MSCPH 511

MSCPH 511
M. Sc. III SEMESTER
NUCLEAR PHYSICS
DEPARTMENT OF PHYSICS
SCHOOL OF SCIENCE
UTTARAKHAND OPEN UNIVERSITY HALDWANI
NUCLEAR PHYSICS
MSCPH511
DEPARTMENT OF PHYSICS
SCHOOL OF SCIENCES
UTTARAKHAND OPEN UNIVERSITY
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Board of Studies
Prof. P. D. Pant Prof. S.R. Jha,

Director School of Sciences School of Sciences, I.G.N.O.U., Maidan
Uttarakhand Open University, Haldwani
Garhi, New Delhi
Prof. P. S. Bisht, Prof. R. C. Shrivastava,
SSJ Campus, Kumaun University, Almora. Professor and Head, Department of Physics,
Dr. Kamal Devlal CBSH, G.B.P.U.A.&T. Pantnagar, India
Department of Physics
School of Sciences, Uttarakhand Open University
Department of Physics (School of Sciences)

Dr. Kamal Devlal, Assistant Professor & Programme Coordinator
Dr. Vishal Kumar Sharma, Assistant Professor Dr. Gauri, Assistant Professor
Dr. Meenakshi Rana, Assistant Professor (AC) Dr. Rajesh Mathpal, Assistant Professor(AC)
Unit Writing and Editing

Editing Writing
Dr. Vishal Kumar Sharma Dr. Vishal Kumar Sharma
Assistant Professor Assistant Professor
Department of Physics Department of Physics
School of Sciences, Uttarakhand Open University School of Sciences, Uttarakhand Open
University
Course Title and Code : Nuclear Physics (MSCPH511)

ISBN :
Copyright : Uttarakhand Open University
Edition : 2022
Published By : Uttarakhand Open University, Haldwani, Nainital- 263139
Printed By :
Contents
Course 11: Nuclear Physics Course code:MSCPH511
Credit: 3
Unit Block and Unit Title Page

Number Number
BLOCK 1: Nuclear Properties and Nuclear Models
1 Nuclear Properties 1-29
2 Nuclear Binding 30-55
3 Nuclear forces I 56-82
4 Nuclear Forces II 83-101
5 Nuclear Models 102-129
BLOCK 2: Radioactivity
6 Radioactivity 130-157
7 Alpha Decay 158-173
8 Beta Decay 174-193
9 Gamma Decay 194-215
BLOCK 3: Nuclear Reactions
10 Nuclear Reactions 216-262
11 Fission and Fusion 263-318

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NUCLEAR PHYSICS MSCPH511
UNIT 1
NUCLEAR PROPERTIES
Structure of the Unit

1.1 Introduction
1.2 Objectives
1.3 Introduction to Nuclear Terminology
1.4 Discovery of Neutron
1.5 Rutherford Scattering and Nuclear Size Estimation
1.5.1 Nuclear Radius
1.5.2 Measurement of Nuclear Radius
1.6 Angular Momentum
1.7 Nuclear Statistics
1.8 Parity and Symmetry
1.9 Magnetic Dipole Moment
1.10 Electric Quadrupole Moment
1.11 Summary
1.12 Glossary
1.13 References
1.14 Suggested Readings
1.15 Terminal Questions
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1.1 INTRODUCTION
In this unit, we will study the basic concepts of the nucleus, constituents of the nucleus and
some basic nuclear properties like nuclear size, nuclear mass, nuclear charge, etc. Before
discussing the unit in detail we must have some knowledge about the atomic structure. Now
question arises, what is the basic building block of all the matters? The answer is an “atom”and
it is the smallest amount of matter that retains all the properties of an element and atomic theory
also suggests that it is composed of smaller particles that no longer have the
same properties as the overall element and consists of two
main components which will be discussed in this unit. Now, when we look into the origins of
nuclear physics, this can be traced back to atomic structure investigations, which began in 1896
with Henry Becquerel's discovery of radioactivity. It's helpful to review how the nuclear atom
came to be in order to grasp the concept of the nucleus. The majority of scientists in the early
nineteenth century agreed that chemical elements are made up of atoms, but they had no idea
about atomic structure. The fact that all atoms contain negatively charged electrons was one of
the first indications. So, in order for an atom to remain neutral, it must possess positively charged
matter of some kind. But how did it come to be like that way? Rutherford discovered this when
he used a thin gold foil to conduct his famous alpha-scattering experiment. The alpha particles
were supposed to pass through the foil with hardly any deflection. This is based on the Thomson
model, which assumes that the electric charge inside an atom is uniformly distributed throughout
its volume. However, they were surprised to find that while the majority of the alpha particles
did not deviate much, a few were distributed over quite wide angles. Some have even become
scattered. This is when the nuclear atom notion was formed, in which the atom is claimed to be
made up of a tiny nucleus carrying all of the positive charge and approximately all of its mass,
with the electrons separated by some distance away.
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1.2 OBJECTIVES
This unit introduces the basic concepts regarding nucleus and their basic properties. After
studying this unit, you should be able to-
• Describe the basic terminology used in nuclear physics like atomic number, atomic mass,
isobars, isotopes etc.
• Define fundamental properties of the nucleus such as nuclear size, nuclear mass and
nuclear charge
• Know about Nuclear spin, magnetic moment and electric quadrupole moment.
1.3 INTRODUCTION TO NUCLEAR TERMINOLOGY
Probing the fundamental particles and their interactions, identifying and interpreting the
features of nuclei, and making technological improvements that benefit society are the three
main components of nuclear physics.
In nuclear physics a nuclear species is characterized by the total amount of positive charge
contained by the nucleus and also by its total number of mass units. The value of the net nuclear
charge is equal to+𝑍𝑒 , where 𝑍 is the atomic number and e denotes the magnitude of the
electronic charge. The fundamental positively charged particle inside the nucleus is the
proton,which is also the nucleus of the simplest atom, Hydrogen. A nucleus with the atomic
number Z almost contains Z protons, and on the other hand an electrically neutral atom must
therefore contain Z negatively charged electrons.
As we know that the mass of the electrons is negligible as compared to the proton mass ( 𝑚𝑃 ≅
2000 𝑚𝑒 ),thereforethe mass of the electron can often be ignored while we have the discussions
about the mass of an atom. The mass number of a nuclear species is denoted by the symbol 𝐴 ,
which is the integer nearest to the ratio between the nuclear mass and the fundamental mass unit.
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In general, Ais greater than Z, nearly for all the nuclei and in most cases, it increases by a factor
of two or more as compared to Z. Thus, there are more massive components in the nucleus.
Before 1932, it was supposed that the nucleus contained 𝐴protons, in order to provide theproper
mass, along with 𝐴 − 𝑍 nuclear electrons to give a net positive charge of𝑍𝑒. One question also
arises here: Does electron exist inside the nucleus? The answer is provided with certain facts that
are not satisfied for the electrons to remain inside the nucleus.
1. The nuclear electrons would have to be held to the protons by a strong force, maybe even
stronger than the Coulomb force. Despite this, there is no evidence for a strong force
between protons and atomic electrons.
2. If we try to confine the electrons in a region of a small space as small as a nucleus
(∆𝑥~10−14 𝑚) the uncertainty principle would require that these electrons have a
momentum distribution with a range ∆𝑝~ ħ⁄∆𝑥 = 20𝑀𝑒𝑉/𝑐 . Electrons which emits
from the nucleus in radioactive β decay have energies value generally less than 1 𝑀𝑒𝑉;
never we have seen a decay of electrons with20 𝑀𝑒𝑉 energies. Thus the existence of
20 𝑀𝑒𝑉 electrons inthe nucleus isnot confirmed by this observation.
3. The total intrinsic angular momentum (spin) of nuclei for which the value of 𝐴 − 𝑍is odd
would disagree with the observed values if 𝐴protons and 𝐴 − 𝑍electrons
were present in the nucleus. For example, consider the nucleus of deuterium with 𝐴 =
2,Z= 1), which according to the proton-electron hypothesis would contain 2
protons and 1 electron. The proton and electron both have intrinsic spin angular
1
momentum2and from the quantum mechanical rules for adding spins of particles would
1 3 1
require that these three spins of 2 combines to give a total of value either2 or 2.But the
observed spin of the deuterium nucleus is 1.

4. Magnetic dipole moments in nuclei containing unpaired electrons should be far higher
than those measured.If a single electron were present in a deuterium nucleus, we would
anticipate the nucleus to have a magnetic dipole moment about the same size as an
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electron, however the observed value of the magnetic moment for the deuterium nucleus
1
is 2000 of the value of the magnetic moment of one electron.
All sorts of above reasons were eliminated by the discovery of neutron in 1932 by Chadwick.
The neutron in general are electrically neutral and it has a mass almost equal to that of proton
mass.Thus a nucleus with 𝑍protons and 𝐴 − 𝑍 neutrons has the proper total mass and charge, without the
need to introduce nuclear electrons.
In general, a nuclear species or nuclide is represented by 𝐴𝑍𝑋, where X represents the chemical
symbol with the number of neutrons 𝐴 − 𝑍, where A is the atomic mass and Z represents the
atomic number which have the same value as that of protons. For example: 11𝐻 , 42𝐻𝑒, 63𝐿𝑖 , 168𝑂,
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26𝐹𝑒 .
The nucleon family consists of two members: neutrons and protons. We use the term nucleons
when we only want to talk about nuclear particles without specifying whether they're protons or
neutrons. As a result, a nucleus with mass number A has A nucleons. Nuclides with the same
proton number but different neutron numbers are called isotopes; for example, the element
chlorine has two isotopes that are stable against radioactive decay,35Cl and 37Cl. It also has many
other unstable isotopes that are artificially produced in nuclear reactions; these are the
radioactive isotopes (or radioisotopes) of Cl. A nuclide with the same N but different𝑍; these are
called isotones. The stable isotones with N = 1are 2Hand3He. Such Nuclides which have the
same mass number A are known as isobars; thus, stable3He and radioactive 3H are isobars.
1.4 DISCOVERY OF NEUTRON
The neutron was not found until 1932, when James Chadwick calculated the mass of this
neutral particle using scattering data. Since the time of Rutherford, scientists have known that the
atomic mass number A of nuclei is somewhat more than double that of most atoms' atomic
number Z, and that the nucleus contains almost all of the atom's mass. Protons and electrons
were thought to be the fundamental particles around 1930, however this required that a number
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of electrons be bound in the nucleus to partially negate the charge of A protons. However, by this
time, the uncertainty principle and "particle-in-a-box" confinement calculations had established
that there just wasn't enough energy available to confine electrons in the nucleus.
By putting the particle's De-Broglie wavelength equal to that dimension, a rough scale of the
energy required for confinement to that dimension can be produced. If we assume a hydrogen
atom has a diameter of 0.2 nm, the equivalent confinement energy is roughly 38 eV, which is the
correct order of magnitude for atomic electrons. However, it takes around 250 MeV of energy to
confine an electron to a nuclear dimension of about 5 fermis.
The maximum confinement energy available from the nucleus's electrical attraction is given by
𝑍𝑘𝑒 2 79(1.44 𝑀𝑒𝑉. 𝑓𝑚)

= ≈ 23 𝑀𝑒𝑉 ≪ 250 𝑀𝑒𝑉 … … … . . (1)
𝑟 5𝑓𝑚
So, from above result there are no electrons in the nucleus.
Bothe and Becker observed in 1930 that bombarding beryllium with alpha particles from a
radioactive source created neutral radiation that was penetrating but non-ionizing, which was an
experimental breakthrough. They assumed it was gamma rays, but Curie and Joliot demonstrated
that when this radiation was used to bombard a paraffin target, it expelled protons with an energy
of roughly 5.3 MeV. As can be seen through momentum and energy analysis, that wasn't the case
with gamma rays:
Fig. 01: Curie and Joliot experiment
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This analysis is similar to that of a head-on elastic collision in which a small particle collides
with a considerably larger one. The required energy for the gamma ray explanation was
significantly higher than any energy measured in the nucleus, thus the neutral radiation must be
some form of neutral particle.
If the neutral particle had a mass comparable to that of the proton, the 5.3 MeV energy of the
expelled protons could be easily explained. This would only require 5.3 MeV from the neutral
particle in head-on collisions, which is within the range of observed nuclear particle emissions.
By bombarding targets other than hydrogen, such as nitrogen, oxygen, helium, and argon,
Chadwick was able to demonstrate that the neutral particle could not be a photon. Not only were
these interactions incompatible with photon emission on energy grounds, but their cross-section
was orders of magnitude larger than that of photon Compton scattering.
Chadwick was left with the problem of estimating the mass of the neutral particle. He decided to
bombard boron with alpha particles and study how the neutral particles interacted with nitrogen.
Figure 02: Discovery of neutron
The masses of boron and nitrogen were well known; thus, these specific targets were chosen.
When energy conservation is applied to the combined interactions, the following formulas result.
1 1 1
𝑚𝛼 𝑣𝛼2 + 𝑚𝛼 𝑐 2 + 𝑚𝐵 𝑐 2 = 𝑚𝑁 𝑣𝑁2 + 𝑚𝑁 𝑐 2 + 2 𝑚𝑛 𝑣𝑛2 + 𝑚𝑛 𝑐 2………….(2)
2 2
Solving for the mass energy of the neutron gives

UTTARAKHAND OPEN UNIVERSITY 7 HALDWANI
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1
𝑚𝛼 𝑣𝛼2 + 𝑚𝛼 𝑐 2 + 𝑚𝐵 𝑐 2 − 𝑚𝑁 𝑐 2
2
𝑚𝑛 𝑐 ≈ 2 … … … … … (3)
𝑣2
1 + 𝑛2
2𝑐
The speed of the neutron is the last remaining unknown on the right hand side of the equation.
Chadwick blasted hydrogen atoms with his created neutrons, assuming that the neutron mass was
near to that of the proton, in order to determine the speed of the protons following the impacts.
He then used the above energy expression to generate a neutron mass of 9381.8 MeV by setting
the neutron speed equal to those proton speeds. Chadwick obtained the initial value for the
neutron mass, which matched the current accepted estimate of 939.57 MeV by using a consistent
set of measurements.
1.5 RUTHERFORD SCATTERING AND NUCLEAR SIZE

ESTIMATION
A Preliminary idea about the nuclear model was first suggested by Rutherford in
1910.The diagram of Rutherford’s experimental set up is shown in the figure 03. Obviously the
apparatus has to be closed in vacuum. A thin film of gold foil is made to strike by a beam of
alpha particles. The scattered alpha particles were detected by a fluorescent screen-telescope
arrangement. Rutherford observed that many of the alpha particles went straight through the foil
or they were deflected by a very small angle. But a few alpha particles were deflected by a very
large angle. Further some were even got scattered back. This observation was not expected on
the basis of idea of atomic structure at that time: that the positive charge is uniformly smeared
over the entire volume of the atom.
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Figure 03: Rutherford’s gold foil experiment
For explanation of these deflections, Rutherford assumed that the positive part of the atom was
concentrated in a very small volume at the center of the atom. This core, which is surrounded by
the cloud of electrons, which makes the entire atom electrically neutral, is now called as nucleus.
This nuclear model of the atom accounts for the scattering of alpha particles at large angles in the
following way. An alpha particle approaching the center of the atom experiences an increasingly
large Coulomb repulsion. Since the atom is mostly empty space , most of the alpha particles do
not go sufficiently close to the nucleus to get sufficiently deflected and so they pass through the
foil without any deviation. However, an alpha particle that passes close to the nucleus is
subjected to a very large Coulomb repulsive force exerted by the massive positive core and is
deflected at a large angle in a single encounter.
1.5.1 Nuclear Radius
From the above discussion we have a basic idea about the nucleus. Now we shall discuss about
the estimation of Nuclear Radius by considering Rutherford’s α-Scattering experiment using
classical approach. Let us consider that the gold nucleus in the scattering experiment has a radius
R. An α-particle trajectory can be specified by its impact parameter b as sown in the figure
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Figure 04: Deflection of an alpha particle by a gold nucleus
For the less coloumb reflection i.e for b is greater than R the deflection of alpha paeticle is also
less. Also when b is less than R, the alpha particle goes through the nuclear positive charge
distribution and so the nuclear charge above the trajectory and the one below the trajectory work
in opposite directions and hence again the deflection produced in the alpha particle trajectory is
small. When the alpha particle just almost touches the nucleus (which is considered to be a
positive charge), this is the case when coulomb repulsion becomes maximum and
correspondingly for b~R, the deflection of alpha particle also becomes maximum. For this
qualitative discussion, let us consider a trajectory corresponding to small deflection. As the
particle approaches near the gold nucleus, it is slowed by the Coulomb repulsion and is speeded
up on its way out i.e away from the nucleus as shown in the figure below
Fig.05: Particle trajectory with impact parameter b and small deflection θ
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The Coulomb repulsion in the region close to the scattering gold nucleus is given by the formula
1 (2𝑒)(𝑍𝑒)
𝐹= ……………..(4)
4𝜋𝜖0 𝑏2
Where 2e represents the charge on α particle. Z represents the atomic number and for gold
(Z=79).
Now this repulsive force can be approximately considered to be operating in the direction
perpendicular to the direction of incidence, over a distance b. If we consider the particle’s
velocity to be v then the time during which the force acts is given by
𝑏
∆𝑡 = … … … … . (5)
𝑣
By this force there is a momentum ∆𝑝 in the direction perpendicular to the incident direction. By
Newton’s law
Δ𝑃
𝐹= Δ𝑡
1 2𝑍𝑒 2 𝑏
∆𝑝 = 𝐹∆𝑡 = ………………….(6)
4𝜋𝜖0 𝑏 2 𝑣
From fig.05 the deflection θ is
∆𝑝 1 2𝑍𝑒 2 ⁄𝑏𝑣
𝜃~ = ………………(7)
𝑝 4𝜋𝜖0 𝑚𝑣
Where m denotes the mass of the α-particle.
So, we have
1 2𝑍𝑒 2 1
𝑏 = ……………(8)
4𝜋𝜖0 𝑚𝑣 2 𝜃
Above equation shows the approximate relation between deflections and impact parameter.
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Rutherford in his experiment observed the maximum deflections were of the order of 1 radian.
This doesn’t mean that large deflections were not observed by Rutherford. Occasionally,
deflections close to π radians were observed by him.
In above equation if we put the values corresponding to maximum deflections i.e θ = 1 and b =
R, we have
1 2𝑍𝑒 2
𝑅~ … … … … … … . . (9)
4𝜋𝜖0 𝑚𝑣 2
If we put other values Z= 79 for gold, mass of α particle 𝑚 = 6 × 10−27 𝑘𝑔
𝑣 ~ 107 m/s for speed of α particle
𝑒 = 1.6 × 10−19 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠
We obtain the radius has order of 10-14m.
This distance is extremely small as compared to the atomic radius value 10-10 m, which is smaller
by a factor of 104.
Since protons are positively charged and the gravitational force is negligible inside the nucleus,
the query then arranges us to how all these protons state together in a nucleus of size having
value 10-14m. There is certainly a nuclear interaction acting between protons and neutrons in
bracket nucleons inside the nucleus over dispenses of about 10-14m that is strong enough to
counteract the massive column repulsion between the closely packed protons. Because the
nuclear interaction's attraction force has no effect on the additional nuclear atomic structure, it
must be a short-range force, effective only across distances of 10-14m. This new force must, by
definition, be more dependent on distance than an inverse square force. With very high energy
particles generated by accelerators, the magnitude of the region where this nuclear short-range
force is considerable (relative to the Coulomb force) may be examined. When such particles are
fired at thin foils, the distribution of scattering angles reveals that the radius R of the nucleus is
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proportional to the cube root of the scattering nuclei's mass number, i.e. R = RoA1/3 , where Ro
has the value of 1.07 x 10-15 m.
1.5.2 Measurement of Nuclear Radius
In the previous section we have estimated the nuclear size by the use of Rutherford’s α-
scattering experiment. There are many experimental methods for measuring nuclear size. In this
section we shall study about the experimental method of measuring the nuclear radius which was
first employed by R. Hofstadter and his colleagues by the important bearing that nuclear size has
our understanding about the nuclear force.
This method is based on measurement of nuclear charge distribution and it assumes that
charge distribution and mass distribution within nuclei are essentially the same. By analyzing
scattering of high-energy electrons from various nuclei we have obtained the charge distribution
in nuclei.
Incident beams of high energy Thin target Beam collector
electrons from accelerator
Detector for scattered electrons
Fig.06: Experimental Setup for nuclear radius measurement
In this experiment electrons are chosen as bombarding particles and its interaction within the
nucleus is well known eletro-magnetic interaction. As shown in the fig.06 the elastically
scattered electrons are observed by a detector as a function of angles.
It is easy to see that the de Broglie wavelength of 200 MeV electrons is about 10 -14 m (by
using the relation𝜆 = ℎ⁄𝑝 ) which is the size of the nucleus. Thus, to get information about
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nuclear charge distribution, the incident electron beam energy has to be the order of 200 MeV.
The assumption that the nuclear charge is uniformly spread over a spherical volume and not a
point charge leads one to expect diffraction effects.
Portions of electron waves incident on different parts of the nucleus will be scattered in a
particular direction, with phase differences resulting in constructor for destructive interference at
some angle. Fig.07 is taken from the work of Hofstadter and his collaborators and shows the
angular distribution of elastically scattered 200 MeV electrons from nuclei assumed to have
spherical uniform charge distribution (having uniform density up to radius r as shown in fig.08.
The actual nuclear charge distribution is not exactly given by a step function, but is shown by
dotted line. One can clearly see in fig.07 that experiments do show diffraction maxima and
minima as expected from a uniform charge distribution. For comparison, the expected result
(which is not obtained experimentally) from a point charge assumption is also shown in fig.07.
For theoretical calculations Hofstadter and his colleagues assumed a charge density of this form
𝜌0
𝜌(𝑟) = ……………….(10)
1+𝑒 (𝑟−𝑅)⁄𝑏
𝜌0
Where 𝜌0 stands for density at the nuclear center, R is the radius at which 𝜌 falls to 2
and 𝑏
measures how rapidly 𝜌 falls to zero at the nuclear surface.
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Fig.07: Shows experimental angular distribution of scattered electrons from three types of
nuclei
Fig. 08: step function shows here the uniform charge distribution
It is clear from the distribution relation given by equation 4 posseses a property that for 𝑅 ≫ 𝑏,
𝜌 is close to 𝜌0 , until (R-r) is a few times b, at which point 𝜌 decreases to very small values
in a distance determined by b, but not by R. If we put the following values, then this charge
distribution agrees with the experimental data involving many nuclei
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𝑛𝑢𝑐𝑙𝑒𝑜𝑛𝑠
𝜌0 ~ 1.65 × 1044 , 𝑎~0.55 𝑓𝑒𝑟𝑚𝑖
𝑚3
and 𝑅~1.07𝐴1⁄3 fermi ………………..(11)
Where 1 Fermi = 1 F = 10−15 m.
If we vary the values of R and b, then we obtain the charge distribution which represents the
observed angular distribution shown in Fig. 07.
The fig. 09 shown below represents that the charge density is essentially the same for almost all
nuclei( except lighter nuclei H and He) decreasing slowly for the increasing values of A. There
exists a general relationship for the nuclear radius, which has been experimentally verified, as
given by equation 11, i.e.
𝑅 = 𝑅0 𝐴1⁄3
Fig. 09: Nuclear charge distribution found by using high-energy electrons as probes in
some nuclei
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Other methods are also there for obtaining the nuclear radius. An important method probes the
extent of the nuclear force rather than the nuclear charge. We have the same general relationship
𝑅 = 𝑅0 𝐴1⁄3 .
Example : 01 : Determine the radii of O16 and Pb206 nucleus.(given that R0= 1.4fm)
Solution :Using the relation R=R0A 1⁄3 and substituting the values of A in the following formula,
we have
R(O16) = 1.41.4 × (16)1⁄3 = 3.33 fm.
R(O16) = 1.4 × (208)1⁄3 = 3.29 fm.
Example : 01 : Determine the stable nucleus that has a radius one third that of Os189.
Solution : By Using the relation R=R0A 1⁄3
1 𝑅 𝐴 1⁄3 𝐴 1⁄3
= = ( ) = ( )
3 𝑅𝑂𝑠 𝐴𝑂𝑠 189
189
𝐴 = =7
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The element with the mass number (A=7) is Lithium(Li7).
1.6 ANGULAR MOMENTUM
The angular momentum of an isolated system is known to be conserved according to elementary

quantum mechanics. Since the nucleus is an isolated system, its angular momentum is a constant
quantity. Most commonly, the total nuclear angular momentum J is denoted by symbol I
corresponding to nuclear spin term.
So, the total angular momentum J of a nucleon is the vector sum of its orbital and spin angular
momenta values
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J = L+S …………………….(13)
where L is orbital angular momentum and S is spin angular momentum.
The magnitude of the total angular momentum of a nucleon is given by
J= j ( j + 1) ……………….(14)
J can take on the half integral values, 1/2, 3/2, 5/2,………
The magnitude of the spin angular momentum is given by
S = s ( s + 1) ……………………….(15)
Where S = ½ is the spin angular momentum quantum number and the magnitude of the orbital
angular momentum is given by
L = l ( l + 1) ………………………….(16)
L can take up integral values 0,1,2,3,
Quantum mechanical considerations show that the total orbital and spin angular moments of the
nucleus is given by
PI2 = I ( I + 1) 2
PL2 = L ( L + 1) 2
………………….(17)
PS2 = S ( S + 1) 2
It is determined during measurement which part of the angular momentum is larger along the
direction of the applied electric or magnetic field. These have the magnitudes I, L, and S,
respectively, for the three examples indicated above.
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The nuclear spin is always zero (I=0) for even Z and even N nuclei, according to measurements
of the ground state spin of nuclei. This demonstrates that the nucleons inside the nucleus have a
tendency to pair off into nucleon-like pairs when their angular momenta are equal and oppositely
aligned.
The measured values of the ground state spins of the nuclei are small integers or half odd
integers; the highest measured value is 9/2, which is small when compared to the sum of the
absolute values of Ii and Si of all the individual nucleons contained in the nucleus. This is an
important point to keep in mind. This is consistent with what was said earlier about pair creation
inside the nuclei. The majority of nucleons of either kind appear to assemble into pairs of zero
spin and orbital angular momentum protons and neutrons, giving the core itself zero total angular
momentum. The nuclear spin, which is thus determined by the few remaining nucleons outside
the core is therefore a small number, integral or half odd integral.
1.7 NUCLEAR STATISTICS
Nuclear statistics can be Either Bose-Einstein statistics or Fermi-Dirac statistics, provide a

description of a system with many particles, such as the nucleus, in terms of quantum mechanics.
Thus, it is possible to divide all fundamental particles into two types
• Bose-Einstein Statistics: The wave function of a system of two identical subatomic

particles is either symmetric or anti-symmetric in the exchange of the two particles'
coordinates in the case of a subatomic particle subject to the laws of quantum mechanics.
If the sign of the wave function is not altered by such an interchange, we have a
symmetric wave function, and for that Bose-Einstein Statistics holds. Bosons, which are
defined as particles with integral spin or zero spin, are subject to Bose-Einstein statistics.
Photon, meson, and deuteron are what they are. Even mass number nuclei are subject to
Bose-Einstein statistics.
• Fermi-Dirac Statistics: But when the exchange results in a change in the wave
function's sign, we get an asymmetric wave function, and the associated statistics are
19
20

known as Fermi-Dirac statistics. Fermions are all particles with half integral spin, which
all correspond to Fermi-Dirac statistics. Protons, neutrons, and electrons also comes into
this category. All odd mass number (A) nuclei correspond to F.D. statistics. No two
fermions can exist in the same quantum state because all fermions obey the Pauli
exclusion principle. On the other hand, Boson particles do not follow to the exclusion
principle and a number of Bosons can exist in the same quantum state.
1.8 PARITY AND SYMMETRY
Another very important property that particles possess is Parity.The particle's parity is a
characteristic of the wave function representing its quantum mechanical state. Positive parity
refers to a wave function that represents a single particle that does not undergo a sign change
upon reflection through the origin, whereas negative parity refers to a wave function that
undergoes a sign change. So we have
 ( x, y, z ) =  ( − x, − y, − z ) =  ( x, y, z ) , for positive parity ...........(18)

 ( x, y, z ) =  ( − x, − y, − z ) = − ( x, y, z ) , for negetive parity ...........(19)
A wave function that describes multiple particles can be expressed as the sum of the wave functions of
each individual particle or as a composite of those wave functions.
Since the parities of the individual particle wave functions are a product, the parity of the entire system
is evidently determined by this. The mass density and charge density of nuclei are symmetric no matter
what the parity is since they are always equal.
Symmetry is an important concept in atomic and nuclear physics. Schrodinger's equation for two
identical, non-interacting particles travelling in the same potential has a straightforward solution that is
 AB =  A (1)  B ( 2 ) ....................(20)
where  A and  B stand for the positions of particles 1 and 2, and are two solutions to the same
one-body wave equation. However, the wave function  AB is not an acceptable function for two
UTTARAKHAND OPEN UNIVERSITY20 HALDWANI
21

identical particles because since it assumes that we can label the particles and distinguish
between those in state A and those in state B. This difficulty can be avoided by writing the wave
function in either one or the other of the two following forms, both of which are solutions of
Schrodinger's equation if Eqn. 20 is a solution
1
S =   A (1)  B ( 2 ) +  A ( 2 )  B (1)  ........(21)
2
1
A =   A (1)  B ( 2 ) −  A ( 2 )  B (1)  ........(22)
2
When labels 1 and 2 are switched in both equations, the probability density is symmetric. As a result,
both are valid solutions to the two-body problem. Because I and 2 can be switched around,  S does
not change and it remains symmetric , while  A is anti-symmetric and sign changes .
1.9 MAGNETIC DIPOLE MOMENT
Now in this unit we shall study about the magnetic dipole moment µ which is associated with a
current loop of area A and having a current I
µ = IA …………………..(23)
Fig. 10: Magnetic dipole with circular loop of area A
21
22

A spinless particle revolving with with electric charge e generates a magnetic dipole moment
which is equal to the number of revolutions per second times the charge times the area enclosed
by the orbit. Now for this circular loop with area A(=π r2) ,we get the value of magnetic dipole
moment by substituting the values of I and A
 ev  2 evr
=  r = ..................(24)
 2 r  2
where v is the velocity of the particle and r is the radius of the orbit. The orbital angular
momentum of this particle is mvr. The ratio of the magnetic dipole moment to the angular
momentum is called the gyromagnetic ratio which is given by
 e
g= = ..................(25)
p1 2m
Naturally, this relationship will hold for the two vectors' components as well as for p, in any
direction, such as the direction of a magnetic field. We know from quantum physics that the
mechanical moment p's component along a specified direction (the z-direction) is m h, which can
be either a positive or negative integer or zero. Consequently, we obtain for the magnetic dipole
moment's z-component
e 
z =   m1 ....................................(26)
 2m 
This expression yields the correct result for the z-component of the magnetic dipole moment that
is due to orbital motion of electrons in an atom. It is believed to give the correct result also for
the orbital motion of protons in a nucleus. The mass in the denominator of Eqn. 26 should then
be the proton mass mp.. In contrary to magnetic dipole moments, we can introduce the nuclear
magneton defined by
e
N = = 5.0505 10−27 Jm2 wb −1 ................(27)
2m p
22
23
 N shows its analogy with the Bohr magneton which is given by the formula 𝑒ħ which is the unit of
2𝑚 𝑒
atomic magnetic moment.  N is much smaller than  B , being only 1/1836 part of the later.
The measured values of the magnetic moments of the electron, proton and the neutron are given
below
e = −1.001145358 B
 p = 2.7927  N
n = −1.9131 N .............................(28)
In the above equation 28 negative sign results from because of the direction of the angular
momentum vector and the magnetic dipole moment vector are opposite to each other. The total
magnetic dipole moment for a nucleus is the sum of the moments associated with the spins of the
protons and neutrons and the moments associated with the orbital motion of the proton. Since the
z-component of the dipole moment is the only component which can be observed through its
interaction with a magnetic field.
1.10 ELECTRIC QUADRUPOLE MOMENT
The nuclear electric quadrupole moment is a parameter which describes the effective shape of
the ellipsoid of nuclear charge distribution. A non-zero quadrupole moment Q indicates that the
charge distribution is not spherically symmetric. By convention, the value of Q is taken to be
positive if the ellipsoid is prolate and negative if it is oblate.
First we have the value of electric potential V of any distribution of electric charges at a distance
R in the Z-direction given by equation 29
1 1 
V= 
4 0  R  R
1
R
1
( )
 dV + 2   zdV + 3   3z 2 − r 2 dV + .......

................(29)
23
24
where,  is the charge density. The integration is performed over the region containing the
electric charge. In the above expression we have a rapidly convergent series and as the value of
R increases, only the first two or three terms become important in this series. The first term
corresponds to the net charge and for larger R this is the only important term. The integral in the
second term is called the dipole moment and the third term is called the quadrupole moment. It
turns out that a nucleus must have a zero electric dipole moment and so the lowest order
contribution to V due to a point charge arises from the quadrupole moment. Two examples of
non-zero electric quadrupole moments are shown in Fig. 11
Fig. 11. Electric Quadrupole Moment
The four-charge (quadrupole) system in the left part of the Fig. 11 has net charge and dipole
moment zero and so the entire electric field is produced by the electric quadrupole moment. An
ellipsoidal charge distribution shown on the right leads to a highly deformed nucleus; it has a
quadrupole moment; however, its dipole moment is zero. The charge density at point r(x, y, z) is
Ze ( r )
given by where Ze is nuclear charge.
The quadrupole moment Q is given by
( )
Q = Z 3 z 2 − r 2  (r )d 3 r
( )
= Z  r 2 3cos 2  − 1  ( r ) d 3 r ...............(30)
24
25
For a spherically symmetric charge distribution

 ( r ) , the quadrupole moment vanishes, for a
prolate nucleus, the charge is concentrated along the z-axis and so Q is positive. Q is negative for
an oblate nucleus as shown in Fig. 12
Fig. 12 Oblate and Prolate shapes
From Eq. 30, it is clear that Q has same dimensions as that of area and is given in terms of m²
or barns (10-24 cm³) or F2 (fermi²).
In a quantum-mechanical definition of the quadrupole moment, the charge density  replaced

by the probability density and the expression is summed over all the protons. In principle, these
calculations are simple and directly yield the quadrupole moment referred to the z-axis. In
classical or semiclassical calculations, one should consider the fact that the nuclear symmetry
axis is not, in general, the space-fixed z-axis to which measured quadrupole moments are
referred. Let us assume that the J-axis can be regarded as a symmetry axis, or at least that the
time average of the charge distribution has a rotational symmetry about the J-axis. It can be
26

defined as a quadrupole moment Q given by Eqn.30 with the z-direction along the J-vector. The
angle è between this vector and the space fixed z-axis in the state for which mJ = J is given by
J J
cos  = = ......(31)
J ( J + 1) ( J + 1)
It can be shown by simple classical calculations that the relationship between Qj and the
observed quadrupole moment Q, which relates to the space-fixed z-axis in the state mJ =J, is
Q=
( J − 12 ) Q ......(32)
( J + 1) J
This shows that a nucleus with J=½ has a zero-quadrupole moment with reference to a space-
fixed axis. The same is true for a nucleus with J=0 for which there are no constraints on the
orientation, and therefore the time-averaged charge distribution is symmetric. An important
aspect of quadrupole moment measurement is that it makes it possible to determine nuclear
deformation. Knowledge of nuclear deformation is important for studying collective behavior of
nuclear particles.
1.11 SUMMARY
In this unit basic properties of Nucleus have been discussed in the simpler manner. In this unit
Rutherford’s scattering and nuclear size estimation have also been explained to understand the
nuclear radius. Learners will also be able to solve the numerical based on nuclear size
determination. The relation between mass and binding energy are discussed. The nuclear wave
electrical properties like nuclear statistics, parity and symmetry have also been explained in this
unit. The concepts of magnetic dipole moment and electrical Quadrupole moment have also been
explained for the nucleus with suitable examples.
26
27
1.12 GLOSSARY
Nucleus the central region of an atom where the majority of the mass is
concentrated.
Isotopes atoms of the same element that have different numbers of neutrons but the
same number of protons and electrons.
Isobars are elements that have the same number of nucleons (sum of protons and
neutrons).
Mirror Nuclei atomic nucleus that contains a number of protons and a number of
neutrons that are mutually interchanged in comparison with another
nucleus.
Scattering a change in the direction of motion of a particle because of a collision
with another particle
Angular Momentum The property of any rotating object given by moment of inertia
times angular velocity.
Parity property important in the quantum-mechanical description of a physical
System
Magnetic dipole moment the measure of the objects tendency to align with the magnetic
field.
Electric Qudrupole moment a parameter which describes the effective shape of the ellipsoid
of nuclear charge distribution
1.13 REFERENCES
1. B.L.Cohen, Concepts of Nuclear Physics; Tata McGraw-Hill Publisher
2. Kenneth S.Krane, Introductory Nuclear Physics, Wiley, New York, 2008.
3. S.B.Patel, Introductory Nuclear Physics, New Age International Publishers.

28

4. S N Ghoshal, Nuclear Physics, S.Chand Publication
5. H.M.Agrawal, Nuclear Physics, Prentice Hall of India
6. A.Beiser, Concepts of Modern Physics, McGraw-Hill, New York, 2003.
7. H.A.Enge, Introduction to Nuclear Physics, Addison-Wesley, London,1966.
8. M.L.Pandya and R.P.S.Yadav, Elements of Nuclear Physics, Kedar Nath Ram Nath, Delhi.
1.14 SUGGESTED READINGS
1. Blin-Stoyle Nuclear and Particle Physics, Chapman and Hall
2. I.Kaplan, Nuclear Physics, Narosa ,2002
3. S.S.M. Wong.,Introductory Nuclear Physics, Prentice Hall of India, New Delhi, 2005
4. B.R.Martin, Nuclear and Particle Physics, Wiley, New York, 2006.
5. J.B.Garg, Nuclear Physics: Basic Concepts, Macmillan, New Delhi, 2011.
6. V.Devnathan, Nuclear Physics, Narosa, New Delhi, 2006.
1.15 TERMINAL QUESTIONS
1. Explain the term nuclear radius.
2. Explain the term nuclear magnetic dipole moment.
3. Define the term statistics and parity for the nucleus.
4. Show that a nucleus a zero electric dipole moment.
28
29

5. Show that nuclear density is constant for all nuclei.
6. Mention various methods for determining the size of the nucleus and describe any one in
detail.
7. Explain the term electric quadrupole moment and find out an expression for quadrupole
moment.
8. Estimate the ratios of the major to the minor axes of 51Sb123. The quadrupole moment is -1.2b.
Take R= 1.5A1/3fm.
9. Explain in brief the different properties associated with the nucleus.
10..Discuss one method for the determination of the size of the nucleus.
11.. Explain the relationship between the nuclear mass with nuclear size.
29
30
UNIT 2
NUCLEAR STABILITY

2.1 Introduction
2.2 Objectives
2.3 Binding Energy
2.3.1 Mass Defect
2.3.2 Binding Energy per Nucleon
2.3.3 Packing Fraction
2.4 Nuclear Reaction
2.5 Types of Nuclear Reactions
2.6 Mass Balance and Energy in Nuclear Reaction
2.7 Q-Value of Equation
2.8 Solution of the Q-Value of Equation
2.8.1 Exoergic Reactions
2.8.2 Endoergic Reactions
2.9 Glossary
2.10 Summary
2.11 References
30
31
2.1 INTRODUCTION
In the previous unit we have obtained the basic information of the nucleus and its properties.
Now, in this unit we shall study about the binding energy which is responsible for the stability of
the nucleus and we know that lighter nuclei are more stable than heavier nuclei in which the
number of neutrons are in excess as compared to the number of protons. In this consequence we
will also study about mass defect and packing fraction. Our current understanding of the nuclear
structure is mostly based on studies in which a chosen nucleus is blasted with various particles,
such as protons, neutrons, and deuterons. When these particles are close enough to the target
nuclei to interact, either elastic or inelastic scattering may occur, one or more completely distinct
particles may be ejected from the nucleus, or the incident particle may be trapped and generate a
gamma ray. After the bombardment, a nuclear reaction is said to have occurred when the mass
number and/or atomic number of the target nuclei changes.
2.2 OBJECTIVES
After studying the unit the learners will be able to understand
• Binding energy, mass defect and packing fraction

• Nuclear Reaction and its types
• Q-value and its solution for nuclear reaction
• Endoergic and Exoergic nuclear reactions
• Solve the numerical problems of finding the Q-value
31
32
2.3 BINDING ENERGY
2.3.1 Mass Defect
Now in this unit we shall study about mass defect which is a very important aspect while we
discuss binding energy responsible for nuclear stability. Mass of an atom is concentrated in the
central part called nucleus, which is made up of neutrons and protons. It has been observed that
the mass of a nucleus is always less than the sum of the individual masses of the protons and the
neutrons, which constitute it. This difference is a measure of the nuclear binding energy which
holds the nucleus together, is known as the mass defect Δ𝑚.
If 𝑚(𝑍, 𝑁) is the mass of the nucleus of an atom consisting of 𝑍 protons and 𝑁 neutrons, the
mass defect Δ𝑚 is given as
Δ𝑚 = 𝑍𝑚𝑝 + 𝑁𝑚𝑛 − 𝑚(𝑍, 𝑁)
Or Δ𝑚 = 𝑍𝑚𝑝 + (𝐴 − 𝑍)𝑚𝑛 − 𝑚(𝑍, 𝑁)………………(1)
where 𝑚𝑝 and 𝑚𝑛 are the masses of a proton and a neutron.
It is convenient to talk in terms of the atomic masses, therefore, adding and subtracting the mass
of 𝑍 atomic electrons on the RHS of the above equation, we get
Δ𝑚 = 𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 − 𝑚(𝑍, 𝑁) − 𝑍𝑚𝑒

Δ𝑚 = 𝑍𝑀𝐻 + (𝐴 − 𝑍)𝑚𝑛 − 𝑀(𝑍, 𝑁) … … … … … … … … … … . (2)
where 𝑚𝑒 is the mass of one electron, 𝑀(𝑍, 𝑁) is the atomic mass and 𝑀𝐻 is the mas of neutral
hydrogen atom.
This missing mass may be regarded as the mass, which would be converted into energy, if a
particular atom is to be formed from the requisite number of electrons, protons and neutrons.
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33

This is also the amount of energy required to break up the atom into its constituents. Therefore,
mass defect is a measure of the binding energy of an atom/nucleus.
Example 1: Calculate the mass defect of deuterium 12 𝐻, which is an isotope of hydrogen known
as heavy hydrogen
Solution: Let us would expect that its mass should be equal to the mass of one neutron in its
nucleus. Thus, one atom, i.e.
𝑚𝑛 + 𝑀𝐻 = 1.008665 + 1.007825𝑢
Therefore, the expected mass of deuterium is 2.016490𝑢.
However, the measured mass of deuterium, is found to be 2.014102𝑢.
Therefore, for deuterium,
the mass defect (Δ𝑚) = 2.016490 − 2.014102𝑢
Δ𝑚 = 0.002388𝑢.
2.3.2 Binding Energy per Nucleon
According to the proton-neutron model of the nucleus, it is obvious that the particles that make
up the stable nucleus are kept together by potent attraction forces, and that effort is required to
break them apart. In other words, the nucleus needs energy to be given in order to be divided into
its component parts. We use the well-known Einstein mass-energy connection E=mc2, where E
and m are the energy and mass of the particle and c is the speed of light in a vacuum, to
investigate what shape this energy can take. Simply said, this shows that energy and mass are
different representations of the same entity.. Therefore, we should anticipate that the mass of the
nucleus as a whole will be less than the sum of the masses of its constituent parts. In fact, a
significant deal of experimentation has led to this observation.
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34

The difference between the energy of the constituent particles and the energy of the entire
nucleus is thus used to determine the binding energy of the nucleus. Take the example of a
nucleus zMA, where the binding energy is given by
𝐵 = [𝑍𝑀𝑃 + 𝑁𝑀𝑁 − 𝑧𝑀 𝐴 ]𝑐 2 …………………………..(3)
where 𝑀𝑃 = Mass of the Proton,
𝑍 = Number of Protons,
𝑀𝑁 = Mass of the Neutron,

𝑁 = Number of Neutrons = (𝐴 − 𝑍)
𝑧𝑀 𝐴 = Measured mass of the neutral atom, [also written as 𝑀(𝑍, 𝐴) ]
The above expression is, now-a-days, generally expressed as
𝐵 = [𝑍𝑀𝐻 + 𝑁𝑀𝑁 − 𝑧𝑀1 ]𝑐 2
where 𝑀𝐻 represents the mass of the neutral hydrogen atom. Since there are A nucleons in the
nucleus, the binding energy per nucleon is also given as
𝐵𝑒 𝑐2
= [𝑍𝑀𝐻 + 𝑁𝑀𝑁 − 𝑧 𝑀 ]……………………….………(4)
𝐴 𝐴
When binding energy fraction 𝐵/𝐴 is plotted against 𝐴, the curve similar to Fig.1 is obtained.
We find from this curve that 𝐵/𝐴 almost remains constant between 𝐴 = 30 and 𝐴 = 100 and
decreases for small and large values of 𝐴.
The coulomb repulsion between the protons, which obviously makes the nuclei less and less
stable, is what causes the reduction for large A. Because only a small number of other nucleons
are attracted to each particular nucleon in light nuclei, their separation distances are greater,
which again affects stability. The drop in B/A for small A is a surface effect. It is obvious that
34
35

the particles at the surface are less tightly bonded than those inside. stronger ties than those in the
interior across. The fraction of constituents at the nuclear surface increasewith the size of the
nucleus. This effect reads B/A for a low A.
The nuclides with an even number of protons and neutrons have greater B/A values than the
nearby odd mass nuclides, according to an analysis of the aforementioned curve. Since there are
no remaining unpaired neutrons in such nuclides, the even number of protons can couple off all
the nucleons, explaining the even-even nucleon's better stability and natural abundance.. This
low mass number nuclides, the stability rule is N = Z, 𝐻 3 , Li2 BH 2 , N14 are examples of odd-odd
examples of odd-odd nuclides which are most stable.
Example 1: Masses of helium nucleus, proton and neutron are 4.0026u, 1.007895u and
1.008665u. Find the energy required to knock out nucleons from the helium nucleus.
Solution: We have the equation
4
2 He → 2𝑝 + 2𝑛
Mass of two protons and 2 neutrons is

35
36

= 2 × (1.007895 + 1.008665)
= 4.03312𝑢
Therefore, mass defect is given as
Δ𝑚 = 4.03312 − 4.0026
= 0.03052𝑢
And the equivalent energy is
= (0.03052𝑢) × 931.49
= 28.42 MeV
2.3.3 Packing Fraction
We have seen that atomic masses are not whole numbers. This divergence of the masses of the
nuclides from whole number was studied by a number of workers like Aston and is expressed in
terms of packing fraction. Packing fraction is defined as
Atomic mass - Mass number

𝑓 =
Mass number
𝑧𝑀 𝐴 − 𝐴
= … … … … … … … … . (5)
𝐴
where 𝑀 𝐴 is the actual mass of a nuclide on the physical atomic mass scale and A mass number.
The quantity 𝑧𝑀 𝐴 − 𝐴 is known as Mass-defect'. If packing fraction is represented as 𝑓, then
𝑧𝑀 𝐴 = 𝐴(1 + 𝑓)………………….……………..(6)
When 𝑓 is plotted against 𝐴, we get the following curve of the shape as shown in fig.2
36
37
Fig 2: Plot of packing fraction(f) versus mass number(A)
The packing fraction for all the elements fall on this curve except He4, C14, and O16.For 𝑂16 the
value of packing fraction is zero. As 𝐴 increases, the falls of the packing fraction becomes
negative, passes through a flat minimum rises gradually becoming positive again at values of 𝐴
above 180 . A a positive packing fraction has an atomic mass 𝑀 greater than its mass nude whilh
This means that the loss of mass due to binding energy requirements, whera, nucleons combine
to form a nucleus, is less than that for oxygen. The packing fractions for light elements like
oxygen, carbon, helium which do not lie on the frac. have positive values but much smaller
(≈ 5 × 10−1 or less) than other light curve like nitrogen, lithium and hydrogen (≈ 80 × 10−4
for hydrogen ) light elements of the former type (O16 , C12 , He4 ) have more stable nuclei’s that
latter.
37
38
Example 1:
The value of fB can be estimated as follows:

For deuteron (H2), since Z=1, N=1,
( )
B.E H 2 = M H + M n − M d
= (1.007825 + 1.008665 − 2.0414102 )  931.5
= 2.224 MeV
( )
fB H 2 =
2.224
2
= 1.112 MeV per nucleon
For alpha-particle (He4), Z=2, N=2,
( )
B.E He 4 = ( 2 1.007825 + 2 1.008665 − 4.002603)  931.5
= 28.3MeV
(
f B He 4 = ) 28.3
4
= 7.075 MeV per nucleon
For (O16),since Z=8, N=8,
( )
B.E O16 = ( 8 1.007825 + 8 1.008665 − 15.994915 )  931.5
= 127.62 MeV
127.62
fB = = 7.98 MeV per nucleon
16
The binding energy fractions of the different nuclei represent the relative strengths of their
binding. Thus H2 is very weakly bound, compared to He4 or O16.
2.4 NUCLEAR REACTION
Typically, an equation representing a nuclear reaction may be written as
38
39

x+ X =Y + y ......(7)
In words this would read like: when an incident projectile x hits the target nucleus X, a nuclear
reaction takes place and as a result there is a new nucleus Y and an outgoing particle The above
reaction can also be written in short form as X (x, y) Y.
In 1919, Lord Rutherford discovered that protons are created when nitrogen is attacked
with polonium particles, and that these protons are capable of piercing 28 cm of air. Rutherford
deserves credit for piercing the "inaccessible armour" of typical non-radioactive nuclei and
causing a transmutation because this was the first nuclear reaction to be set off un a laboratory.
You might think of the nuclear reaction as
14
7 N ( , p ) 178 O ......(8)
In 1930 Cockcroft and Walton used artificially accelerated protons and produced the following
nuclear reaction.
7
3 Li + p → 24 He + 
7
3 Li ( p,  ) 24 He ......(9)
Thousands of nuclear reactions have previously been examined by various laboratories

throughout the world, and the number is constantly increasing. In this chapter, we'll look at the
scenario when the target nucleus is at rest before to the collision and all of the particles can be
considered non-relativistically. Finding out when a specific nuclear reaction becomes
energetically feasible is what we're interested in learning. By using the equations of momentum
and energy conservation, we may determine this.
2.5 TYPES OF NUCLEAR RECATION

Nuclear reactions are classified on the basis of the projectile used, the particle detected and the
residual nucleus.
39
40
i. Scattering: In the scattering reaction, the projectile and the detected (outgoing) particle
are the same. The scattering is elastic when the residual nucleus is left in the ground
state. When the residual nucleus is left in an excited state, the scattering is called
inelastic.
ii. Pickup reactions: When the projectile gains nucleons from the target, the nuclear
reaction is referred to as pickup reaction, e.g.
16
8 (
O 12 H , 13H ) 15
8 O
iii. Stripping reactions: In this type, the projectile loses nucleons to the target nucleus, e.g.
16
O
8 ( 4
2 He, 12 H ) 18
9 F
In a pickup or stripping reaction it is assumed that the nucleon involved in the process
enters or leaves (the shell of) the target nucleus without disturbing the other nucleons.
These reactions are therefore referred to as direct reactions. In contrast, we have the
following type compound nuclear reactions. Of
iv. Compound nuclear reactions: Here the projectile and target form a compound nucleus
which has a typical life span of 10−16 sec. When this time is compared with a typical
nuclear time, i.e., the time taken by the projectile to traverse the target nucleus ( 10−22
sec.) as in the case of direct reactions, we can conclude that the decay compound nucleus
does not depend on the way it was formed. of A
This situation is often described as: the compound nucleus does not "remember" how it was
formed. Usually, the same compound nucleus is given rise to by a number of nuclear
reactions. This compound nucleus can decay in a number of ways or channels. This is
64
illustrated by taking for example, the compound nucleus 30 Zn formed in an excited state
( )
*
64
30 Zn by two different methods and then decays as given below:
40
41
( )
*
p + 2963Cu → 64
30 Zn → 3063Zn + 01n
→ 3062 Zn + 01n + 01n
→ 62
29 Zn + 01n + p
( )
*
+ 60
20 Ni → 64
30 Zn → 63
30 Zn + 01n
→ 62
30 Zn + 01n + 01n
→ 62
29 Cu + 01n + p .....(9)
These reactions were experimentally studied by S.N. Ghoshal [Phys. Rev. 80939 (1950)]
Figure 4 depicts how a compound nucleus in an excited state can have different modes of decay,
giving rise to different residual nuclei and detected particles.
Fig 4: Modes of decay in a compound nucleus in an excited state
41
42

The compound nucleus model has been successfully applied for targets of A>10 and projectile
energies up to about 15 MeV. For projectile energies between 15 MeV and 50 MeV the direct
reaction model and the optical model are found to be successful. In this unit, our main emphasis
is on understanding energies of nuclear reactions.
2.6 THE BALANCE OF MASS AND ENERGY IN NUCLEAR

REACTIONS
Consider the nuclear reaction,
x+ X =Y + y
Since the total mass and energy is conserved, we have
(E x ) ( ) (
+ mx c2 + M X c2 = EY + MY c2 + Ey + my c 2 ) ......(10)
Where,
E x = kinetic energy of the projectile
mx c 2 = rest energy of the projectile
And similarly, EY and M Y c 2 , E y , my c 2 and M X c 2
The target nucleus X is assumed to be at rest.
The Q value is expressed as,
Q = EY + Ey − Ex ......(11)
i.e., it is the change in the total kinetic energy.
42
43

This change in the total kinetic energy in a nuclear reaction is clearly the nuclear
disintegration energy.
From Eq. (10) and Eq. (11) becomes,
Q = EY + E y − Ex
= ( M X + mx ) − ( M Y + m y )  c 2 .......(12)
Thus, we see that Q is also change in the total rest mass. We have an exoergic nuclear reaction if
Q is positive. The nuclear reaction is an endoergic reaction if the Q value is negative.
From Eq. (12) it is clear that to determine the Q value or nuclear disintegration energy for
a nuclear reaction, we must know the kinetic energy of the particles.
E y , the kinetic energy of the residual nucleus, is usually small and hard to measure.
In the following section, we will see how E y can be eliminated and how the Q equation
can be set up.
2.7 THE Q EQUATION

The analytical relationship between the kinetic energy of the projectile and outgoing particle and
the nuclear disintegration energy Q is called as the Q equation.
To understand the dynamics of two body nuclear reactions in laboratory coordinate

system, refer to Fig.5
43
44
Fig. 5: Two body nuclear reaction in lab system
Conservation of mass-energy gives
Q = EY + Ey − Ex
As remarked before, E y is small and hard to measure and is therefore eliminated.
Conservation of linear momentum along the direction of projectile x gives,
2mx Ex = 2my Ey cos  + 2M Y EY cos  ......(13)
Angles  and  are coplanar as linear momentum perpendicular to the  plane is equal to zero.
Therefore, conservation of linear momentum normal to projectile direction (in the  ,  plane)
gives,
0 = 2MY EY sin  − 2my Ey sin  .............(14)
From these equations E y and  are eliminated. Squaring and adding Eqs. (13) and (14),
M Y EY = mx Ex + my Ey − 2 mx my Ex Ey cos 
44
45

But EY = Q − Ey + Ex
mx my 2 mx m y E x E y
Q = E y − Ex + Ex + Ey − cos 
MY MY MY
 my   mx  2 mx m y E x E y
=E y  1 +  − Ex 1 − − cos  ......(15)
 M Y   M Y  M Y
This is the standard form of the Q equation.
It is interesting to note that one can also obtain Eq.15 by solving the momentum triangle and
substituting for E y . This is indicated below:
Fig 6: Momentum triangle, using the notation of fig. 05
We see that,
PY2 = px2 + p y2 − 2 px p y cos  ......(16)
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46
2MY EY = 2mx Ex + 2my Ey − 2.2 mx my Ex Ey cos 
giving,
mx my 2 mx my Ex E y
EY = Ex + Ey − cos  ......(17)
MY MY MY
From Eq. (6)
EY = Q − Ey − Ex
Substituting this in Eq (12) we get,
 my   mx  2 mx my Ex E y
Q =E y 1 +  − Ex  1 − − cos 
 MY   MY  MY
Which is the Q Equation (10).
The kinetic energy E x , E y and  all are measured in laboratory system. Since the Q
equation is based on mass-energy conservation in a nuclear reaction, it holds for all types of
reactions. The exact masses can be replaced by the corresponding mass numbers, in many
applications without significant error. For very accurate calculations, the neutral atomic masses
are used. Let us see how we can use the isotopic masses (neutral atomic masses) to obtain Q
value in alpha and beta decay reactions. For convenience, let the element symbol represent the
isotopic mass. We recall that Q = ( mx + M X ) − ( my + MY ) where the masses are the nuclear
masses and Q is in mass units.
i. Consider the reaction

U→
238
92 Th + 24 He
234
90
Q = (U − 92e ) − (Th − 90e ) + ( He − 2e )

46
47

Where e stands for electronic mass.
i.e.,
Q = (U − 92e ) − (Th + He − 92e ) 
= U − (Th + He ) 
∴ Isotopic masses can be used to evaluate Q.
ii. Consider the reaction,

C → 147 N + −10e + 00
14
6
Q = ( C − 6e ) − ( N − 7e ) + e 
= ( C − 6e ) − ( N − 6e )
=C−N
iii. Consider the reaction,

63
30 Zn → 2963Cu + +10e + 00
Q = ( Zn − 30e ) − ( Cu − 29e ) + e 

= ( Zn − 30e ) − ( Cu − 30e + 2e )
= Zn − ( Cu + 2e )
Thus, we see that the isotopic masses can be used to compute Q value in a positron decay
reaction, provided two electron masses are included with that of the product particle.
47
48
2.8 SOLUTION OF THE Q EQUATION
We are often interested in E y , the energy of the detected particle and its variation with E x , the
energy of the bombarding projectile, for a fixed Q. For this purpose, the Q equation (Eq. 10) can
be regarded as a quadratic in Ey . Then its general solution can be conveniently put in the form,
Ey =    2 + w ......(18)
Where,
mx m y E x
= cos  ......(19)
my + M Y
And,
M Y Q + E x ( M Y − mx )
w= ......(20)
my + M Y
When Ey is real and positive the reaction is energetically possible.
Emission of m y becomes energetically impossible when Q value is negative, ( M Y − mx ) is
negative (i.e., a heavy projectile) and a large angle of observation  , making cos negative.
Equation (13), for a various energy E x of the following particle, tells us about the types of
nuclear reactions which can occur.
Let us now consider energies of exoergic and endoergic reactions.
2.8.1 Exoergic Reactions: Here Q > 0.
a. Very low energy projectiles: e.g., thermal neutrons.

48
49

We can take E x  0
MY Q
v = 0 and w =
my + M Y
And so,
MY
Ey = Q ......(21)
my + M Y
i.e., E y is same for all angles 
An example of such reaction is,
C ( n,  ) 49 Be
12
6
b. Finite energy projectiles: For most of the reactions, mx M Y and so, w  0 for all E x .
Thus, it can be seen from Eq. (13) that E y is single valued for all E x , and is given by,
Ey = v + v2 + w ......(22)
An example of such a reaction is,
10
5 B ( , p ) 136 C and Q = +4.0 MeV
c. Double values of Ey: In a number of reactions E y is no single valued. For example, in
the reaction,
15
7 N ( d , n ) 168 O, Q  10 MeV
16
If the observed particle m y is chosen to be the residual nucleus O , then,
my mass number , AY = 16 and M Y − AY = 16 (for neutron).
Equation (15) gives,

49
50

Q − Ex
w=
17
Which is negative for all deuteron bombarding energies greater than,
Ex = Q = 10 MeV
Thus, for Ex > 10 MeV , we have two real positive values of E y at  = 0 .
That is, in the forward direction, there are two monoenergic groups of 16O nuclei. Physically this
implies that in the center of mass system, these two groups are projected forward and backward.
2.8.2 Endoergic Reactions: Q < 0 .
For every nuclear reaction with a positive Q value, the inverse reaction has a negative Q value of
the same magnitude. Thus, for example, 168 O ( n, d ) 157 N has Q = −10 MeV , or 136 C ( p,  ) 105 B has
Q = −4 MeV
a) Very low energy projectiles:

Ex  0, Q  0
v 2 + w2  0 and so...
Ey is imaginary, implying that no reaction occurs. ( E x is insufficient to start the
reaction.)
b) Threshold energy ( Ex )thresh : In an endoergic reaction, the energy -Q is needed to excite the
compound nucleus sufficiently so that it will break up. The bombarding particle must
supply this energy in the form of kinetic energy. However, not all of that kinetic energy is
available for excitation because some is used to give momentum to the compound
nucleus which, is distributed among the products of the reaction. Thus, the bombarding
particle (projectile) must supply some energy in addition energy in addition to - Q so that
the energy -Q becomes available for the excitation of the compound nucleus.
51

The smallest value of projectile energy (bombarding energy) at which an endoergic
reaction can take place is called the threshold energy for that reaction.
From Eq. 13, it can be seen that the reaction first becomes possible when E x is large
enough to make,
v 2 + w2 = 0
E x has its minimum possible value at  = 0 , which is the ( Ex )thresh .
This condition gives:
mx my Ex M Y Q + Ex ( M Y − mx )
+ =0
( my + M Y ) my + M Y
2
Which gives,
 m + MX 
( Ex )thresh = −Q  x  ......(23)
 MX 
In obtaining Eq. (18), we have made use of the fact that the value of Q is much less than the
c2
masses of particles in a nuclear reaction, and so
my  mx + M X − MY ......(24)
Equation (18) can be also obtained in a straightforward way by considering M c and Vc as mass
and velocity of the compound nucleus. Then we have,
mx vx = M cVc
mx
Or Vc = vx
Mc
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52

Now, the part of kinetic energy of the projectile needed for the excitation of the compound
nucleus is,
1 1
Q= mx vx2 − M cVc2
2 2
2
1 1 m 
= mx vx2 − M c  x  vx2
2 2  Mc 
1  m 
= mx vx2  1 − x 
2  Mc 
But
M c = mx + M X
1  Mx 
Therefore, −Q = mx vx2  
2  mx + M X 
The threshold energy is then,
1  m + MX 
( Ex )thresh = mx vx2 = −Q  x 
2  MX 
Which is Eq.23.
2.9 GLOSSARY
Kinetic Energy - the energy possessed by a body due to its motion.
Magnetic Moment - a vector quantity that provides a measure of the torque exerted on a
magnetic system (as a bar magnet or dipole) when placed in a magnetic field.
52
53

Mass Defect - the difference between the mass of an atomic nucleus and the sum of the masses
of its constituent particles.
MeV (Mega Electron Volt) – the energy possessed by a particle with one electronic charge in
passing through a potential difference of one million volts.
Nuclear Binding Energy – the minimum energy required to separate an atomic nucleus into it
constituent particles i.e., protons and neutrons.
Nuclear magneton - a unit of magnetic moment, used to measure proton spin and approximately
equal to 1/1836 times Bohr magneton.
Q value - the Q value for a reaction is the amount of energy absorbed or released during the
nuclear reaction.
Exoergic(exothermic)reaction - a nuclear reaction that releases energy and have positive Q-

value.
Endoergic (or endothermic)reaction – a nuclear reaction that requires an input of energy to take
place with negative Q-value.
2.10 SUMMARY
In this unit, we have explained various important terms like binding energy, mass defect, binding
energy per nucleon and packing fraction with suitable examples. In this consequence we have
defined the nuclear reaction and also its various types has also been explained. In this unit we
have also learnt that to determine the Q value or nuclear disintegration energy for a nuclear
reaction, we must know the kinetic energy of the particles. We have an exoergic nuclear reaction
if Q is positive. The nuclear reaction is an endoergic reaction if the Q value is negative. The
analytical relationship between the kinetic energy of the projectile and outgoing particle and the
nuclear disintegration energy Q is called as the Q equation has also been explained
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54

mathematically. In addition to this the solution of the Q equation has also been established in this
unit for better understanding of the topic.
2.11 REFERENCES
1 Nuclear Physics - Problem-based Approach including MATLAB, Hari M. Agrawal. PHI

Learning, Dehi
2 Modem Physics, Stephen T. Thorthon - Cengage Learning India, New Delhi
3 S.B.Patel, Introductory Nuclear Physics, New Age International Publishers.
4 Kenneth S.Krane, Introductory Nuclear Physics, Wiley, New York, 2008.
5 B.R.Martin, Nuclear and Particle Physics, Wiley, New York, 2006.
1 B.L.Cohen, Concepts of Nuclear Physics; Tata McGraw-Hill Publisher

2 S N Ghoshal, Nuclear Physics, S.Chand Publication
3 H.A.Enge, Introduction to Nuclear Physics, Addison-Wesley, London,1966.
4 M.L.Pandya and R.P.S.Yadav, Elements of Nuclear Physics, Kedar Nath Ram Nath,
Delhi.
5 I.Kaplan, Nuclear Physics, Narosa ,2002
6 S.S.M. Wong.,Introductory Nuclear Physics, Prentice Hall of India, New Delhi, 2005
1. What do you meant by nuclear binding energy.

2. Explain the terms mass defect and packing fraction.
54
55

3. Plot the general shape of the binding energy curve (Binding energy per nucleon versus
mass number A. Explain fission and fusion on the basis of this plot.
4. What do you mean by Q value of a nuclear reaction?
5. Write down the expression for Q value in the class of α decay.
56
6. Find the Binding Energy per nucleon for Fe.(Given that mass of Fe atom= 55.93493
amu).
7. Find the B/A for deuterium atom . Mass of deuterium atom(1H2) is 2.01402 amu.
8. Find the B/A for 7Li. Mass of lithium atom is 7.016004 amu.
9.The B/A for Deuterium and Helium are 1.1 MeV and 7 MeV respectively. What would be the
energy released when two deutrons are fused together to form a He nucleus.
55
56
UNIT 3
NUCLEAR FORCES I

3.1 Introduction
3.2 Objectives
3.3 Basic Understanding of Nuclear Forces
3.4 Basic properties of deuteron
3.4.1 Binding Energy
3.4.2 Size
3.4.3 Spin of Deuteron
3.4.4 Magnetic Dipole Moment
3.4.5 Quadrupole Moments
3.5 Existence of excited states of deuteron

3.6 n-p scattering at low energies with specific square well potential
3.6.1 What is Scattering
3.6.2 Neutron – Proton Scattering at Low Energies
3.7 Results of Low Energy n- p and p-p Scattering
3.8 Glossary
3.9 Summary
3.10 References
56
57
3.1 INTRODUCTION
In the previous unit we have studied the basic information about binding energy which refers to
the stability of nucleons. Now in this unit we shall study about the forces that binds the nucleon
together i.e. nuclear force.in this consequence we known that large number of nuclei, available in
nature, are stable. Now the natural question is what bounds the nucleons together in the nucleus
to make it stable? It is further known that most of the nuclei consist of more than one proton –
the positive charged particle; it implies that there exists in the nucleus forces which are strong
enough to overpower the coulomb repulsion and hold the nucleons together. The forces which
hold the nucleons together are commonly called nuclear forces and are short range forces as it is
evident from the fact that binding energy is proportional to the number of constituent nucleons.
3.2 OBJECTIVES
The main aim of studying this unit is to understand the nuclear forces which is responsible for
binding the nucleons. For understanding the nuclear force, we will study the Basic properties of
deuteron viz, its binding energy, its size, spin, magnetic and quadrupole moments etc. After
going through this unit learners should be able to:
• Understand the various properties of deuteron.

• Analyze the existence of excited states of deuteron with the solution of spherically
symmetric square well potential for higher angular momentum states.
• Learn n-p scattering at low energies with specific square well potential.
• Comparatively study the results of low energy n-p and p-p scattering.
3.3 BASIC UNDERSTANDING OF NUCLEAR FORCES

The nuclear forces cannot be of electromagnetic origin because nuclear forces involve
uncharged particle while electromagnetic forces do not. The purely magnetic forces cannot
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58

account the binding energy of about 8 MeV per nucleon (1.1 MeV per nucleon in deuteron) in
nuclei in general and it is clear that magnetic forces are some hundred times smaller than nuclear
forces.
The gravitational forces too cannot explain the existence of such huge forces because they are
about 1036 times smaller. It implies that nuclear forces cannot be of any of the type discussed so
far.If we compare the stable nucleus with the atom then we find that stability of the atom is
governed by predominant central particle but in the case of nucleus there is no predominant
central particle. The forces which hold together the different nucleons should have mutual forces
between the individual nucleons in the ensemble. It turns out that the nuclear force are strange
and of intriguing nature.
Now let us turn to the nuclear interaction which according to Yukawa’s theory, may be conceived as due
to the exchange of a relatively massive particle - the π- meson or pion with a mass approximately 270
times that of an electron. Thus, nuclear interaction is about 1038 times stronger than the gravitational
interaction and about 1000 times than the electromagnetic interaction and so comes under what are
called ‘strong interaction’.
• We then infer that none of the only two interactions, encountered previously, is able to
account for the existence of nuclei.
• The only way out then is to recognize the existence of another fundamental interaction –
the nuclear interaction (force).
• Since nuclei are composed of protons and neutrons only which are packed very densely
within the small volume of the nucleus, the heavier nuclei will be subjected to very strong
Columbian repulsive force – the one acting between the positively charged protons,
which tends to tear the nucleus apart.
• The fact that nuclei stay as bound systems even in the take of these strong repulsive.
• Columbian forces, is a sufficient proof of the great strength of the nuclear forces and
that at distances of the order of nuclear dimensions, it should be attractive in nature .
As a general rule, the wave nature of matter (quantum mechanical principles) is relevant where the de
Broglie wavelength of the particles is of the order of the size of the system to be studied.
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59

So let us compare nuclear size with the wavelength of a nucleon of energy 10 MeV.
h h 6.6256  10=27
= = = ……….(1)
( 2M N E )
1/2
MNv  2  1.67  10−24  10  1.6  10−19 
(
= 9.3 10−23 cms. = 9.3Fermi 1Frtmi = 10−13cms . )
Which is obviously of the size of the nuclei and hence quantum mechanical considerations are
indeed relevant to the study of nuclei.
Having ascertained that nuclei are quantum mechanical systems composed of nucleons, it is quite
plausible to study the nuclear forces under the simplest possible conditions.
➢ The simplest case in which the nuclear force is effective is when there are only two
experimentally achievable situations:
1- When the two nucleons are bound together
Of the three possible bound states of a two – nucleon system, di-neutron (nn), di-proton
(pp) and deuteron (np), nature has provided us with only the deuteron and the other two are
unstable.
2- When the two nucleons are in free state and one is made to impinge on the other, i.e.
The scattering processes.
In practice, it is not possible to make a neutron target and therefore scattering experiments are
limited only to neutron proton (np) scattering and proton-proton (pp) scattering.
3.4 BASIC PROPERTIES OF DEUTRON
The deuteron consists of a neutron and proton, having charge equal to proton +e, mass
2.014735 atomic mass units and it obeys the Bose-Einstein statistics.
The experimentally measured properties of deuteron are:

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60
3.4.1 Binding energy
Binding energy . (Experimental) = 2.225±.003 Mev.
The binding energy of deuteron can be determined from a number of experiments.
The easiest one comprise of allowing slow neutrons to be captured by protons in a

material containing hydrogen i.e. hydrogenous substances such a paraffin, plastic etc. and
measuring the energy of the emerging  rays. The reaction is called (n-p) capture
reaction and may be written as
0 n1 +1 H 1 →1 H 2 + 
Because the neutron carries no charge, the nuclear force binding the deuteron cannot be
electrical. It can also not be to MN = 1.67×10-24 gm. to provide a 2.225 MeV binding
energy. Therefore the binding force is of nuclear origin .
3.4.2 Size:
Deuteron radius: The root- mean – square value of deuteron radius is 2.1 Fermi.
3.4.3 Spin:
Spin. 1 (in units of h).
3.4.4 Magnetic dipole moment:
The magnetic dipole moment of deuteron is d = 0.85735 ± 00003 nuclear magneton.
In a structure made up of particles, one expects the total magnetic moment to be the vector sum
of the magnetic moments due to the orbital motion of the charged particles.
60
61

Applying quantum mechanics to describe deuteron we may reasonably assume the ground state
of deuteron to be an S state for which the angular momentum L = 0. With L=0, the wave function
 is spherically symmetrical and for the S state the angular momentum quantum number l=0,
no contribution from orbital motion is expected to spin.
• The nucleons are half spin particles and deuteron is known to have a spin equal to unity
which implies that the proton and neutron spins are parallel.
• In such a case the proton and neutron magnetic moments should also add up.
• Experimental measurements for nucleon magnetic moments give the following values:
▪ Magnetic moment of proton  p =2.79281 ±0.00004 nm.
▪ Magnetic moment of neutron  n =-1.913148±0.000066 nm.
▪ Sum of the two moment ( p + n ) = 0.879662±0.00005 nm.
Thus deuteron is expected to have a magnetic moment of 0.8797±.00015 nm. However from
experimental measurements deuteron magnetic moment is found to have a value 0.85733 ± .0002
nm between the expected and the measured values which is difficult to explain.
The simplest interpretation being the at deuteron possesses some orbital motion and that our
previous assumption of l=0 in the ground state is not correct.
Even this approximate agreement is valid only for the S state . Others give values very much
different from the measured value.
Thus magnetic moment measurements of deuteron establish the following important conclusions.
(1) In the ground state of deuteron, the proton and neutron spins are parallel (triple
state 3S1.)
(2) Neutron is a half spin particle.
In the ground state of deuteron, the orbital angular momentum is zero
(l=0, S=1 state ).

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62
3.4.5 Quadrupole moments:
The electric quasrupole moment of deuteron as measured by Rabi et al in a radio frequency

molecular bean method is
Qd = 2.82×10-27 cm.2 or 0.00282 barn,
Which although small but is not zero.
Alternately it can be put as to give
average z2 for proton
1 z2
(1.14) = 2 ……………..(2)
3 r
average r2 for proton
It implies that charge distribution in the ground state is not spherically symmetric because
a spherical chare distribution needs a value for Qd=0
z2 1
Or for the = .
r2 3
The result also indicate that charge distribution is of prolate shape, i.e., elongated along
the z axis. The electric quadruple moment and the magnetic moment discrepancy cannot
be explained by assuming the state to have some other value of l.
➢ This suggests that the wave function contains a mixture of l values. Since total angular
momentum is equal to the vector sum of the orbital and spin angular moments i.e.
J = L + S ………………..(3)
62
63

1
In a system like deuteron which consists of one proton and one neutron, each having a spin ,
2
the spin quantum number S be given by :
1 1
𝑆 = |2 ± 2|= 1 or 0………….(4)
For L=1 and a maximum value of S=1. from equation l can have only the values 0, 1 and 2 . But
the conservation of parity demands that even and odd values of l should not be simultaneously
present in the same wave function and therefore with l=0 , only l=2 can be present.
The wave function then may be written as:
 = a0 1s + a2 1d ……………(5)
This means that the system spends a fraction |𝑎0 |2 of its time in l=0 state and a2 2
of its time in
an l=2 state. Therefore the ground state may be taken to be a mixture of 3S1 and 3D1 states.
The magnetic moment and electric quadrupole moment discrepancies can be fully accounted for
with a2 = 0.96 and a2 = 0.04 . This means that deuteron spends 96% of the time in an l = 0
2 2
state and only 4% of the time in an l = 2 state.
We therefore infer that the deuteron is not in a purely spherically symmetric state. However, we
shall assume that the ground state function is spherically symmetric one .
3.5 EXISTENCE OF EXCITED STATE OF DEUTRON
Extending the calculations of the bound state to cases where the orbital angular quantum number
l is greater than zero leads to a result that deuteron cannot exist in these states.
For the extreme case, binding energy EB  0 , kr0 is still only slightly greater than  / 2 , since the
binding energy EB of the ground state has already been found negligible compared to the
potential well depth V0 . For the first excited state kr0 would have to be greater than 3 / 2 , since
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64

the wave function u(r) would have to have a radial node inside the well. But from equation: kr0
must certainly be less than  for all positive values of binding energy.
We shall here prove that for (l  0) no bound state exists. It shall be assumed that the potential is
central and of square well type in this case.
The differential equation used for this case (l  0) , is which through the substitution
u (r ) = r (r ) takes the form as shown below.
𝑑2 𝑢(𝑟) 𝑀 𝑙(𝑙+1)ℎ2
+ ℎ2 [𝐸 − 𝑉(𝑟) − ] 𝑢(𝑟) = 0. ………………(6)
𝑑𝑟 2 𝑀𝑟 2
V(r)
Fig. 1. Deuteron wave function of the first excited states
When we compare these equations, we find that it is analogous to an S-wave radial equation with
potential form
l (l + 1)h 2
Ve ff (r ) = V (r ) + ……………………..(7)
Mr 2
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65

Here the second term on R.H.S. is called the centrifugal potential as its space derivative gives the
classical centrifugal force. This potential is repulsive, there forces ‘l’ increases, the binding
energy of the lowest bound state decreases.
Coming back to equation(6) and setting l= 1, the next acceptable value of l after 0, we have ,
𝑑2 𝑢(𝑟) 𝑀 2ħ2
+ ħ2 [𝐸 − 𝑉(𝑟) − 𝑀𝑟 2 ] 𝑢(𝑟) = 0 ……….(8)
𝑑𝑟 2
Now E=-EB’ the binding energy of deuteron in the P- state (l=1) and using a square well
potential V (r) = V0’ for r < r0, for the p- state , equation may be written as
𝑑2 𝑢(𝑟) 𝑀 2ħ2
+ ħ2 [𝑉0 ′ − 𝐸𝐵 ′ − 𝑀𝑟 2 ] 𝑢(𝑟) = 0 for r  r0 ……(9)
𝑑𝑟 2
and
𝑑2 𝑢(𝑟) 𝑀 2ħ2
+ ħ2 [𝐸𝐵 + 𝑀𝑟 2 ] 𝑢(𝑟) = 0 for r  r0 ……..……..(10)
𝑑𝑟 2
Now assuming
𝑀 𝑀𝐸′𝐵
𝑘′ = √[ħ2 (𝑉0 ′_𝐸𝐵 ′)] and 𝛾′ = √( ). …….......(11)
ħ2
The above equation (11) may be written as
𝑑2 𝑢(𝑟) 2
+ [𝑘′2 − 𝑟 3] 𝑢(𝑟) = 0 for r  r0 …………..(12a)
𝑑𝑟 2
and
𝑑2 𝑢(𝑟)
− [𝛾′2 +
2
] 𝑢(𝑟) = 0 for r  r0 …………..(12b)
𝑑𝑟 2 𝑟3
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The least well depth, just repaired to produce this bound state, is the one for which the binding
energy EB’ is just equal to zero, i.e., when 𝛾′ = 0 and 𝑘′ = √(𝑀𝑉0 ′/ℎ2 ) = 𝑘0 .
If we put k0r = x, the wave equation reduces to
𝑑2 𝑢(𝑟) 2𝑢(𝑟)
+ 𝑢(𝑟) − =0 for 𝑥 < 𝑘0 𝑟0 ……………….(13a)
𝑑𝑥 2 𝑥2
𝑑2 𝑢(𝑟) 2𝑢(𝑟)
and − =0 for 𝑥 > 𝑘0 𝑟0 ……………….(13b)
𝑑𝑥 2 𝑥2
The solution of equation(13b) with the correct boundary condition becomes
𝑢(𝑟) = 𝐴𝑥 −1 for 𝑥 > 𝑘0 𝑟0 , …………………………………...(14)
To solve equation(13a), we make the substitution v = xu(r), so that
𝑑𝜐 𝑑𝑢(𝑟)
=𝑥 + 𝑢(r)
𝑑𝑥 𝑑𝑥
𝑑2 𝜐 𝑑2 𝑢(𝑟) 𝑑𝑢(𝑟)
and =𝑥 +2
𝑑𝑥 2 𝑑𝑥 2 𝑑𝑥
and equation (13a) can then be re-written as follows
𝑑2 𝜐 2 𝑑𝜐
− 𝑥 𝑑𝑥 + 𝜐 = 0 for 𝑥 < 𝑘0 𝑟0 .
𝑑𝑥 2
Differentiating this equation with respect to x, we get
𝑑3 𝜐 2 𝑑2 𝜐 2 𝑑𝜐 𝑑𝜐
− 𝑥 𝑑𝑥 2 + 𝑥 2 𝑑𝑥 + 𝑑𝑥 = 0for 𝑥 > 𝑘0 𝑟0.
𝑑𝑥 3
Dividing this equation by x throughout , we get
1 𝑑3 𝜐 2 𝑑2 𝜐 2 𝑑𝜐 1 𝑑𝜐
− 𝑥 2 𝑑𝑥 2 + 𝑥 3 𝑑𝑥 + 𝑥 𝑑𝑥 = 0 for 𝑥 < 𝑘0 𝑟0. ………………(15)
𝑥 𝑑𝑥 3
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𝑑2 1 𝑑𝜐 1 𝑑3 𝜐 2 𝑑2 𝜐 2 𝑑𝜐
Now since ( ) = 𝑥 𝑑𝑥 3 − 𝑥 2 𝑑𝑥 2 + 𝑥 3 𝑑𝑥′
𝑑𝑥 2 𝑥 𝑑𝑥
then the equation(15) may be re-written as
𝑑2 1 𝑑𝜐 1 𝑑𝜐
( ) + 𝑥 𝑑𝑥 = 0 for 𝑥 < 𝑘0 𝑟0. …………. …..(16)
𝑑𝑥 2 𝑥 𝑑𝑥
Now since u(r)=vx-1, must vanish for x = 0, the solution of above equation is found to be
1 𝑑𝜐
= 𝐴1 sin x for x < k0ro.
𝑥 𝑑𝑥
Integrating it, we get
𝜐 = 𝑥𝑢(𝑟) = 𝐴1 (sin x-x cos x) for x < k0r0, …..(17)
Equation (14) and (17) respectively provides the solution for outside and inside the well. To
satisfy continuity condition at the boundary (r=r0 or x=k0r0), these solutions should be matched
at the boundary. Which yields,
𝑑
(sin x-x cos x ) = 0 at x = k0r0
𝑑𝑥
Or x sin x = 0 at x = k0r0
Or k0r0 sin k0r0 = 0 at x = k0r0 ………………………………(18)
The smallest positive root of this equation is k0r0=  . Hence a bound state of the deuteron for l
 can exist only if k0r0<  and this contradicts the previous statement that k0r0<  . Therefore
we conclude that no bound states exist for deuteron when l  , i.e., deuteron does not possess any
excited state.
_________________________________________________________
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3.6 n-p SCATTERING AT LOW ENERGIES WITH SPECIFIC

SQUARE WELL POTENTIAL
3.6.1 What is Scattering
When an intense and collimated beam of nucleons is bombarded on target nuclei the interactions
between incident nucleus and target nuclei takes place.
As a result we may observe the following two possibilities:
CASE I :-
(i) The interactions does not change the incident particles, i.e., incoming and
outgoing particles are the same.
(ii) The change is in the path of incoming nucleons, i.e., they are deviated from
their original path. This process is known as scattering,
(iii) In scattering processes the outgoing particles may have same energy as that of
incident particles or may have the changed energy value. The former is known
as elastic scattering and latter is known as inelastic scattering.
CASE II:-
The second possibility is that the outgoing particles are different from the incident particles.
Then the interaction process is known as nuclear reaction.
In nuclear reaction we have two alternatives:
(a) The incident material particles are fully captured by the target and instead  -radiations (  -
photons) are emitted. The situation is termed as radiative capture (e.g., n-  -reaction).
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69

(b) In the second alternative the outgoing particles are either charged particles or some other
material particles which are the product of the process itself, then the process is known as
nuclear reaction.
It should be remembered that any of the above alternative may occur, either alone or with other
competing processes. Among the nucleon-nucleon scattering, neutron proton (n-p) scattering is
the simplest one, because here the complication due to coulomb forces are not present.
• In (n-p) scattering neutron proton system is analyzed in the state of positive energy, i.e.,
in a situation when they are free.
• In the experiment, a beam of neutrons from an accelerator is allowed to impinge on a
target containing many essentially free protons.
• The simplest substance is hydrogen gas but in some cases other substances like thin nylon
sheet and paraffin are used. Hence, it is natural to think that in target protons are not free
but are bound in molecules.
• The molecular binding energy is so small about 1eV, therefore, for the impinging
neutrons of energy greater than 1eV, protons are treated as free.
• The presence of electrons also do not affect the process because they are too light to
cause any appreciable trouble to incoming neutrons.
• When neutrons impinge on protons, some of them are captured to form deuteron and
balance of energy is radiated in the form of  rays; but the great majority of neutrons
undergo elastic scattering.
• In the process, the interactions between two nucleons is of such a order that the neutrons
changed their velocities in magnitude as well as in direction.
The proton - proton (p-p) scattering is due to the presence of coulomb repulsion between two
protons.
• The presence of coulomb repulsion increases the change of direction the account of
which is made in estimating the nuclear forces.
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70

• It appears that nuclear force between the protons is not sufficiently strong to bind the
protons against coulomb repulsion. (It is supported by the fact that no bound state with
two protons He3 exists).
• It will be seen later that p-p system remains unbound even in the absence of coulomb
repulsion.
The neutron neutron (n-n) scattering is not practically possible because of the non availability of
neutron target (because neutron decays into proton in a few minutes).
However, their are evidences to support if n-n forces are similar to p-p forces, a bound state for
two neutrons cannot exist.
3.6.2 Neutron – Proton Scattering at Low Energies

In the low energy range, most of the measurements of scattering cross section are due to
Melkonian and Rainwater et.al. A beryllium target bombarded at by deuterons accelerated in a
cyclotron, provided the neutron beam which was shot at a target containing free protons.
Fig. 2. n-p scattering cross section
• These results show that the scattering cross section depends very much on the energy of
the incident neutrons.
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• At low energies below 10 MeV, the scattering is essentially due to neutrons having zero
angular momentum (l=0) and hence in the Centre of mass system, the angular distribution
of scattered neutrons is isotropic.
• In order to avoid complications due to Coulomb forces we shall consider the scattering of
neutrons by free protons viz. those not bound to molecules.
• However in practice the protons are of course bound to molecules but the molecular
binding energy is only about 0.1 eV.
• Therefore if the incident neutrons have an energy greater than about 1eV. The protons
can be regarded as free.
• In describing elastic scattering events like the scattering of neutrons by free protons it is
more convenient to use the center of mass system.
The quantum mechanical problem describing the interaction between two particles, in the
center of mass system, is equivalent to the problem of interaction between a reduced mass
such as the system.
Although while wording out the following theory we shall think in terms of a neutron being
scattered by a proton but it applies equally well to spin less, reduced mass particle which is
being scattered by a fixed force center.
Let us suppose that the neutron and the proton interact via a spherically symmetric force field
whose potential function is V (r), where r is the distance between the particles .
The Schrodinger equation for a central potential V (r) in the center of mass system , for the n-p
system is
𝑀
[𝛻 2 + ħ2 {𝐸 − 𝑉(𝑟)}] 𝜓 = 0 ………. …(19)
Where M is the reduce mass of the n-p system .
To analyze the scattering event, we have to solve this equation under proper boundary
conditions.
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72

• The action in the immediate region of the scattering centre will be violent and difficult to
describe.
• Things will be easier, though, if the experimenter waits for the scattered particles at a
sufficient distance from the scattering centre.
• For scattering, the boundary condition states that the wave should have two components
when it is far from the scattering centre.
(i) an incident plane wave that describes the un scattered particles and superimposed
upon it,
(ii) an outgoing scattered spherical wave which emanates from the scattering center.
To solve in asymptotic form,
 =  inc + sc …………………(20)…
The wave function that describes an incident plane wave (a beam of particles ) moving in the
positive z-direction is
 inc = e ikz = e ikz cos  , …………………..(21)
 ME 
Where k =  
 h 
2
Which is a solution of the wave equation with V (r) set equation zero,
 2 ME 
 +  inc = 0,
 h2 

……………….(22)
eliminate scattering so that the total wave function becomes identical with the incident wave
function  inc .
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73

• The wave function represents one particle per unit volume since the square of the wave
function is equal to unity.
• Having known the form of the incident wave function, the next problem is to devise a
suitable form for the scattered wave function .
This obviously is
e ikr
 sc = f ( ) ,
……………(23)
r
• For large ‘r’ f (  ) in this expression indicate amplitude of the scattered wave in the
direction  . This wave function is a necessary consequence of the assumption that the
scatterer simply scatters the particles and does not absorb them at all.The probability
density and hence the number of scattered particles per unit volume shall be proportional
to  sc .
2
• If scattering is considered to be isotropic, the density (number per unit volume) of

scattered particles through a large spherical shell of radius r is inversely proportional to r2
since the volume of the spherical shell, being given by 4r 2 dr, is proportional to r2 and
density therefore is proportional to 1/r2 which is also proportional to  sc Hence 1/r2

2
dependence of  sc .
Therefore the wave function  , in a form we are actually interested viz. asymptotic, may be
written as.
eikr
 =  inc + sc = e + f ( ) .
iks
………(24)
r
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74

Now, in Fourier analysis we often expand as arbitrary function into a sense of harmonic
functions of various frequencies.
So we expand the incident plane wave function eikz in terms of Legendre Polynomials Pl (cos  )
and write.
𝜓𝑖𝑛𝑐 = 𝑒 𝑖𝑘𝑟𝑐0𝑠𝜃 = ∑∞
𝑙=0 𝐵𝑙 (𝑟)𝑃𝑙 (𝑐𝑜𝑠 𝜃)……………….(25)
Where l is an integer representing the various partial waves. This particular way of writing the
wave function is termed as the partial wave expansion.
The radial functions Bl (r) in the above equation are given by
𝐵𝑙 (𝑟) = 𝑖 𝑙 (2𝑙 + 1)𝑗𝑙 (𝑘𝑟), …………………(26)
Where Jl(kr) is the Spherical Bessel function which is related to the ordinary Bessel function
through the formula
𝜋 1/2
𝑗𝑙 (𝑘𝑟) = (2𝑘𝑟) 𝐽𝑙+1/2 (𝑘𝑟) ……………(27)
and can be represented as
1 𝑑 𝑙 𝑠𝑖𝑛 𝑘𝑟
𝑗𝑙 (𝑘𝑟) = (−𝑘𝑟)𝑙 [𝑘𝑟 𝑑(𝑘𝑟)] ( ) ………….(28)
𝑘𝑟
Whence asymptotically
𝑙𝜋
𝑠𝑖𝑛(𝑘𝑟− )
𝑗𝑙 (𝑘𝑟)𝑟→∞ → 2
……………(29)
𝑘𝑟
Asymptotically, Bl(r) from is given by
𝑙𝜋
𝑠𝑖𝑛 (𝑘𝑟 − 2 )
𝑙
𝐵𝑙 (𝑟)𝑟→∞ → 𝑖 (2𝑙 + 1)
𝑘𝑟
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75

1 −𝑖(𝑘𝑟−𝑙𝜋/2)
≅ 2𝑖𝑘𝑟 𝑖 𝑙 (2𝑙 + 1). [𝑒 𝑖(𝑘𝑟−𝑙𝜋/2)−𝑒 ]. ………..(30)
The Spherical Bessel function Jl(kr) for various values of l are
3 1 3 𝑐𝑜𝑠(𝑘𝑟)
𝑗2 (𝑘𝑟) = [(𝑘𝑟)3 − 𝑘𝑟] 𝑠𝑖𝑛( 𝑘𝑟) − (𝑘𝑟)2
…………….(31)
These functions are plotted in the Fig 3.
Similarly f (  ) can also be expanded in terms of the Legendre Polynomials as follows
𝑖
𝑓(𝜃) = 2𝑘 ∑∞
𝑙=𝑜 𝑓1 (2𝑙 + 1)𝑃1 (𝑐𝑜𝑠 𝜃). ……………(32)
Substituting the values in equation(24) we have
𝑒 𝑖𝑘𝑟
𝜓 = 𝜓𝑖𝑛𝑐 + 𝜓𝑠𝑐 ≈ ∑∞ 𝑙
𝑙=0 [𝑖 (2𝑙 + 1)𝑗𝑙 (𝑘𝑟) + 𝑓𝑙 ] 𝑃1 (𝑐𝑜𝑠 𝜃). ………….(33)
𝑟
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Fig.3. Variation of Bessel function with orbital angular momentum quantum number.
Due to the fact that each term in the sum [equation 33] with a particular value of the orbital
angular momentum quantum number 'l' reflects a solution of the wave equation in spherical polar
coordinates for constant potential energy. Because the majority of the scattering occurs at
energies below 10 Mev and the number of partial waves is severely constrained in this situation,
it is sufficient to study the scattering only for l=0 particles, or S-waves, the expansion classifies
the particles in the beam according to their angular momenta.
From equation (30) by putting l=0, we have
𝑠𝑖𝑛(𝑘𝑟) (𝑘𝑟)2
𝐵0 (𝑟) = ≈1− +. . . . . .. …………..(34)
𝑘𝑟 6
And for l=1
𝑠𝑖𝑛( 𝑘𝑟) 𝑐𝑜𝑠( 𝑘𝑟)

𝐵1 (𝑟) = 3𝑖 [ − ]
𝑘𝑟 𝑘𝑟
 kr (kr )3 
 3i  − + ...... .
 3 30 
B1 (r )
  (kr) 2 .
B0 (r )
Since Bl2(r) determines the probability density, we instead determined the ratio of the square (r)
B1 ( r )
instead of just . Consider a neutron with energy 1 MeV in the L-system compared to 0.5
B0 ( r )
MeV in the C-M system to get a sense of the magnitude of this ratio. then, neutron momentum is
 2 1.67 10−24 1.6 10−6 

1
p = (2ME ) = 2
 ………..(35)
 2 
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77
= 1.6310−15 gm.cm. / sec.

Now the wave number number is given by
p 1.6310−15 −1
k= = = 1.55  1012
cm.
h 1.054510−27
If we assume the nuclear forces to have a range r = 2 Fermi, then
2
B1 (r )
B0 (r )
( )
 (kr) 2 = 1.55 1012  2 10−13 = (0.31) ,
2
= 0.0961,
i.e., only 9% of the scattering at 1 Mev is caused by neutrons with l=1. Similar calculations
increase this amount to roughly 49% for a neutron with an energy of 10 MeV. In the energy
range below 10 Mev, thus. The major scattering type is S-wave (l=0).
3.7 RESULTS OF LOW ENERGY n- p and p-p SCATTERING
The theory for the scattering cross-section that was constructed in the previous section is actually
a theory for the phase shift, which in turn depends on the assumptions about the characteristics of
the scattering potential V. (r). To provide an example, we now carry out the computations using
the same rectangular potential well that was presupposed in earlier sections for the deuteron
ground state.
Equation inside and outside the nuclear square potential well, which is the radial Schrodinger
equation for l=0, can be represented as
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𝑑2 𝑢(𝑟) 𝑀
+ ħ2 (𝐸 + 𝑉0 )𝑢(𝑟) = 0 for r < r0,
𝑑𝑟 2
𝑑2 𝑢(𝑟) 𝑀
+ 2 𝐸𝑢(𝑟) = 0 for r > r0 (36)
𝑑𝑟 2 ħ2
Since the negative binding energy is substituted in the current case of n-p scattering by a low
positive energy E that is substantially smaller than the well-depth V0. These calculations can be
expressed as
𝑑2 𝑢𝑖
+ 𝐾 2 𝑢𝑖 = 0 for r < r0, ……………….(37a)
𝑑𝑟 2
𝑑2 𝑢0
+ 𝑘 2 𝑢0 = 0 for r > r0, ………………(37b)
𝑑𝑟 2
Where ui is the wave function inside the well and u0 that outside the well and
𝑀(𝐸+𝑉0) 𝑀𝐸
𝐾2 = , 𝑘2 = . ………(37c)
ℎ ℎ2
Equation(37a) has the solution
ui = A sin Kr …….……..(38)
Equation(37b) has the solution
u0 = C sin kr + D cos kr, ………………..(39)
Which may be written as
u0 = B sin(kr +  0 ). …………………..(40)
To understand the significance of the phase shift  0 , the Schrodinger equation would V(r) set
equal to zero, the solution of which is of the form
u ( r ) = sin kr. ……………. (41)
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Because it must disappear when r = o. the only viable solution outside of the well. This is how
turning on the scattering potential causes the phase shift at great distances.
We now require that the solution (38) and (40) join smoothly at r = r0 i.e. the logarithmic
derivative must be continuous at r=r0 viz,
1 dui 1 du0
r = r0 =
ui dr u0 dr r = r0
This condition, with equation (38) and (40) gives
K cot Kro = k cot(kr0 +  0 ) ………… ( 42)
By comparing this result with the continuity condition equation for the ground state of the
deuteron viz.
√{𝑀(𝑉0 − 𝐸𝐵) }
√{𝑀(𝑉0 − 𝐸𝐵 )}
𝑐𝑜𝑡 𝑟0 = −𝛾
ħ ħ
([ ] )
We assume that inside the well, the scattering wave function is not significantly different from
the deuteron wave function in order to simplify the matching condition in the case of n-p
scattering. This seems quite acceptable given that the sole difference between the two scenarios
is that while the total energy E in this case is little, it is positive, whereas the deuteron binding
energy EB is also small, but it is negative. We therefore suppose that the value of the logarithmic
derivative of the ground state wave function of the deuteron, -γ, might be used to estimate the
logarithmic derivative K cot Kr0 of the inside wave function for scattering. Hence from (42)
𝑘 𝑐𝑜𝑡( 𝑘𝑟0 + 𝛿0 ) = −𝛾 ……………..(43a)
At this point we introduce another approximation that r0 is very small (possibly zero) compared
to 𝑘 = √(𝑀𝐸/ħ) so that kr0 may be neglected in the above equation and then
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𝑘𝑐𝑜𝑡𝛿0 = −𝑦 or cot 0 = − y / k ……………..(43b)
Now the total scattering cross-section for l=0 from equation is given by
4𝜋 4𝜋 1 4𝜋 1
𝜎𝑠𝑐,0 = 2
𝑠𝑖𝑛2 𝛿0 = 2 . = 2.
𝑘 𝑘 (1 + 𝑐𝑜𝑡 𝛿0 ) 𝑘 (1 + 𝑦 2 /𝑘 2 )
2
4𝜋 4𝜋ℎ2
= 𝑘 2 +𝑦 2 = 𝑀(𝐸+𝐸 ). …………………(44)
𝐵
Where we have substituted the values of k2 and y2 from equations (37). The relation was first
arrived at by E.P. winger which although agrees with experimental results at high energies but
fails miserably at low energies.
3.8 GLOSSARY
Nucleon: The proton and the neutron, constituting atomic nuclei.
Bound state: A quantum state of a particle subject to a potential such that the particle has a
tendency to remain localized in one or more regions of space.
Binding Energy: Amount of energy required to separate a particle from a system of particles or
to disperse all the particles of the system.
Magnetic moment: Magnetic Moment is defined as magnetic strength and orientation of a

magnet or other object that produces a magnetic field.
Electric Quadrupole Moment: A parameter which describes the effective shape of the ellipsoid
of nuclear charge distribution.
Scattering : In physics, a change in the direction of motion of a particle because of a collision

with another particle
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3.9 SUMMARY
After going through this unit, you would be able to achieve the following objectives. Now we
recall what we have discussed so far.
• We have learnt the basic properties of deutron, its charge (+e), mass (~2.014 amu), its
radius (2.1 fermi), its binding energy (=2.225  .003 Mev), Spin (1  ) and statistics
(Bose-Einstein) and the electric quadrupole moment Qd =0.00282 barn .
• The study of deuteron problem, although hopelessly limited in as much as deutron
possesses only the ground state and no-excited states exist for the bound neutron-proton
system, gives invaluable clues about the nature of the nuclear force.
• We learnt that neutron and proton can form stable combination (deuteron) only in the
triplet state means when the n & p spins are parallel. The singlet state, i.e. a state of anti
parallel n-p spins being unbound.
• The existence of non-zero magnetic moment and electric quadrupole moment for deutron
suggests that at least a part of the neutron proton force acting in deutron is non-central.
• The nuclear forces are spin dependent i.e., nuclear forces not only depend upon the
separation distance but also upon the spin orientations of two nucleons. They are
independent of the shape of nuclear potential.
3.10 REFERENCES
1. Elementary Nuclear Theory, 2nd ed. by Bethe and Morrison, Wiley: New York.
2. The Atomic Nucleus by R D. Evans, McGraw-Hill Publications.
3. Atomic and Nuclear Physics by Brijlal and Subhraininyan.
4. Nuclear Physics by D. C Tayal.
5. Nuclear Physics by Irving Kaplan, Narosa Publishing House

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1.”The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of Physics, Berlin: Springer
Ver.
2. “Nuclear Two Body Problems and Elements of Nuclear Forces” Experimental
Nuclear Physics by N. F. Ramsey, Wiley: New York.
3. Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum
Press, New York.
1. Discuss the n-p and p-p scattering at low energies.

2. Solve the Schrodinger equation for the deuteron in a S-state under the assumption of
square well potential.
3. Show that deuteron has no excited state.
4. Write short note on scattering length.
5. Explain clearly how the properties of the deuteron indicate the presence of spin dependent
force and tensor force between two nucleons.
6. Briefly explain the properties of nuclear forces.
7. If the binding energy of the deutron were 12 MeV , what would be roughly the depth of
the potential well, assuming it to be square?
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UNIT 4
NUCLEAR FORCES II

4.1 Introduction
4.2 Objectives
4.3 Spin Dependence
4.4 Scattering Length
4.5 Pauli Principle and Antisymmetrisation
4.6 Isospin
4.7 Two Nucleon system
4.8 Generalized Pauli Exclusion Principle
4.9 Meson Theory of Nuclear Forces:(Yukawa Theory)
4.9.1 Estimation of mass of meson using uncertainty principle
4.9.2 Yukawa Potential
4.10 Summary
4.11 References
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4.1 INTRODUCTION
The nucleons that make up the nucleus have two important properties. Firstly, you cannot
distinguish one proton/neutron from another proton/neutron. All protons are indistinguishable
and all neutrons are indistinguishable. We can extend this to also say that, as far as strong or
nuclear interaction properties are concerned, no nucleon is distinguishable from any other
nucleon. This charge independence symmetry is at the root of the concept of isospin. This is a
seemingly obscure quantity but it allows simplicity and transparency when specifying
wavefunctions and has some useful properties giving rise to useful selection rules
4.2 OBJECTIVES
After studying the unit learners will be able to

• Know the spin dependence and scattering length.
• Explain meson theory of exchange forces.
• Understand Pauli exclusion principle
• Differentiate between symmetric and antisymmetric wave functions
• Describe Isospin
• Solve questions based on Isospin
• Explain the Yukawa potential
4.3 SPIN DEPENDENCE
E. P. Wigner postulated that spin effects the internucleon forces. Since the neutron and proton
are both spin particles, their spins can either be parallel or antiparallel during n-p scattering. The
neutron and proton spins are parallel in the bound state of the n-p system in deuterium, which has
a binding energy of EB, suggesting that this equation may remain true in the parallel spin case.
The triple state of parallel spins has a statistical weight of three, which corresponds to the three
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permitted orientations of the angular momentum vector in the presence of an outside magnetic
field. Because a vector with zero length cannot be oriented, the state of antiparallel spins is a
singlet state with a statistical weight.
1- The singlet and triplet states of the n-p system will occur in scattering experiments in
proportion to the statistical weight factors for these states, which are and respectively. This is
because neutron and proton spins are generally randomly oriented in scattering experiments, as
are the spins of neutrons in the incident beam. As a result, the overall scattering cross-section
will consist of two components: the scattering cross-section for the triplet state and the scattering
cross-section for the singlet state, as shown below.
3 1
σ0 = 4 σt + 4 σs ……….(45)
From a naive perspective, the two spins are equally likely to be parallel and antiparallel in a
random distribution of spins, such as in n-p scattering, giving the two states identical statistical
weight. The phrase "spin pointing up" merely indicates that the spin vector is pointing
somewhere in a cone that is centered on the vertical direction because, according to quantum
mechanics, the spin direction cannot be defined as uniquely as a vector in space. The four
equally likely configurations for the respective spins of the two particles are schematically
shown in the following figure.
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(1)
(2)
(3)
(4)
Fig. 5 Spin dependence
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The total spin unity shown in Figures (1) and (4) corresponds to the magnetic quantum number
values +1 and -1, respectively. Since the spins are not aligned along the z-axis, the z-components
in examples (2) and (3) may add up to zero, resulting in a singlet state, or they may add up to a
total spin perpendicular to the z-axis, resulting in a triplet state.
4.4 SCATTERING LENGTH
Fermi and Marshall introduced a very useful concept the ‘scattering length a’ for the discussion
of nuclear scattering at very low incident neutron energy.
 ME 
[i.e.E → 0 and hence k =  2  → 0
 h  ……..(46)
Which may be defined as follows:
𝑠𝑖𝑛 𝛿0
𝑎 = 𝐿𝑖𝑚 (− ); ……………..(47)
𝑘→0 𝑘
By this definition, equation which gives the total scattering cross section for S-wave (l=0) can be
written for very low incident neutron energy as
4𝜋 𝑠𝑖𝑛2 𝛿0
𝐿𝑖𝑚(𝜎𝑠𝑐 , 𝑜) = 𝐿𝑖𝑚 ( ) = 4𝜋𝑎2 ………….(48)
𝑘→0 𝑘→0 𝑘2
Equation therefore shows that "a" has the dimensions of length, hence the name scattering
length, and has the geometric significance of being the radius of a hard sphere around the
scattering center from which neutrons are scattered.
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Now, it should be noticed from equation that as the energy E of the incident neutron approaches
0, it must also approach either 0 or else the cross-section at zero neutron energy would become
infinite, which is physically absurd. Therefore, at very low incident neutron energies ( E → 0 ),
equation reduces to
0
a=− ………….(49)
k
Then at very low incident neutron energies, the wave function outside the range of nuclear force
as expressed by equation may be written as
 sin(kr +  0 
Lim U (r ) = Lim(r 0 ) = Lim e i 0 
k →0 k →0 k →0
 k 
The scattering length is then simply represented graphically by the equation. The scattering
length 'a' is the intercept on the r-axis and this equation depicts a straight line for U (r). Fig. 6
illustrates this.
The relevance of positive or negative scattering length is that it informs us what is the
significance of attaching a positive or negative sign with at the scattering length, an inquisitive
reader may inquire quite naturally after we have defined the scattering length using equations.
whether a bound or unbound state exists in the system.
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Fig. 6. Graphical interpretation of scattering length
It is evident from Fig. 6 that a positive scattering length denotes a bound state while a negative
scattering length denotes an unbound or virtual state.
Since the wave function for the bound state of the n-p system, known as the deuteron wave
function, must curve towards the r-axis in order to match the exponentially decaying solution
(c.f. equation), r>r0 will result in a positive intercept on the r-axis, indicating a positive scattering
length. When extrapolating U(r), the wave function for an unbound state must match an
increasing solution outside of the range r0, and this must result in a negative intercept on the r-
axis, signifying a negative scattering length
4.5 PAULI PRINCIPLE AND ANTISYMMETRISATION

The nucleons that make up the nucleus have two important properties. Firstly, you cannot
distinguish one proton/neutron from another proton/neutron. All protons are indistinguishable
and all neutrons are indistinguishable. We can extend this to also say that, as far as strong or
nuclear interaction properties are concerned, no nucleon is distinguishable from any other
nucleon. This charge independence symmetry is at the root of the concept of isospin.
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Nucleons are fermions and therefore should obey the Pauli principle. We can guarantee this if we
construct wavefunctions which are antisymmetric with respect to particle exchange.
Consider for now a simple two-particle system with no interactions between the two identical
particles. Under these conditions the Hamiltonian is simply the sum of the Hamiltonians for the
two particles alone:
𝐻ˆ = 𝐻ˆ1 + 𝐻ˆ2
You can easily show by substitution that a simple product wavefunction, Ψ = 𝜑𝑎 (𝑟1 )𝜑𝑏 (𝑟2 ),
satisfies the Schrodinger equation, 𝐻ˆ 𝜓 = 𝐸𝜓, if 𝐸 = 𝐸1 + 𝐸2 . Here the a and 𝑏 represent the
quantum numbers of the two single-particle states, and 𝑟1 and 𝑟2 are the spatial coordinates of the
two nucleons.
There are problems with writing the wavefunction as 𝜓 = 𝜑𝑎 (𝑟1 )𝜑𝑏 (𝑟2 ). It suggests that you can
label the first nucleon and distinguish it from the second; if you exchange the first for the second
nucleon you get a different state 𝜓 = 𝜑𝑎 (𝑟2 )𝜑𝑏 (𝑟1 ). Also if the two single-particle states are the
same, 𝑎 = 𝑏, the wavefunction becomes 𝜓 = 𝜑𝑎 (𝑟2 )𝜑𝑎 (𝑟1 ), rather than zero as expected by the
Pauli principle.
We can construct something which obeys Pauli by forming the wavefunction:
1
𝜓= [𝜑𝑎 (𝑟1 )𝜑𝑏 (𝑟2 ) − 𝜑𝑎 (𝑟2 )𝜑𝑏 (𝑟1 )]
√2
By substitution, show that this wavefunction satisfies the Schrodinger equation for the two-
particle system and that it is properly normalised.
It is clear that in this particular case if 𝑎 = 𝑏, the wavefunction is zero and therefore the Pauli
principle is obeyed. If you perform a particle exchange operation on this wavefunction, where
you swap the first for the second (1 → 2) nucleon and the second for the first (2 → 1) nucleon,
you end up with the final answer −Ψ. This wavefunction is therefore antisymmetric under
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particle exchange. Generally, wavefunctions must be antisymmetric to be consistent with the
Pauli principle.
In general, the Hamiltonian depends on both spatial and spin components. Consider a simple
separation of spin and spatial coordinates can be performed such that the wavefunction is written
as a product of a spatial and a spin part, Ψ𝑋. In order to be consistent with the Pauli principle, it
is the overall wavefunction that must be antisymmetric. This means that either the spatial part is
antisymmetric combined with a symmetric spin part, or the spatial part is symmetric and the spin
part is antisymmetric.
Consider the possible spin wavefunctions of two identical independent nucleons. They both have
spin-1/2 and so the system can couple to give and overall spin of 𝑆 = 0 or 1.
Let's take the 𝑆 = 1 possibilities first.
To have 𝑆 = 1, we must have wavefunctions with 𝑚𝑆 = +1,0 and −1.
𝑀𝑆 = 𝑚𝑠1 + 𝑚𝑠2 .
For 𝑀𝑠 = 1 we have 𝑚𝑠1 and 𝑚𝑠2 = +1/2.
For 𝑀𝑠 = −1 we require 𝑚𝑠1 and 𝑚𝑠2 = −1/2.
If we interchange 1 for 2 and visa versa for these wavefunctions we would see that they are
symmetric under particle exchange since each nucleon has the same quantum numbers. The
𝑀𝑆 = 0 wavefunction could be made from either 𝑚𝑠1 = 1/2 and 𝑚𝑠2 = −1/2, or 𝑚𝑠2 = +1/2
and 𝑚𝑠1 = −1/2.
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4.6 ISOSPIN or ISOTOPIC SPIN
.Protons and neutrons are identical as far as the operation of the strong nuclear force is
concerned. Here are a few observations that indicate that we could start to treat them as two
different types of the same particle:
(a) The proton-neutron mass difference is very small.
(b) In low-energy ( <5MeV ) np and pp scattering, the cross section for the 1S channel is equal
to within a few percentage if Coulomb effects are corrected.
(c) If we exchange all the protons in a nucleus to neutrons, and all the neutrons to protons, and
generate its mirror system. Apart from small Coulomb effects and the n-p mass difference, the
energy levels are remarkably similar. The nuclear interactions are therefore unchanged if n - n
forces are exchanged for p p forces and the nuclear force must be CHARGE SYMMETRIC.
(d) If we exchange protons, one by one for neutrons gradually and generate a sequence of
nuclei of the same mass but different numbers of protons and neutrons. States in nuclei in this
isobaric sequence have analogues in the other members of the sequence. The nuclear interactions
between n-p and p-p and n-n are therefore the same and the nuclear force must be CHARGE
INDEPENDENT.
30
An example was discussed before in Si,30P and 30
S where analogue states were seen in all
30
three. Some states though were unique to P and we will see that these are states which are
allowed in an np system but forbidden in the pp or nn cases by Pauli.
This is the origin of the name nucleon. Heisenberg was the first to treat protons and neutrons as
two substates of the same particle, the nucleon back in 1932. This is a two substate system, rather
like intrinsic spin- 1/2 which can have substates m_s=+1/2 or -1/2, hence the name of this new
characteristic, isospin.
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A nucleon has isospin I=1/2; a proton has Iz=-1/2 and a neutron has Iz=+1/2. This z doesn't refer
to real space anymore; isospin operates in a fictitious new space called isospin space. All the
quantum mechanical mathematics of spin operators, Pauli spin matrices, angular momentum
coupling and raising/lowering operators can be applied to isospin in a completely analogous way.
4.7 TWO-NUCLEON SYSTEM
Now we take the two nucleons and couple their isospin together. Each nucleon carries I=1/2, so
we can couple them together to make I =0 or 1 . The third components couple according to Iz=
Iz1+ Iz2. A system of states is then generated completely analogous to the spin components
discussed above. The T=0 state is a singlet with Iz=0, and it is antisymmetric under particle
exchange. The I=1 coupling is a triplet of states with Iz = +1,0, and -1 and is symmetric under
particle exchange. What does all this mean?
We have two nucleons so we have generated nuclei with A=2. The third component tells us
something important. Protons individually have Iz=-1/2 and neutrons Iz=+1/2, so the third
component is all wrapped up in what electrical charge the system has. The quantity, -(Iz-1/2),
tells us the charge on a nucleon. In a multinucleon system, the overall charge can be found by
summing -(Iz-1/2) over all the nucleons. Doing this gives you, Z=-(Iz-A/2). For the A=2 nuclei,
Iz=+1,0, and -1 must therefore correspond to Z=0,1 and 2 , or equivalently two neutrons (a
dineutron), the deuteron and two protons ( a diproton). Notice that in the dineutron and diproton,
there is no I=0 state since it is prevented by the Pauli principle. But in the deuteron, both I=0 and
I=1 states are available.
In addition, for the A=2 case we would expect that the spatial wavefunction is going to be s(1/2)
for both nucleons. The spatial wavefunction in this case cannot be antisymmetric, and only a
symmetric version is non-zero. Prove this by looking at the spatial two-particle wavefunctions
above and substitute in the same single-particle states. To get an overall antisymmetric
wavefunction, if the spin part is symmetric, the isospin part must be antisymmetric, or visa versa.
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So the I=1 (symmetric) state must have J=0 (antisymmetric). The T=0 state must have J=1 for
the same reason.
Combining this with the knowledge of the deuteron we have from before, namely the ground
state is J=1 and all other states are unbound and at higher energies, we can draw a diagram
summarising all of this.
We've used charge independence to tell us that, apart from Coulomb effects and pn mass
differences, the 𝐼 = 1 states should be at the same energy.
We would expect that the diproton and the dineutron are not bound nuclei. Two identical
nucleons cannot stick together under nuclear forces.
Notice that for the deuteron two isospin couplings are possible, 𝐼 = 0 and 𝐼 = 1. We see
experimentally that the 𝐼 = 0 coupling is lowest. It turns out that for most nuclei, many isospin
couplings are available, but usually the one that is lowest in energy is 𝐼 = 𝐼𝑧 .
Take a nucleus composed of 𝐴 nucleons, with 𝑍 protons and 𝑁 neutrons. The 𝐼𝑧 of the nucleus is
(𝑁 − 𝑍)/2. Many values of the isospin are available by coupling together the isospin of each
1
nucleon i.e. couple 2 , 𝐴 𝑡𝑖𝑚𝑒𝑠. The maximum value of the isospin is the A/2. I cannot be lower
than Iz.
30
For example, S has 𝐴 = 30, 𝑍 = 14 and 𝑁 = 16
𝐼𝑧 = 1/2(𝑁 − 𝑍) = +1
Maximum 𝐼 = 15
Minimum 𝐼 = 1
Ground-state 𝐼 = 1
4.8 GENERALIZED PAULI EXCLUSION PRINCIPLE

The Pauli exclusion principle says that no two identical fermions can simultaneously occupy the
same quantum state. Generalized Pauli Exclusion Principle All fermions and particles derived
from fermions, such as protons and neutrons, obey Fermi-Dirac statistics; this includes obeying
the Pauli exclusion principle. Quarks (up and down) and leptons (electrons, electron neutrinos,
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muons, muon neutrinos, taus, and tau neutrinos) are all fermions. Particles obeying the exclusion
principle have a characteristic value of spin, or intrinsic angular momentum; their spin is always
some odd whole-number multiple of one-half.
The Pauli exclusion principle does not apply to bosons: these are particles that obey Bose-
Einstein statistics; they all have integer values of spin. Photons, gluons, gravitons, and the W, Z
and Higgs bosons are all bosons
Quantum numbers should be considered as address of each electron within an atom, each address
has four components, and no two electrons can have the exact same address. The Pauli exclusion
principle does not apply to bosons: these are particles that obey Bose-Einstein statistics; they all
have integer values of spin. Photons, gluons, gravitons, and the W, Z and Higgs bosons are all
bosons. This Pauli Exclusion principle states that no two electrons in an atom can have the same
four quantum numbers. If two electrons occupy the same orbital, they must have different spins.
Pauli’s Exclusion Principle limits the numbers of electrons that a shell or a subshell may contain.
4.9 MESON THEORY OF NUCLEAR FORCES:(YUKAWA

THEORY)
The range and power of the nuclear force were topics covered in the previous section. The
experimental understanding of the deuteron served as the foundation for this information. Instead
of providing us with a theoretical representation of the nuclear force, it informs us about some of
its characteristics. Any theory regarding it hinges on whether the deuteron system's spin is 0 or 1
and whether the orbital momentum is even or odd. Why should the nuclear force be dependent
on these? The well-known Yukawa theory of the exchange of mesons by nucleons offers the
solution. This meson exchange produces exchange forces. Let's examine the fundamental aspects
of Yukawa's concept.
(1) Nuclear force originates from exchange of a particle with a non -zero rest mass called meson.
(2) Each nucleon is surrounded by a meson field as a result it continuous emits and absorbs
mesons. This exchanger of meson gives rise to the nuclear force rand exchange of meson leads to
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constant transfer of momentum from one nucleon to other and hence the force is exerted between
them.
(3) The emission of meson ought to reduce the mass of nucleon. However this is not observed.
Thus, exchange of meson must takes place in such a short time that the uncertainty in the energy
is in consistence with the Heisenberg's uncertainty principle Δ𝐸 ⋅ Δ𝑡 ≈ ℏ
This makes the detection of meson exchange impossible and so the meson in this exchange
process is referred to as virtual mesons.
(4) The meson responsible for strop g nuclear forces are 𝜋-mesons (𝜋 + , 𝜋 0 , 𝜋 − ). The force
between p n is due to 𝜋 + and 𝜋 −
𝑝 → 𝑛 + 𝜋4
}
𝑛 → 𝑝 + 𝜋−
𝑝 → 𝑝 + 𝜋0
} ………………(50)
𝑛 → 𝑛 + 𝜋0
4.9.1 Estimation of mass of meson using uncertainty principle
If a nucleon emits a virtual meson 𝜋 of rest mass 𝑀𝜋 , it will loose an amount of energy 𝑀𝜋 𝑐 2
Δ𝐸 = 𝑀𝜋 𝑐 2 𝑝 → 𝑝 + 𝜋0 ……………(51)
Now the proton mass in above transition must decrease which is not observed in the experiment
This violation of conservation of energy is understood if the meson is absorbed again in a short
time Δ𝑡 such thai Δ𝐸. Δ𝑡 ≈ ℏ. So, the complete reaction is given as
𝑝 → 𝑝 + 𝜋 0 → 𝑝, …………………….(52)
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Thus, the emission and absorption of 𝑎 particle with rest energy 𝑀𝜋 𝑐 2 is consistent with
uncertainty principle. If the particle encounters the same or another nucleon in time Δ𝑡 given by
ℏ
(Δ𝑡 ≃ 𝑀 2
). Even if we assume that meson travels at the maximum possible speed that is the
𝜋𝑐
speed of light (c),then the maximum distance it can cover in this time is 𝑐.Δ𝑡 .
ℏ
𝑅 = 𝑐.Δ𝑡 = ………… (53)
𝑀𝜋 𝐶
Mass of meson is
ℏ
𝑀𝜋 =
𝑅𝐶
6.64 ×10−34 ⁄2𝜋

𝑀𝜋 = 2 𝜋 ×1.2 ×10−15 ×3×108
≈ 270 𝑚𝑒 ………………..(54)
Table 01: Comparison of masses, spin and magnetic moment for pions
𝝅+ 𝝅𝟎 𝝅−
𝑴𝝅 273.2 𝑚𝑒 264.4 𝑚𝑒 273.2 𝑚𝑒
Spin 0 0 0
Magnetic 0 0 0
Moment
From the equation (54), a mass of about 270𝑚0 (where 𝑚0 is the trest mass of an electron) for
the meson mass 𝑀𝜋 .One can estimate the experimentally observed value of 𝑅 by putting rest
mass 𝑀𝜋 of the meson.
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ℏ
𝑅∼ ∼ 10−15 m = one Fermi
2 × (270𝑚0 )𝑐
So, we have estimated roughly that the nuclear force is due to the process of meson exchange
then its range is of the order of a few fermis, which agrees well with the experiments.
4.9.2 Yukawa Potential
The discovery of 𝜋-meson predicted by Yukawa was a triumph of the meson theory and an
important landmark in our quest to understand the nuclear force. Yukawa was also able to write
an approximate expression for the potential energy of interaction between two nucleons:
𝑒 −𝑟/𝑅
𝑉 = −𝐶 ………………..(55)
𝑟
where 𝐶 is a constant, 𝑟 is the internucleon distance and 𝑅 is the range of force.
The Yukawa potential has been found to be successful in the discussion of deuteron problem and
also in understanding the low energy nucleon scattering data. The constant 𝐶 occupies a place in
the Yukawa meson theory, similar to that of charge 𝑒 in the electromagnetic theory.
In other words, the electromagnetic potential energy 𝑉 = 𝑒𝜙 and in an analogous manner the
Yukawa potential energy is 𝑉 = √𝐶∅. We can also show the meson filed equation which can be
written in the same way as electromagnetic field equations are
1
∇2 ∅ − ∅=0 ………..(56)
𝑅2
With the solution
𝑒 −𝑟⁄𝑅
∅ = √𝐶 ………………..(57)
𝑟
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Which have the potential
𝑒 −𝑟⁄𝑅 𝑒 −𝑟⁄𝑅
∅ = √𝐶 =𝐶 ………………..(58)
𝑟 𝑟
Fig.7: Yukawa Potential.
The Yukawa mesons were predicted to interact with nuclei quite strongly. It took around 12
years for physicists at Berkeley, University of California, to make the discovery of the mesons
responsible for the strong nuclear interaction. These are the referred to mesons (pi mesons). It
has been discovered that there are three different kinds of mesons or pions
Salient Features of Nuclear Forces:
(1) It is short range attractive force.
(2) It is in general a non-central force.
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(3) They have property of saturation only i.e each nucleon interacts only with its nearest
neighbors and not with all the constituents in the nucleus.
(4) They are charge independent i.e. 𝑛 − 𝑛, 𝑛 − 𝑝, 𝑝 − 𝑝 have same nuclear forces.
(5) They are spin dependent.
(6) They are exchange forces proposed by Yukawa.
4.10 SUMMARY
After going through this unit, you would be able to achieve the following objectives
• Know the spin dependence and scattering length.

• Understand Pauli exclusion principle
• Differentiate between symmetric and antisymmetric wave functions
• Describe Isospin
• Solve questions based on Isospin
• Explain the Yukawa potential .
• Studied the meson theory of nuclear forces as proposed by Yukawa with experimental
verified results predicted for nuclear forces.
4.11 REFERENCES
3. Atomic and Nuclear Physics by Brijlal and Subhraininyan.
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1.”The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of Physics,Berlin:

SpringerVer.
2. “Nuclear Two Body Problems and Elements of Nuclear Forces” Experimental Nuclear
Physics by N. F. Ramsey, Wiley: New York.
3. Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum Press, New
York.
1. Explain meson theory of exchange forces in detail..

2. What do you understand by Pauli exclusion principle?
3. Differentiate between symmetric and antisymmetric wave functions with examples.
4. Describe Isospin and discuss its utility.
5. Explain the Yukawa potential
6. Show that deuteron has no excited state.
7. Write short note on scattering length.
8. Explain the nature of Yukawa Potential.
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UNIT 5
NUCLEAR MODELS
5.1 Introduction
5.2 Objectives
5.3 Liquid drop model
5.4 Semi Empirical Mass Formula
5.4.1 Formula for the total binding energy of a nucleus
5.4.2 Weizsacker’s Semi Empirical formula
5.5 Nuclear Shell Model
5.5.1 Experimental evidences of Shell Model
5.5.2 Predictions about Shell Model
5.5.3 Spin-Orbit Coupling
5.5.4 Achievements and Applications of Shell model
5.6 Collective Model of the Nucleus
5.7 Glossary
5.8 Summary
5.9 Reference
5.11 Terminal Question
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5.1 INTRODUCTION
When we use any nuclear model, there are numerous significant facts that need to be explained.
How well a model's predictions are supported by experiments determines how valuable it is.
Understanding the nature of the internucleon reaction is essential for comprehending the
observable features of the nucleus. According to Yukawa's hypothesis, there is a highly powerful
short-range force that acts between nucleons when the distance between the points is less than
the interaction's range. This force is based on the exchange of pions between two nucleons.It
should be noted that even if the precise nature of the inter-nucleon interaction had been
understood, it would have been extremely challenging to construct a suitable explanation of the
structure of the nucleus because the Schrodinger equation cannot be precisely solved for a
system with that many bodies. Different models for the nucleus have been developed, each of
which can explain some of the attributes of various categories of nuclei, due to the
aforementioned challenges in constructing a satisfactory theory of the nucleus' structure.
The liquid drop model and collective model are based on the similarity to a drop of liquid, and
the Fermi gas model and shell model are based on the similarity to a weakly interacting gas.
Some limiting characteristics of the nucleus can be explained by the various nuclear models that
have been periodically offered. The nucleons are thought to influence and interact significantly
only with their close neighbors in the liquid drop model of the nucleus.
Thus, the nuclear binding energy found to vary with mass number and heavy nuclei fission may
be explained by the liquid drop model. The vast spacing of low-lying excited states in nuclei,
however, could not be accounted for by the liquid drop model. This and other characteristics of
the nucleus would force us to think about how individual nucleons move inside a potential well,
leading to the formation of a nuclear shell structure like to the electronic shells in an atom.
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It is very likely that the nucleons in the nuclei are grouped in distinct shells, just as it is the case
with the binding of the electrons in the atoms.
The nuclear shell model, developed in 1949 by Maria Mayer and separately by Haxel, O. Jensen,
J.H.D., and Suess, H.E., was able to explain all the magic numbers under the assumption that
individual nucleons had strong spin orbit forces. The development of the j-j coupling scheme
was a crucial step that led to a series of independent particle states that matched the empirically
discovered magic numbers.
The shell model is quite effective at predicting the excited states of nuclei and explaining the
extent to which many nuclei are non-spherical, in addition to providing proof of the magic
numbers. This suggests that, despite the shell model's reasonable representation of a nucleus, it is
still a simplified model. A nuclear model, however, becomes an effective plan when it can
explain certain experimental data.
5.2 OBJECTIVES
After studying this unit, you will be able to
• Understand liquid drop model

• derive the semi empirical formula
• Find the stable nucleus
• discuss the role of shell model
• understand magic number s
• Solve numerical based on shell model
5.3 LIQUID DROP MODEL
The liquid drop model of the nucleus was put forth by Neils Bohr and Wheeler in their theory of
nuclear fission, which is based on Gamow's idea of a potential barrier. According to this model,
an electrically charged, incompressible liquid drop with a mass that varies but has a very high
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density of 107 kg/m3 can be used to represent an element's nucleus. Additionally, it is assumed
that the nucleons in the nucleus are free to interact with their nearest neighbors, similar to how
liquid molecules are free to move about a given intermolecular distance from their nearest
molecules.
The nuclear forces have a number of characteristics, including a short range and a tendency to
saturate. These forces, which resemble the characteristics of the forces that keep the liquid drop
together, were inferred from the linear relationship between the binding energy and the volume
of the number of particles in the nucleus. This theory has been used to support the analogies
between the molecules in the liquid drop and the nucleons in the nucleus, which have been
shown to be comparable.
These are the following assumptions on which liquid drop model is based upon
(i) surface tension force of a liquid is analogous to the nuclear forces.
(ii) the constant binding energy per nucleon is analogous to the latent heat of vaporization.
(iii) the disintegration of nuclei by the emission of particles is analogous to the evaporation of the
molecules from the surface of liquid.
(iv) the energy of nuclei corresponds to internal thermal vibrations of liquid drop molecules and
(v) the formation of compound nucleus and absorption of bombarding particles correspond to the
condensation of liquid drops.
5.4 SEMI EMPIRICAL MASS FORMULA (SEMF)
The mass and binding energy of the nucleus have received a considerable deal of attention, thus
a formula that would enable the computation of these quantities would be extremely useful and
informative. Weizascker devised such a formula, which is also known as the semi-empirical
mass formula or the semi-empirical binding energy formula. Bohr and Wheeler were able to
establish the stability limit against spontaneous fission using the Weizsacker formula.
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5.4.1 Total binding energy formula of the nucleus
The total binding energy of a nucleus can be constructed by the following five energy terms
(i) Volume energy

(ii) Surface energy
(iii) Coulomb energy
(iv) Asymmetry energy and
(v) Pairing and shell energy.
Various contributions to the binding energy of a nucleus are shown below.
(i) Volume energy
This first term that contributes mainly to the binding energy of the nucleus comes from a term
which is proportional to the mass A, which can be shown as
𝐸𝑉 = 𝑎𝑉 𝐴 ………(1)
Since the volume of the nucleus is also proportional to A, may be regarded as volume energy. In
Eqn.(1) 𝑎𝑉 is a constant called volume co-efficient.
(ii) Surface Energy
The nucleons on the surface of the drop interact only with half as many other particles as that
done by the particles in the interior of the nucleus So the binding energy is reduced because this
nuclear surface. In Eqn.(1) it has been assumed that all nucleons are being attracted uniformly
to all sides and hence it is necessary to subtract a term proportional to the surface area of the
nucleus.
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The surface energy is proportional to the surface area of the nucleus, which is assumed to be
spherical shape i.e.,
𝐸𝑆 𝛼 4𝜋𝑅 2 ……(2)
where R is the nuclear radius.
𝑅 = 1.07𝐴1/3 fermi
1 fermi = 10−15 m
By substituting the above values the expression for the surface energy of the nucleus
𝐸𝑆 = -a𝑆 𝐴2/3 (3)
Where, aS is called the surface co-efficient and can be evaluated using empirical data.
(iii) Coulomb Energy
The binding energy tends to decrease due to the Coulomb energy between the protons, and this
tendency is represented by a term with a minus sign. Numbers represent the amount of work
required to construct a nucleus against Coulomb repulsion. Considering the nucleus to be a
uniformly charge spherical shell with a constant charge
 Ze 
density  =  3
upto their radius r.
 4 / 3 r 
The work dW which is required to bring the spherical shell up to its radius r is given by
1 4
𝑑𝑊 = [ × 4𝜋𝑟 2 𝑑𝑟 𝜌] ……….(4)
4𝜋𝜀0 𝑟 3𝜋𝑟 3 𝜌
Now the Coulomb repulsion energy Ec can be calculated by integrating the Eqn.(4) between the
limits r=0 and r = R. Thus Eqn.(4) can be written after integrating and substituting for the value
of  . W e have
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3/5𝑍 2 𝑒 2
𝑊 = 𝐸𝐶 = ......(5)
4𝜋𝜀0 𝑅
the total Coulomb energy of a nucleus of charge Z is given by
−3/5𝑍(𝑍 − 1)𝑒 2
𝐸𝐶 = ......(6)
4𝜋𝜀0 𝑅
The negative sign here shows the disruptive nature of this term Eqn.(6) now takes the form
(𝑍 − 1)
𝐸𝐶 = −4𝑎𝑐
𝐴1/3
3𝑒 2
where, 𝑎𝐶 =
5(4𝜋𝜀0 )𝑅0
where  0 represent the permittivity of free space.
(iv) Asymmetry energy
This term arises because in heavy nuclei, the number of neutrons (N) are greater as compared to
the number of protons (Z).On the other hand in light nuclei N = Z and these nuclei are highly
stable.
Without the Coulomb effect, a deviation from Z = A/2 would typically result in instability and a
lower binding energy value.
As the number of neutrons rises, the nucleus becomes more asymmetrical, which brings about a
force that lowers the volume energy.
The size of this effect might be expressed as a term proportional to the square of the neutron
excess (A-2Z) over protons. The symmetry effect is also inversely proportional to A, as
established by a detailed analysis of it. Now
(𝑁 − 2𝑍)2
𝐸𝐴 𝛼 ……(7)
𝐴
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Or
-a𝐴 (𝑁 − 2𝑍)2
𝐸𝐴 = ......(8)
𝐴
where aA. called the asymmetry co-efficient.
(v) Pairing Energy
Nucleons have a tendency to exist in pairs. Hence the nuclei with even number of protons and
even number of neutrons are the most stable and the most abundant in nature. Nuclei with odd
numbers of both protons and neutrons are the least stable, while nuclei for which either the
proton or neutron number is even are not so much stable.
This pairing effect can be shown by
𝐸𝑝 =-a𝑝 𝐴−3/4 ......(9)
It was empirically determined by Fermi. Eqn.(9) is called the pairing coefficient.
5.4.2 Weizsacker's Semi-Empirical Mass formula
Total binding energy (B.E) can be expressed by combining all the terms which are contributing towards
the binding energy of the nucleus
B.E. = 𝐸𝑉 + 𝐸𝑆 + 𝐸𝐶 + 𝐸𝐴 + 𝐸𝑃
𝑎𝐶 𝑍(Z-1) 𝑎𝐴 (𝐴 − 2𝑍)2
B.E. = 𝑎𝑉 𝐴 − 𝑎𝑆 𝐴2/3 − − − 𝑎𝑃 𝐴−3/4 …….(10)
𝐴1/3 𝐴
Above equation can also be expressed in terms of the nuclear (or atomic) mass, since the mass and
binding energy are related by the relation
𝐵. 𝐸. = (𝑍𝑀𝑝 + 𝑁𝑀𝑛 − 𝑀)𝑐 2

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or
𝐵. 𝐸.
𝑀 = 𝑍𝑀𝑝 + (𝐴 − 𝑍)𝑀𝑛 − ……(11)
𝑐2
where Mp, Mn and M are the masses of proton, neutron and nucleus respectively. By substituting
Eqn.(11) in Eqn.(10), we get
𝑎𝑣 𝐴 4𝑎𝑐 𝑍(𝑍 − 1) 𝑎𝑠
𝑀 = 𝑍𝑀𝑝 + (𝐴 − 𝑍)𝑀𝑛 − 2
+( 2 )× 1/3
+ ( 2 ) 𝐴2/3
𝑐 𝑐 𝐴 𝑐
𝑎𝑎 (𝐴 − 2𝑍)2
+ ( 2) × + 𝑎𝑝 𝐴−3/4 ..........(12)
𝑐 𝐴
Eqn.(12) is known as Weizsacker's semi-empirical mass formula.
Fig.1 summarizes the effect of different energy terms in deciding B.E/A of a nucleus.
Fig. 1: Schematic representation of the various terms in the mass formula as a function of A
(source: https://astarmathsandphysics.com )
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The values of the co-efficient can be determined by a combination of theoretical observed
calculations and adjustments to fit experimental values of the masses.
Typical values of the co-efficients are:
av = 14 MeV
as = 13MeV
ac = 0.60 MeV
aa = 19 MeV
a p = −33.5MeV for even-even nuclei
= 0 for even-odd nuclei
= +33.5MeV for odd-odd nuclei
Examples 1: By Using the semi empirical binding energy formula calculate binding energy of
40
20Ca .
Solution : The semi empirical formula for binding energy is
( A − 2Z ) + a A−3/4
2
Z2
B.E. = av A − as A − ac 1/3 − aa
2/3
p
A A
where, av = 14 MeV ; as = 13MeV ; ac = 0.60 MeV
aa = 19 MeV and a p = −34 MeV
as A is even = 40 and Z is even =20.

av A = 14  40 = 560 MeV
as A2/3 = 13  402/3 = 13 11.696 = 152MeV

Z2 400
ac 1/3
= 0.60  2/3 = 0.60 11.696 = 70MeV
A 40
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( A − 2Z )
2
aa = aa  0 = 0
A
a p A3/4 = −34  40−3/4 = −34  0.063 = 2.14 MeV
B.E. = 560 − 152 + 70 + 2.14 = 335.86 MeV
Example 2. Calculate the atomic number of the most stable nucleus for a given mass number
based on the liquid drop model. Hence explain why out of 2He6, 4Be6 and 3Li6 only the last one is
stable.
Stable nucleus. According to the liquid drop model the binding energy B.E. is given by
( A − 2Z ) + a A3/4
2
Z2
B.E. = av A − as A 2/3
− ac 1/3 − aa p
A A
The most stable nucleus for a given mass number A is that which has the maximum value of
binding energy i.e., for which
(  B.E ) = −2a A−2/3 Z + 4a

( A − 2Z ) A−1 = 0
z
c a
Or
4aa − 8aa A−1Z = 2ac A−1/3Z
Or
( )
Z 4aa + ac A2/3 = 2aa A
A
Z=
a
2 + c A2/3
2aa
Substituting the value of ac = 0.5053 and aa = 23.702 MeV, we get
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A
Z=
2 + 0.015 A2/3
For light nuclei having small A the term 0.015A2/3 can be neglected as it is very small for
stability Z = A/2 this result is confirmed experimentally.
Out of 2He6, 4Be6 and 3Li6 it is only 3Li6 for which A= 6 and Z=3
i.e Z=A/2
Therefore it is the only stable nucleus out of the given three.
Example 3: For "mirror" nuclei which have N and Z differing by one unit, determine the mass
difference. Consider A to be odd.
Solution : Mirror nuclei to be considered have the same odd value of A but the values of N and Z are
interchanged such that they differ by one unit.
The (A-2Z) value in the asymmetry term in mass formula can be written as,
A - 2Z = N + Z - 2Z = N - Z so that,
If N and Z differ by one unit,
N − Z = 1
M z +1 − M z = ( M p − M n ) ( Z + 1) − Z  +
ac 
1/3 (
Z + 1) 
2
A 
=M p − M n + ac
( 2Z + 1)
A1/3
M Z +1 − M Z = M p − M n + ac A2/3
Since A = 2Z+1 for these mirror nuclei
Example 4: The masses of N15 and O15 are 15.000108u and 15.003070u respectively. Using this data,
determine the Coulomb coefficient ac in the semi-empirical mass formula.
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Solution : N15 and O15 are mirror nuclei, Therefore
M Z +1 − M Z = M p − M n + ac A2/3
Using the respective data
( 2.96 10 ) u = ( −0.000844 ) + a (15)

−3
c
2/3
ac = 3.542 6.08 MeV

ac = 0.58 MeV
5.5 NUCLEAR SHELL MODEL
This model is similar to the Bohr model for electrons in the extra nuclear space. By analogy with
the closed sub-shells and shells in the case of atoms it is assumed that nucleons also form similar
closed sub-shells and shells within the nucleus.
5.5.1 EVIDENCES THAT LED TO SHELL MODEL
The following are the facts, that are favoring the shell model
1. Pairs of nucleons frequently form inside. A single unpaired nucleon can be taken out of
the nucleus more easily than a paired one. Two protons and two neutrons combine to
form a very stable nucleus. The fact that a particle has a significant binding energy of
approximately 28.3 MeV lends support to this.
2. The graph between binding energy per nucleon and atomic number exhibits numerous
kinks, one of which is illustrated as an example in the range A= 126 to 150 in Fig.2.
These kinks are associated with a sharp increase in the binding energy per nucleon.
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Fig. 2: Graph between binding energy per nucleon vs atomic number
These kinks or discontinuities have been found to occur whenever either the Neutron
number or the proton number or both take the values 2,8,20,50,82 and 126. Nuclei
containing 2,8,20,50,82 and 126 nucleons of the same kind known as Magic numbers,
have a very high stability. For example, 2He4 with Z=N=2 and 8O16 with Z = N = 8 are
highly stable. Similarly, the nuclei with 14,28 and 40 nucleons (semi-magic numbers) are
slightly less stable but are more stable than the rest.
3. The most numerous nuclei are those with even numbers of both protons and neutrons; the
least abundant are those with odd numbers of both protons and neutrons; and the
intermediate types are those with odd numbers of one type and even numbers of the other
type. High natural abundance is, of course, related to stability. Brown provided
information regarding the relative abundances of nuclei in 1949 using information about
the elements that make up the sun, the earth, and the stars.
The relative abundances of naturally occurring isotopes with nuclei that contain magic
numbers of neutrons or protons are typically larger than 60%. For instance, the relative
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abundances of the isotopes 88Sr (N = 50), 138Ba (N = 82), and 140Ce (N = 82) are
82.56%, 71:66%, and 88.48%, respectively. The stable end product of the natural
radioactive series, Lead 82Pb208 has underline Z= 82 and N=126 both magic numbers.
4. In contrast to other elements, an element with a magic number of protons typically has a
higher number of stable isotopes. For instance, whereas argon (Z = 18) and titanium (Z =
22) have 3 and 5 stable isotopes, respectively, calcium (Z = 20) has 6 stable isotopes.
Again, the greatest number of stable isotopes is found in tin with Z = 50. This value is 10,
as opposed to 8 for tellurium (Z = 52) and cadmium (Z = 48).
5. Compared to the nearby isotones, the number of naturally occurring isotones with the
magic numbers of neutrons is typically high. For instance, at N = 82 there are seven
stable isotones as opposed to three and two at N = 80 and two at N = 84, respectively.
Similar circumstances exist at N = 20, 28, and 50, which have 5, 5, and 6 isotones,
respectively. These numbers are larger than in the cases of the nearby isotones.
6. The nuclei with the magic numbers of neutrons typically have low neutron capture cross-
sections. The probability of these nuclei acquiring an extra neutron is low because their
neutron shells are already full, as seen in Fig. 3. Similar to this, the cross sections for
proton capture are small for nuclei with the magic proton numbers.
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Fig.3: Neutron capture
7. If the heavy nuclei's disintegration energies are plotted as functions of mass number A for
a given Z, a regular fluctuation is typically seen up until the magic neutron number N =
126, at which point there is a sharp discontinuity. This demonstrates the neutron number
126's magical properties.
8. β-emitters exhibit discontinuities at the magic proton or neutron values
9. The three lead isotopes, which all have the same magic number Z = 82 of protons in their
nuclei, are the stable end products of all three naturally occurring radioactive series.
10. The earliest excited states of nuclei with magic numbers of neutrons or protons occur at
energies higher than those of the nearby nuclei.
5.5.2 PREDICTIONS OF SHELL MODEL
Shell model predictions are given as:
1. Within the nucleus, nucleons arrange themselves into closed sub shells in a manner
akin to that of atoms' electrons. Thus, shell model is the name given to it.
2. There is some sort of shell structure to the nucleons that make up the nucleus. Protons
and neutrons in the proper quantity are used to complete the shells.
3. The extra nuclear electrons spin around in the nucleus' Coulomb electric field, which is
meant to be heavy and far away. Just a few allowed orbits are used by electrons to
rotate. Similar to this, each nucleon in the nucleus moves freely in a defined orbit under
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the influence of the central potential created by the average interaction amongst the
remaining (A - 1) nucleons in the nucleus.
4. Each nucleon is predicted to have a spin angular = s ( s + 1) where s is the spin
quantum number = 1/2 and an orbital angular momentum= l ( l + 1) where I is the
orbital angular momentum quantum number having values 0, 1, 2, 3,... etc.
These facts have served as the foundation for the development of a nuclear shell model. The
theory is comparable to the Bohr model for additional nuclear space electrons. It is expected that
nucleons will likewise create similar closed sub shells and shells within the nucleus by similarity
with the closed sub shells and shells in the case of atoms.
The protons and neutrons that make up the nucleus are meant to be arranged in a shell structure,
and these shells are supposed to close with the right number of protons and neutrons. The excess
nuclear electrons are predicted to spin under the nucleus' heavy and far-reaching Coulomb
electric field. The electrons only rotate in certain specified orbits.
The theoretical understanding of the origin of the nuclear shell model is based on the assumption
of the existence of a dominant spherically symmetric central field potential force governing the
motion of the individual nucleons in the nuclei. From the compelling similarity of stability
between the magic nuclei and the inert gases, it is predicted in the shell model that each nucleon
moves independently inside the nucleus in a fixed orbit under the influence of a central field of force or
a central potential V(r) produced by the average interaction between all the remaining (A - l) nucleons in
it.
It is assumed that the nucleons move in an average harmonic oscillator potential given by
1 1
𝑉(𝑟) = 𝑘𝑟 2 = 𝑚𝜔2 𝑟 2 ......(13)
2 2
where m is the mass of the nucleon and  the oscillator frequency. The 3-dimensional
Schrodinger's wave equation for the harmonic oscillator can be solved using rectangular
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Cartesian co-ordinates by a generalization of the method. However, for the present purpose it is
more appropriate to use spherical polar co-ordinates.
If the potential given by Eqn. 13 is substituted in the 3-dimensional Schrodinger equation, then
the following radial equation can be obtained by solving it through the method of variables
1 𝑑 2 𝑑𝑅𝑡 2𝑚 𝑙(𝑙 + 1)ℏ2

(𝑟 ) + 2 {𝐸 − 𝑉(𝑟) − } 𝑅𝑡 =0 ......(14)
𝑟 2 𝑑𝑟 𝑑𝑟 ℏ 2𝑀𝑟 2
where R1(r) is the radial function. The term l(l + 1) is the centrifugal potential. The angular part
of the wave function is the spherical harmonic Y1m ( ,  ) so that the total wave function is
𝜓=R 𝑛𝑙 (𝑟)𝑌1𝑚 (𝜃, 𝜙) ......(15)
By employing the quantum mechanical approach, the 3-D harmonic oscillator problem can be
solved. Thus, from the solution for the Eqn. 14 we find that the various energy levels are given
by
𝐸 = (𝜆 + 3/2)ℏ𝜔 ......(16)
Further it can be shown that angular part of the wave function  requires that the oscillator
quantum number  is related to the orbital quantum number l and the radial part of the quantum
number (similar to the principal quantum number of the electronic orbit), referred to as the radial
quantum number n, by the relation
𝜆 = 2(𝑛 − 1)+l=2n+l-1 ......(17)
The levels of different azimuthal quantum numbers l are designed by the usual symbols used in
atomic spectroscopy, as given below:
l:0123456
Spectroscopic notation: s p d f g h i
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The nucleons are designed by marking their n values followed by spectroscopic notation giving
the l value. The lowest energy level of the harmonic oscillator is the Is level with n = 1, l = 0 its
3
energy being  which is the zero-point energy. The level structures for the neutrons and the
2
protons are similar.
5.5.3 Spin Orbit Coupling
In order to explain the disagreement at the higher magic numbers, Mayer and independently
Haxel and Jensen suggested that a spin-orbit interaction term should be added to the central
potential V(r) in Eqn.13.
Taking the spin-orbit interaction into account, the sequence of the energy levels is designated as
follows:
1s1/2 ;1 p3/2 ,1 p1/2 ;1d5/2 ,1d3/2 ;....2s1/2 ; 2 p3/2 , 2 p1/2 ;.....2 f7/2 , 2 f5/2 ; etc.
In accordance with the Pauli exclusion principle, each sub level of a given j can accommodate a
maximum of (2j + 1) nucleons of either kind for which the magnetic quantum numbers m, are
different. The possible values are n j =j, j-1,...-j when a sub level of given j is completely filled
up with (2j + 1) nucleons of particular kind, the extra nucleons of the same kind must go the next
higher state of different j,
The group of sub levels (n,l,J) having energy values close to one another now constitute a shell.
The number of nucleons required to fill up the shell is the sum of the nucleon numbers (2j + 1).
Using this scheme, the number of nucleons required to complete a sub shell can be calculated as
follows:
The lowest level  = 0 , according to this scheme is 1s1/2 with j = 1/2 which contains
( 2  1 2 + 1) or 2 nucleons.
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The next higher level with  = 1 is a combination of two sub levels 1p3/2 and 1*p1/2The maximum
number of nucleons which can occupy these sub-levels are 4 i.e ( 2  3 2 + 1) and 2 i.e
( 2  1 2 + 1) respectively, so that the total number of nucleons in the group of sub-levels is (4+2)
or 6. So the shell closure takes place in this case with (2+6) or 8 nucleons.
Fig. 04 : Representation of energy levels in shell model with magic numbers (https://nuclear-
power.com)
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5.5.4 Achievements of the Shell Model
The shell model in atomic physics can be used to predict the order and relative positions of
excited states in addition to providing a good description of the atomic ground state. Using the
nuclear shell-model, we may try to forecast the locations, angular moments, and parity of nuclear
energy levels.
Magic numbers are explained by the shell model. By referring to Table 10.1 and adding the
nucleons from below, it can be determined that the magic numbers 2, 8, 20, 50, 82, and 126
correspond to neutrons or protons when the shell ends.
The magnetic moments and ground state spins of the nuclei are successfully explained by the
shell model. In order for the mechanical and magnetic moments of the neutrons and protons with
opposing spin to cancel, they pair off. Thus, the odd or left proton or neutron that could not pair
with one another contributes to the overall nucleus's spin and magnetic moment.
The j-values of the individual nucleons in it are added to get the total angular momentum
quantum number l, also known as the spin quantum number, under the typical quantum
conditions.
For example, 1H2 has Z=1 i.e, one proton and N = A - Z = 1 neutron. Each nucleus will,
therefore, be in the ground state l = 0 and will have a j-value ½ for either. The j's of the proton
and the neutron are oriented parallel to each other so that l = 1 which agrees with the
experimentally observed value of spin.
For 1H³ and 2He3 nuclei we have two nucleon of the same kind which are paired off and one
nucleon of the other kind with j value ½. Thus, the l value of each of these two nuclei is 1/2.
The j values of the odd nucleon in the nuclei are used to explain the observed nuclear magnetic
moments. Due to spin, the odd neutron has a negative magnetic moment, whereas the odd proton
has a positive magnetic moment. Because the two extra neutrons are paired off and have no
influence on the single odd nucleon that contributes to the nucleus' spin and magnetic moment,
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the spins and magnetic moments of isotopic nuclei with odd A that differ in their A values by
two are the same. The nuclear spin and magnetic moment of 47Ag107,109 and 55Cs133,135,137isotopes,
for instance, are equal.
On the basis of closed shells, it is possible to explain the extremely high stability and low
binding energy of nuclei. For instance, the first shell of 2He4 is completely filled with protons
and neutrons (2 of each), resulting in l = 0. Additionally, the closure of the shell gives it an
extremely high binding energy of 28.3 MeV. The á particle 2He4 nucleus has an extremely
stable structure as a result. The case of 8016 is also comparable, with the first two shells of
neutrons and protons completely occupied with 2 and 6 nucleons, for a total of 8 nucleons,
respectively.
For maximum stability, the proton and neutron levels should be filled equally when adding
nucleons to the nucleus. As a result, the number of neutrons and protons is almost equal for low
values of mass number 4. The most stable nuclides will once more develop as the number of
nucleons increases when the lowest energy levels accessible to protons and neutrons are full. The
number of neutrons will exceed the number of protons as the space between proton levels
increases.
Due to the large variation in the nuclear spins of the isomeric states of these nuclei and the fact
that their A values are close to the magic numbers needed to complete the various nuclear shells,
the phenomenon of nuclear isomerism, or the existence of isobaric, isotopic nuclei in different
energy states, has been explained on the basis of the shell model.
5.6 THE COLLECTIVE MODEL OF A NUCLEUS
Let's examine some of this model's features. Collective motions of all nucleons are taken into
consideration in addition to the shell model characteristics. In essence, this combines elements of
both the liquid drop model and the shell model. Consequently, a substantially greater amount of
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experimental data may be explained by the collective model. Nuclear deformation resulting in
collective modes of excitation, such as collective oscillations and rotations, are two forms of
collective effects that stand out in particular. Centrifugal and Coriolis forces are responsible for
the less evident impacts.
Large quadrupole moments, which are seen in many nuclei, cannot be explained by the shell
model, which necessitates large nuclear deformations. The nuclear quadrupole moment, which
we defined before, is a measurement of the nuclear shape's departure from a sphere.
As a result, the experimentally observed quadrupole moment suggests the existence of a

distorted, non-spherical nucleus. It was brought up during the debate of the shell model that the
quadrupole moment of the nuclei corresponding to magic numbers is almost zero. In the region
between closed shells, nuclei have significant quadrupole moments. They cannot be explained by
a single particle shell model where the electromagnetic moments are determined by the final odd
nucleon. Due to its spherical nature, the core does not contribute to the quadrupole moment.
( )
Q = Z  3z 2 − r 2  (r )d 3 r ……….(18)
The quadrupole moment Q has dimensions of area.
If a homogeneously charged ellipsoid with charge Ze and semi-axes a and b is considered (b

along z-axis), then,
Q=
2
5
(
Z b2 − a 2 ) …………(19)
a+b
For small deviation from spherical symmetry, we can talk about an average radius, R0 
2
R
and R = b − a . Defining a deformation parameter  = , the quadrupole moment becomes,
R0
4 2
Q= ZR0  ……….(20)
5
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125

It is common to introduce a reduced quadrupole moment
Q
Qreduced = ………(21)
ZR02
For a uniformly charged ellipsoid, from
4
Qreduced =   deformation parameter  .
5
In wave mechanics,  ( r ) the probability density in Eq.18 has to be replaced by  *j j where, j is
the spin quantum number. The nuclear spin is taken along the z-direction so that j=m, the
magnetic quantum number
Thus,
( )
Q = Z  * 3z 2 − r 2  d 3r …………….(22)
To calculate Q within the framework of the single particle shell model, the wave function  is
taken to be a single particle wave function  n / m . It can be for example, the wave function of a
particle in a harmonic oscillator potential well. It can be shown that,
2 j −1 2
Q=− r ……………..(23)
2 ( j + 1)
Where j=angular momentum quantum number of the single particle ( j  1/ 2 ) .
r 2 = expectation value of the square of the radius of the single particle orbit.
For a single proton, j = 3/2, 5/2, 7/2, etc. and so, for order of magnitude calculation, for higher
2 j −1
values of j,  1 . Also, if we take, r 2 = R02 , then,
2 ( j + 1)
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126
Q  − R02
Q
Qreduced ( single-particle ) = ………….(24)
ZR02
1
−
Z
175
Now, experimentally the measured quadrupole moment of middle of the shell nuclei like 71 Lu is
Qreduced ( 175
71 )
Lu = +0.25
On the other hand, the single-particle estimate is (from Eq. 24).
Qreduced ( single-particle ) ( 175

71 )
Lu = −0.014
Giving,
Qreduced ( Expt.)
 −18
Qreduced ( cal. single-particle )
The single-particle estimate of quadrupole moment is not only wrong in magnitude but also
wrong in sign. On the other hand, for a doubly magic plus one proton nucleus like
209
83 Bi (83 protons and 126 neutrons) ,
Qreduced ( Expt.)  −0.014
And
1
Qreduced ( cal. single-particle )  − = −0.012 ……..(25)
Z
These values are in good agreement.
Obviously, the shell model is not able to explain the observed large quadrupole moments for
middle of the shell nuclei. For the nuclei like 175Lu, the core is not spherical (as assumed by shell
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127

model). It is permanently deformed by the nucleons in the outermost shell. Above equation give
us an estimate of the deformation that produces the large quadrupole moment.
Because orbits are oriented arbitrarily in a closed shell configuration, the deforming effects of
multiple nucleons balance out. The nucleus often changes its surface so that it aligns with the
density distribution of the nucleons in the unfilled shell. The nucleus takes on a spheroidal
equilibrium shape when there are enough nucleons outside the closed shells. Contradictory
forces resulted in this equilibrium shape. The individual nucleons outside the closed shells that
make up the core, on the other hand, attempt to change this core, which enables them to move in
a deformed nuclear potential. The pairing forces, on the other hand, couple two identical
nucleons in order to decrease their net angular momentum to zero.
This second effect has the tendency to yield a spherical symmetrical condition. More nucleons
outside the closed shell core certainly has a positive impact on the first effect. In other words, the
deformation of a nucleus rises as we move away from closed shells, leading to the observation of
large quadrupole moments. Do other experimentally observed phenomena also exhibit nuclear
deformation? Yes, it is clearly noticeable in the nuclei's rotational and vibrational spectra.
5.7 GLOSSARY
Nucleon: The proton and the neutron, constituting atomic nuclei.
Binding Energy: Amount of energy required to separate a particle from a system of particles or
to disperse all the particles of the system.
Magnetic moment: Magnetic Moment is defined as magnetic strength and orientation of a

magnet or other object that produces a magnetic field.
Electric Quadrupole Moment: A parameter which describes the effective shape of the ellipsoid
of nuclear charge distribution.
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128
5.8 SUMMARY
In this unit liquid drop model of the nucleus explained. Semi empirical mass formula has also
been explained by using binding energy formula . Along with it some numerical problems has
also been solved for the convenience of learners. Shell model and its role in significance of
magic numbers is explained by considering magic numbers the basis of spin orbit interaction
theory. Some achievements of the shell-model is discussed in this unit as well. The collective
model which is the combined form of the liquid drop model which is totally built on the
foundation that the collective nucleon motion dominates over individual nucleon motion with we
have shell model where nucleons are assumed to move completely independently. These models
based on apparently opposite assumptions, are found to be very useful in understanding what
actually goes on with in the nucleus. The evidence of the existing magic numbers is beyond of
the range of the liquid drop model of nucleus.
5.9 REFERENCES
3. Elements of Nuclear Physics by M.L.Pandya,R.P.S.Yadav
4. Nuclear Physics An Introduction by S.B.Patel
6. https://astarmathsandphysics.com/a-level-physics-notes/nuclear-physics/2909-the-liquid-
drop-model-of-the-nucleus.html
7. https://nuclear-power.com/wp-content/uploads/Shell-model-of-nucleus.png
8. Nuclear Physics by S.N.Ghoshal
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129
1.”The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of Physics, Berlin:
Springer Ver.
3. Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum
Press, New York.
1. Derive an expression for the semi-empirical formula of Weizsacker.
2 Draw the graph and show that the binding energy per nucleon is the sum of volume, surface,
Coulomb and asymmetry energies.
3. Calculate the binding energies and atomic mass of the nuclides of Ca40, Sn120and Pb208
4. Calculate the binding energies for the mirror nuclei K39 and Ca39 from the semi empirical
binding energy equation. What do you infer from the energies about the relative stabilities?
5. Give an account of the assumptions of the Liquid drop model proposed by N.Bohr and Wheeler.
6.Obtain an expression for the total binding energy of a nucleus based on the liquid drop model.
7. Calculate the binding energies of the nuclei - Be9, Al27, Cu63 Mo98, Xe130 W184, U238, from the semi
empirical binding energy formula and compare the results with the experimental values.
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130
UNIT 6
RADIOACTIVITY
6.1 Introduction
6.2 Objectives
6.3 Properties of Radioactive Rays
6.4 The Law of Radioactive Decay
6.4.1 Unit of Activity
6.5 Radioactive Growth and Decay
6.6 Ideal Equilibrium
6.7 Transient Equilibrium and Secular Equilibrium
6.8 Radioactive Series
6.9 Determination of the Age of the Earth
6.10 Carbon Dating-Archaeological Time Scale
6.11 Summary
6.12 References
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131
6.1 INTRODUCTION
In this unit learners will study about the basic properties of Radioactivity As we know that
information about atomic nuclei started as early as 1896 with the discovery of radioactivity.
In 1895, Roentgen discovered the X-rays. The French scientist Becquerel got interested in
Roentgen's work. Becquerel was seized of the fact that the production of 𝑋-rays was always
accompanied by fluorescence from the material of the X-ray tube (glass). He thought that Xrays
existed whenever there was fluorescence. To investigate this problem, Becquerel took uranium
sulphate, which fluoresces under the action of sunlight. He found that fluorescent uranium
sulphate did give out rays, which could affect a photographic plate even when wrapped in thick
black paper. Becquerel argued that the fluorescent salt had given rise to X-rays, which had
penetrated the black paper and affected the photographic plate.
But he soon saw that he was mistaken. During one such experiment the sky happened to be
overcast and the uranium salt was hardly fluorescent. On developing the photographic plate,
Becquerel was surprised to see a dark spot on it, as before. He had obviously stumbled on some
new kind of rays (1896) which could penetrate the thick wrapper and affect the photographic
plate. It was soon established that any salt of uranium emits Becquerel rays. Unlike the X-rays,
which appear in an 𝑋-ray tube only under special conditions, the Becquerel rays are emitted in a
spontaneous manner.
Marie. Curie found that pitchblende, the ore from which uranium is extracted, emits Becquerel
rays with a much stronger intensity than what its uranium content would. After a long and
laborious process of chemical separation, Marie Curie and her husband Pierre Curie discovered
two new elements, polonium and radium,which emitted Becquerel rays.
They gave the name 'radioactive' to all substances capable of emitting, Becquerel rays and
phenomenon itself came to be known as 'radioactivity'. the phenomenon itself came to be known
as 'radioactivity'.
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132

The discovery of radium was a great event, as it was found to be about a million times more
radioactive than uranium. This power of radium radiation made it possible to study
radioactivity systematically.
6.2 OBJECTIVES
In this unit learners will study about
• Properties of Radioactive Rays

• The Law of Radioactive Decay
• Radioactive Growth and Decay
• Ideal Equilibrium
• Transient Equilibrium and Secular Equilibrium
• Determination of the Age of the Earth
• Carbon Dating-Archaeological Time Scale
6.3 PROPERTIES OF RADIOACTIVE RAYS
(1)By a calorimetric experiment, Curie estimated that one gm of radium liberates 140 calories in
one hour. Though small, this energy is released continuously over a very long period of time.
(2) Radioactive rays ionize the surrounding air and affect photographic plates.
(3) Radioactive rays act differently on different cells and tissues. Cells that multion are most
readily destroyed by these rays. This outstanding discovery made radidm and invaluable aid to
physicians to fight tumours, particularly discovery made radium an
(4) Fluorescence is produced in substances like zime surticularly cancerous growths. adding
minute quantities of radium to, say, zinc sulphide, we can get a compounde, by continuously
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133

luminous in the dark. This can be used to produce get a compound that is gun sights and
instrument pointers (coating) for thosed to produce luminous watch dials, be read in the dark.
(5) Rutherford found that a beam of radioactive rays from radium sample splits into three
components in a strong magnetic or electric field. See Fig radium sample splits into three
Fig. 1. Deflection of 𝛼, 𝛽, 𝛾 rays
(a) The 𝛼 − 𝑟𝑎𝑦𝑠 (particles) are the nuclei of helium atoms. This identification was made by
Rutherford and Royds in 1909. By the spectroscopic method, they found traces of helium in an
originally pure sample of radon gas, which is an alpha emitter. When an electric discharge was
passed through a tube containing pure radon gas, initially only characteristic radon lines
appeared.
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134

After a day Rutherford found that, Radon lines became somewhat weaker and new lines started
appearing. These new lines were identified with helium spectrum. As the days passed, the radon
spectrum began weakening while the helium lines grew brighter. Thus, for the first time, people
'saw' decay of an element (radon) and 'birth' of a new element, helium. Such a transformation in
which a parent element gives rise to a new element-called the daughter product-by emitting
radioactive rays is called a radioactive transformation.
For the above example of radon, we can write the radioactive transformation equation as:
222 218
86 Rn → 42 He + 84 Po
Parent element → 𝛼-particle + daughter product
During a radioactive transformation, the mass number and total charge is conserved.
So we can say that radioactivity is a nuclear phenomenon. In other words, the radioactive rays
come out of the atomic nucleus.
A few examples of alpha decay are,
238
92 U → 42 He + 234
90 Th
226
Ra → 2 He + 222
4
86 Rn
218 4 214
88 Po → 2 He + 82 Pb
Alpha rays can be stopped by a thin sheet of paper. On the other hand, they cause intense
ionization in air. Most 𝛼-particles are emitted with velocities between ∼ 1.5 × 107 m/s and ∼
2.2 × 107 m/s. Any group of 𝛼-particles emitted from the same type of nuclei always has a
definite velocity and hence a definite energy.
The alpha particles cover a definite distance in a given material, practically without any loss of
intensity and then suddenly in a small distance are absorbed completely. The definite distance
they travel within a given material is called their range in that material.
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135

(b) The 𝛽 rays are identical with electrons. A 𝛽-particle therefore has a mass (1/1836) of mass
of a proton.
A few examples of 𝛽 decay are:
234
90 Th → 234 0
91 Pa + −1 𝑒
210
Bi → 210 0
84 Po + −1 𝑒
14 14 0
6 C → 7 N + −1 𝑒
Notice that the mass number and charge are conserved and the daughter product moves one place
up in the periodic table, as loss of a negative charge by a nucleus would imply gain of a positive
charge. Beta rays cause much less ionization in air, but are ∼ 100 times more penetrating than 𝛼-
rays. They can penetrate a sheet of aluminium a few mm thick.
The velocities of 𝛽 -particles emitted from various nucleii range up to 0.99c , where 𝑐 =
3 × 108 m/s , is the velocity of light. A particular 𝛽 -active element emits 𝛽 -particles with
energies varying between zero and a certain maximum. This maximum energy is called the end-
point energy.
(c) The 𝛾-rays are part of the electromagnetic spectrum. They have wavelengths smaller than
those usually associated with the X-rays. Thus, usually 𝛾-ray photons (energy of a photon, 𝐸 =
ℎ𝑓 = ℎ𝑐/𝜆) are more energetic than the X-ray photons and are even more penetrating than the
X-rays. (They are ∼ 100 times more penetrating than 𝛽-rays.) The wavelength of 𝛾-ray photons
ranges between ∼ 1.7 × 10−8 cm and ∼ 4 × 10−6 cm.
The ionization due to 𝛾-rays is a photoelectric effect. Owing to their large energies, the 𝛾-ray
photons can dislodge electrons not only from outer orbits (valence orbits on conduction bands) of
the atoms but also from the inner orbits. Besides this photoelectric effect, 𝛾-rays lose energy by
(i) Compton scattering, in which the 𝛾 -photon collides with an electron and gets
scattered with a shift in wavelength [Δ𝜆 = ℎ/𝑚0 𝑐(1 − cos 𝛼)].
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136

(ii) (ii) pair production in which a 𝛾-photon is converted into a pair consisting of an
electron and a positron. For this, the energy of the 𝛾-ray has to be > 1.02MeV.
6.4 THE LAW OF RADIOACTIVE DECAY
When a nucleus disintegrates by emitting a particle (𝛼 or 𝛽) or a gamma ray, or by capturing an

electron from the atomic shell ( 𝐾-capture), the process is called radioactive decay. This decay is
spontaneous.
By using a Geiger Counter, it is possible to study how radioactive decay depends on time. Let us
take a radioactive sample containing 𝑁0 nuclei at time 𝑡 = 0, i.e., at the beginning. We wish to
calculate the number 𝑁 of these nuclei left after time 𝑡.
The number of nuclei of a given radioactive sample disintegrating per second is called the
activity of that sample.
𝑑𝑁
∴ = Rate of decrease of nuclei with time
𝑑𝑡
= Activity at time 𝑡.
Experimentally, it is found that the activity at any instant of time 𝑡 is directly proportional to the
number 𝑁 of parent type nuclei present at that time.
𝑑𝑁
− 𝑑𝑡 ∝ 𝑁
𝑑𝑁 ………..(1)
− 𝑑𝑡 = 𝜆𝑁
where 𝜆 > 0, is the proportionality constant.
The negative sign indicates that 𝑁 decreases as 𝑡 increases.
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For unit time, from Eq.1, we get,
𝑑𝑁
𝜆 = (− )
𝑁
i.e., 𝜆 is fractional change in 𝑁 per sec. and we see that 𝜆 is not a mere proportionality constant,
but it gives us the probability of decay per unit interval of time. Hence 𝜆 is called the probability
constant, the decay constant or the disintegration constant.
Notice that 𝑑𝑁 is the number of parent nuclei that decay between times 𝑡 and 𝑡 + 𝑑𝑡, and that we
have taken 𝑁 as a continuous variable.
From Eq.1, by integration we get,
𝑁 𝑡
𝑑𝑁
∫ = − ∫ 𝜆𝑑𝑡
𝑁0 𝑁 0
𝑁 = 𝑁0 𝑒 −𝜆𝑡
Here, 𝑁̀0 = Number or radioactive nuclei at 𝑡 = 0.
Thus we see that the law of radioactive decay is exponential in character.
Notice that only half the amount of radon present initially remains after 3.83 days; only one-
fourth of it remains after 7.66 days and only one-eighth remains after 15.32 days.
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138
Fig. 2. Decay curve for radon, ( 222

88 Rn).
The plot shows that in a fixed time interval (say, 3.83 days for radon) a fixed fraction (1/2 for
radon) of the amount of radioactive substance at the beginning of the interval decays. This
fraction is independent of the amount of radioactive substance or whether you have a freshly
prepared sample or not and instead depends only on the interval of time. This is a characteristic
of the exponential nature of the law of radioactive decay.Figure 2 depicts a typical experimental
decay curve for radon.
The decay constant 𝜆 is a characteristic of radioactive substance and it depends in no way on the
amount of the substance present.
Half Life (𝑻) : It is convenient to define a time interval during which half of a given sample of
radioactive substance decays. This interval is called the half life or half value period of that
substance, denoted by 𝑇.
𝑁 1
= 2 = 𝑒 −𝜆𝑇
𝑁0
𝜆𝑇
𝑒 =2 ……….(2)
𝜆𝑇 = log 𝑒 2 = 0.693
0.693
𝑇 = 𝜆
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139

Mean Life (𝝉) : Individual radioactive atoms may have life spans between zero and infinity.
Hence it is meaningful to talk about the average or mean life 𝜏, defined as,
Total life time of all nuclei in a given sample

𝜏= ………..(3)
Total number of nuclei in that sample
To evaluate this we have from Fig.3.
Fig.3: Curve shows how 𝑑𝑁 number of nuclei decay in time 𝑑𝑡.
From the curve, one can see that each of the 𝑑𝑁 number of radioactive nuclei has lived a life of 𝑡
seconds, i.e., the total life span of 𝑑𝑁 nuclei is (𝑑𝑁 ⋅ 𝑡) seconds.
∴ We can put Eq. 3 as,
0
∫𝑁 𝑡𝑑𝑁
0
𝜏= 0
∫𝑁0 𝑑𝑁
∞
−𝑁0 𝜆 ∫0 𝑡𝑒 −𝜆𝑡 𝑑𝑡
= , ∵ 𝑑𝑁 = −𝜆𝑁𝑑𝑡 = −𝜆𝑁0 𝑒 −𝜆𝑠 𝑑𝑡
−𝑁0
∞
= 𝜆 ∫ 𝑡𝑒 −𝜆𝑡 𝑑𝑡,
0
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which on integration by parts becomes,
1 1
𝜏 = 𝜆 ⋅ 𝜆2 = 𝜆 …………..(4)
Example 01: According to measurements by Rutherford and Geiger, one gram of radium emits
in one second 3.7 × 1010 alpha particles. Estimate the half life of radium.
Solution: The decay constant of radium is,
3.7 × 1010
𝜆Ra = sec −1
2.7 × 1021
where 2.7 × 1021 = Number of radium atoms in one gram of radium.
( ∵ There are 6.02 × 1023 atoms [Avogadro's number] in one gm-atom of radium-226)
∴ 𝜆Ra = 1.37 × 10−11 sec −1
0.693
∴ Half life, 𝑇Ra = ≅ 5 × 1010 sec ≅ 1600 yrs.
𝜆Ra
6.4.1 Unit of Activity
The most commonly used unit is the curie. It was originally based on the rate of decay of a gram
of radium. Experiments have yielded the result that there are-about 3.7 × 1010 disintegrations
per second per gram of radium. This number is taken as a standard and is called the curie. Thus
by definition,
disintegrations
One curie = 1Ci = 3.7 × 1010 sec
This is applicable to all types of nuclear disintegrations.
A Curie of activity is a very strong source of radiation.

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141

Thus, one has
and
1 millicurie = 1 mCi = 10−3 Ci

1 microcurie = 1𝜇Ci = 10−6 Ci.
Sometimes one uses another unit for activity, called the rutherford.
disintegrations
1 rutherford = 1rd = 106
sec.
1mrd = 10−3 rd
1𝜇rd = 10−6 rd
Activity can also be defined in terms of N as
𝑑𝑁 0.693
Activity = | | = 𝜆𝑁 = 𝑁
𝑑𝑡 𝑇
Thing to remember is that a very short-lived substance gives rise to large activity, even if it is
present in minute quantities.
226
Example 02 : Calculate the activity of (i) One gram of radium 88 Ra, whose half life is 1622
200
years and (ii) 3 × 10−9 kg of active gold, 79 Au, whose half life is 48 mins. Solution:
1gm − mole atoms

𝑁 = 1gm ( ) (6.03 × 1023 )
(i) 226gm gm-mole
= 2.66 × 1021 atoms
0.693
and 𝜆 = 𝑇
0.693
= 𝑠𝑒𝑐 −1
1622 × 365 × 24 × 60 × 60
= 1.355 × 10−11 𝑠𝑒𝑐 −1

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142

disintegrations
∴ Activity = 𝜆𝑁 = 3.604 × 1010
sec
= 0.974Ci
∼ 1Ci
1gm−mole atoms
(ii) 𝑁 = (3 × 10−6 gm) ( ) (6.03 × 1023 )
200gm gm-mole
= 9.04 × 1015 atoms.
0.693
and 𝜆 = 𝑇
0.693
= sec −1
48 × 60
= 2.406 × 10−4 sec −1
disintegrations
Activity = 𝜆𝑁 = 2.18 × 1012 .
sec
= 58.9Ci
6.5 RADIOACTIVE GROWTH AND DECAY
Let us consider decay of the type,
𝐴 → 𝐵 → 𝐶 (stable)
Our aim is to find out the abundance of substance 𝐵 if 𝐴 decays to 𝐵 and 𝐵 decays to 𝐶.
Let 𝑁𝐴 be the number of nuclei of 𝐴 type at any instant. Assume that originally only 𝐴 type was
present and the initial number of nuclei of 𝐴 type be 𝑁0 . is zero.
Let 𝑁𝐵 be the number of nuclei of 𝐵 type at the same instant 𝑡. At 𝑡 = 0, the initial number 𝐵 is
formed as a result of decay of 𝐴.
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143

Therefore, the number of nuclei entering the 𝐵 category is,
𝑑𝑁𝐴
− = 𝜆𝐴 𝑁𝐴 …………(5)
𝑑𝑡
where 𝜆𝐴 is decay constant of 𝐴 type.
The number of nuclei leaving the 𝐵 category is,
= 𝜆𝐵 𝑁𝐵
where 𝜆𝐵 is decay constant of 𝐵 type.
∴ The net change in number of nuclei (per sec) of the 𝐵 category is,
𝑑𝑁𝐵
𝜆𝐴 𝑁𝐴 − 𝜆𝐵 𝑁𝐵 = ……..(6)
𝑑𝑡
Now,
𝑁𝐴 = 𝑁0 𝑒 −𝜆𝐴𝑡
𝑑𝑁𝐵
= 𝜆𝐴 𝑁0 𝑒 −𝜆𝐴𝑡 − 𝜆𝐵 𝑁𝐵
𝑑𝑡
Multiplying by the integrating factor, 𝑒 𝜆𝐵 𝑡 𝑑𝑡
𝑒 𝜆𝐵 𝑡 𝑑𝑁𝐵 = 𝜆𝐴 𝑁0 𝑒 (𝜆𝐵 −𝜆𝐴)𝑡 𝑑𝑡 − 𝜆𝐵 𝑁𝐵 𝑒 𝜆𝐵 𝑡 𝑑𝑡 …….(7)
L.H.S. can be now integrated by parts.
∫ 𝑒 𝜆𝐵 𝑡 𝑑𝑁𝐵 = 𝑒 𝜆𝐵 𝑡 𝑁𝐵 − ∫ 𝜆𝐵 𝑁𝐵 𝑒 𝜆𝐵 𝑡 𝑑𝑡 ………(8)
Integrating Eq. 7, and substituting Eq.8, we get,
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144

𝜆𝐴
𝑒 𝜆𝐵 𝑡 𝑁𝐵 =𝜆 𝑁0 𝑒 (𝜆𝐵 −𝜆𝐴)𝑡 + 𝐶
𝐵 −𝜆𝐴
𝑁𝐵 = 0, 𝑡 = 0
𝜆 𝑁 …..
𝐶 = − (𝜆 𝐴−𝜆0 )
𝐵 𝐴
𝑁0 𝜆𝐴
𝑁𝐵 =𝜆 (𝑒 −𝜆𝐴𝑡 − 𝑒 −𝜆𝐵 𝑡 ) … … … … . . (9)
𝐵 −𝜆𝐴
Which is the desired result, giving the number of nuclei of 𝐵 type present at any time 𝑡. The
decay of 𝐴 type and growth and decay of 𝐵 type is shown in Fig.4.
Fig.4: Decay and growth of radioactivity.
Example 3: Show that if there are initially 𝑁0 radioactive nuclei of the parent present, the time at
which the number of radioactive daughter nuclei is maximum is:
log (𝜆𝐵 /𝜆𝐴 )

𝑡max =
𝜆𝐵 − 𝜆𝐴
Solution: We know that,

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145

𝑁0 𝜆𝐴
𝑁𝐵 = (𝑒 −𝜆𝐴𝑡 − 𝑒 −𝜆𝐵 𝑡 )
For 𝑁𝐵 to be maximum,
𝑑𝑁𝐵 𝑁0 𝜆𝐴
=0= [−𝜆𝐴 𝑒 −𝜆𝐴𝑡max + 𝜆𝐵 𝑒 −𝜆𝐵 𝑡max ]
𝑑𝑡 𝜆𝐵 − 𝜆𝐴
log (𝜆𝐵 /𝜆𝐴 )
𝑡max =
6.6 IDEAL EQUILIBRIUM
From Eq.6, at 𝑡max we have
𝜆𝐴 𝑁𝐴 = 𝜆𝐵 𝑁𝐵 ……………(10)
since
𝑑𝑁𝐵
=0
𝑑𝑡
Thus at time 𝑡max , (and only at 𝑡max ), the activity of parent and activity of accumulated daughter
are equal.
At 𝑡max (as considered in example above), we have:
𝜆𝐴 𝑁𝐴 = 𝜆𝐴 𝑁0 𝑒 −𝜆𝐴 𝑡max
𝜆
log ( 𝐵 )
𝜆𝐴
−𝜆𝐴
= 𝜆𝐴 𝑁0 𝑒 𝜆𝐵 −𝜆𝐴
𝜆𝐴 𝜆𝐴
= 𝜆𝐴 𝑁0 ( ) 𝑒 𝜆𝐵 −𝜆𝐴
𝜆𝐵
𝑇𝐵 𝑇𝐵
= 𝜆𝐴 𝑁0 ( ) 𝑒 𝑇𝐴−𝑇𝐵 … … … … … (11)
𝑇𝐴
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146

When the activities of parent and daughter are, equal, the situation is called ideal equilibrium.
Note that this situation exists only at the moment when, time = 𝑡max .
By Fig. 4 where 𝑑𝑁𝐵 /𝑑𝑡 is positive between times, 𝑡 = 0 and 𝑡 = 𝑡max . This implies that parent
activity in this time range always exceeds daughter activity. Conversely between times 𝑡 = 𝑡max
and 𝑡 = ∞, 𝑑𝑁𝐵 /𝑑𝑡 is negative, implying that the daughter activity exceeds the activity of its
parent.
6.7 TRANSIENT EQUILIBRIUM AND SECULAR EQUILIBRIUM
Let us proceed by considering, the daughter shorter-lived than the parent. i.e
𝑇𝐴 > 𝑇𝐵 .
From Eq. 9 the activity of 𝐵 type is,
𝜆𝐵
𝜆𝐵 𝑁𝐵 = 𝑁0 𝜆𝐴 (𝑒 −𝜆𝐴𝑡 − 𝑒 −𝜆𝐵𝑡 )
Since
𝜆𝐴 𝑁𝐴 = 𝜆𝐴 𝑁0 𝑒 −𝜆𝐴𝑡 ,
𝜆𝐵
𝜆𝐵 𝑁𝐵 = (𝜆̀𝐴 𝑁𝐴 ) (1 − 𝑒 −(𝜆𝐵 −𝜆𝐴)𝑡 ) … … … … . (12)
𝜆𝐵 𝑁𝐵 𝑇𝐴
= (1 − 𝑒 −[(𝑇𝐴−𝑇𝐵)/𝑇𝐴]𝜆𝐵𝑡𝑡 ) … … … … … . (13)
𝜆𝐴 𝑁𝐴 𝑇𝐴 − 𝑇𝐵
For large 𝑡, Eq.13 becomes,
𝜆𝐵 𝑁𝐵 𝑇𝐴
=𝑇 …………(14)
𝜆𝐴 𝑁𝐴 𝐴 −𝑇𝐵
𝑇𝐴
When the ratio, 𝑇 is greater than one, from Eq.13, it is clear that for large 𝑡,
𝐴 −𝑇𝐵
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147

Eq.14 should hold. Since we have 𝑇𝐴 > 𝑇𝐵 such is the case. When the Eq.14 holds, we say that
transient equilibrium exists between parent and the daughter. Learners can also see that
according to Eq. 14 the ratio of daughter and parent activities is constant.
A prominent example of transient equilibrium is the following decay:
228 228
89 Ac has half life of ∼ 6 hours and promptly decays into 90 Th
Thus, we can ignore the presence of 228Ac as an intermediate product. From Eq.14; the activity
ratio corresponding to equilibrium is,
228
𝜆𝐵 𝑁𝐵 Activity of Th
=
𝜆𝐴 𝑁𝐴 Activity of 228 Ra
𝑇𝐴
=
𝑇𝐴 − 𝑇𝐵
6.7
=
6.7 − 1.9
= 1.39 … … … . . (15)
228
i.e., it is > 1. of Th.
228
Thus it is an example of transient equilibrium. Figure 5 shows decay of Ra and growth of
228
Th.
Notice from the Fig. 5 that at larger values of time, the activity of the daughter approaches its
transient equilibrium value of 1.39 times the remaining parent activity (as given by Eq.15).
147
148

If the daughter is longer-lived than the parent, i.e., if 𝑇𝐴 < 𝑇𝐵 , from Eq.13 it can be seen that, the
𝜆 𝑁
ratio 𝜆𝐵 𝑁𝐵 continuously increases as time 𝑡 increases.
𝐴 𝐴
228 228 228

Fig.5: Transient equilibrium between Ra parent and Th daughter. ( source is pure Ra)
148
149
In other words, after sufficient time, the activity of the daughter becomes independent of the
residual activity of the parent, and there can be no equilibrium between them. This is indicated in
Fig. 6 for the case of
210
𝛽 210
𝛼 206
83 Bi 84 Po 82 Pb
𝑇𝐴 = 5.1 days 𝑇𝐵 = 138.4 days
Let us now consider the important case of the daughter being much shorter-lived than parent. Here,
and Eq. 12 becomes
𝜆𝐴 ≪ 𝜆𝐵 (𝑇𝐴 ≫ 𝑇𝐵 )
𝜆𝐵 𝑁𝐵 = 𝜆𝐴 𝑁𝐴 (1 − 𝑒 −𝜆𝐵 𝑡 ) ………..(16)
210 210
Fig.6: There is no equilibrium between Bi(5.1𝑑) and Po(138.4𝑑).
that is the daughter activity is controlled only by its own decay constant.
It can be at once seen that for,
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150
𝑡 ≫ 𝑇𝐵 ,
0.693
the exponential, 𝑡
𝑒 −𝜆𝐵 𝑡 =𝑒 𝑇𝐵
≅0
and so
𝜆𝐵 𝑁𝐵 = 𝜆𝐴 𝑁𝐴 ……………(17)
It should be kept in mind that Eq. 17 is valid only if 𝑇𝐴 ≫ 𝑇𝐵 and 𝑡 ≫ 𝑇𝐵 .
Thus in these cases, we have seen that when the daughter activity equals the parent activity and
the equilibrium is called secular equilibrium.
For a very long-lived parent, on physical grounds, the result given by Eq. 17 is expected. In this
case, the activity of the parent can be taken as almost constant and so the rate of production of the
daughter too is constant. As the daughter quantity grows (accumulates), its rate of decay increases
and finally catches up, after sufficient time, with the rate of its production. When the two rates
become equal, we say that the daughter is in secular equilibrium with the parent, and,
giving,
𝜆𝐵 𝑁𝐵 = 𝜆𝐴 𝑁
𝜆𝐵 𝑇 𝑁 ……….(18)
= 𝑇 𝐴 = 𝑁𝐴
𝜆𝐴 𝐵 𝐵
Therefore in the case of secular equilibrium, the ratio of the number of parent nuclei and daughter
nuclei is constant, and is equal to the ratio of their half lives.
This explains why the percentage of radium contained in uranium was always experimentally found
to be the same-an average of one gram of radium per 3.2 tons of pure uranium.
From Eq. 18
𝑁Ra 𝑇Ra 1620yrs.

= = (𝑇 ≫ 𝑇Ra )
𝑁𝑈 𝑇𝑈 4.5 × 109 yrs. 𝑈
= 3.6 × 10−7
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151
giving the experimentally observed ratio of 1gm to 3.2 tons. Actually, radium is not the immediate
daughter of uranium and we have to consider a radioactive series.However, uranium has the longest
half life in the series and hence all other products which are successively formed (except the end-
product) are in secular equilibrium with uranium and their per cent content in a given sample of
uranium is constant, equal to the ratio of their half lives as given by Eq. 18.
Another example of secular equilibrium, (which is not a part of a naturally occurring radioactive
series) is,
90
𝛽− 𝛽− 90
38 Sr 90 Y 40 Zr (stable).
28 yrs. 39 64.8 hrs.
6.8 RADIOACTIVE SERIES
Each series is formed by successive daughter products, all ultimately derived from a single parent.
Radioactive nuclei found in nature are said to exhibit natural radioactivity. There are now a few
thousand radioactive isotopes which have been produced in the laboratory, mostly by neutron
bombardment. These are said to exhibit artificial radioactivity. Compared to this, the number of
naturally occurring radioactive nuclides is quite small, about 70 .
When naturally occurring radioactive isotopes were studied, scientists found that they could be
divided into four series (isotopes between 𝑍 = 81 and 𝑍 = 92 ).
The reason that there are exactly four series is a consequence of the fact that alpha decay reduces
the mass number of nucleus by 4 .
Therefore, the nuclei whose mass numbers are all given by
𝐴 = 4𝑛.
where 𝑛 is an integer, can decay in descending order of mass number.
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152
6.9 DETERMINATION OF THE AGE OF THE EARTH
Radioactivity is the best clock we can employ to estimate the absolute age of the earth, as it is
totally unaffected by environmental change or natural upheavals like earthquakes, storms etc. The
determination of geological ages is done very often by the so-called lead methods.
It involves the following nuclear processes:
238 206
92 U → emission of 8𝛼-particles → 82 Pb
235 207
92 U → emission of 7𝛼-particles → 82 Pb
232 208
90 Th → emission of 6𝛼-particles → 82 Pb
238
To illustrate how an estimate of time can be made, consider the 92 U series. The stable end-product
206 238
is 82 Pb, and so we can take 𝜆Pb = 0. The half life of U is 4.5 × 109 years. Hence after
sufficient time, (say, a billion years) the only elements present in any appreciable amount will be
uranium and lead. This is because, all elements in the uranium series will be in secular equilibrium
238 206
with the parent U and only Pb will not be in equilibrium, and hence quantity of Pb will
continuously go on increasing while uranium will continuously get depleted. This is indicated in
Fig. 2.13 which depicts the flow of water from a reservoir into a collector, through a chain of other
tanks in between. The diameters of the connecting pipes determine the 'probability' of decay (rate
of flow). The situation is analogous with decay of a radioactive series, finally ending in a stable
product.
Therefore, it is possible for us to apply Eq.9 not only to the first and second elements (𝐴 → 𝐵) but
238 206
also to the first and last. ( U and Pb in the case referred).
Equation 9 is:
152
153
𝑁0 𝜆𝐴
𝑁𝐵 = (𝑒 −𝜆𝐴𝑡 − 𝑒 −𝜆𝐵 𝑡 )
𝜆𝐴 = 𝜆𝑈
𝜆𝐵 = 𝜆Pb = 0 as Pb is stable.
𝑁𝐵 = 𝑁Pb
𝑁0 = 𝑁𝑣
𝑁Pb = −𝑁𝑣 (𝑒 −𝜆U 𝑡 − 1)
= 𝑁𝑣 (1 − 𝑒 −𝜆U 𝑡 ) … … … . (19)
238 206 206

U ore always contains Pb, which can be assumed to be of radioactive origin Pb is the
end-product).
∴ Present number of Pb atoms + Present number of U atoms
= Number of 𝑈 atoms originally present.
i.e.,
𝑁pb + 𝑁U = 𝑁𝑣 ………(20)
Eqs. 19 and 20 can be solved simultaneously to give:
1 𝑁Pb +𝑁U
𝑡 = 𝜆 log ( ) ………..(21)
U 𝑁U
Thus, by spectro-chemically analyzing a sample and knowing its uranium and lead content, it is
possible to estimate the age of that sample. Of course, the most important factor is to ensure that no
helium or uranium has escaped the rock sample during its lifetime.
The oldest surface rocks have been found to have an age of about 3 × 109 years.
When the same method is applied to determine the age of meteorites, it is found that oldest of these
are about 4.5 × 109 years. This corresponds to the age of the earth and is different from the
segregation of surface rocks.
206
Another method may be mentioned. It consists of using the Pb/ 207 Pb isotopic ratio of
radiogenic lead.
153
154
Using Eq. 19 it is easy to see that
206 238 U(1−𝑒 −𝜆𝑈𝑡 )

Pb
207
= 235 U(1−𝑒 −𝜆𝑈′ 𝑡 )
………..(22)
Pb
where
238
𝜆𝑈 = 𝜆 of 𝑈
𝜆′𝑈 = 𝜆 of 235
𝑈.
From Eq. 22, 𝑡 can be evaluated.
238
The main disadvantage of this method is that radon which forms in the decay of U has a half life
of 3.82 days and being gaseous, might escape from the system.
Thus for this method or the first method, the rock samples have to be taken from massive
pitchblende, deeply buried. In such a case, the gas would get trapped in rock cavities and radon
(T = 3.82 days ) cannot be expected to diffuse very far from the dense rocks.
6.10 CARBON DATING-ARCHAEOLOGICAL TIME SCALE

14
Cosmic rays continuously form 6 C in atmosphere. The nuclear reaction is
1 14 14
0𝑛 + 7 N→ 6 C + 11 H
14
6 C is a neutron-rich (6 protons and 8 neutrons) isotope of carbon, which is 𝛽 − active.
14 14
6 C → 6 N + 𝛽 − + antineutrino
14
Half life of 6 C is 5730 years.
14
Thus 6 C would not have been present in the atmosphere, had it not been continuously replenished.
14
C gets combined with hydrogen and oxygen and eventually finds its way into all organic matter.
154
155
14
Since C has a half life of 5730 years it is an ideal radioactive isotope for studying the age of
civilisations and is extensively used in archaeology and anthropology.
14
When an animal or plant dies, its intake of carbon stops and from that moment decay of C is the
only process that continues. Coal and petroleum are organic in nature, but they are so old that there
14
is no trace of C in them. This has been established by experiments. In a young animal or tree,
14
C activity is same as atmospheric carbon which is about 15 disintegrations per gm per min.
14
By carefully measuring the C activity of a fossil or dead tree, it is possible to estimate its age.
This is shown in the problem given below.
EXAMPLE 04: In an archaeological expedition, charcoal from an ancient fire-pit was excavated.
14 14
This sample showed a C activity of 11.3 counts per gm per min. The absolute activity of C in a
living tree is independent of species and it is ∼ 15.3 counts per gm per min. Estimate the age of the
charcoal sample.
Solution: We have,
11.3 = 15.3𝑒 −𝜆𝑡
where
0.693 0.693
𝜆= =
𝑇 5730 years
1.354 = 𝑒 𝜆𝑡 = 𝑒 0.000121𝑡 where 𝑡 is in year

𝑡 = age of charcoal sample in years
log 𝑒 1.354
= years
0.000121
= 2504.65 years.
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156
6.11 SUMMARY
In this unit learners have learned about the basic Radioactive α -,β-, and γ- rays and its properties.
The Law of Radioactive Decay have also been discussed along with Unit of Activity. Radioactive
Growth and Decay has been graphically explained with suitable example. Ideal Equilibrium,
Transient Equilibrium and Secular Equilibrium with examples has been discussed. How the Age of
the Earth can be determined with the help of Radioactive phenomenon has also been discussed in
this unit.
6.12 REFERENCES

1.”The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of Physics, Berlin: Springer
Ver.
2. “Nuclear Two Body Problems and Elements of Nuclear Forces” Experimental Nuclear Physics by N. F.
Ramsey, Wiley: New York.
3. Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum Press, New York.
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157
1. Explain the properties of the radioactive decay.
2. Discuss the law of radioactive decay.
3. Give examples to illustrate the generality of the law of exponential decay and growth.
4. Differentiate between the transient equilibrium and secular equilibrium.
5. Discuss how the age of the earth can be determined by using the phenomenon by radioactivity.
157
158
UNIT 7
ALPHA DECAY
7.1 Introduction
7.2 Objectives
7.3 The α-Spectrum and Fine Structure
7.3.1 Long Range α- Particles
7.4 Q- value in the α-decay
7.5 Range of α particles
7.6 Range-Energy (velocity) relationship
7.7 Geiger-Nuttal law.
7.8 Gamow's theory of α -decay
7.9 Selection rules in Alpha Decay
7.10 Summary
7.11 References
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159
7.1 INTRODUCTION
The alpha decay is one of the earliest nuclear processes and a typical type of radioactivity in heavy
nuclei. A Coulomb barrier of more than 20 MeV is experienced by an alpha particle when it
collides with a heavy nucleus like uranium, whereas alpha particles produced by uranium have
energies of less than 5 MeV. Energy conservation laws appear to be broken in the area of the
potential barrier by this. The close link between half-life and corresponding decay energy, or Q
value, is another significant aspect of the -decay. For understanding the above observed
phenomena, a theory has to be developed because the classical theory was not able to explain the
alpha emission process.
7.2 OBJECTIVES
After studying the unit learners will be able to:
• explain the Range of α particles

• derive Range-Energy (velocity) relationship
• describe Geiger-Nuttal law.
• Explain Gamow's theory of α -decay
• Describe  -particle energies and selection rules
7.3 THE  -SPECTRUM AND FINE STRUCTURE
When alpha particles are emitted from one distinct energy level of the parent nucleus to another
distinct energy state of the product nucleus, they are said to be mono-energetic. When it comes to
some nuclei, it has been noticed that the same radioactive material emits many groups of particles,
each with a distinct energy. The transitions from the various discrete energy states of the parent
nucleus to the various discrete energy states of the daughter nucleus make this possible. The
emission of particles is caused by an energy transition between two distinct nuclear energy states;
the parent nucleus' starting energy state and the daughter nucleus' final energy state are responsible,
159
160
and as a result, the emission spectrum is a line spectrum. These  particles emission give rise to
the so called-fine structure' of a spectrum. We can group such  transitions into two types. As an
228
example, the decay of 90 Th as shown in Fig.1.
Fig. 1 Five groups of  - particles, indicated by slanting lines, emitted by 228Th.
The transitions in this case are from a single, clearly excited energy level of the parent nucleus
(ZXA) to many excited states of the product nucleus (Z-2XA-4). Particles are divided into five groups
according to their energy. Out of them, fifth groups of 5.42 MeV energy -particles leave the
daughter product in ground state, whereas four groups of α-particles leave the daughter nucleus in
excited state. By emitting rays, which are shown by vertical lines, the daughter nucleus
reaches from its excited states to the ground state.
160
161
7.3.1 Long Range α- Particles
The usual or main category of α-particles are those that are emitted by a parent nucleus in ground
state and leave the daughter nucleus in ground state. Long-range α-particles are those released by
an excited parent nucleus that leave the daughter product in the ground state, as in the instance of
212Pb's emission, which is illustrated in Fig.
2.
Fig. 2 Long Range a-particles emitted by 212 Po
This is because the energy of excitation becomes available to the  -particle, when the daughter
nucleus reaches its ground state. The  - emitter is known to emit a few  -particles (  1 ,  2 ) with
much greater energies than those of main group (  0 ) as shown in Fig.2
This is due to the fact that when the daughter nucleus reaches its ground state, the energy of
excitation becomes available to the α-particle. According to Fig. 2, the α- emitter is known to emit
a few α -particles( α1, α2 ) with energies that are significantly higher than those of the main group
(α0).
161
162
Remember, in case of normal emission, the transition is from ground state of parent to ground state
f daughter. In the case of long-range  -particle emission, the transitions are from excited states of
parent nucleus to ground state of daughter nucleus directly. The emission of long-range  particles
provides information about energy levels of the parent nucleus. The emission of short range
 particles due to transition from ground state of parent nucleus to excited states of daughter
nucleus (Fig.1) tells about the energy levels of daughter nucleus.
7.4 Q- VALUE IN THE  -DECAY
The Q value in the  -decay process is the total energy released in the decay process and is called
"disintegration energy" which can be calculated by considering the conservation of momentum and
the energy Q-value is given by
𝑚𝛼
𝑄𝛼 = 𝐾𝛼 ( + 1) ......(1)
𝑚𝑑
with
Ka = Kinetic energy of  particle
ma = mass of  particle
md = mass of the daughter nucleus
It is quite reasonable to replace the ratio of masses with the ratio of the mass numbers. The new
expression for Q can be written as
𝐴
𝑄𝛼 = 𝐾𝛼 ( ) ......(2)
𝐴−4
with A = mass number of parent nucleus
Since usually A is large and so Q  K , i.e. the  particle carries away most of the disintegration
energy.
162
163
7.5 RANGE OF α-PARTICLES
The α-particles released by naturally occurring radioactive elements are quickly absorbed by
matter. They are able to pass through thin foils made of mica or aluminium as well as thin papers.
They can only get through the top few layers of such material, though. As they go through stuff,
they lose energy. After a given distance in matter, the mono energetic particles lose all of their
energy by subsequent collisions (substance). The term "range" refers to the maximum distance a
particle in a substance can go before losing all of its energy. It is expressed in unit of
length(cm).The range is seen to be very small in solids and liquids ( 10-3 mm for energy of a few
MeV) and large in gases (few centimetres). Increases in temperature and pressure have opposite
effects on the range of the gas, respectively.
By measuring the ionisation that particles cause along their passage through the medium at various
points along their range, they can be identified. Another approach is to take photos of their tracks in
the Cloud Chamber.
7.6 RANGE-ENERGY (VELOCITY) RELATIONSHIP
Rutherford pointed out the fact that there appears to be a systematic relationship between the half-
period and the range (R) of emitted -particles. If the half-period is shorter, it seems like the particle
velocity (energy) will be higher (large decay constant). The following mathematical relationship
between the measured values of the ranges and energy of the -particles has been established.
3
𝑅 = 𝛼𝐸 2 ......(3)
This empirical relationship, valid in a limited energy range (3 to 7 MeV) is known as Geiger's law.
For the energy (E) in MeV and the range (R) in meter, the value of the constant a = 3.15 x 10 -3
m/MeV.
If v is the velocity of  particles then
we have
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164
𝑅 = 𝑏𝑣 3 ......(4)
With b = 9.416 x 10-24 unit.
In the case of a solid, the range RS (in meter) is related to the range RS in air. As
1
0.312𝐴2
𝑅𝑆 = ......(5)
𝜌
where  is the density of the solid of mass number A.
For higher and lower energies, more appropriate relationships are give by Eqs. 4 and 5 respectively.
𝑅 = 𝑣4
and 𝑅 𝛼 𝑣 3/2 ………( 6)
7.7 GEIGER-NUTTAL LAW
Geiger and Nuttal had discovered an empirical relationship between the decay constant (λ) of
naturally occurring radioactive elements and the range (R) of α-particles. Geiger-Nuttal law, which
may be written as, describes this relationship between R and λ as
𝑙𝑜𝑔 𝜆 = 𝐴 + 𝐵 𝑙𝑜𝑔 𝑅 ......(7)
with A and B as constants:
According to this law, the a-particles emitted by elements having larger disintegration constant (i.e.
shorter half-period) have longer ranges and vice-versa. Geiger - Nuttal law tells that the graph of
log λ versus log R is a straight line with slope B.
As illustrated in Fig.3, distinct straight parallel lines are obtained for various radioactive series,
suggesting that the constant B has the same value across the spectrum while the constant A varies.
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The Geiger-Nuttall law can also be expressed as, given that the range is related with energy as
indicated by Eq.1
Fig.3 Variation of log  with log R
𝑙𝑜𝑔 𝜆 = 𝐶 + 𝐷 𝑙𝑜𝑔 𝐸 ......(8)
where C and D are other new constants.
Fig.4. Variation of log  with E
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Since the half-life (T) is inversely proportional to decay (  ) constant, Geiger - Nuttall law gives
relation between half-life and range or energy. The variation of half-life with energy is shown in
Fig.4.
It shows that for a small increase of energy (by a factor of 2) the half-life decreases enormously.
Remember (by a factor of 1024). This enormous variation in half-life cannot be explained by the
classical theory.
7.8 GAMOW'S THEORY OF  - DECAY
In 1928, G. Gamow, R. W. Gurney, and E.U. Candon independently and nearly simultaneously
provided an acceptable explanation of decay using the principles of quantum mechanics. The α-
particle is believed to be trapped inside the nucleus by the Coulomb barrier in a basic form of this
theory, but it only has a small chance of exiting the nucleus.
The disintegration of heavy nuclei reveals that two protons and two neutrons can occasionally
group together to form a particle within the nucleus. The strong, short-range, and attractive nuclear
forces must act on a α-particle when it is inside the nucleus. However, as it leaves the nucleus, the
Coulomb repulsion caused by the remaining positive nucleus (Z-2)e, where +Ze is the charge of the
parent nucleus, acts on it.
Outside the nucleus the Coulomb potential energy V(r) at a distance r of  -particle from the centre
of the nucleus is given by
2(𝑍 − 2)2
𝑉(𝑟) = ......(9)
4𝜋𝜀0 𝑟
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Fig.5. Potential well and Coulomb potential barrier of a nucleus. The  particle wave function is
oscillatory nature inside the nucleus (r < R) and at large distances s(r>b) .
Fig.5 depicts a schematic representation of the Coulomb potential barrier for disintegration (the
potential energy between the particle and daughter nucleus). The -particle is bound and cannot
escape at all if the decay Q-value is positive.
The Coulomb potential given by equation, (r > R) V(r), describes what happens outside the nucleus
eq. (9). If we neglect the recoil energy of the daughter nucleus, the kinetic energy of the -particle
asymptotically equals Q. The -particle encounters the attractive nuclear forces close to the nuclear
surface and develops a kinetic energy that is dependent on the depth of the nuclear potential well
U). In Fig. 5, the nuclear potential is shown as a square well with a sharp edge and a width R
(nuclear radius).
The value of V(r) turns out to be 34 MeV in the case of 88Ra226 emitting -particles that produce
83Ra223 as a daughter nucleus. However, Q = 4.88 MeV, which is much lower than V, is the
energy of a- disintegration (r) According to classical mechanics, a particle must have an energy at
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least equal to V(r) in order to leave or enter the nucleus, which is contradictory to the energy levels
that have been measured.
However, such a classical forbidden occurrence is possible in quantum mechanics. According to

the Schrödinger wave equation, the particle is represented as a wave. By substituting the relevant
particle potentials, we can write the wave equation down for the various regions. An initial particle
inside the nucleus (rR) has a finite probability of leaving the nucleus when these equations are
solved with the proper boundary conditions. We can imagine a particle escaping from the nucleus
as if there were tunnels in the potential barrier. Thus, the phenomenon is known as the "quantum
mechanical tunnel effect." The initial energy and the half-life of the dissolving nucleus can be
calculated according to the quantum mechanical theory (parent nucleus). As a result, this quantum
mechanical barrier penetration theory provides an explanation of Geiger -Nuttall law.
The probability that a particle exists as a recognizable entity inside the nucleus prior to emission is
a significant aspect of the decay problem. The structure of the states of the parent and daughter
nuclei strongly affects the formation probability (P), which is what it is called. It is a measure of
how much the daughter nucleus in its final state plus a particle "looks like" the initial state.
Although there are certain errors in determining the barrier penetration factor, it is theoretically
possible to derive it from experimental measurement. P is difficult to calculate, so we assume it is
equal to unity for the sake of simplicity.
The  -particle wave function is oscillatory inside the nucleus (r<R) and also at large distance (r>
b) as shown in Fig.5. Within the barrier region (R r b), it is a decreasing function of r. The barrier
can be considered to be a set of narrow rectangular barriers of width dr, which the  particle has to
penetrate in order to escape form the nucleus. For r>R, V(r) varies as 1/r, therefore R/b = Q/H
where H being the barrier height.
Given that for existence of  -particle we can imagine it moving back and forth inside the nucleus
with a speed 'v' given by its kinetic energy and reaches the barrier with a frequency f given by
𝑣
𝑓≈ ......(10)
2𝑅
where R is the radius of the nucleus.

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The value of f for  particle is
1.7 107
f = =8.5 1020s −1
2 10−14
As mentioned earlier, quantum mechanically, the α-particle in a nucleus is represented by a wave

function  . Inside the nucleus  is oscillatory with a wave number
√2𝑚(𝑄 + 𝑈)
𝐾= ......(11)
ℏ2
with (Q+U) as the total kinetic energy of a-particle inside nucleus.
and m is the particle's mass. If we assume that any reflected wave from the barrier's outer edge at
r=b is negligible, the wave function within the barrier region (R r b) is not zero and takes the shape
of a falling exponential. The wave function is once more oscillatory outside of the nucleus, with a
wave number at very long distances.
By treating the barrier as a series of narrow rectangular barriers in one dimension and assuming that
the element's barrier height is given below , we can estimate the penetration probability.
2𝑚𝑄
√
ℏ2
2𝑍𝑒 2
𝑉(𝑟) = ......(12)
4𝜋𝜀0 𝑟
where Z is the atomic number of the daughter nucleus. The wave function at each barrier has the
form e −  r
2𝑚[𝑉(𝑟)−𝑄]
where 𝛽=√ ......(13)
ℏ2
−  ( r + dr )
and after the barrier the wave function is reduced to have the form e
The transmission probability through a single element of the barrier is e −2  dr .
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Now, the total of these infinitesimal probabilities, calculated for the set of barrier elements between
r = Rand and r = b, is the probability of breaking through the entire barrier. The entire
"transmission probability" (T) can be expressed as because the product of such infinitesimal
exponentials is equal to the exponential of the integral of the exponents, i.e., it equals the
exponential of the sum of their exponents.
𝑏
𝑇 = 𝑒𝑥𝑝 [−2 ∫ 𝑅(𝑑𝑟)]
𝑅
= 𝑒𝑥𝑝(−2𝐺) ......(14)
where G is called the 'Gamow factor' which can be evaluated, with V(r) given by equation (11) and
obtained as
(2𝑚) 2𝑍𝑒 2
𝐺=√ 2 ( ) [𝑐𝑜𝑠 −1 √𝑥 − √𝑥(1 − 𝑥)] ......(15)
ℏ 𝑄 4𝜋𝜀0
with x = r/b = Q/H ......(16)
Combining the results of Eqns.10 and 14 with the preformation probability, we can write an
expression for the  - decay rate or the decay constant (  ) as
𝜆 = 𝑝𝑓 𝑒𝑥𝑝(−2𝐺) ......(17)
In order to estimate the dependence of the half-life on the decay Q-value, as shown in Fig.4, we
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take U as the parent nucleus and calculate the penetrability factor (transmission probability)
given by Eqn 14 for Q = 4.268 MeV (which is the actual  decay Q-value of 238
U and Q = 5.268
MeV i.e. for two values that differ by 1 MeV.
Now slightly rewriting Eqn 13 as
2𝑚𝑐 2 𝑄 𝑄 𝑄
𝐺 = 2𝛼𝑧√ [𝑐𝑜𝑠 −1 √ − √( ) (1 − )] ......(18)
𝑄 𝐻 𝐻 𝐻
where the quantity  = e2 4 0 C is the usual fine-structure constant and is equal to 1/137.
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In the case of 238

U we take the following data Z = 90, the rest-mass energy me of  particle equal
to 3727 MeV, the Coulomb barrier height H = 27.87 MeV. (Obtained using Eqn 10) with r = R =
1.2 (A11/3,A21/3) = 9.3 fm, for the effective separation distance between the daughter nucleus with
A1 = 234 and the  -particle with A2 = 4, at the peak of the barrier.
Substituting this data into Eqn 16, we obtain the G = 44.391 and 36.047 for Q = 4.268 and 5.268
MeV, respectively.
1
By using Eqn 15 and since half-life is related to decay constant as, t1/2 

we obtain
t1 2 ( 5.268) / t1 2 ( 4.628) = exp ( −16.69 ) = 5.7 10−8
This suggests that an increase of 1 MeV in Q-value changes the half-life more by seven orders in
magnitude.
We can utilize on another alternate method by using the following data for the  -decay.
Barrier height = H = V(r) = 15MeV, Q= 5MeV, m = 4 x 1.66 x 10-27 kg. From equation 13 we have
2𝑚
𝛽=√ [𝑉(𝑟) − 𝑄]
ℏ2
= 1.388 ×10+15
Then the value of T = 2.55 x 10-24 s
Accordingly, the probability of the α-particle colliding with the barrier wall and passing through it
is around 2.5 in 1024 collisions.
 = fT = 8.5 1020 s −1  2.55 10−24 = 2.17 10−3 s −1
So, the mean life time against  decay is
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1 1
m = = = 461sec = 7 min 41sec
 1.17 10−3
7.9 SELECTION RULES IN ALPHA DECAY
. When the spins of the initial and final states differ, it can be shown that the -particle loses some
orbital angular momentum, which prevents the emission. However, it often has little effect on the
decay probability. There are two key elements that control the intensity variations. The first is
caused by the substantial energy dependence of barrier penetrability, which falls off as the final
state's energy rises. The second component, which may affect the chance of preformation, is the
structural variations between the initial and final states. Now, the effect caused by a particle losing
a few units of angular momentum is minimal in comparison to these two effects.
The parity of the wave function is (-1)L, where L is the orbital angular momentum quantum
number, and there is a parity selection criterion for decay that depends on angular momentum. The
parities of the initial and final nuclear states will differ if L is odd and remain the same if L is even
due to the conservation of parity in α- decay. As a result, this rule forbids some decays.For an
example, a o+ parent state cannot decay into odd parity state with even spin or an even parity state
with odd spin.
7.10 SUMMARY
After studying the unit learners have learnt about
• The α-Spectrum and Fine Structure

• Long Range α- Particles
• Q- value in the α-decay
• Range of α particles
• Range-Energy (velocity) relationship
• Geiger-Nuttal law.
• Gamow's theory of α -decay
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• Selection rules in Alpha Decay
7.11 REFERENCES
 Nuclear Physics by Irving Kaplan, Narosa Publishing House

 The Atomic Nucleus by R D. Evans, McGraw-Hill Publications.
 Elements of Nuclear Physics by M.L.Pandya,R.P.S.Yadav
 Nuclear Physics An Introduction by S.B.Patel
 Nuclear Physics by D. C Tayal.
 Nuclear Physics by S.N.Ghoshal
1. “The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of Physics, Berlin: Springer Ver.
2. “Nuclear Two Body Problems and Elements of Nuclear Forces” Experimental Nuclear Physics by N. F.
Ramsey, Wiley: New York.
3. Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum Press, New York.
1. Discuss the quantum mechanical tunnelling of a particle.
2. Write about 'fine structure of a spectrum and its importance.
3. Write a note on the selection rules for a particle emission.
4. Explain how the range and energy of the  -particle and the half-life period are related through
Geiger-Nuttall Law.
5. Discuss the Gamow theory of  -decay and how it explains the main features of  particle
emission process.
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UNIT 8
BETA DECAY
8.1 Introduction
8.2 Objectives
8.3 What is Beta Decay?
8.3.1What happens in Beta Decay?
8.3.2 Types of Beta Decay
8.4 Q-Value in Beta Decay
8.5 Neutrino Hypothesis
8.6 Fermi's Theory of β -Decay
8.7 Fermi-Kurie Plots
8.8 Selection Rules for  -Decay
8.9 Parity Violation in Beta Decay
8.10 Summary
8.11 References
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8.1 INTRODUCTION
The emission of electrons is associated with several nuclear processes. Some nuclei have been seen
to emit positrons. In some nuclei, electron capture—in which the nucleus captures an orbital
electron, usually from the K-shell—occurs. The term "beta decay" refers to all three closely related
phenomena.
The process by which an unstable nucleus modifies itself to achieve the best possible neutron-to-
proton ratio for a given mass number is known as β-decay. In contrast to α-decay, which results in a
4 unit decrease in mass number,β -decay has no change in the mass number (A = 0). The β-
spectrum, one of the nuclear spectra, has the unusual quality of being "continuous," whereas the α-
spectrum is a line spectrum. Every β- spectrum contains electrons (β- particles) with energies
ranging from zero to a specific maximum. This radionuclide's unique property is its maximum
energy value
Fermi proposed a successful theory of the shape of the spectrum and the lifetime of emitters in
1934. Many of the fundamental issues relating to -particle emission were resolved by 1950.
This continuous nature of  -spectrum gave rise to serious difficulties, in understanding the  -
decay process. The chief problem is that, just like the  - decay,  -decay also an energy transition
between two definite energy states as shown in Fig.1
Fig.1  -decay transition between two definite energy states.

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8.2 OBJECTIVES
After studying this unit, learners will be able to
• Explain what is Beta Decay?

• Describe the types of Beta Decay
• Evaluate the Q-Value in Beta Decay
• Discuss neutrino hypothesis
• Describe Fermi's Theory of β -Decay
• Understand the various selection rules for  -Decay
• Explain parity violation in Beta Decay
8.3 WHAT IS BETA DECAY?
Beta Decay is a type of radioactive decay in which a proton is transformed into a neutron or vice
versa inside the nucleus of the radioactive sample. Processes like beta decay and alpha decay allow
the nucleus of the radioactive sample to get as close as possible to the optimum neutron/ proton
ratio. While doing so, the nucleus emits a beta particle which can either be an electron or positron.
Remember that there either a proton can turn into a neutron or a neutron into a proton. Electron and
the positron are generated to obey the law of conservation of charge. Beta-decay occurs via weak
interaction.
8.3.1What happens in Beta Decay?
Beta decay is a radioactive decay in which a beta ray is emitted from an atomic nucleus. During
beta decay, the proton in the nucleus is transformed into a neutron and vice versa. If a proton is
converted to a neutron, it is known as β+ decay. Similarly, if a neutron is converted to a proton, it is
known as β– decay. Due to the change in the nucleus, a beta particle is emitted. The beta particle is
a high-speed electron when it is a β– decay and a positron when it is a β+ decay. Beta particles are
used to treat health conditions such as eye and bone cancer and are also used as tracers.
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Example:
One of the examples of beta decay is the β– decay of the carbon atom. Here, a neutron of carbon is
converted into a proton, and the emitted beta particle is an electron. Similarly, the β+ decay of
carbon-10 can be represented by an equation as follows:
10 10
6𝐶 → 5𝐵 + 01𝑒 + ……………..(1)
Here, the proton of the carbon atom is converted into a neutron, and the emitted beta particle is a
positron.
8.3.2 Types of Beta Decay
There are two types of beta decay types:
(i)beta minus (β–) and
(ii) beta plus (β+).
(i)Beta-Minus Decay
• In beta minus, a neutron is transformed to yield a proton, causing an increase in the atom’s
atomic number. The neutron is neutral, but the proton is positive.
• To maintain the conservation of charge, the nucleus in the process also produces an electron
and an antineutrino.
• Antineutrino is the antimatter counterpart of neutrino. Both of these are neutral particles
with negligible mass. They interact with matter very weakly and can even pass through the
entire earth without being disturbed.
• In beta minus decay, the change in atomic configuration is:
𝐴 𝐴
𝑍𝑋 → 𝑍+1𝑌 + 𝑒 − + 𝜈̅
𝑛 → 𝑝 + 𝑒 − + 𝜈̅ ……………(2)
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14
Examples of beta minus decay include the decay of C into 14N and it usually occurs in neutron-
rich nuclei.
(ii)Beta-Plus Decay
• In beta plus decay, the proton disintegrates to yield a neutron causing a decrease in the
atomic number of the radioactive sample. The nucleus experiences a loss of proton but
gains a neutron.
• Again, conservation of charge is important. The beta plus decay in order to obey the
conservation law also yields a positron and a neutrino.
• A positron is the antimatter equivalent of an electron; the same in all aspects except that a
positron has a positive charge.
• A neutrino’s behaviour is the same as the antineutrinos. As expressed in the equation, it is:
𝐴 𝐴
𝑍𝑋 → 𝑍−1𝑌 + 𝑒+ + 𝜈
𝑝 → 𝑛 + 𝑒 + + 𝜈 ……………(3)
Beta plus decay can happen only if the daughter nucleus is more stable than the mother nucleus.
This difference goes into the conversion of a proton into a neutron, a positron and a neutrino. There
is no increase in mass number because a proton and a neutron have the same mass.
The below image depicts the example of beta minus (β–) decay and beta plus (β+) decay.
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Fig 1: Beta decay types (https://cdn1.byjus.com)
8.4 Q-VALUE IN BETA DECAY
In nuclear and particle physics, the energetics of nuclear reactions is determined by the Q-
value of that reaction. The Q-value of the reaction is defined as the difference between the sum
of the rest masses of the initial reactants and the sum of the masses of the final products in
energy units (usually in MeV).
Consider a typical reaction in which the projectile a and the target A give place to two products, B
and b. This can also be expressed in the notation we have used so far, a + A → B + b, or even in a
more compact notation, A(a,b)B.
The Q-value of this reaction is given by:
𝑄 = [𝑚𝑎 + 𝑚𝐴 – (𝑚𝑏 + 𝑚𝐵)]𝑐2 ……………(4)
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When describing beta decay (reaction without projectile), the disintegrating nucleus is usually
referred to as the parent nucleus and the nucleus remaining after the event as the daughter
nucleus. The emission of a beta particle, either an electron, β–, or a positron, β+, changes the
atomic number of the nucleus without affecting its mass number. The total rest mass of the
daughter nucleus and the nuclear radiation released in a beta disintegration, m Daughter +
mRadiation, is always less than that of the parent nucleus, m parent.
The mass-energy difference,
Q = [mparent – (mDaughter + mRadiation)]c2 …………….(5)

appears as the disintegration energy, liberated in the process. For example, the Q-value of typical
beta decay is:
Fig.3: Q-value in beta decay (http://nuclear-power.com)
In the process of beta decay, either an electron or a positron is emitted. This emission is
accompanied by the emission of antineutrino (β- decay) or neutrino (β+ decay), which share the
energy and momentum of the decay. The beta emission has a characteristic spectrum. This
characteristic spectrum is caused by the fact that either a neutrino or an antineutrino is emitted
with the emission of a beta particle. The shape of this energy curve depends on what fraction of
the reaction energy (Q value-the amount of energy released by the reaction) is carried by the
massive particle. Beta particles can therefore be emitted with any kinetic energy ranging from 0 to
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Q. After an alpha or beta decay, the daughter nucleus is often left in an excited energy state. To
stabilize itself, it subsequently emits high-energy photons, γ-rays
8.5 NEUTRINO HYPOTHESIS
Pauli, in 1930, postulated the existence of a new particle called neutrino denoted by  and should
be emitted in the  -decay process. This new particle carries away an amount of energy equal to the
difference between the observed  - particle energy and the endpoint energy r(Emax) of the
continuous spectrum, in order to satisfy the conservation of energy. So, there can be 3-ways of
energy sharing between the  -particle and neutrino as mentioned below:
(i)  -particle can carry the maximum energy ( E  = Emax) and neutrino carrying zero energy ( E =
0)
(ii)  -particle carrying some medium energy ( E  <Emax) and neutrino carrying low energy ( E <
E  ). So that E + E = Emax
(iii)  -particle with very low energy ( E  << E) and neutrino carrying maximum energy ( E > E  )
So that E + E = Emax
Now, in order to satisfy the conservation of angular momentum, charge and nuclear statistics, the
following properties are assigned to neutrino. Remember, the existence of neutrino is observed
experimentally, after about 25 years of its proposal.
(i) Neutrino has zero charge and hence negligible magnetic moment.
(ii) The mass of neutrino is zero or almost zero
(iii) Neutrino has a spin of ½
(iv) Neutrino must be a fermion in order to obey nuclear statistics.
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Just like other elementary particles like electron, neutrino should have an antiparticle, called anti-
neutrino denoted by  and has same properties as that of neutrino. The only difference between
neutrino and antineutrino particles is the following. A neutrino has spin always antiparallel to its
momentum, where as anti-neutrino has its spin parallel to its momentum. In other words, neutrino
is a left-handed particle while antineutrino is a right-handed particle, as shown in Fig.4..
To describe the handedness of the neutrino, the idea of helicity, (which is presently out of the scope
of this course) is used. Remember that the helicity (H) is -1 for neutrino and +1 for anti-neutrino
Fig.4: Neutrion and Antineutrino
8.6 FERMI'S THEORY OF  -DECAY
Now, let us understand the features of Fermi's theory to explain the  -decay process.
According to Fermi's theory, the electron and neutrino are produced during the process of -particle
emission. Keep in mind that the electron is created, not present, in the nucleus. The neutrino is also
referred to as the -particle's "birth partner." The basic concept In order to use the perturbation
theory, Fermi's hypothesis requires that the interaction responsible for the decay process be "weak."
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The Hamiltonian that causes the decay provides the formula for the transition rate between the
initial and final states in quantum mechanics. The likelihood of the decay is affected by a number
of variables, including the energy released during the process, the angular momentum of the
particles released, the nuclear charge, and the characteristics of the beginning and final nuclear
states The Fermi's Golden Rule, which is a foundation of quantum mechanics and is written as the
transition probability as
2𝜋
𝑑𝑊𝑓𝐼 = |< 𝑓|𝐻𝑊 |𝑖 >|2 𝜌(𝐸) ......(6)
ℏ
where Hw is the interaction Hamiltonian, /P(E) is called the 'density of states factor' or 'phase space
factor', f and i indicate the final and initial states.
The expression for density of states  ( E ) given by
𝑑𝑁
𝜌(𝐸) = ......(7)
𝑑𝐸
where N is the number of states of particles involved in the decay that can be put in a given volume
and E is the total energy of the particles, dN is the number of final states with energy in the interval
E to (E + dE).
The phase-space factor is mainly responsible for how the energy produced during the transition is
transformed into the kinetic energy of the particles. The distribution of energy is also influenced by
the Coulomb field. For the sake of simplicity, it is assumed that the particles are released when the
Coulomb field is absent.
If the kinetic energy of the remaining heavy nucleus is neglected, the neutrino and the β- particle
each receive a portion of the total energy released during the transition, i.e. Emax = E + E
The density of final states  ( E ) completely determines the form of  -spectrum. In order to
compute  ( E ) , we have to consider 'phase-space'. The graph of position of the particle(x) against
its momentum (  x ) constitute a phase-space as shown in Fig.5
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Fig.5: Phase space in quantum mechanics.
A particle in a phase-space is represented by a 'cell' and not by a point. The volume of the cell is
always same and equal to h.
The transition rate (  ) between an initial state (i) and a final state (f) is given by
2𝜋 𝑑𝑁
𝜆= |𝑀|2 ......(8)
ℏ 𝑑𝐸
With M is the matrix element of the transition (it gives an energy times volume) and is equal to
|<f|Hw|i>| given in Eqn.2.3 and is independent of energy of the electrons (  -particles) emitted and
can be treated as constant for most of the  - transitions.
The number of energy states available to a  -particle with momentum between pẞ and (pẞ+ dpẞ)
and constrained to a volume V is given by
4𝜋𝜌𝛽2 𝑑𝜌𝛽 𝑉
[𝜌(𝑃)𝑑𝑃]𝛽 = ......(9)
ℎ3
A similar formula applies to the particle neutrino, so we can write
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4𝜋𝜌𝜈2 𝑑𝜌𝜈 𝑉
[𝜌(𝑃)𝑑𝑃]𝜈 = ......(10)
ℎ3
Assuming that the phase-space available for either particle is independent of the other i.e. there is
no angular correlation between their directions of emission, the total number of energy states
available for the two particles is
(4𝜋)2 𝑉 2 𝜌𝛽2 𝑑𝜌𝛽 𝜌𝛽 𝑑𝜌𝛽

𝑑𝑁 = ......(11)
ℎ6
The energy of neutrino can be written as
𝐸𝜈 = 𝜌𝜈 𝑐 = 𝐸𝑚𝑎𝑥 − 𝐸𝛽 ......(12)
For a fixed ẞ particle energy (Eẞ) we can write
𝑑𝐸
𝑑𝜌𝜈 = ......(13)
𝐶
Substituting Eqn.12 and 13 into Eqn.11 gives an expression for the phase-space factor in terms of
the energy and momentum of the  -particle as
𝑑𝑁 2
𝛼 𝜌𝛽2 (𝐸𝑚𝑎𝑥 − 𝐸𝛽 ) 𝑑𝜌𝛽 ......(14)
𝑑𝐸
Now, we have to consider a correction factor for the effect of Coulomb field on the emitted
particles. This correction factor is called "Fermi function', denoted by F (Z, E  ) where Z is the
atomic number of the final (daughter) nucleus. This function has been evaluated for many different
situations and tabulated values are available elsewhere.
Now, substituting the result of Eqn14, including the Fermi function, into Eqn.6, we obtain finally
an expression, for the differential  decay probability per unit time for  -particle momenta in the
range pẞ to (pẞ+ dpẞ) as
2
dW𝑓𝐼 = d𝜆(𝜌𝛽 ) = C|𝑀|2 𝐹(𝑍, 𝐸𝛽 )𝜌𝛽2 (𝐸𝑚𝑎𝑥 − 𝐸𝛽 ) 𝑑𝜌𝛽 ………(15)
where C is a constant
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By integrating the equation (15), the total decay rate can be obtained and we get
2
𝜆 = 𝐶|𝑀|2 ∫ 𝐹(𝑍, 𝐸𝛽 ) 𝜌𝛽2 (𝐸𝑚𝑎𝑥 − 𝐸𝛽 ) 𝑑𝜌𝛽 ………….(16)
2
with f(Z, 𝐸𝑚𝑎𝑥 ) ∫ 𝐹(𝑍, 𝐸𝛽 ) 𝜌𝛽2 (𝐸𝑚𝑎𝑥 − 𝐸𝛽 ) 𝑑𝜌𝛽 ………(17)
Accurate calculations can be made for the function (Z, Emax). Thus, it is convenient to describe the
β-decay results in terms of the product of f(Z, Emax) and the transition half-life (t). The
"comparative half-life" is the result of ft. According to the Eqn16, which separates the phase-space
1
effects from the transition's important details, the product ft is proportional to to 2
. It is more
M
convenient to describe the value of log10ft instead of ft with t in seconds because the value of ft
changes over many orders of magnitude (103 to 1020 sec) for different transitions.
In general, the largest matrix elements and hence the smallest ft values result from the most
fundamental β-decays (of low z elements). Since the matrix element produces energy multiplied by
volume, as was already established, the interaction energy necessary for theβ -decay process is on
the order of 10-4 MeV. This suggests that the interaction of decay is weak.
8.7 FERMI-KURIE PLOTS
Eqn15 can be written as below gives a staright line graph plotted against electron (  -particle)
energy E  .
1/2
𝑑𝑊𝑓𝐼
[ ] = (Constant)|𝑀|(𝐸𝛽𝑚𝑎𝑥 − 𝐸𝛽 ) … … … … . (18)
𝐹(𝑍, 𝐸𝛽 )𝜌𝛽2 𝑑𝜌𝛽
The function is called 'Kurie function' (K) written as
1/2
𝑑𝑊𝑓𝐼
𝐾=[ ] ......(19)
𝐹(𝑍, 𝐸𝛽 )𝜌𝛽2 𝑑𝜌𝛽
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The Fermi-Kurie plot is the graph that K is plotted on and is found to be a straight line over a
wide energy range. Experimentally, the transition dwfi is the quantity of β -particles per energy
interval, and the particles may be found and counted using a Geiger counter. A magnetic
spectrograph may be able to determine the energy (or momentum) interval.
The Fermi-Kurie plot's linear nature supports the notion that leptonic particles do not carry orbital
angular momentum. The linear dependence is not present in Fermi-Kurie graphs for decays in
which the leptons take orbital angular momentum.
8.8 SELECTION RULES FOR  -DECAY
Because the matrix element determines the probability of a transition and, hence, the order of
transition. So, a sequence of terms can be used to express the matrix element (M). When they don't
disappear, these terms get smaller and smaller. It's important to focus on the first non-vanishing
term. The transitions that result from the first term being unity and energy-independent are known
as "permitted transitions."
The first term in the expansion of the matrix element becomes zero for some transitions because of
the initial and final nuclear wave functions. These transitions are known as "first forbidden"
transitions, which is what the expansion's second term refers.
We must use the third non-vanishing term, which is much smaller, if the first and second terms
both disappear during the expansion of the matrix element. Second forbidden transitions are these
transitions.
For β-decay, the criteria that link the transition's properties, such as changes in angular momentum
and parity, with the transition order are referred to as "selection rules." It is possible to determine
the type of interaction producing the β-decay by comparing the experimentally found order of -
transition with the theoretical values order. There are two types of selection rules to be applied for
 -decay process
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CASE I : Li and Lf are the nuclear spin vectors for the initial and final states in the transition, then
the change in angular momentum (L) is given as
𝛥𝐿 = 𝐿𝑓 − 𝐿𝑖 ......(20)
where i and f refer to initial and final states respectively
The orbital angular momentum carried away by the electron-neutrino pair must be equal to the
change L. of the nucleus in  -transition. So, for allowed transition 𝛥𝐿 = 0 The particles can be
aligned either parallel or anti-parallel to each other, so that the total spin angular momentum carried
away by them can be either S = 0 (parallel) or S=1 (anti-parallel) when S = 0, the spin change 𝛥𝑆 =
𝛥𝐿 = 0 and this rule is called 'Fermi Selection Rule'.
When S = 1, then |  S|=| I | = 1 (in Unit of ) and this rule is called 'Gamow-Teller selection
Rule."
CASE II: The vector difference between initial and final angular momentum of the nucleus will be
equal to 1. So, the selection rule can be written as 𝛥𝐿 = ±1. But there cannot be transition from
zero spin angular momentum states to zero angular momentum states i.e. 𝐿𝑖 = 𝐿𝑓 = 0. This is
excluded in 'Gamow-Teller section rule."
Example: Check whether the transition 𝑛 → 𝑝+ 𝑒 − is allowed by both the type of selection rules.
Solution: Since parity associated with orbital angular momentum is (-1) L, and in allowed
transitions (L = 0) there will be no parity change.
The selection rules forbidden transitions are less simple than the rules for allowed transitions.
Here the leptons carry away an additional spin angular momentum of one unit.
Here, there will be a parity change if the orbital angular momentum quantum number L is odd and
no change in parity if L is even.
For a first forbidden transition L = (1) there will be a parity change.
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8.9 PARITY VIOLATION IN BETA DECAY
β-decay exhibits a further peculiarity. This was discovered in 1957 by C.S. Wu who observed the
decay of radioactive cobalt into nickel
60 Co → 60 Ni + e− + ν̄ ……….(21)
The cobalt sample was kept a low temperature and placed in a magnetic field so that the spin of
the cobalt was pointing in the direction of the magnetic field.
Fig.6: Beta emission from Cobalt
She discovered that most of the electrons emerged in the opposite direction from the applied
magnetic field. If we write s for the spin of the parent nucleus and pe for the momentum of an
emitted electron, this means that the average value of the scalar product s · pe was negative. In order
to balance the momentum the antineutrinos are usually emitted in the direction of the magnetic field,
so that the average value of s · pν¯ was positive.
Under the parity operation

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r → −r
and
p → −p
but angular momentum which is defined as a vector product
L = r × p,
is unchanged under parity

L → L.
Spin is an internal angular momentum and so it also is unchanged under parity.
But this means that the scalar product s · pe does change under parity
s · pe → −s· pe
so that the fact that this quantity has a non-zero average value (or expectation value in quantum
mechanics terms) means that the mechanism of β-decay violates parity conservation
If we view the above diagram in the corner of a mirrored room so that all the directions were reversed
the spin would point in the same direction, but the electron direction would be reversed so that in
that world the electrons would prefer to emerge in the direction of themagnetic field.
60
The spin of the daughter nucleus
2 Ni28is 4 (it is produced in an excited state) whereas that of the
parent 60 Co was275, so that in order to compensate for unit of angular momentum lost (in the direction
of the magnetic field) the angular momentum the antineutrinos and electrons have their spins in the
1
2 This means that the antineutrinos have a spin component + in their
direction of the magnetic field. 2
1
direction of motion (in units of h̄) whereas the electrons have a spin component − 2 in their direction
of motion. The sign of the component of the spin of a particle in its direction of motion is called
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the “helicity” of the particle. Neutrinos always
2
have
2
negative helicity (antineutrinos always have
1 1
positive helicity). An electron can have component of spin either + 2 𝑜𝑟 − 2
in its direction
of motion (either positive or negative helicity). However, the electrons emitted in β-decay usually
have negative helicity (positrons emitted in β-decay usually have positive helicity). This means that the
mechanism responsible for β-decay (called the “weak interaction”) distinguish between positive and
negative helicity and therefore violate parity.
8.10 SUMMARY
After studying this unit, learners have learnt about how to
• Explain what is Beta Decay?

• Describe the types of Beta Decay
• Evaluate the Q-Value in Beta Decay
• Discuss neutrino hypothesis
• Describe Fermi's Theory of β -Decay
• Understand the various selection rules for  -Decay
• Explain parity violation in Beta Decay
8.11 REFERENCES

4. The Atomic Nucleus (Tata McGraw-Hill Publishing Company Ltd.)
R. D. Evans
5. Atomic and Nuclear Physics Vol. II (S. Chand & Company Ltd.)
S. N. Ghoshal
6. Nuclear Physics (An Introduction) (Wiley Eastern Ltd.)
S. B. Patel
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192

7. Nuclear Physics (Narora Publishing Ltd.) :
Iriving Kaplan
8. https://cdn1.byjus.com/wp-content/uploads/2021/05/Radioactivity-Beta-Decay.png
9. http://nuclear-power.com/wp-content/uploads/2018/08/Beta-Decay-Beta-
Radioactivity-definition.png
10. http://nuclear-power.com/wp-content/uploads/2018/08/beta-decay-q-value-
example.png
1. “The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of Physics,

Berlin: Springer Ver.
2. “Nuclear Two Body Problems and Elements of Nuclear Forces” Experimental Nuclear
Physics by N. F. Ramsey, Wiley: New York.
3. Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum Press, New
York.
1. Discuss the postulates of Pauli's neutrino hypothesis.

2. What are Fermi-Kurie plots? Discuss their importance.
3. What are allowed and forbidden  - transitions?
4. Give the Fermi theory of  decay discuss how it explains the important features of  -
spectrum.
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5. State and explain the selection rules of  -emission.
6. Discuss the parity violation in beta decay.
7. Discuss the types of beta decay with suitable examples.
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UNIT 9
GAMMA DECAY
9.1 Introduction
9.2 Objectives
9.3 Gamma Decay
9.4 COMPTON SCATTERING OF GAMMA RAYS
9.5 PAIR PRODUCTION AND ANIHILATION
9.6 K-CAPTURE or ELECTRON CAPTURE
9.7 INTERNAL CONVERSION
9.8 MULTIPOLRAITY IN GAMMA TRANSITIONS
9.9 SELECTION RULES
9.10 Summary
9.11 References
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9.1 INTRODUCTION
An excited or unstable nucleus transforms into a more stable structure and releases extra energy as
radiation. An excited state is typically left over after a nucleus decay by alpha or beta emission.
When emitting a particle from a nucleus is energetically impossible, the nucleus decays through
electromagnetic interaction. Because -emission can be measured between more pairs of nuclear
states, -spectroscopy can provide more about the structure of the nucleus. Gamma rays rarely absorb
or scatter in air, which is crucial for determining whichever states are involved in the nuclear
transition. The -rays can be easily detected with sensors having good energy resolution.
9.2 OBJECTIVES
After studying this unit, the learners will be able to how to
• Explain Gamma Decay

• Discuss Compton Scattering of Gamma Rays
• Describe Pair Production and Annihilation
• Explain K-capture or Electron Capture
• Describe Internal Conversion
• Discuss Multipolarity in Gamma Transitions
• Explain Selection Rules
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9.3 GAMMADECAY
Gamma decay or γ decay represents the disintegration of a parent nucleus to a daughter through
gamma rays (high energy photons) emission. This transition (γ decay) can be characterized as:
Fig.1: Gamma decay(http://nuclear-power.com)
As can be seen, if a nucleus emits a gamma-ray, atomic and mass numbers of the daughter nucleus
remain the same, but the daughter nucleus will form a different energy state of the same element.
Note that nuclides with equal proton number and mass number (thus making them by definition the
same isotope) but in a different energy state are known as nuclear isomers. We usually indicate
isomers with a superscript m, thus: 241mAm or 110mAg.
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Fig. 02: Decay scheme of Iodine 131(http://nuclear-power.com)
In most practical laboratory sources, the excited nuclear states are created in the decay of a parent
radionuclide. Therefore, gamma decay typically accompanies other forms of decay, such as alpha or
beta decay. After a beta decay (isobaric transition), nuclei usually contain too much energy to be in
their final stable or daughter state.
Gamma rays are high-energy photons with very short wavelengths and thus very high frequency.
Gamma rays from radioactive decay are in the energy range from a few keV to ~8 MeV,
corresponding to the typical energy levels in nuclei with reasonably long lifetimes. As was written,
they are produced by the decay of nuclei as they transition from a high energy state to a lower state.
Since the gamma rays are in substance only very high-energy photons, they are very penetrating
matter and are thus biologically hazardous. Gamma rays can travel thousands of feet in the air and
easily pass through the human body.
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In contrast to alpha and beta radioactivity, gamma radioactivity is governed by an electromagnetic
interaction rather than a weak or strong interaction. As in atomic transitions, the photon carries
away at least one unit of angular momentum (the photon, described by the vector electromagnetic
field, has a spin angular momentum of ħ), and the process conserves parity.
9.4 COMPTON SCATTERING OF GAMMA RAYS

Compton scattering involves the scattering of photons by charged particles where both energy and
momentum are transferred to the charged particle while the photon moves off with a reduced energy
and a change of momentum. Generally, the charged particle is an electron considered to be at rest
and the photon is usually considered to be an energetic photon such as an X-ray photon or gamma
ray photon. In this experiment gamma rays from a cesium-137 source are used for the source of
photons that are scattered and each photon has an energy of 0.662MeV when incident on the target
scatterer. The charged particle is assumed to be an electron at rest in the target. While the theory
here is applied to gamma rays and electrons, the theory works just as well for less energetic photons
such as found in visible light and other particles.
The theory of Compton scattering uses relativistic mechanics for two reasons. First, it involves the
scattering of photons that are massless, and secondly, the energy transferred to the electron is
comparable to its rest energy. As a result the energy and momentum of the photons and electrons
must be expressed using their relativistic values. The laws of conservation of energy and
conservation of momentum are then used with these relativistic values to develop the theory of
Compton scattering.
From the special theory of relativity, an object whose rest mass is 𝑚0 and is moving at a velocity 𝑣
will have a relativistic mass 𝑚 given by
𝑚0
𝑚= ……………(1)
√1−(𝑣/𝑐)2
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199

The relativistic momentum 𝑝⃗ is defined as 𝑚𝑣⃗ so that squaring both sides of Equation (1) and
multiplying by 𝑐 4 leads to
𝑚2 − 𝑚2 (𝑣/𝑐)2 = 𝑚02 ,
𝑚2 𝑐 4 − 𝑚2 𝑣 2 𝑐 2 = 𝑚02 𝑐 4 ,
(𝑚𝑐 2 )2 − (𝑝𝑐)2 = (𝑚0 𝑐 2 )2 ,
𝐸 2 − (𝑝𝑐)2 = 𝐸02 ,
and
(𝑝𝑐)2 = 𝐸 2 − 𝐸02 . ………………….(2)
Equation (2) then relates the magnitude of the relativistic momentum p

⃗⃗ of an object to its
relativistic energy 𝐸 and its rest energy 𝐸0 . From this equation it is readily seen that the
magnitudes of momentum and energy of a massless particle such as a photon are related by
𝑝𝑐 = 𝐸. ………….(3)
or
𝐸 ℎ𝑣 ℎ
𝑝= = = 𝜆. …………….(4)
𝑐 𝑐
Figure3 illustrates the scattering of an incident photon of energy 𝐸 = ℎ𝑣 moving to the

ℎ𝑣 ℎ
right in the positive x direction with a momentum 𝑝 = = 𝜆 and interacting with an
𝑐
electron at rest with momentum 𝑝𝑒 = 0 and energy equal to its rest energy, 𝐸0 = 𝑚0 𝑐 2 .
The symbols, ℎ, 𝑣 , and 𝜆 , are the standard symbols used for Planck's constant, the
photon's frequency, and its wavelength. 𝑚0 is the rest mass of the electron. In the
interaction, the gamma ray is scattered in the positive x and y directions at an angle 𝜃
ℎ𝑣 ′ ℎ
with momentum of magnitude 𝑝′ = = 𝜆′ and energy 𝐸 ′ = ℎ𝑣 ′ . The electron is
𝑐
scattered in the positive x-direction and negative y-direction at an angle 𝜙 with respect to
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1
the positive x-direction with momentum 𝑝𝑒′ = 𝑐 √𝐸𝑒2 − 𝐸02 and energy 𝐸𝑒 = 𝑚𝑐 2 where
𝑚 is the relativistic mass of the electron after the interaction.
Figure 3:. Compton scattering diagram
From the law of conservation of energy, the energy of the incident gamma ray, ℎ𝑣, and
the rest mass of the electron, 𝐸0 , before scattering is equal to the energy of the scattered
gamma ray, ℎ𝑣 ′ , and the total energy of the electron, 𝐸𝑒 , after scattering, or
ℎ𝑣 + 𝐸0 = ℎ𝑣 ′ + 𝐸𝑒 . ……………..(5)
From Equation (2), the relationship between the total energy, 𝐸𝑒 , of the electron after
scattering, its rest mass, 𝐸0 , and its relativistic momentum, 𝑝𝑒 , is given by
𝐸𝑒2 = (𝑝𝑒′ 𝑐)2 + 𝐸02 ……………(6)
and
𝐸𝑒 = √(𝑝𝑒′ 𝑐)2 + 𝐸02 . ……………(7)

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Substituting Equation (7) into Equation (5) yields
ℎ𝑣 + 𝐸0 = ℎ𝑣 ′ + √(𝑝𝑒′ 𝑐)2 + 𝐸02 . ……………..(8)
Using the relationship between the energy of a photon (massless particle) and its momentum
from Equation (4) gives
𝑝𝑐 + 𝐸0 = 𝑝′ 𝑐 + √(𝑝𝑒′ 𝑐)2 + 𝐸02 . …………….(9)
Rearranging gives
(𝑝 − 𝑝′ )𝑐 + 𝐸0 = √(𝑝𝑒′ 𝑐)2 + 𝐸02 ………………(10)
and
(𝑝 − 𝑝′ )2 𝑐 2 + 𝐸02 + 2(𝑝 − 𝑝′ )𝑐𝐸0 = (𝑝𝑒′ 𝑐)2 + 𝐸02 ………..(11)
that results in the following expression based on conservation of energy
2(𝑝−𝑝′ )𝐸0
𝑝2 + 𝑝′2 − 2𝑝𝑝′ + 𝑐
= 𝑝𝑒′2 . (12)
Equation (12) is then an expression relating the momentum 𝑝𝑒 of the electron given to it
by a scattered gamma ray whose initial momentum was 𝑝 and whose final momentum is
𝑝 '. The electron was assumed to be initially at rest and it was also assumed to be given
enough energy for relativistic mechanics to apply. Equation (12) is solely based on the
law of conservation of energy, but another independent expression for the momentum 𝑝𝑒
can be found based on the law of conservation of momentum.
In the scattering process momentum must be conserved so that
Total Momentum Before = Total Momentum After . ………..(13)
Since momentum is a vector quantity,
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Total Momentum in X-Direction Before = Total Momentum in X-Direction After ……..(14)
and
Total Momentum in Y-Direction Before = Total Momentum in Y-Direction After . ………(15)
For an electron at rest, its initial momentum is zero and has no x and y components. For
an incident gamma ray photon moving in the positive x direction and interacting with an
electron at rest, the initial x-component of momentum is 𝑝 and the y-component is zero
so that
𝑝 = 𝑝′ cos 𝜃 + 𝑝𝑒′ cos 𝜙 ……………….(16)
and
0 = 𝑝′ sin 𝜃 + (−)𝑝𝑒′ sin 𝜙 ……………….(17)
where 𝑝′ and 𝑝𝑒′ are the momenta of the scattered gamma ray and electron after interacting.
Rearranging Equations (16) and (17) and squaring both sides of each produces
𝑝𝑒′ cos 𝜙 = 𝑝 − 𝑝′ cos 𝜃 , … … … … . (18)

𝑝𝑒′ sin 𝜙 = 𝑝′ sin 𝜃 , … … … … … (19)
𝑝𝑒′2 cos2 𝜙 = 𝑝2 + 𝑝′2 cos 2 𝜃 − 2𝑝𝑝′ cos 𝜃 , … … … . (20)
and
𝑝𝑒′2 sin2 𝜙 = 𝑝′2 sin2 𝜃. …………….(21)
Adding Equations (20) and (21) yields
𝑝𝑒′2 (sin2 𝜙 + cos 2 𝜙) = 𝑝2 + 𝑝′2 (sin2 𝜃 + cos 2 𝜃) − 2𝑝𝑝′ cos 𝜃 ………….(22)
That can be simplified using the indentity cos 2 𝑥 + sin2 𝑥 = 1 to further yield
𝑝𝑒′2 = 𝑝2 + 𝑝′2 − 2𝑝𝑝′ cos 𝜃. …………(23)
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Equation (23) is then an expression based on the law of conservation of momentum that
relates the momentum given to the electron from its rest position by the incident gamma
ray of momentum 𝑝 interacting with the electron so that it is scattered off at angle 𝜃 with
momentum 𝑝′ .
Equating Equations (12) and (23), one based on conservation of energy and the other on
conservation of momentum, gives
2(𝑝−𝑝′ )𝐸0
𝑝2 + 𝑝′2 − 2𝑝𝑝′ + = 𝑝2 + 𝑝′2 − 2𝑝𝑝′ cos 𝜃 …………(24)
𝑐
that reduces to
2(𝑝−𝑝′ )𝐸0
= 2𝑝𝑝′ − 2𝑝𝑝′ cos 𝜃, ………(25)
𝑐
and
1 1 𝑐
𝑝′
− 𝑝 = 𝐸 [1 − cos 𝜃]. ……………..(26)
0
Using the relationships for momentum, energy, wavelength, and frequency for photons,
ℎ ℎ𝑣 𝐸
𝑝=𝜆= = 𝑐 , Equation (26) can be transformed into
𝑐
1 1 1
𝐸′
− 𝐸 = 𝐸 [1 − cos 𝜃] ……..(27)
0
that relates the energy of a scattered photon 𝐸 ' to the energy of the incident photon 𝐸 and
the scattering angle 𝜃.
Equation (27) is a simple equation that can be used to verify the theory for the Compton
Effect. The energy of incident gamma rays 𝐸 can be easily measured with a
scintillatorphotomutiplier detector and multichannel analyzer system. The energy of the
scattered gamma rays 𝐸 ′ as a function of 𝜃 can also be easily measured with the same
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1 1
system. A plot of measurements of − 𝐸 versus measurements of [1 − cos 𝜃] should
𝐸′
1
result in a linear graph whose slope is the inverse of the electron's rest energy 𝐸 .
0
9.5 PAIR PRODUCTION AND ANIHILATION

Pair production is a process in which a gamma ray of sufficient energy is converted into
an electron and a positron. A fundamental law of mechanics, given by Newton, is that in
any process total linear (as well as angular) momentum remains unchanged. In the pair-
production process a third body is required for momentum conservation. When that body
is a heavy nucleus, it takes very little recoil energy, and therefore the threshold is just
twice the rest energy of the electron; i.e., twice its mass, m, times the square of the
velocity of light, c2, or 2mc2.
Pair production also can occur in the field of an atomic electron, to which considerable
recoil energy is thereby imparted. The threshold for such a process is four times the rest
energy of an electron, or 4mc2. The total pair-production cross section is the sum of the
two components, nuclear and electronic. These cross sections depend on the energy of the
gamma ray and are usually calculated in an electron theory proposed by the British
physicist P.A.M. Dirac through a method of approximation that is a simplification of a
method (a “first approximation”) devised by the German physicist Max Born (i.e., a “first
Born approximation”). The process is envisaged by Dirac as the transition of an electron
from a negative to a positive energy state. Corrections are required for these cross
sections at high energy, at high atomic number, and for atomic screening (the intrusion of
the field of the electrons in an atom); these are normally made via numerical procedures.
The fraction of residual energy, symbolized by the Greek letter alpha, unexpended in
conversion of energy to mass, that appears in any one particle (e.g., the electron) is thus
given by the kinetic energy of that electron Ee minus its rest energy mc2 divided by the
energy of the gamma ray hν (i.e., the product of Planck’s constant and the frequency of
the gamma ray) minus twice the rest energy of the electron 2mc2, or α = (Ee -mc2)/(hν -
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2mc2). Because the same equation applies to each of the two electrons that are formed, it
must be symmetric about the condition that each of the particles has half the residual
energy, symbolized by the Greek letter alpha, α (in excess of that conveyed to the “third
body”); i.e., that α = 0.5. Below an energy of about 10,000,000 eV for the gamma ray, the
probability for pair production (i.e., the pair-production cross section) is almost
independent of the atomic number of the material, and, up to about 100,000,000 eV of
energy, it is also almost independent of the quantity α. Even at extremely high energies
the probability that a certain fraction of the total available energy will appear in one
particle is almost independent of the fraction as long as energy is comparably distributed
between the two particles (excepting in cases in which almost all energy is dumped into
one particle alone). Typical pair-production cross sections at 100 MeV (million electron
volts) are approximately 10-24 to 10-22 square centimetre, increasing with atomic number.
At high energies, approximately equal to or greater than 100 MeV, pair production is the
dominant mechanism of radiation interaction with matter.
Clearly, as the photon energy increases, the dominant interaction mechanism shifts from
photoelectric effect to Compton scattering to pair production. Rarely do photoelectric
effect and pair production compete at a given energy. Compton scattering, however, at
relatively low energy competes with the photoelectric effect and at high energy competes
with pair production. Thus, in lead, interaction below 0.1 MeV is almost exclusively
photoelectric; between 0.1 MeV and 2.5 MeV both photoelectric and Compton processes
occur; and between 2.5 MeV and 100 MeV Compton scattering and pair production share
the interaction. In the pair process the photon is annihilated, and an electron–positron pair
is created. On the other hand, an electron or positron with energy approximately equal to
or greater than 100 MeV loses its energy almost exclusively by production of high-
energy bremsstrahlung (X rays produced by decelerating electric charges) as the result of
interaction with the field of a nucleus. The cross section for bremsstrahlung production is
nearly independent of energy at high energies, whereas at low energies the dominant
energy-loss mechanism is by the creation of ionizations and excitations. A succession of
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bremsstrahlung and pair-production processes generates a cascade or shower in the
absorber substance. This phenomenon can be triggered by an electron, a positron, or a
photon, the triggering mechanism being unimportant as long as the starting energy is
high. A photon generates a pair through pair production, and the charged particles
generate photons through bremsstrahlung, and so on repeatedly as long as the energy is
kept sufficiently high. With penetration into the substance, the shower increases in size at
first, reaches a maximum, and then gradually decreases. Loss of particles by degradation
to lower energies (in which the yield of bremsstrahlung is low), ionization loss, and
production and absorption of low-energy photons eventually reduce the size of the
cascade. The mathematical theory of cascades has been developed in great detail.
X rays and gamma rays
When light of sufficiently high frequency (or energy equal to hν), independent of its
source, is absorbed in a molecular system, the excited molecular state so produced, or
some excited state resultant from it, may either interact with other molecules or
decompose to produce intermediate or ultimate products; i.e., chemical reactions ensue.
Study of such processes is encompassed in the subject of photochemistry (see below
Molecular activation).
Electromagnetic waves of energy greater than those usually described as ultraviolet light
(see Figure 2) are included in the classes of X rays or gamma rays. X-ray and gamma-ray
photons may be distinguished by definition on the basis of source. They are
indistinguishable on the basis of effects when their energy is absorbed in matter.
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9.6 K-CAPTURE or ELECTRON CAPTURE
The capture of electrons is a nuclear process. An atom's inner electron is absorbed by a

nucleus rich in protons. The nucleus emits high-energy gamma rays as part of this
process. The weak force causes electron capture, which is a relatively minor decay
phase.If the number of protons in a nucleus exceeds the number of neutrons, electron
capture occurs. The nucleus of an atom absorbs an electron from the atom's K-shell or L-
shell. The absorbed electron and one proton then unite to generate a neutron, bringing the
number of protons and neutrons in the nucleus back to a healthy balance.
The capture of a 1s electron by the nucleus of an unstable isotope is known as K-electron

capture. When the nucleus of an unstable isotope captures an inner-orbital electron, it is
known as electron capture. A proton combines with an electron to make a neutron, and an
X-ray is produced as a result of the process.
The atomic number decreases by one unit, but the mass number stays the same. Because
they are nearest to the nucleus, the captured electron usually comes from the 1sor
2sorbitals.
The process is known as K-electron capture if the electron arrives from the 1s level (the
K-shell).The mass number of a nucleus does not change when it undergoes an electron
capture reaction, but the atomic number lowers by one. The total number of protons and
neutrons inside the nucleus is known as the mass number, while the number of protons or
electrons in an atom is known as the atomic number. The electron is grabbed by the
nucleus in an electron capture reaction.
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9.7 INTERNAL CONVERSION
Internal conversion is another electromagnetic process which can occur in the nucleus
and which competes with gamma emission. Sometimes the multipole electric fields of the
nucleus interact with orbital electrons with enough energy to eject them from the atom.
This process is not the same as emitting a gamma ray which knocks an electron out of the
atom. It is also not the same as beta decay, since the emitted electron was previously one
of the orbital electrons, whereas the electron in beta decay is produced by the decay of a
neutron.
Fig.4: Internal Conversion
203 203
An example used by Krane is that of Hg, which decays to Tl by beta emission,
leaving the 203Tl in an electromagnetically excited state. It can proceed to the ground state
209

by emitting a 279.190 keV gamma ray, or by internal conversion. In this case the internal
conversion is more probable. Since the internal conversion process can interact with any
of the orbital electrons, the result is a spectrum of internal conversion electrons which
will be seen as superimposed upon the electron energy spectrum of the beta emission.
The energy yield of this electromagnetic transition can be taken as 279.190 keV, so the
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ejected electrons will have that energy minus their binding energy in the Tl daughter
atom.
The diagram above is of course conceptual only and not to scale since the nuclear radius
of thallium is modeled to be about 0.7x10-14 m and the radius of the atom is about
1.76x10-10m, a factor of about 25,000 larger! And of course the planetary type orbits of
the electrons are unrealistic since the wave properties of the electrons lead to charge
distributions that give a finite probability that the K electron shown above will actually
extend inside the nucleus so that the nucleus can interact with it and hand off its excess
energy. An examination of the electron distribution for the simplest atom, hydrogen, can
give the perspective that the electron has a small but finite probability of extending into
the nucleus. From the table of binding energies below, you can see that the binding
energy of the K-shell electron is over 85,000 electron volts compared to 13.6 eV for the
hydrogen electron, or over 6,000 times larger.
Electron emissions from the Hg-203 to Tl-203

decay, measured by A. H. Wapstra, et al., Physica
20, 169 (1954).
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At higher resolution, the internal conversion

electrons from the L, M and N shells can be
resolved. Z. Sujkowski, Ark. Fys. 20, 243 (1961).
At even higher resolution, the three L shells can be

resolved. From C. J. Herrlander and R. L. Graham,
Nucl. Phys. 58, 544 (1964).
The resolution of the electron detection is good enough that such internal conversion
electron spectra can be used to study the binding energies of the electrons in heavy atoms.
In this case, the measured electron energies can be subtracted from the transition energy
as indicated by the gamma emission, 279.190 keV.
Fig.05: Internal conversion electron spectrum
In addition to information from the internal conversion electrons about the binding
energies of the electrons in the daughter atom, the relative intensities of these internal
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conversion electron peaks can give information about the electric multipole character of
the nucleus.
9.8 MULTIPOLRAITY IN GAMMA TRANSITIONS
Transitions between excited states (or excited states and the ground state) of a nuclide
lead to the emission of gamma quanta. These can be classified by their multipolarity.
There are two kinds: electric and magnetic multipole radiation. Each of these, being
electromagnetic radiation, consists of an electric and a magnetic field.
The conservation of angular momentum and parity between the emitting system and the
radiation field forms the basis for categorising the various photon emission processes.
The potential (VL) resulting from an electric multiple of an order (L) at the origin is often
described by
1
𝑉𝐿 = ∑ 𝑞𝑖 𝑟𝑖2 𝑃𝐿 (𝑐𝑜𝑠 𝜃𝑖 ) ......(28)
4𝜋𝜀0 𝑅0𝐿+1
Where R0 is the position vector of the field-point, PL (cos) is the Legendre polynomial of
order L, qi is the ith charge, ri is the position of ith charge from origin, and there is an
angle between ri and Ro. In Eqn.28, the various terms represent the electric vectors of the
radiation fields resulting from various orders of multipoles. The radiation caused by an
electric monopole, which does not radiate energy, is represented by the first term. The
second term is the electric dipole's (E1) time-varying moment's radiation field. The
following word denotes the radiation field of the magnetic dipole (M1) of the time-
varying moment as well as the radiation field of the electric quadruple (E2) of the time-
varying moment.
In general, VL is equivalent to the radiation field of a magnetic 2L-1 pole or an electric

2L pole (EL) with time-varying moments. L is a measurement of the radiation's
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multipolarity and the amount of angular momentum that was transmitted during the
radiative transition. The orbital angular momentum transferred in the case of particle
emission is equivalent to this L.
The multiple order of a transition is referred to as 2L, as was already mentioned earlier in
this section. L stands for the multipolarity of photons, and multipole radiation and
transition are referred to
(i) dipole ( L=1) i.e., Δ I = L = 1 or 21 = 2
(ii) quadrupole ( L=2 ) i.e., Δ I = L = 2 or 2 2 = 4
as
(iii) octupole ( L=3 ) i.e., Δ I = L = 3 or 23 = 8 and higher order
transitions
For any order, the transitions can be either Electric or Magnetic, in nature.
9.9 SELECTION RULES
In simple definition we have
|𝐼𝑖 − 𝐼𝑓 | ≤ 𝐿 ≤ |𝐼𝑖 + 𝐼𝑓 | ......(29)
So, we have the following selection rules for the  -ray emission process.
i) the  -photon must carry away at least one unit of angular momentum, and
ii) in  -decay process, the polarity must always be conserved.
The change in the magnitude of the angular momentum is given by
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𝐿 = 𝛥𝐼 = |𝐼𝑖 ± 𝐼𝑓 | ......(30)
Suppose, the nuclear spin changes from 4 to 2 during a transition. Then I is simply the
scalar difference 2. However, I which is the vector change in angular momentum
(  I ) can have values i.e. Ii − I f to| Ii + I f 2,3,4,5 and 6 in this case, since any
integral value form 4-2 to 4+2 is possible. Remember L = 0 y-emission is not allowed.
The parity change for the nucleus determines the native of the multipole transition.
The  -transitions are classified as electric 2L pole and magnetic 2L pole according to
 = ( −1) for electric 2L pole radiation

L
 = ( −1)
L +1
for magnetic 2L pole radiation
The radiation field can have even or odd parity for a given L, depending on whether the
nature is electric or magnetic.
The electric multipole radiation, denoted by E1, E2,...EL has parity (-1)L and magnetic
multipole radiation M1, M2......ML. has parity (- 1)L+1 The electric and
magnetic/multipoles of the same order L have opposite parity. The electric and magnetic
dipole moment are shown in Fig.
Fig.6 (a) Electric dipole moment (b)Magnetic dipole moment

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An electric dipole, as shown in Fig6 (a) is a displacement of charge q and is of the form
which under the parity operation of inversion transforms into - and therefore, has 'odd'
(negative) parity. A magnetic dipole like a circulatory charge q which forms a current
loop (Fig.6(b)), has a magnetic moment proportional to (q r x v), which under inversion,
transforms into [q (- r) x (- v)] i.e. it does not charge sign and so has even (positive)
parity.The emission rate strongly depends on photon energy as well as on multipolarity.
This means that normally, radiation will be emitted with lowest multipolarity, allowed by
the selection rules.
9.10 SUMMARY
After studying this unit, the learners have learnt about, how to
• Explain Gamma Decay

• Discuss Compton Scattering of Gamma Rays
• Describe Pair Production and Annihilation
• Explain K-capture or Electron Capture
• Describe Internal Conversion
• Discuss Multipolarity in Gamma Transitions
• Explain Selection Rules
9.11 REFERENCES
1. The Atomic Nucleus (Tata McGraw-Hill Publishing Company Ltd.) by R.D. Evans
2. Nuclear Physics (An Introduction) (Wiley Eastern Ltd.) by S. B. Patel
3. Atomic and Nuclear Physics: Vol. II (S. Chand & Company Ltd.) by S. N. Ghoshal
4. Atomic and Nuclear Physics (Books & Allied (P) Ltd A.B. Gupta and Dipak Ghosh
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215

5. Nuclear Physics (Principles and Applications) by John Lilly.(John Wiley & Sons Ltd.)
6.Arthur H. Compton, "A Quantum Theory of the Scattering of X-rays by Light Elements," The
Physical Review, Vol. 21, No. 5, (May, 1923).
7. http://nuclear-power.com/wp-content/uploads/2018/07/Iodine-131-decay-scheme.png
8. http://nuclear-power.com/wp-content/uploads/2018/08/Gamma-Decay-Gamma-
Radioactivity-definition.png
9.https://www.britannica.com/science/radiation/Pair-production
1. R. M. Eisberg, Fundamentals of Modern Physics (John Wiley & Sons, New York, 1961)
2. R. M. Singru, “ Introduction to Experimental Nuclear Physics”, Wiley Eastern Pvt. Ltd.,
1972, Chs. 1, 4, and 7.
3. G. F. Knoll, “Radiation Detection and Measurement” John Wiley & Sons, 1979, Chs. 2,
10, 16, 17, and 18.
4. S. S. Kapoor and V. S. Ramamurthy, “Nuclear Radiation Detector”, Wiley Eastern Ltd.,
1986, Ch V, page 176, 185.
5. GDM 10, Handbook and other related manuals.
1. Explain Gamma decay.

2. Discuss the multipolarity of a gamma transition and explain the selection rules.
3. Describe Compton theory of scattering in detail.
4. What is pair production and annihilation in gamma rays?
5. What do you understand by electron capture in gamma decay?
6. Write a note on Internal Conversion.
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UNIT 10
NUCLEAR REACTIONS
10.1 Introduction
10.2 Objectives
10.3 Types of Nuclear Reaction
104. Conservation Laws for Nuclear Reactions
10.5 Charged Particle Induced Nuclear Reactions
10.5.1 Proton Induced Nuclear Reactions.
10.5.2 Deuteron Induced Nuclear Reactions:
10.5.3 α-particle induced reactions
10.6 Neutron Induced Reactions
10.7 Photo-disintegration (Nuclear Reactions Induced by Photons or γ-rays)
10.8 Reaction Energetics - the Q-value Equation
10.9 Nuclear Reaction Cross-Sections
10.10 Partial Wave Analysis of Nuclear Reaction Cross-sections
10.11 Summary
10.12 References
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10.1 INTRODUCTION
We have preliminary idea that our present-day knowledge about the nuclear structure
involves many experiments involving bombardment of heavy nuclei by means of lighter
energetic particles. Rutherford used particles from a radioactive source to create the first
nuclear reaction in his lab in 1919. The artificial transmutation of elements was
discovered as a result of the experiment that generated artificial nuclei disintegration. A
composite system was created in Rutherford's experiment as a result of a very high
velocity particle colliding head-on collision with nitrogen nuclei N14. This composite
system disintegrated in matter of 10-15 seconds, generating a high-velocity proton and a
small amount of remnant nuclear material. An equation analogous to the equation for a
chemical reaction that is given below can be used to represent the nuclear transmutation
process mentioned below
2He
4
+ 7N14 → 9F18 → 8O17 + 1H1 ………(1)
In Eqn.1 the composite system 9F18 is known as compound nucleus.
.Any such collision, in which the characteristic of the struck particle is changed from
original identity , is known as a nuclear reaction.
In general, a nuclear reaction can be represented by
ZX
A
+ x → ZYA + y ………(2)
or in compact notation it can be written as X (x, y) Y
A compound nucleus is created when the incident particle (or projectile) x collides with
the target nucleus X. This compound nucleus quickly disintegrates to form a residual
nucleus Y and a product particle y. Most nuclear reactions involve light particles like
neutrons, protons, deuterons, and particles as the incident and product particles. Nuclear
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reactions offer a testing ground for theories regarding nuclear forces as well as a crucial
function in elucidating the structure of nuclei. Nuclear spectroscopy puts a high value on
the information that nuclear reactions convey regarding the energy levels and decay
patterns of the excited product nucleus.
10.2 OBJECTIVES
After studying this unit learners will be able to how to
• Describe the Types of Nuclear Reaction

• Explain the Conservation Laws for Nuclear Reactions
• Discriminate between Charged Particle Induced Nuclear Reactions
• Understand the Photo-disintegration process
• Evaluate the Reaction Energetics - the Q-value Equation
• Explain Nuclear Reaction Cross-Sections
• Describe Partial Wave Analysis of Nuclear Reaction Cross-sections
10.3 TYPES OF NUCLEAR REACTION
Various types of nuclear reactions are possible when a target nucleus is bombarded by means of
energetic projectile particles, the bombardment process may give rise to the following events
singly or jointly.
(i) Elastic Scattering :It is a process in which the incident and the outgoing particles
are the same and the kinetic energy is conserved. Thus, if the incident particles and the
target nuclei are simply scattered by each other without any change in their relative
energy, the process is aid to be elastic scattering.
If the incident particle is denoted by ' 𝑎 ' and the target nucleus by 𝑋, then the elastic
scattering process may by symbolically written as
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e.g.,
𝑎+𝑋 →𝑋+𝑎 … … (3)

𝑒 + 6 C12
−
→ 6 C12 + 𝑒 −
1 238
0 𝑛 + 92 U → 92 U 238 + 0 𝑛1
Sometimes the charge exchange reactions like the one given below, are also referred to as
elastic scattering.
𝜋− + 𝑝 → 𝑛 + 𝜋0
(ii) Inelastic Scattering: The bombardment process may result into a change in the
relative energy without change in the internal structure of the target nucleus. This type of
process is known as inelastic scattering and in it although the incident and the outgoing
particles are the same but kinetic energy is not conserved during the process. In this
process the target nucleus, after the interaction, is not left in its initial state but is raised to
an excited state and so the emergent particle, although the same as the incident particle,
emerges with an energy lower than the energy of incident particle by an amount equal to
the excitation energy given to the target nucleus. Symbolically inelastic scattering is
represented as
𝑎 + 𝑋 → 𝑋∗ + 𝑎 …..(4)
where * indicates that the target nucleus is in an excited state. To quote specific examples
1
0𝑛 + 82 Pb208 → 82
Pb208∗ + 0 𝑛1
Subsequently 82 Pb200∗ emits 𝛾-rays.
1
1H + 7 N14 → 7 N14∗ + 1 H1 ; 7 N14∗ → 7 N14 + 𝛾
(iii) Nuclear Reactions: Although in the broadest sense, nuclear reactions may be
taken to include all types of processes which result from the bombardment of target
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nuclei by means of an energetic particle but the term nuclear reaction is generally used to
indicate the processes in which the incident and the emergent particles are not the same
but are different i.e. an initial state involving nucleons is converted into a different final
state involving nucleons. Therefore, a nuclear reaction is a process in which a change in
the composition (internal structure) and/ or energy of the target nucleus is brought about
through bombardment with a nuclear projectile. The projectiles may be nucleons,
complex nuclei, photons, electrons or any particle with a life time >> 10−8 sec. The
target also may be a single nucleon or a complex nucleus. Symbolically, a nuclear
reaction is written as
𝑎+𝑋 → 𝑌+𝑏 ……(5)
To quote a particular example;
1
0𝑛 + 82 Pb208 → 81 Tl
208
+ 1 H1
It is a standard practice to designate a nuclear reaction symbolically by enclosing the

incoming and outgoing particles in parentheses separated by a comma and writing the
target nucleus on the left of the parentheses and product nucleus to the right of it. Thus,
the reactions (11.3) may be written as
𝑋(𝑎, 𝑏)𝑌 ……(6)
or in particular 82 Pb208 (𝑛, 𝑝)81 Tl208 )
Ever since the discovery of the first artificial nuclear reaction by Rutherford [ 2 He4 +
7 N14 → 8 O17 + 1 H1 ], experimental nuclear physicists have been concerned with the
investigations of various kinds of nuclear reactions. As we have seen above, a nuclear
reaction is usually initiated by exposing a suitable target material to a collimated beam of
mono-energetic lighter nuclear particles (protons, neutrons, 𝛼-paricles etc.). While using
charged particles as projectiles, it is be noted that the projectile must have sufficient
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energy to overcome the Coulomb potential barrier projectile must have sufficient energy
to overcome the Coulomb potential barrier surrounding the target nucleus. This
necessitated the development of charged particle accelerators to accelerate the projectiles
to high enough energies. Therefore, with the development of particle accelerators in 1932
onwards and now nuclear reactions may be produced with any of the element of the
periodic table. Studies on the cross-sections of nuclear reactions and the angular
distributions of the disintegration fragments coupled with the various theoretical led to a
better understanding of the nature of nuclear forces.
The richness of phenomena encountered in the study of nuclear reactions is revealed by

the colorful terminology in current use such as, the cloudy crystal ball model of the
nucleus, boil off processes, stripping 2 He4 + 8 O16 → 9 F18 + 1 H 2 ) i.e. projectile losing
nucleons to the target and pick up reactions, ( 1 H 2 + 8 O16 → 8 O15 + 1 H 3 ) i.e. projectile
gaining nucleons from the target, knock out, spallations (nuclear reaction in which
several particles are emitted) etc. The study of nuclear physics covers one of the largest
areas of nuclear and subnuclear physics and in a brief survey (as in this chapter) with any
pretensions to coherences it is impossible therefore to make detailed mention of all the
phenomena : Ideally, the task of a comprehensive theory of nuclear fractions is to predict
the cross-sections for individual energetically allowed processes that can be initiated by
the bombardment of a given target nucleus with a given projectile. In the search for such
a theory, we shall be limiting ourselves to a modest goal of predicting the absorption and
scattering cross-sections of the nuclear reactions in this unit.
The study of nuclear reactions involves the measurement of the following quantities:
(i) Intensity, energy and identity of the incident beam of particles.
(ii) The number of particles which are emitted from the target per unit time.
(iii) The energy and identity of the emitted particles and of the residual (product) nucleus.
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(v) Induced activity of the product nucleus if any. type:
Nuclear reactions at low excitation energies (below 10MeV ) are mostly of the
𝑎+𝑋 → 𝑌+𝑏+𝑄 ……(7)
where 𝑄 is the energy released or absorbed during the reaction. It is equal to the total
rest-mass of the left-hand side minus the total rest mass of the right-hand side of the
equation, multiplied by 𝑐 2 . In these reactions the major share of available kinetic energy
is carried away by the emitted particle. If the product nucleus 𝑌 happens to be in an
excited state after the emission of the light particle hitherto designated by 𝑏 , it
subsequently decays through one or more 𝛾-emissions. The product nucleus may also be
beta unstable. In such a case, it decays by electron or positron emission at some later
time, possibly also followed by 𝛾-emission.
Nuclear reactions at low excitation energies include following types of reactions:
(i) (𝑝, 𝛾), (𝑖𝑖)(𝑝, 𝑛), (𝑖𝑖𝑖)(𝑝, 𝛼), (𝑖𝑣)(𝑛, 𝛾), (𝑣)(𝑛, 𝑝), (𝑣𝑖)(𝑛, 𝛼),
(vii) (𝑑, 𝑝), ( viii) (𝑑, 𝑛), (𝑖𝑥)(𝛼, 𝑛) …
An approximate classification of nuclear reactions may be done on the basis of energy

and wavelength of the incident particle. If the wavelength of incident particle is large in
comparison with the size of the nucleus, the scattering is found to be insensitive to the
detailed structure of the nucleus and the nucleus as a whole act coherently. As the energy
of the-incident particle increases, its wavelength decreases, the interaction takes place
with a local cluster of nucleons only and not with the nucleus as a whole. When the
energy of the incident particle becomes so high that the wavelength associated with it
becomes about the size of a nucleon, the primary interaction takes place with a nucleon
only although secondary interactions may raise the residual nucleus to an excited state
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which may evaporate nucleons and decay to ground state electro-magnetically. At an
energy of several hundred MeV and greater, meson production predominates.
The wavelength associated with an incident nucleon of kinetic energy 𝐾 is,
ℎ ℎ ℎ ℎ2
𝜆 = 𝑝 = 𝑚𝑣 = and hence 𝐾 = 2𝑚𝜆2
√2𝑚𝐾
If 𝜆 = 10−14 meters Nuclear Reactions
(6.625 × 10−34 Joule − Sec)2

𝐾 =
2 × (1.67 × 10−27 Kg)(10−14 m)2 (1.6 × 10−19 J/eV)
≈ 8 × 106 eV = 8MeV
So, we find that up to an energy ≈ 10MeV, the nucleus as a whole act as one entity. If the
wavelength becomes 1/10 of this value 𝑖. 𝑒. 𝜆 = 10−15 meter, the kinetic energy has to
increase by a factor of 100 as kinetic energy is inversely proportional.
As the energy of the projectile particle is increased, the type and variety of the possible
type of nuclear reactions also increases. For example, if the bombarding energy of the
nucleons lies between 20 and 30MeV, two light particles instead of just one for 10MeV
projectile energies, may be emitted. Examples of such reactions are (𝑝, 𝑝𝑛), (𝑛, 𝑝𝑛)
reactions etc.
As the projectile energy is further increased and reaches beyond about 200 MeV which is
the threshold energy for the creation of pions ( 𝜋-mesons), we enter a new field - the field
of sub-nuclear Physics. Thus, the distinctive boundary between nuclear and sub-nuclear
physics is set by the threshold energy, for the creation of pions. Below this threshold
energy we deal only with nuclear reactions in the strict sense of the word and above this
threshold, mesic and particle phenomena become predominantly important until
projectile energy reaches multiBeV region. In this region nuclear features become of
secondary importance and the interactions between sub-nuclear particles like pions,
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nucleons, 𝐾-mesons etc. become most prominent features, while nuclear composition of
targets becomes of secondary importance.
We can sum up these consideration as follows:
1 When the energy of the incident particle lies below about 10MeV, the nucleus as a
whole act as one entity.
2 When the energy of the incident particle is of the order of few hundreds of MeV,
interactions with local clusters of nucleons predominate.
3 At bombarding energy of several hundred MeV, meson production predominates

and nuclear structure becomes of little importance.
10.4 CONSERVATION LAWS FOR NUCLEAR REACTIONS
Nuclear reactions like several other physical phenomena and like chemical reactions are
governed by certain conservation laws which provide very valuable information about the
nuclear reactions. These conservation laws represent certain features of the nuclear
reactions which are common to all and are of great importance in analyzing the
experimental data. We list below these conservation properties.
(i)Conservation of Electric Charge: The total electric charge is conserved in all

nuclear reactions. This means for example that a proton can change itself into a neutron
but a positron or a positive meson must appear or all electron must disappear in the
process. A most straight forward confirmation of electric charge conservation comes
from the creation of electron-positron pair.
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225
(ii) Conservation of Nucleons : So far no nuclear reaction has been observed

in which the total number of nucleons i.e. the value 𝐴 is not conserved. On the
contrary in all nuclear reactions, nucleon number is always conserved.
(iii) Conservation of the Constants of motion of Classical and

Quantum Mechanics :In all nuclear reactions, at least to an order of accuracy
beyond experimental interest, the total energy (sum of the rest mass energy and the
kinetic energy), the linear and angular momenta, spin angular momentum, isotopic spin
to be defined later in chapter 13 and parity (provided weak interactions like 𝛽 and meson
decay are excluded but not electro-magnetic processes), are conserved. The conservation
of energy belps us to calculate 𝑄 -value of the reaction. The conservation of linear
momentum guarantees that the incoming particle and the product nucleus move in a plane
if only two product nuclei result.
(iv) Conservation of statistics: The conservation of statistics is also followed by

all nuclear reactions. This law asserts that the spin character of a closed system cannot
change.
10.5 CHARGED PARTICLE INDUCED NUCLEAR

REACTIONS
Charged particle induced nuclear reactions are those which are induced by charged
particles like protons, deuterons, 𝛼-particles etc. With the invention of charged particle
accelerators capable of accelerating particles, to the energy range of several hundred BeV,
the scope and variety of such reactions has become so large that even a brief mention of
these all, will cover a very large volume. For a systematic study, these nuclear reactions
are classified therefore in accordance with the energy of the incident particle as follows:
225
226

(i) Low energy reactions (0 < 𝐸 𝑖 < 100eV)
(ii) Moderate energy reactions (1keV < 𝐸 𝑖 < 500keV)
(iii) High energy reactions (0.5MeV < 𝐸 𝑖 < 10MeV)
(iv) Very high energy reactions (10MeV < 𝐸 𝑖 < 50MeV)
(v) Ultra high energy reactions (𝐸 𝑖 > 50MeV)
Target nuclei with 𝐴 < 25, are called light nuclei, those with 25 < 𝐴 < 80, intermediate
nuclei and those with 80 < 𝐴 < 250. are called heavy nuclei.
Under the charged particle induced reactions, we first put forth nuclear reactions induced
by protons.
10.5.1 Proton Induced Nuclear Reactions.
Proton induced reactions include(𝑝, 𝛼), (𝑝, 𝑛), (𝑝, 𝛾) and (𝑝, 𝑑) reactions.
(i) (𝐩, 𝜶) reactions. (𝑝, 𝛼) nuclear reactions can in general be represented as
𝐴
𝑧X + 1 H1 → [𝑧 + 1C𝐴+1 ] → 𝑍−1 Y
𝐴−3
+ 2 He4 ……….(8)
If we recollect, this type of reaction has the status of being the first artificially induced
reaction, studied by Walton and Cock-croft in 1932. In the reaction studied by Walton
and Cockcroft 3 Li7 target was bombarded by means of 0.7MeV protons to produce two
𝛼-particles as follows.
7
3 Li + 1 H1 → [ 4 Be8 ] → 2 He4 + 2 He4
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227

The reaction is important on account of another historical background in as much as it
was this reaction which provided the first quantitative verification of Einstein's mass
energy relation (𝐸 = 𝑚𝑐 2 ). Further examples of reactions of this type are :
6 1
3 Li + 1H → [ 4 Be7 ] → 2 He3 + 2 He4
9 1
4 Be + 1H → [ 5 B10 ] → 3 Li6 + 2 He4
27 1
13 Al + 1H → [ 14 Si28 ] → 12 Mg 24 + 2 He4
11 1
5 B + 1H → [ 6 C12 ] → 4 Be8 + 2 He4
In the last reaction 4 Be8 nucleus is unstable and therefore breaks up giving rise to two 𝛼-
particles.
8
4 Be → 2 He4 + 2 He4
Thus the final reaction reads
5 B11 + 1 H1 → [ 6 C12 ] → 4 Be8 + 2 He4 → 3 2 He4
(ii) (𝐩, 𝐧) reactions : In general scheme such reactions can be written as
𝐴
𝑍X + 1 H1 → [Z + 1 C𝐴+1 → Z 2 Y𝐴 + 0 𝑛1 ……..(9)
It is clear from the above equation that these type of reactions produce a nucleus which is
one atomic number higher than the original (target) nucleus and as such the nucleus rises
higher up (to the right) in the periodic table.
Examples of these reactions are :
5 B11 + 1 H1 → [ 6 C12 ] → 6 C11 + 0 𝑛1

58
28 Ni + 1 H1 → [ 29 Cu59 ] → 29 C58 + 0 𝑛1
65
29 Cu + 1 H1 → [ 30 Zn66 ] → 30 Zn65 + 0 𝑛1
All the (𝑝, 𝑛) reaction are usually endoergic reactions.
227
228

(iii) (𝐩, 𝜸) reactions: The general scheme of such reactions is
𝐴
𝑍X + 1 H1 → [𝑧 + 1C𝐴+1 ] → 𝑍+1 Y
𝐴+1
+𝛾 ……..(10)
In this case also the product nucleus has an atomic number greater by 1 than the target
nucleus and mass number also increases by unity.
Examples of this type are
7
3 Li + 1 H1 → [ 4 Be8 ] → 4 Be8 + 𝛾
19
9 F + 1 H1 → [ 10 Ne20 ] → 10 Ne20 + 𝛾
27
13 A + 1 H1 → [ 14 Si28 ] → 14 Si28 + 𝛾
The 𝛾-rays which are produced in these reactions are very energetic and can be used to
induce nuclear reactions.
(iv) (p, d) reactions: The general scheme of these reaction is
𝐴
𝑧X + 1 H1 → [𝑧 + 1CA+1 ] → 𝑧−1 Y
𝐴−1
+ 1H2 ……..(11)
Note that in this case, the atomic number of the product nucleus is one less than that of
the target nucleus and therefore the product nucleus goes one place down (to the left) in
the periodic table.
9 1 10 8 2
4 Be + 1 H → [ 5 Be ] → 4 Be + 1 H
7 1 8 6 2
3 Li + 1 H → [ 4 Be ] → 3 Li + 1 H
10.5.2 Deuteron Induced Nuclear Reactions:
These reactions include (𝑑, 𝛼), (𝑑, 𝑝) and (𝑑, 𝑛) types of nuclear reactions.
On general scheme, all these reactions may be written as;
228
229

𝐴
𝑍𝑋 + 1 H 2 → [𝑧 + 1C𝐴+2 ] → 𝑍−1 Y𝐴−2 + 2 He4 (𝑑, 𝛼) reaction
→ 𝑍 X 𝐴+2 + 1 H1 ((𝑑, 𝑝) reaction
………..(12)
𝐴+1 1
→ 𝑍+1 Y + 0 𝑛 (𝑑, 𝑛) reaction
.
Examples: (i) (𝒅, 𝜶) reactions
6
3 Li + 1 H 2 → [ 4 Be8 ] → 2 He4 + 2 He4
16 2 18 14
8 O + 1 H → [ 9 Fe ] → 7 N + 2 He4
27
13 Al + 1 H 2 → [ 14 Si29 [→ 12 Mg 25 + 2 He4
(ii) (𝐝, 𝐩) reaction

12
6C + 1 H 2 → [ 7 N14 ] → 6 C13 + 1 H1
23 2 25 24 1
11 Na + 1 H → [ 12 Mg ] → 11 Na + 1 H
31
15 P + 1 H 2 → [ 16 S33 ] → 15 P32 + 1 H1
(iii) (𝐝, 𝐧) reaction 12

+ 1 H 2 → [ 7 N14 ] → 7 N13 + 0 𝑛1
6C
9 2 11 10
4 Be + 1 H → [ 5 B ] → 5 B + 0 𝑛1
7 2 9 8 1
3 Li + 1 H → [ 4 Be ] → 4 Be + 0 𝑛
However most interesting which the target itself contains deuterons, e.g.
2
1H + 1 H 2 → [ 2 He4 ] → 1 H 3 + 1 H1
2 2 4 3 1
1 H + 1 H → [ 2 He ] → 2 He + 0 𝑛
The 𝑄 values for (𝑑, 𝑝) reactions are generally positive and so they are exoergic
reactions.
10.5.3 𝜶-particle induced reactions
(i) (𝜶, 𝐧) reactions: General scheme of such reactions is

230

𝐴
2X + 2 He4 → [𝑧 + 2C𝐴+4 ] → 𝑧+2 Y
𝐴+3
+ 0 𝑛1 ………..(13)
Particular examples are
7
3 Li + 2 He4 → [ 5 B11 ] → 5 B10 + 0 𝑛1
23
11 Na + 2 He4 → [ 13 Al25 ] → 13 Al6 + 0 𝑛1
40 4 44 43
18 A + 2 He → [ 20 Ca ] → 20 Ca + 0 𝑛1
(ii) (𝜶, 𝒑) reactions. General scheme of such reactions is
𝐴
𝑧X + 2 He4 → [𝑧 + 2 C𝐴+4 ] → 𝑧+1 Y
𝐴+3
+ 1 H1 ………(14)
Particular Examples
7 N14 + 2 He4 → [ 9 F18 ] → 8 O17 + 1 H1

27
13 Al + 2 He4 → [ 15 P 31 ] → 14 Si30 + 1 H1
32
16 S + 2 He4 → [ 18 A36 ] → 17 Cl35 + 1 H1
39
19 K + 2 He4 → [ 21 Se43 ] → 21 Ca42 + 1 H1
10.6 NEUTRON INDUCED REACTIONS
Neutron induced nuclear reactions are the most interesting ones. Since neutrons are
uncharged particles, they do not have to overcome the Coulomb potential barrier
surrounding a target nucleus and therefore even slow neutrons are very much effective in
inducing nuclear reactions. Because of this property more very much effective in
inducing nuclear reactions. Because of this property more induced reactions. Neutron
induced reactions because of their use in nuclear reactors, have a special status. Types of
the neutron induced reactions are given as (𝑖)(𝑛, 𝛼), (𝑖𝑖)(𝑛, 𝑝), ( iiii )(𝑛, 𝛾) and (𝑛, 2𝑛).
230
231
(i) (𝐧, 𝜶) reactions: (𝑛, 𝛼) reaction

𝐴
𝑍X + 0 𝑛1 → [ 𝑍 C𝐴+1 ] → 𝑍−2 Y
𝐴−3
+ 2 He4 … … … … (15)
Particular examples are
6
3 Li + 0 𝑛1 → [ 3 Li7 ] → 1 H 3 + 2 He4
10
5 B + 0 𝑛1 → [ 5 B11 ] → 3 Li7 + 2 He4
(ii) (n. p) reactions.
𝐴
𝑧X + 0 𝑛1 → [ 𝑧 C𝐴+1 ] → 𝑧−1 Y
𝐴−1
+ 2 0 𝑛1 …………..(16)
Particular examples:
7 N14 + 0 𝑛1 → [ 7 N15 ] → 6 C14 + 1 H1

27
13 Al + 0 𝑛1 → [ 13 Al28 ] → 12 Mg 27 + 1 H1
64
30 Zn + 0 𝑛1 → [ 30 Zn65 ] → 29 Cu64 + 1 H1
(iii) (𝐧, 𝟐𝐧) reactions.
𝐴
2𝑋 + 0 𝑛1 → [ 2 𝐶 1+1 ] → 2 𝑋 Λ−1 + 2 0 𝑛1 ………..(17)
Particular examples:
27
13 Al + 0 𝑛1 → [ 13 Al28 ] → 13 Al
26
+ 0 𝑛1
The 𝑄-value of (𝑛, 2𝑛) reactions is always negative and therefore can be produced only
by the fast neutrons.
Another type of reaction induced by neutrons is the so-called radiative capture of

neutrons, in which photons ( 𝛾-rays) are emitted out and the product nucleus is an isotope
of the target nucleus with a mass number one unit greater.
General scheme of such reactions is

231
232

𝐴
𝑧X + 0 𝑛1 → [ 𝑧 C𝐴+1 ] → 𝑧 X 𝐴+1 + 𝛾
Typical examples are
27
13 Al + 0 𝑛1 → [ 13 Al28 ] → 13 Al28 + 𝛾
115
49 In + .0 𝑛1 → [ 49 In116 ] → 49 In116 + 𝛾
238
92 U + 0 𝑛1 → [ 92 U 239 ] → 92 U 239 + 𝛾
10.7 PHOTO-DISINTEGRATION (NUCLEAR REACTIONS

INDUCED BY PHOTONS OR 𝜸-RAYS)
Photo-disintegration is a process in which disintegration is produced by bombardment

through high energy photons. Photons possess zero rest-mass energy and therefore are
capable of delivering only their kinetic energy and for the ejection of a particle from the
target nucleus, the kinetic energy of the photon must exceed or be at least equal to the
binding energy of the target nucleus. This means that photo-disintegration reactions are
endoergic with threshold energies of the order of 10MeV.
Examples of photo-disintegration reactions are
2
1H +𝛾 → [ 1 H 2 ] → 1 H1 + 0 𝑛1
9
4 Be + 𝛾 → [ 4 Be9 ] → 4 Be8 + 0 𝑛1
31
15 p +𝛾 → [ 15 p31 ] → 15 p30 + 0 𝑛1
These are the examples of [𝛾, 𝑛] reactions. Another type of photo-disintegration reactions
may be (𝛾, 𝑝) reactions. The (𝛾, 𝑝) reactions need photons of energies higher than those
used in (𝛾, 𝑛) reactions.
232
233
10.8 REACTION ENERGETICS - THE Q-VALUE EQUATION
The fundamental principle of conservation of energy and momentum has a decisive

influence upon the characteristics of all physical processes and so also of nuclear
reactions whose energetics, we are going to study in this unit.
Most nuclear reactions involve two-body interactions of the type depicted in figure 01
[Symbolically written as 𝑀𝑇 (𝑀𝑖 , 𝑀𝑒 )𝑀𝑅 , in which a projectile or incident particle of rest
mass 𝑀𝑖 impinges with a velocity 𝑣 𝑖 upon a stationary target (𝑣 𝑇 = 0) of rest mass 𝑀𝑇
and as a result of this interaction a lighter particle of rest mass 𝑀𝑒 is emitted out with a
velocity 𝑣 𝑒 at an angle 𝜃 with the incident direction and the residual nucleus of rest mass
𝑀𝑅 moves off with a velocity 𝑣 𝑅 at an angle 𝜙. From energy conservation principle, the
total energies of the initial and final systems must be equal.
𝑖 𝑇 𝑒 𝑅
i.e. 𝐸tot + 𝐸tot = 𝐸tot + 𝐸tot ……(18)
or [𝐾 𝑖 + 𝐸0𝑖 ] + [𝐾 𝑇 + 𝐸0𝑇 ] = [𝐾 𝑒 + 𝐸0𝑒 ] + [𝐾 𝑅 + 𝐸0𝐾 ] ……(19)
where 𝐾 𝑖 , 𝐾 𝑇 , 𝐾 𝑒 , 𝐾 𝑅 refer to the kinetic energies and 𝐸0𝑖 , 𝐸0𝑇 , 𝐸0𝑒 , 𝐸0𝑅 , refer to the rest
mass energies, of the incident particle, the target, the emergent particle and the residua!
nucleus respectively. It is to be noted that 𝐾 𝑇 , the kinetic energy of the target, is zero for
a target at rest in the 𝐿-system.
An important parameter of all the nuclear reaction experiments is the measurement of the
energy released in the reaction which appears as kinetic energy of the residual nucleus
and of emitted particles or as the energy of the photons etc., set free at the expense of the
internal energy of the colliding systems. This energy is called the 𝑄-value of the nuclear
reaction. Equation (19) can be rearranged to read.
[𝐸0𝑖 + 𝐸0𝑇 ] − [𝐸0𝑒 + 𝐸0𝑅 ] = [𝐾 𝑒 + 𝐾 𝑅 ] − [𝐾 𝑖 + 𝐾 𝑇 ] ……(20)
233
234

The 𝑄-value is then defined as Nuclear Reactions
or also
𝑄 = [𝐾 𝑒 + 𝐾 𝑅 ] − [𝐾 𝑖 + 𝐾 𝑇 ]
… . . (21)
𝑄 = [𝐸0𝑖 + 𝐸0𝑇 ] − [𝐸0𝑒 + 𝐸0𝑅 ]
Now, since the target is stationary in the laboratory system so 𝐾 𝑇 = 0 and from (21), we
get another useful relation viz.
𝑄 = 𝐾𝑒 + 𝐾𝑅 − 𝐾𝑖 ……(22)
The 𝑄-value as defined in equation (21), equals the difference between the rest mass
energies of the pre-reaction quantities (incident particle and target) and the post-reaction
quantities (emergent particle and the residual nucleus). Using Einstein's mass energy
relations (𝐸 = 𝑚𝑐 2 ), equation (21) can also be written in terms of the rest masses of the
participants and the products of the nuclear reactions.
Thus
𝑄 = 𝑐 2 [𝑀𝑖 + 𝑀𝑇 ] − (𝑀𝑒 + 𝑀𝑅 ) ……(23)
Every nuclear reaction possesses a characteristic 𝑄-value and is termed as Exoergic or

Exothermic if 𝑄 is +𝑣𝑒 i.e. (𝑄 > 0) meaning thereby that energy is released during the
nuclear reaction and Endo-ergic or Endo-thermic, if 𝑄 is -ve i.e. (𝑄 < 0) i.e. if energy is
absorbed during the reaction. An exoergic reaction can take place (at least in principle)
spontaneously but an endoergic reaction cannot take place unless the incident particle
possesses a kinetic energy that exceeds the threshold energy of the nuclear reaction. This
threshold energy is equal so the absolute 𝑄-value in the centre of mass system. Exactly at
threshold, the reaction can just commence.
234
235

The 𝑄-value of the reaction can be calculated by means of either of the equations (22) or
(23) : If we use equation (21), then in the centre of mass system, measurements of 𝐾 𝑖
and 𝐾 𝑒 would suffice as 𝑄 = 𝐾 𝑖 − 𝐾 𝑒 . Then 𝐾 𝑅 and 𝐾 𝑇 could be calculated from the
momentum conservation law. But measurements are made in L-system and the
experimental observables which are simplest to record, are the values of 𝐾 𝑖 and 𝐾 𝑐 . In
the 𝐿-system 𝐾 𝑇 is zero as the target is stationary in this system. The kinetic energy of
the recoil nucleus 𝐾 𝑅 is usually very small and hard to measure and therefore is estimated
by expressing it in terms of other measured quantities as follows:
Fig.01: Schematic diagram depicting kinematics of a nuclear reaction.
The momentum conservation law in vector form may be written as
𝐩𝑖 = 𝐏 𝑒 + 𝐩𝐑 ……(24)
The relativistic relationship between momentum and kinetic energy is*
𝑝2 = 2𝑚0 𝐾 + 𝐾 2 /𝑐 2 ……(25)
Then from the momentum triangle (fig.1) we have
2
(𝑝𝑅 )2 = (𝑝𝑖 ) + (𝑝𝑒 )2 − 2𝑝𝑖 𝑝𝑒 cos 𝜃 ………..(26)
235
236

Substituting from equation (25) with appropriate suffixes we get
2
𝑅
(𝐾 𝑅 )2
𝑅 𝑖 𝑖
(𝐾 𝑖 ) 𝑒 𝑒
(𝐾 𝑒 )2
2𝑀 𝐾 + = 2𝑀 𝐾 + + 2𝑀 𝐾 +
𝑐2 𝑐2 𝑐2
(𝐾 𝑖 )2 (𝐾 𝑒 )2
-2 𝑐𝑜𝑠 √[{2𝑀𝑖 𝐾 𝑖 + } × {2𝑀 𝑒𝐾𝑒 + }] ......(27)
𝑐2 𝑐2
(i) Non-Relativistic Q-value Equation
For low energy reactions, the term (𝐾 𝑅 )2 /𝑐 2 shall be very small and then we can find an
approximate solution of equation (27). It is to be noted that neglecting 𝐾 2 /𝑐 2 term in the
expression (25) for relativistic momentum gives,
𝑝2 = 2𝑚0 𝐾 ……(28)
which is the classical expression for the momentum i.e. the expression for the momentum
under non-relativistic limits. Under this case, equation (27) reduces to ;
2𝑀𝑅 𝐾 𝑅 = 2𝑀𝑖 𝐾 𝑖 + 2𝑀𝑒 𝐾 𝑒 − 2cos 𝜃 √[(2𝑀𝑖 𝐾 𝑖 )(2𝑀𝑒 𝐾 𝑒 )]

= 2𝑀𝑖 𝐾 𝑖 + 2𝑀𝑒 𝐾 𝑒 − 4cos 𝜃√(𝑀𝑖 𝑀𝑒 𝐾 𝑖 𝐾 𝑒 )
𝑅
𝑀𝑖 𝑖 𝑀𝑒 𝑒 √(𝑀𝑖 𝑀𝑒 𝐾 𝑖 𝐾 𝑒 )
𝑜𝑟 𝐾 = 𝑅 𝐾 + 𝑅 𝐾 − 2 cos 𝜃 … … (29)
𝑀 𝑀 𝑀𝑅
Now for a stationary target in 𝐿-system, 𝐾 𝑇 = 0 and from equation (22), we have
𝑄 = 𝐾𝑒 + 𝐾𝑅 − 𝐾𝑖
Substituting the value of 𝐾 𝑅 in this equation from equation (29), we get
𝑀𝑖 𝑖 𝑀𝑒 𝑒 2cos 𝜃√(𝑀𝑖 𝑀𝑒 𝐾 𝑖 𝐾 𝑒 )
𝑒
𝑄 =𝐾 + 𝑅𝐾 + 𝑅𝐾 − 𝑅
− 𝐾𝑖
𝑀 𝑀 𝑀
236
237
𝑀𝑖 𝑀𝑒 2cos 𝜃√(𝑀𝑖 𝑀𝑒 𝐾𝑖 𝐾𝑒 )
or 𝑄 = (𝑀𝑅 − 1) 𝐾 𝑖 + (𝑀𝑅 + 1) 𝐾 𝑒 − …………..(30)
𝑀𝑅
This is what is called the 'non-relativistic 𝑄-value equation'. All the masses appearing in
this equation are the nuclear rest masses. With 𝐾 𝑖 , 𝐾 𝑒 and 𝜃 measured in a Laboratory
System, 𝑄 of the reaction can be calculated from this equation. When 𝜃 = 𝜋/2, i.e. when
the emergent particles are observed at right angles to the incident beam of particles, the
last term in equation (30) disappears and the equation reduces to,
𝑀𝑖 𝑀𝑒
𝑄 = ( 𝑅 − 1) 𝐾 𝑖 + ( 𝑅 + 1) 𝐾 𝑒
𝑀 𝑀
The formula is independent of the mechanism by which the nuclear reaction proceeds as
also of any particular nuclear model and so can be applied to all types of two body non-
relativistic nuclear reactions.
For example, we consider some particular processes:
(i) Elastic Scattering. In this process:
𝑀𝑖 = 𝑀𝑒 and 𝑀𝑇 = 𝑀𝑅 ,
hence from equation (23) viz. 𝑄 = [(𝑀𝑖 + 𝑀𝑇 ) − (𝑀𝑒 + 𝑀𝑅 )]𝑐 2 we have,
𝑄 = 0. ……(31)
(ii) Inelastic Scattering. In this process,
𝑀𝑖 = 𝑀𝑒 and 𝑀𝑇 = 𝑀𝑅
𝑏𝑢𝑡 𝑄 = −𝐸 ∗ … … (32)
where 𝐸 ∗ is the excitation energy imparted to the target nucleus.

238

(ii) Relativistic Q-value Equation: For finding out the relativistic 𝑄 -equation, the
equation (27) is to be solved as such i.e. without the appoximation (𝐾 2 /𝑐 2 = 0). This
adds a correction term 𝛿rel to the non-relativistic 𝑄-equation (30). The relativistic 𝑄 -
value equation may then be written as:
𝑀𝑒 𝑀𝑖 √𝑀𝑖 𝑀𝑒 ⋅𝐾𝑖 𝐾 𝑒 ]
𝑄 = (1 + 𝑀𝑅 ) 𝐾 𝑒 + (𝑀𝑅 − 1) 𝐾 𝑖 − 2cos 𝜃 + 𝛿𝑟𝑒𝑙 …(33)
𝑀𝑅
1 2
where 𝛿rel = 2𝑀𝑅𝑐 2 [(𝐾 𝑖 ) + (𝐾 𝑒 )2 − (𝐾 𝑅 )2
𝐾𝑖 𝐾𝑒
−cos 𝜃√ (𝑀𝑖 𝑀𝑒 ⋅ 𝐾 𝑖 𝐾 𝑒 ) ⋅ (𝑀𝑖 + 𝑀𝑒)] …(34)
Mostly 𝐾 𝑅 is small as compared with 𝐾 𝑖 and 𝐾 𝑐 and therefore while evaluating 𝛿rel we
can set 𝐾 𝑅 = 0. It is worthwhile to remark here that in all the relativistic formulae above,
the case in which some particle denoted by the suffix 𝑖 is a photon, can be obtained by
𝑖 𝑀𝑖 𝑐 2
replacing the total energy 𝐸tot = (1−𝑣2/𝑐 2) by the quantum energy ℎ𝑣 and neglecting 𝑀𝑖
wherever it occurs, in comparison with factor 1/(1 − 𝑣 2 /𝑐 2 )
Solution Of Non-Relativistic Q-value Equation.
The variation of emergent particle kinetic energy 𝐾 ′′ for a fixed 𝑄-value can be studied
by regarding the non-relativistic 𝑄-value equation (30) as a quadratic equation in √𝐾 𝑒 .
We can re-write this equation as follows,
𝑀𝑒 𝑒
𝑀𝑖 𝑖
2√𝑀𝑖 𝑀𝑒 𝐾 𝑖 𝐾 𝑒
𝑄 = (1 + 𝑅 ) 𝐾 + ( 𝑅 − 1) 𝐾 − cos 𝜃
𝑀 𝑀 𝑀𝑅
or 𝑄𝑀𝑅 = (𝑀𝑅 + 𝑀𝑒 )𝐾 𝑒 + (𝑀𝑖 − 𝑀𝑅 )𝐾 𝑖 − [2√(𝑀𝑖 𝑀𝑒 𝐾 𝑖 )cos 𝜃] √(𝐾 𝑒 )
or (𝑀𝑅 + 𝑀𝑒 )𝐾 𝑒 − [2√(𝑀𝑖 𝑀𝑒 𝐾 𝑖 )cos 𝜃] √𝐾 𝑒

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−[𝑄𝑀𝑅 + (𝑀𝑅 − 𝑀𝑖 )𝐾 𝑖 ] = 0 ……..(35 )
which is a quadratic equation in √𝐾 𝑐 . The solution of this equation can be written

as,
2√(𝑀𝑖 𝑀𝑒 𝐾 𝑖 )cos 𝜃 ± √4𝑀𝑖 𝑀𝑒 𝐾 𝑖 [cos2 𝜃

+4(𝑀𝑒 + 𝑀𝑅 ){𝑄𝑀𝑅 + (𝑀𝑅 − 𝑀𝑖 )}𝐾 𝑖 ]
√𝐾 𝑒 =
2(𝑀𝑅 + 𝑀𝑒 )
√(𝑀𝑖 𝑀𝑒 𝐾 𝑖 )cos 𝜃 𝑀𝑖 𝑀𝑐 𝐾 𝑖 cos2 𝜃
= ± √
(𝑀𝑅 + 𝑀𝑐 ) (𝑀𝑅 + 𝑀𝑒 )2
{𝑄𝑀𝑅 + (𝑀𝑅 − 𝑀𝑖 )𝐾 𝑖
+ ]
𝑀𝑅 + 𝑀 𝑒
or √𝐾 𝑒 = 𝑝 ± √(𝑝2 + 𝑞) (say)
√(𝑀𝑖 𝑀𝑒 𝐾𝑖 ) [𝑄𝑀𝑅 +(𝑀𝑅 −𝑀𝑖 )𝐾𝑖 ]

where 𝑝 = (𝑀𝑅 +𝑀𝑐 )
cos 𝜃, 𝑞 = (𝑀𝑅 +𝑀𝑒 )
…….(36)
Energetics Of Nuclear Reactions. We can now discuss energetics of nuclear

reactions. Like chemical reactions, nuclear-reactions can also be of two types (i) Exo-
ergic Reactions and (ii) Endo-ergic Reactions, depending upon
whether 𝑄-value for the reaction is respectively positive or negative.
We discuss their energetics below :
(i) Exo-ergic Reactions : For reactions of this type, 𝑄 -value for the reaction is
necessarily positive. These reactions can take place spontaneously (at least in principle)
i.e. even when the kinetic energy of the incident particle is zero(𝐾 𝑖 = 0). We first study
the variation of the kinetic energy of the emergent particles and their angular distribution
as the kinetic emergent particles gradually increased from zero.
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(ii) Zero Projectile Energy : When kinetic energy of the incident particle 𝐾 𝑖 → 0, 𝑝 as
defined in equation also tends to zero
𝑄𝑀𝑅
i.e. 𝑝 → 0 and 𝑞 → 𝑀𝑅 +𝑀𝑒. Hence
or
√𝐾 𝑒 = [𝑝 ± √(𝑝2 + 𝑞)] → √𝑞 or 𝐾 𝑒 → 𝑞
𝑄𝑀𝑅
𝐾𝑒 → … … … (37)
It is evident from this expression for 𝐾 𝑒 that kinetic energy of the emergent particle viz.
𝐾 𝑒 is independent of 𝜃, i.e. it is the same in all directions if the projectile energy is zero.
Finite Projectile Energy. We have defined 𝑞 as follows.
𝑄𝑀𝑅 + (𝑀𝑅 − 𝑀𝑖 )𝐾 𝑖
𝑞=
Since in all nuclear reactions, rest mass of the recoil nucleus is always greater than the
rest mass of the incident particle (projectile) i.e. 𝑀𝑅 > 𝑀𝑖 , hence 𝑞 is positive for all
values of 𝐾 𝑖 (kinetic energy of the incident particle). Therefore, only one of the two roots
of the 𝑄-value equation. 𝑣𝑖𝑧. 𝑝 + √(𝑝2 + 𝑞), shall be acceptable. Thus in this case 𝐾 𝑐 is
single valued for all possible values of incident particle energy 𝐾 𝑖 is single valued for all
possible values of incident particle energy 𝐾 𝑖 and is given by,
𝐾 𝑒 = 𝑝 + √(𝑝2 + 𝑞) ………(38)
(ii) Endo-ergic Reactions : For every nuclear reaction with a positive 𝑄 -value, the
inverse reaction has always a negative 𝑄-value of exactly the same absolute magnitude.
Reactions having negative 𝑄 -values are termed as "endo-ergic reactions". When
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projectile (indicent) particle encrgy is zero, viz. 𝐾 𝑖 = 0, the term (𝑝2 + 𝑞) becomes
negative since 𝑄 is necessarily negative. Therefore √𝐾 𝑒 becomes "imaginary" i.e. the
reaction can not take place even in principle. As the kinetic energy of the incident particle
𝐾 𝑖 is increased from zero, one finds a minimum value of the kinetic energy of the
incident particle at which the reaction is just initiated. This minimum value of the kinetic
energy of the incident particle 𝐾 𝑖 , required to initiate a necessarily endo-ergic reaction is
called the "threshold energy" which is different for different endo-ergic reactions.
From our above analysis, it is at once clear that the reaction becomes first possible when
the kinetic energy of the incident particle 𝐾 −𝑖 is large enough to make the term (𝑝2 + 𝑞)
equal to zero 𝑖. 𝑒.
𝑝2 + 𝑞 = 0
or
𝑀𝑖 𝑀𝑖 𝐾 𝑖 cos2 0 𝑄 ⋅ 𝑀𝑅 + (𝑀𝑘 − 𝑀𝑖 )𝐾 𝑖
+ =0
(𝑀𝑅 + 𝑀𝑖 )2 (𝑀𝑘 + 𝑀𝑖 )
or
[𝑀𝑖 𝑀𝑒 cos2 𝜃 + (𝑀𝑅 + 𝑀𝑒 )(𝑀𝑅 − 𝑀𝑒 )𝐾 𝑖 = −𝑄𝑀𝑅 𝑀𝑅 + 𝑀𝑒 )
or
𝑖
−𝑄𝑀𝑅 (𝑀𝑅 + 𝑀𝑒 )
(𝐾 )𝜃 = 𝑖 𝑒 2
𝑀 𝑀 cos 𝜃 + (𝑀𝑅 + 𝑀𝑒 )(𝑀𝑅 − 𝑀𝑖 )
−𝑄𝑀𝑅 (𝑀𝑅 + 𝑀𝑒 )
(𝐾 𝑖 )𝜃 = 𝑖 𝑒 2
𝑀 𝑀 cos 𝜃 + (𝑀𝑅 )2 − 𝑀𝑅 𝑀𝑖 + 𝑀𝑅 𝑀𝑒 − 𝑀𝑖 𝑀𝑒
−𝑄(𝑀𝑅 + 𝑀𝑒 )
(𝐾 𝑖 )𝜃 = … … … . . (39)
𝑅 𝑒 𝑖 𝑀𝑖 𝑀𝑒 2
𝑀 +𝑀 −𝑀 − sin 𝜃
𝑀𝑅
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𝑀𝑖 𝑀𝑒
If the emergent particle is observed at 𝜃 = 0, then the term sin2 𝜃 becomes zero and
𝑀𝑅
in this case (𝐾 𝑖 )𝜃=0 assumes its minimum possible value (𝐾 𝑖 )𝜃=0 , called "threshold
energy" which is given by,
−𝑄(𝑀𝑅 +𝑀𝑐 )
(𝐸𝑡ℎ )𝜃=𝑛 = (𝐾 𝑖 ) = …….(40)
𝜃=0 𝑀𝑅 +𝑀𝑒 −𝑀𝑖
Now, we know that the general relationship between 𝑄-value and rest masses of reaction
participants and products is given by equation (23) which is reproduced below,
or
𝑄 = [(𝑀𝑖 + 𝑀𝑇 ) − (𝑀𝑅 + 𝑀𝑒 )]𝑐 2

𝑄
𝑀𝑖 + 𝑀𝑇 = 𝑀𝑅 + 𝑀𝑒 + 2
𝑐
𝑄
but as is usually the case, 𝑀𝑇 >> 𝑐 2, then as a very good approximation, we can write
from the above relation,
𝑀𝑖 + 𝑀𝑇 = 𝑀𝑅 + 𝑀𝑒 ………….(41)
𝑀𝑅 + 𝑀𝑒 − 𝑀𝑖 = 𝑀𝑇
Substituting these values in equation (40), we obtain
−𝑄(𝑀𝑖 +𝑀𝑇 )
(𝐸𝑡ℎ )𝜃=0 = ……….(42)
𝑀𝑇
This is the threshold energy for 0 = 0∘
At the threshold of the reaction, product particles first emerge in the direction 𝜃 = 0 and
we can calculate the kinetic energy of these emergent particles as under,
We know √𝐾 𝑐 = 𝑝2 + √(𝑝2 + 𝑞)
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but at threshold, 𝑝2 + 𝑞 = 0, hence in the direction 0 = 0,
𝑀𝑖 𝑀𝑒 (𝐾 𝑖 )𝜃−0
𝑒 2
𝐾 =𝑝 = =0
(𝑀𝑅 + 𝑀𝑒 )
where we nave substituted for 𝑝 from equation (36).
𝑀𝑖 𝑀𝑒
𝐾 𝑒 = (𝐸𝑡ℎ )𝜃=0 (𝑀𝑅 ……….(43)
+𝑀𝑒 )2
As 𝐾 𝑖 increases from the threshold value, product particles start emerging in the
directions 𝜃 > 0 also. At 𝜃 = 90∘ , 𝑝 = 0 and then
√𝐾 𝑒 = √𝑞 or 𝐾 𝑒 = 𝑞
𝑄𝑀𝑅 +(𝑀𝑅 −𝑀𝑖 )(𝐾𝑖 )𝜃=90∘ ………..(44)
𝐾𝑒 = (𝑀𝑅 +𝑀𝑒 )
Now at Threshold, 𝑝2 + 𝑞 = 0 and if 𝜃 = 90∘ , from equation (36), it can be seen that 𝑝
is also zero. Hence, we conclude that 𝑞 = 0
𝑄𝑀𝑅 +(𝑀𝑅 −𝑀𝑖 )(𝐾𝑖 )𝜃=90∘

(𝑀𝑅 +𝑀𝑒 )
=0
𝑄𝑀 + (𝑀 − 𝑀𝑖 )(𝐾 𝑖 )𝜃=90∘ = 0
𝑅 𝑅 ………..(45)
−𝑄𝑀𝑅
(𝐾 𝑖 )𝜃=90∘ = (𝐸th )𝜃=90∘ 𝑀𝑅 −𝑀𝑖
which is the threshold energy for 90∘ .
Threshold Energy: (CASE I): The minimum kinetic energy that an incident particle
must possess in order to bring about a necessarily endo-ergic reaction, is called the
Threshold Energy 𝐸𝑡ℎ for the nuclear reaction. Threshold energy just suffices to initiate
the nuclear reaction process in which the reaction products are formed with zero mutual
velocity in the 𝐶𝑀 system. It therefore follows that at threshold, the final net amount of
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kinetic energy in the 𝐶𝑀-system is zero. However, in the 𝐿-system 𝐾 𝑒 and 𝐾 𝑅 are not
zero because the velocity of the 𝐶𝑀 itself is not zero in 𝐿-system.
In the 𝐿-system the target is stationary and therefore the total kinetic energy of the
reaction participants is given by 𝐾 𝑖 only.
In the 𝐶𝑀-system both the incident particle and the target nucleus are moving towards
each other such that total linear momentum in 𝐶𝑀-system is zero.
i.e.
𝑀𝑖 𝑣 𝑖 = 𝑀𝑇 𝑣 𝑇
𝑀𝑖 𝑣 𝑖 …………..(46)
𝑣 = 𝑀𝑇
Since the target is stationary in the 𝐿-system and hence the total kinetic energy in the 𝐿-
system is equal to the kinetic energy of the incident particle only viz. 𝐾 𝑖 and total kinetic
energy in the 𝐶𝑀-system is
[𝐾]𝐶𝑀 = [𝐾 𝑖 ]𝐶𝑀 + [𝐾 𝑇 ]𝐶𝑀 ……..(47)
The relationship between the kinetic energy of a particle in the 𝐶𝑀-system and in 𝐿-
system is
𝑀𝑇
(𝐾)𝐶𝑀 = 𝐾 𝑖 (𝑀𝑖 +𝑀𝑇 ) ………..(48)
The net final kinetic energy is obtained by adding 𝑄 to this, which gives us what is called
threshold condition
𝑀𝑇
[𝐾 𝑖 ( ) + 𝑄] = 0
𝑀𝑖 + 𝑀𝑇 𝑖ℎ
Therefore, the positive threshold energy 𝐸𝑡ℎ in the 𝐿-system is

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with 𝑄 < 0 i.e. negative.
𝑀𝑖 +𝑀𝑇
𝐸𝑡ℎ = [𝐾 𝑖 ]𝑡ℎ = (−𝑄) ( ) ………(49)
𝑀𝑇
The threshold energy can be determined experimentally and the result can be used to
calculate the 𝑄-value from the above equation. The equation (49) holds good in quite a
general case. However, if the incident particles are 𝛾-rays, then 𝑀𝑖 = 0 and in that case,
threshold energy will be just equal to −𝑄.
The masses entering the 𝑄-value equation (31) and the threshold energy equation are the
nuclear rest masses. However, in actual calculations, they may be replaced by the masses
of the neutral atoms. The electrons that are added to the nuclei to form the neutral atom,
cancel on the two sides of the nuclear reaction equation as the number of electrons on the
two sides of the equation is the same.
CASE II : If the incident particle of mass 𝑀𝑖 strikes the target with a velocity 𝑣 and
forms a compound nucleus of mass 𝑀𝑐 which moves off with a velocity 𝑣 𝑐 , then from
momentum conservation principle, we have
𝑀𝑖 𝑣 𝑖 = 𝑀𝑐 𝑣 𝑐 = (𝑀𝑖 + 𝑀𝑇 )𝑣 𝑐 since 𝑀𝑒 = 𝑀𝑖 + 𝑀𝑇
𝑐
𝑀𝑖 𝑣 𝑖
∴ 𝑣 = 𝑖 … … … … . . (50)
𝑀 + 𝑀𝑇
Now, the energy balance equation can be written as:
Since
𝑄 = 𝐾𝑐 − 𝐾𝑖
1 1 2
𝑄 = 𝑀𝑐 (𝑣 𝑐 )2 − 𝑀𝑖 (𝑣 𝑖 ) … … … … (51)
𝑜𝑟 2 2
2
1 (𝑀𝑖 + 𝑀𝑇 )(𝑀𝑖 𝑣 𝑖 ) 1 𝑖 𝑖 2 1 𝑖 𝑖 2 𝑀𝑖
𝑄= − 𝑀 (𝑣 ) = 𝑀 (𝑣 ) [ − 1]
2 (𝑀𝑖 + 𝑀𝑇 )2 2 2 𝑀𝑖 + 𝑀𝑇
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where we have substituted the value of 𝑣 𝑐 from equation (50)
or
1 𝑖 𝑖 2 𝑀𝑇
𝑄 = 𝑀 (𝑣 ) (− 𝑖 )
2 𝑀 + 𝑀𝑇
or
1 𝑖 𝑖 2 𝑀𝑖 + 𝑀𝑇
𝑀 (𝑣 ) = (−𝑄) ( )
2 𝑀𝑇
∴ Threshold energy
1 2 𝑀𝑖 +𝑀𝑇
𝐸𝑇ℎ = 2 𝑀𝑖 (𝑣 𝑖 ) = (−𝑄) ( ) …………….(52)
𝑀𝑇
Binding Energy And Q-value: The 𝑄-value of a nuclear reaction can be expressed
in terms of the binding energy of the interactting nuclei. If 𝑊 represents the mass of the
constituents of an atom in free state, then
𝑊 = 𝑍𝑚𝑝 + (𝐴 − 𝑍)𝑚𝑛 + 𝑍𝑚𝑒
where 𝑍 = Atomic number, 𝐴 = mass number and 𝑚𝑝 , 𝑚𝑛 and 𝑚𝑒 refer to the proton,
neutron and electron mass respectively.
The binding energy is then defined to be the energy equivalent of the difference of
masses of the constituent particles of the atom in the free state and the mass of bound
atom (𝑀).
∴ B.E. (binding energy) = (𝑊 − 𝑀)𝑐 2
∴ 𝑀𝑖 = 𝑍 𝑖 𝑚𝑝 + (𝐴𝑖 − 𝑍 𝑖 )𝑚𝑛 + 𝑍 𝑖 𝑚𝑒 − (B.E. )𝑖 /𝑐 2
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𝑀𝑇 = 𝑍 𝑇 𝑚𝑝 + (𝐴𝑇 − 𝑍 𝑇 )𝑚𝑛 + 𝑍 𝑇 𝑚𝑒 − ( B.E. )𝑇 /𝑐 2
𝑀𝑒 = 𝑍 𝑒 𝑚𝑝 + (𝐴𝑒 − 𝑍 𝑒 )𝑚𝑛 + 𝑍 𝑐 𝑚𝑒 − ( B.E. )𝑒 /𝑐 2
and 𝑀𝑅 = 𝑍 𝑅 𝑚𝑝 + (𝐴𝑅 − 𝑍 𝑅 )𝑚𝑛 + 𝑍 𝑅 𝑚𝑒 − (B.E.) 𝑅

/𝑐 2
where the raised suffixes 𝑖, 𝑇, 𝑒, 𝑅 refer to incident particle, target, emitted particle and
the residual nucleus respectively.
∴𝑄 = [(𝑀𝑖 + 𝑀𝑇 ) − (𝑀𝑒 + 𝑀𝑅 )]𝑐 2 = −[( B.E. )𝑖 + ( B.E. )𝑇 ]

+[( B.E. )𝑒 + ( B.E. )𝑅 ]
= [Σ( B.E. )final − Σ( B.E. )initial ]
10.9 NUCLEAR REACTION CROSS-SECTIONS
We know that for all types of nuclear reactions, certain conservation laws are satisfied. It
is a very difficult task to ascertain whether a particular projectile will initiate a particular
nuclear process in a given target or not? But somehow we can predict with calculable
accuracy the fraction of the incident particles that induces a particular nuclear process i.e.
we can determine the probability of occurrence of a particular nuclear process. This then
introduces the concept of nuclear reaction cross-section.
Cross-section, for a particular nuclear reaction,
Number of given type of events per unit time per nucleus

𝜎=
Number of projectile particle per unit area per unit time
The total cross-section 𝜎𝑡𝑜𝑡 may be written as
𝜎𝑡𝑜𝑡 = 𝜎𝑠𝑐 + 𝜎𝑎 ……………..(53)
with 𝜎𝑠𝑐 = scattering cross-section
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and 𝜎𝑎 = absorption cross-section
Cross-section has the dimensions of area, and is consequently measured in a unit called
'barn' ( 1 barn = 10−24 cm2 = 10−28 m2 ). The results of all the experimental studies
may be expressed quite conveniently in terms of the concept of cross-section.
Level Width (Γ): The compound nucleus, on the average, remains in a given excited
state for a certain time before decaying either through particle or 𝛾-ray em ission. The
reciprocal of this mean life time 𝜏, is called the decay constant which represents the
probability per unit time of the emission of a particle or 𝛾-ray. In the study of energy
states excited by nuclear reactions, it is customary to use, instead of disintegration
constant, a quantity proportional to it which is called the level width and defined as
ℏ
Γ = = ℏ𝜆 …………..(54)
𝜏
where 𝜆 is the decay constant.
Level width is expressed in energy units and the basis for its use is the uncertainty
principle. In case of energy-time measurements, the uncertainty principle may be written
as :
Δ𝐸 ⋅ Δ𝑡 ≈ ℏ …………..(55)
The mean life time of an excited state of the compound nucleus may be identified with
the time uncertainty Δ𝑡 corresponding to an energy uncertainty Δ𝐸. The level width Γ is
then identified with this energy uncertainty Δ𝐸. It is at once clear from equation (11.40)
that a state having a short mean life time, has a large level width and is thus very poorly
defined in energy whereas a long-lived excited state is quite sharply defined in energy.
Different decay modes possess different level widths. The total width may be written as
the sum of the partial level widths
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Γ = Γ𝑑 + Γ𝛼 + Γ𝑛 + Γ𝛾 + ⋯ ………..(56)
where Γ𝑑 , Γ𝛼 , Γ𝑛 , Γ𝛾 etc. are the partial level widths for the emission of deuteron, 𝛼 -
particle, neutron, 𝛾-rays etc.
The concept of level width is useful in that it can be obtained from the measurement of
resonances. By knowing the total level width, the mean life time can be calculated from
the relation:
ℏ( Joule-sec ) 1 ⋅ 106 × 10−34

𝜏(sec) = = sec.
Γ( Joule ) Γ(eV) × 1 ⋅ 6 × 10−19
6 ⋅ 9 × 10−16
= sec. … … … … (57)
Γ(eV)
Experimental values of the average life time of the compound nucleus are found to be of
the order of 10−14 sec. Comparing this life time with the natural nuclear time which is the
time required for a nuclear projectile to traverse a target nucleus, it is found that the mean
life time is considerably longer than the natural nuclear time. To get an estimate, if we
take a nuclear diameter of ∼ 10−15 meter and a velocity 107 m/sec . for absorbed
projectile, then the natural time is ≈ 10−15 /107 sec ≅ 10−22 sec. This means that the
mean life of the compound nucleus is about 108 times as long as the natural nuclear time.
Example: The 5 B10 (𝛼, 𝑝) 6 C13 reaction shows among others a resonance for an
excitation energy of the compound nucleus of 13.23MeV. The width of this level as
found experimentally is 130KeV . Calculate the mean life of the nucleus for this
excitation.
Solution: The reaction is
5 B10 + 2 He4 → ( 7 N14 )∗ → 6 C13 + 1 H1
The mean life time is given by :

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ℎ 6.625 × 10−34 Joule-sec.
𝜏 = =
2𝜋Γ 2(3 ⋅ 14) × 130 × 103 eV × 1 ⋅ 6 × 10−19 Joule/eV
= 5 × 10−21 sec.
Example: Neutron capture of slow neutrons by 𝑈 235 shows a resonance for an excitation
energy of 0.29eV . The compound nucleus can become de-excited either through 𝛾 -
emission or by tission into larger nuclear fragments. The mean life time of the compound
nucleus was found to be 4.7 × 10−15 sec. and the partial width for 𝛾 -emission Γ𝛾 =
3.4 × 10−2 eV. Calculate the partial fission width 𝑇𝑓 .
Solution: Total width
ℎ (6.625 × 10−34 Joule/sec)

Γ= =
2𝜋𝜏 2 × 3.14 × (4.7 × 10−19 sec)(1.6 × 10−19 Joule/eV)
= 0.14eV
Now since Γ = Γ𝛾 + Γ𝑓
∴ Γ𝑓 = Γ − Γ𝛾
= 0 ⋅ 14eV − 0.034eV
= 0.106eV
10.10 PARTICAL WAVE ANALYSIS OF NUCLEAR

REACTION CROSS-SECTIONS
When a beam of particles strikes against a target, the individual nuclei in the target
behave as independent interaction centres. The incident particle beam is partly scattered
and partly absorbed and therefore the total cross-sectioin of the process, 𝜎𝑡𝑜𝑡 , is
composed of the scattering cross-section 𝜎𝑠𝑐 and the reaction (absorption) cross-section
𝜎𝑎 .i.e.
𝜎𝑡𝑜𝑡 = 𝜎𝑠𝑐 + 𝜎𝑎 …………(58)

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The free particles wave equation describing the process may be written as
∇2 𝜓 + 𝑘 2 𝜓 = 0 …………(59)
The beam of incident particles may be thought of or described by a plane wave moving in
the 𝑍-direction and for particles of constant momentum, the wave function describing the
beam of incident particles as
𝜓𝑖𝑛𝑐 = 𝑒 𝑖𝑘𝑧 = 𝑒 𝑖𝑘𝑟cos 𝜃 ………….(60)
The wave number 𝑘 is given by
1
𝑀𝑣 1
𝑘= = ℏ [2𝑀𝐸]2 ..............(61)
ℏ
where 𝑀 is the reduced mass of the incident particle and target nucleus and 𝑣 is the
velocity of the particle. The incident plane wave can be expressed as superposition of
waves with different orbital angular momenta as follows
𝜓𝑖𝑛𝑐 = 𝑒 𝑖𝑘𝑟cos 𝜃 = ∑𝑙=∞

𝑙=0 𝐵𝑙 (𝑟)𝑌𝑙,0 (𝜃) ……………..(62)
The expansion is obtainable in this form because the expression for the wave function is
independent of the azimuthal angle 𝜙 and therefore the only spherical harmonics
entering the expansion are
𝑌𝑙,0 (0) i.e. with 𝑚 = 0
The coefficients 𝐵𝑙 (𝑟) can be expressed in terms of Spherical Bassel functions 𝑗𝑙 (𝑘𝑟) as
follows :
𝐵𝑙 (𝑟) = 𝑖 𝑙 √[4𝜋(2𝑙 + 1) ∣ 𝑗𝑙 (𝑘𝑟) ……………..(63)
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The spherical Bassel function 𝑗𝑙 (𝑘𝑟) is related to the ordinary Bessel function 𝐽𝑙+12 (𝑘𝑟)
through the formula
𝜋 1/2
𝑗l (𝑘𝑟) = (2𝑘𝑟) 𝐽l+1/2 (𝑘𝑟) ………….(64)
and can be represented as
1 𝑑 ′ sin 𝑘𝑟
𝑗doqn 10𝑙(𝑘𝑟) = (−𝑘𝑟)𝑙 [𝑘𝑟 ⋅ 𝑑(𝑘𝑟)] [ ] ………….(65)
𝑘𝑟
whence asymptotically.
(𝑘𝑟)𝑙
𝜆|𝑘𝑟| << 𝑙 (2𝑙+1)!!
𝑗𝑙 (𝑘𝑟) = { ………….(66)
1 𝑙𝜋
𝑘𝑟 ∣>> 𝑙 sin (𝑘𝑟 − 2 )
𝑘𝑟
where (2𝑙 + 1)!! = (2𝑙 + 1)(2𝑙 − 1) … 7.5.3.1.
𝜓𝑖𝑛𝑐 = 𝑒 𝑖𝑘𝑟cos 𝜃 = ∑𝑙=∞ 𝑙

𝑙=0 𝑖 √[4𝜋(2𝑙 + 1)]𝑗𝑙 (𝑘𝑟) ⋅ 𝑌𝑙,0 (𝜃) …………(67)
At large distances from the interaction centre i.e. |𝑘𝑟| >> 𝑙, the expansion (62) of the
incident wave function with the aid of equation (66), may be written as
𝑙=∞
𝑗𝑘𝑟cos 𝜃
1 𝑙𝜋
𝜓𝑖𝑛𝑐 =𝑒 = ∑ √[4𝜋(2𝑙 + 1)] ⋅ sin (𝑘𝑟 − ) 𝑌𝑙,0 (𝜃)
𝑘𝑟 2
𝑙=0
𝑙=∞
𝜋1/2
= ∑ √(2𝑙 + 1)𝑖 𝑙+1 [𝑒 −𝑖(𝑘𝑟−𝑙𝜋/2) − 𝑒 𝑖(𝑘𝑟−𝑙𝜋/2) ]𝑌𝑙,0 (𝜃) … (68)
𝑘𝑟
𝑙=0
where 𝑙 is the orbital momentum quantum number which designates each incoming
partial wave. Equation (68) contains both the incoming and the outgoing spherical
waves. The first of the exponential terms 𝑣𝑖𝑧. 𝑒 −𝑖[(𝑙𝑘−(𝑙𝜋/2)] represents a series of
incoming spherical waves impinging at the interation centre i.e. 𝑧 = 0 and second of the
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exponential terms viz. 𝑒 𝑖[(𝑘𝑟−(𝑙𝜋/2)] represents outgoing spherical waves. The above
equation (68) represents undistored waves in the absence of any absorbing or scattering
centre. If however a nucleus is located at the origin (𝑧 = 0), the amplitudes of the
outgoing spherical waves from the origin, shall undergo changes.
The stream of particles in the above picture shall have all sorts of impact parameters with
the target nucleus. The impact parameter is the separation at which the collision partners
(i.e their centers) pass one another in the 𝐶𝑀 system if there were no interaction between
them is given by
𝑚𝑣𝑏 = lℏ …………..(69)
𝑙, is orbital quantum number and can take up values 0,1,2,3, ….
From the above equation, the impact parameter is then given by:
𝑙ℏ
𝑏 = 𝑚𝑣 = 𝑙 ƛ …………..(70)
where ƛ (read as lambda cross) is the rationalised de Broglie wavelength for the non-
relativistic particle. The impact parameter b could also be conceived of as quantized Then
only certain discrete could be assumed by the incident particle with respect to the
interacting nucleus e.g.
particle waves with 𝑙 = 0 (S-waves) have impact paraneter 𝑏 = 0.

particle waves with 𝑙 ≠ 1(𝑃-waves ) have impact parameter 𝑏 = ƛ
particle waves with 𝑙 = 2 ( 𝐹-waves) have impact parameter 𝑏 = 2ƛ and so on..
To understand the significance of the angular momenta in a plane wave, the incident
beam can be conceived of as arranged in concentric cylindrical zones, each having
discrete value of the orbital angular momentum. The inner-most zone then contains
particles with impact parameter less than ƛ i.e. ( 𝑆-waves). The next zone contains all
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particles with impact parameters between ƛ and 2ƛ. The 𝑙𝑡ℎ zone contains all particles
with impact parameters between lƛ and (𝑙 + 1)ƛ. The particles in the lth zone, then have
angular momenta berween lℏ and (𝑙 + 1)ℏ.
Fig. 1. Illustration of the quantization of the incident particle orbital angular momentum.
The cross-sectional areà of the lth zone
= 𝜋(𝑙 + 1)2 (ƛ)2 − 𝜋𝑙 2 (ƛ)2 = (2𝑙 + 1)𝜋(ƛ)2
i.e., the reaction cross-section = (2𝑙 + 1)𝜋(ƛ)2 ………….(71)
Since no more particles can be removed from the incident beam by the interaction centre
than are originally present in it and therefore the upper limit to the reaction cross-section
is,
𝜎𝑎,l ≤ (2𝑙 + 1)𝜋(ƛ)2 ……………….(72)
where 𝜎𝑎,𝑙 stands for the reaction cross-section for the lth partial wave.
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. The interaction can change the phase and amplitude of the outgoing wave only and not
of the incoming wave. To make provision for these changes in the outgoing wave, we
modify the wave by introducing a complex multiplicative scattering coefficient 𝜂𝑙 , where
𝑙 refiers to the orbital angular momentum. Naturally then |𝜂𝑙 | ≤ 1, since the number of
outgoing particles cannot exceed the number of incoming particles. Incorporating this
factor in equation ( 68), the new wave function is
𝑙=∞
𝜋1/2 𝑙𝜋
𝜓(𝑟) = ∑ √ (2𝑙 + 1)𝑖 𝑙+1 [𝑒 −𝑖|𝑘𝑟−( 2 )| + 𝜂𝑙 𝑒 𝑖∣𝑘𝑟−(𝑙𝜋/2)] ] 𝑌𝑙,0 (𝜃) … … … (73)
𝑘𝑟
𝑙=0
= 𝜓𝑖𝑛𝑐 + 𝜓𝑜𝑢𝑡 = 𝜓𝑖𝑛𝑐 + 𝜓𝑠𝑐 … … … (74)
This equation (73) holds good for neutrons only. If the incident particles are charged, the
equation must be replaced by an appropriate Coulomb wave function.
The elastically scattered wavefunction 𝜓𝑠𝑐 is the difference of the new total wave
function and the original total wave function i.e. 𝜓𝑠𝑐 = 𝜓(𝑟) − 𝜓𝑖𝑛𝑐
Substituting for 𝜓(𝑟) and 𝜓𝑠𝑐 from equation (73) and (68), we get,
𝜋 1/2
𝜓𝑠𝑐 = ∑𝑙=∞
𝑙=0 √ (2𝑙 + 1)𝑖
𝑙+1 (1
− 𝜂1 )𝑒 𝑖[𝑘𝑟−(𝑙𝜋/2)] 𝑌𝑙,0 (𝜃) …………(74)
𝑘𝑟
The differential elastic scattering cross-section in the direaction 𝜃, which is defined as
outgoing scattered flux of particlesthrough the solid

angle 𝑑Ω in a direction 𝜃
𝜎𝑠𝑐 (𝜃) =
incident flux of particles × 𝑑Ω
To obtain 𝑁𝑠𝑐 (𝜃)𝑑Ω, we enclose the interaction centre by a large sphere of radius 𝑟0 . The
flux corresponding to 𝜓𝑠𝑐 , through a sphere of radius 𝑟0 , according to the rules of
quantum inechanics, is calculated as follows
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The number of particles scattered per second into the solid angle 𝑑Ω at an angle 0 , is
equal to the number of particles scattered through an area 𝑑𝐴 = 𝑟02 𝑑Ω, which is given by
𝑁𝑠𝑐 (𝜃)𝑑Ω = 𝑗𝑠𝑐 𝑑𝐴 ………..(75)
where 𝑗𝑠𝑐 = scattered probability current density and 𝑑𝐴 = 𝑟02 𝑑Ω
∗
ℏ ∂𝜓𝑠𝑐 ∂𝜓𝑠𝑐
∴ 𝑁𝑠𝑐 (θ)𝑑Ω = ∗
(𝜓𝑠𝑐 − 𝜓𝑠𝑐 ) ⋅ 𝑟 2 𝑑Ω
2𝑀𝑖 ∂𝑟 ∂𝑟 𝑟=𝑟0 0
ℏ ∂𝜓𝑠𝑐
= ∗
(𝜓𝑠𝑐 ) ⋅ 𝑟02 𝑑Ω
𝑀𝑖 ∂𝑟 𝑟=𝑟0
ℏ𝑘
= |𝜓 (𝑟 , 𝜃)|2 𝑟02 𝑑Ω
𝑀 𝑠𝑐 0
ℏ𝑘
= 𝑣 ∣ 𝜓𝑠𝑐 (𝑟0 , 𝜃)⌋2 𝑟02 ⋅ 𝑑Ω, Since = 𝑣 … … … … … (76)
𝑀
With the help of equation (74), this equatiion may be written as,
∞ 2
𝜋
𝑁𝑠𝑐 (𝜃)𝑑Ω = 𝑣 ⋅ 2 2 |∑ √ (2𝑙 + 1)𝑖 𝑙+1 (1 − 𝜂𝑙 )𝑒 𝑖(𝑘𝑟−𝑙𝜋/2) 𝑌𝑙,0 (𝜃)| × 𝑟02 𝑑Ω
𝑘 𝑟0
𝑙=0
∞ 2
𝜋
= 𝑣 2 |∑ √ (2𝑙 + 1)𝑖 𝑙+1 (1 − 𝜂𝑙 )𝑒 𝑖(𝑘𝑟−𝑙𝜋/2) 𝑌𝑙,0 (𝜃)| 𝑑Ω … … … … . (77)
𝑘
𝑙=0
And, the incident flux is given by
ℏ ∗
∂𝜓𝑖𝑛𝑐
𝑁= (𝜓𝑖𝑛𝑐 )
𝑀𝑖 ∂𝑟
Ѱ𝑖𝑛𝑐 = 𝑒 𝑖𝑘𝑧
ℏ −𝑖𝑘𝑧 ∂ 𝑖𝑘𝑧
∴ 𝑁 = [𝑒 (𝑒 )]
𝑀𝑖 ∂𝑧
ℏ𝑘
= =𝑣 ……………(78)
𝑀
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Hence
𝑁𝑑Ω = 𝑣 = 𝑑Ω
𝑁𝑠𝑐 (𝜃)𝑑Ω
∴ 𝜎𝑠𝑐 (𝜃) =
𝑣𝑑Ω
∞ 2
𝜋
= 2 |∑ √ (2𝑙 + 1)𝑖 𝑙+1 (1 − 𝜂𝑙 )𝑒 𝑖(𝑘𝑟−𝑙𝜋/2) 𝑌𝑙,0 (𝜃)|
𝑘
𝑙=0
∞ 2
𝜋
= 2 |∑ √ (2𝑙 + 1)(1 − 𝜂𝑙 )𝑌𝑙,0 (𝜃)| … … … . . (79)
𝑘
𝑙=0
The total elastic scattering cross-section is accordingly given by
𝜋
𝜎𝑠𝑐 = ∫ 𝜎𝑠𝑐 (𝜃)𝑑Ω = 𝑘 2 ∑∞
𝑙=0 (2𝑙 + 1)|1 − 𝜂𝑙 |
2
………..(80)
Now since 𝜎𝑠𝑐 = ∑∞

𝑙=1 𝜎𝑠𝑐,𝑙 ……….(81)
where 𝜎𝑠𝑐 = total scattering cross-section and 𝜎𝑠𝑐,𝑙 is the scattering cross-section for the
𝑙 𝑡ℎ partial wave of angular momentum corresponding to 𝑙.
From equations (80) and (81), we have
∞ ∞
𝜋
∑ 𝜎𝑠𝑐,𝑙 = 2 ∑ (2𝑙 + 1)|1 − 𝜂𝑙 |2
𝑘
𝑙=0 𝑙=0
𝜋
∴ 𝜎𝑠𝑐,𝑙 = 2 (2𝑙 + 1)|1 − 𝜂𝑙 |2 … … … … . (82)
𝑘
Reaction Cross-Section: We can also calculate the reaction cross-section by taking

account of particle loss. In the case of absorption (reaction), the number of particles
coming out of the interaction centre is less than the number coming in and so |𝜂𝑙 | < 1.
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Therefore, total absorption cross-section 𝜎𝑎 can be obtained by calculating the number
𝑁𝑎 of the particles removed from the beam per sec. and dividing it by the number 𝑁 of
the incident particles per unit area per second.
Then 𝑁𝑎 is the number of particles that enter this sphere without leaving it again through
the entrance chancel It is therefore equal to the total net flux into this sphere 𝜓(𝑟) as
contained in equation (73).
Thus 𝑁𝑎 = − ∬ 𝑗𝑎 (𝑟0 , 𝜃)𝑟02 𝑑Ω
where 𝑗𝑎 refers to the absorbed probability current density.
ℏ ∂𝜓(𝑟) ∂𝜓 ∗ (𝑟)
∴ 𝑁𝑎 =− ∗
∬ (𝜓 (𝑟) − 𝜓(𝑟) ) × 𝑟02 𝑑Ω
2𝑀𝑖 ∂𝑟 ∂𝑟 𝑟=𝑟0
ℏ ∂𝜓(𝑟)
=− ∬ (𝜓 ∗ (𝑟) ) × 𝑟02 𝑑Ω
𝑀𝑖 ∂𝑟 𝑟=𝑟0
Making use of equation (11 ⋅ 109), and carrying out integration.
∞
ℏ 𝜋
𝑁𝑎 = − ∑ (2𝑙 + 1)[exp {𝑖(𝑘𝑟 − 𝑙𝜋/2)}
𝑀𝑖 𝑘 2
𝑙=0
−𝜂𝑙∗ exp {−𝑖(𝑘𝑟 − 𝑙𝜋/2)}][(−𝑖𝑘)exp {−𝑖(𝑘𝑟 − 𝑙𝜋/2)}
−(𝑖𝑘)𝜂𝑙 exp {𝑖(𝑘𝑟 − 𝑙𝜋/2)}]
On account of the orthonormality of 𝑌𝑙,0 (𝜃)
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∞
ℏ𝜋
𝑁𝑎 = ∑ (2𝑙 + 1)(1 − |𝜂𝑙 |2 )
𝑀𝑘
𝑙=0
∞
ℏ𝑘 𝜋
𝑁𝑎 = ⋅ ∑ (2𝑙 + 1)(1 − |𝜂𝑙 |2 )
𝑀 𝑘2
𝑙=0
∞
𝜋
𝑁𝑎 = 𝑣 ∑ (2𝑙 + 1)(1 − |𝜂𝑙 |2 )
𝑘2
𝑙=0
Hence absorption cross-section,
∞
𝑁𝑎 𝜋 (2𝑙 + 1)(1 − |𝜂𝑙 |2 )
𝜎𝑎 = = 𝑣 2 ∑
𝑁 𝑘 𝑣
𝑙=0
∞
𝜋
𝜎𝑎 = 2
∑ (2𝑙 + 1)(1 − |𝜂𝑙 |2 ) … … . . (83)
𝑘
𝑙=0
𝜎𝑎 = ∑∞
𝑙=0 𝜎𝑎 , 𝑙 ……..(84)
where 𝜎𝑎 = total absorption (reaction) cross-section and 𝜎𝑎 , 𝑙 is the absorption cross-

section for the 𝑙 𝑡ℎ partial wave. From equations (83) and (84)
𝜋
∴ 𝜎𝑎,𝑙 = 𝑘 2 (2𝑙 + 1)(1 − |𝜂𝑙 |2 ) …………(85)
The cross-section (scattering and absorption) given by equations (82) and (85), are valid
for neutron scattering and absorption only as we have not included Columb wave
function in our analysis.
Total Cross-Section: The total crosssection is the sum of the elastic scattering
cross-section 𝜎𝑠𝑐 and the reaction cross-section 𝜎𝑎 and is given by
𝜋
𝜎𝑡𝑜𝑡 = 𝜎𝑠𝑐 + 𝜎𝑎 = 𝑘 2 ∑∞ 2 2
𝑙=0 (2𝑙 + 1)(|1 − 𝜂𝑙 | + 1 − |𝜂𝑙 | ) ……….(86)
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To evalute the above expression we know that if 𝑍1 and 𝑍2 are two complex quantites,
then
|𝑍1 − 𝑍2 |2 = |𝑍1 |2 − 2Re (𝑍1 𝑍2∗ ) + |𝑍2 |2

… … … . (86)
∴ |1 − 𝜂𝑙 |2 = |1|2 − 2Re (𝜂𝑙∗ ) + |𝜂𝑙 |2
Substituting it in the above equation (86), we get
∞
𝜋
𝜎𝑡𝑜𝑡 = 2 ∑ (2𝑙 + 1)[1 − 2Re (𝜂𝑙∗ ) + |𝜂𝑙 |2 + 1 − |𝜂𝑙 |2 ]
𝑘
𝑙=0
∞
2𝜋
= ∑ (2𝑙 + 1)[1 − Re (𝜂𝑙∗ )]
𝑘2
𝑙=0
∞
= 2𝜋(ƛ) ∑ (2𝑙 + 1)[1 − Re (𝜂𝑙∗ )]

2
𝑙=0
∗
or also 𝜎𝑡𝑜𝑡 = 𝜋(ℏ)2 ∑∞ ∗
𝑙=0 (2𝑙 + 1)[2 − (𝜂𝑙 + 𝜂𝑙 )] The symbol Re (𝜂𝑖 ) stands for the
real part of 𝜂𝑖∗ . The total cross-section for the 𝑙 th partial wave may then be written as
𝜎𝑡𝑜𝑡,𝑙 = 𝜋(𝜆)2 (2𝑙 + 1)[𝜂𝑙 + 𝜂𝑙 ∗ ]

∞
… … … … (87)
𝜎𝑡𝑜𝑡 = ∑ 𝜎𝑡𝑜𝑡,𝑙
𝑙=0
Conclusions. We can infer that
(i) If 𝜂𝑙 = −1, 𝜎𝑎,𝑙 = 0 but 𝜎𝑠𝑐,𝑙 is at its maximum and is given by
[𝜎𝑠𝑐 , 𝑙]max = 4𝜋(ℏ)2 (2𝑙 + 1)
This shows that scattering can take place without absorption.
(ii) If 𝜂𝑙 = 0, 𝜎𝑎,𝑙 is at its maximum and is given by,
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𝜎𝑠𝑐,𝑙 = [𝜎𝑎,𝑙 ]max = 𝜋(ℏ)2 (2𝑙 + 1)
meaning thereby that absorption is always accompanied by scattering, although scattering

may and may not be accomponied by absorption. Thus the scattering and absorption
cross-sections for the 𝑙 th partial wave, lie between the limits :
0 ≤ 𝜎𝑠𝑐,𝑙 ≤ 4𝜋(ℏ)2 (2𝑙 + 1)

0 ≤ 𝜎𝑎,𝑙 ≤ 𝜋(ℏ)2 (2𝑙 + 1)
10.11 SUMMARY
After studying this unit learners have learnt about how to
• Describe the Types of Nuclear Reaction
• Explain the Conservation Laws for Nuclear Reactions
• Discriminate between Charged Particle Induced Nuclear Reactions
• Understand the Photo-disintegration process
• Evaluate the Reaction Energetics - the Q-value Equation
• Explain Nuclear Reaction Cross-Sections
• Describe Partial Wave Analysis of Nuclear Reaction Cross-sections
10.12 REFERENCES

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262

1“The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of

Physics, Berlin: Springer Ver.
2.“Nuclear Two Body Problems and Elements of Nuclear Forces” Experimental
3.Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum
Press, New York.
4.Elementary Nuclear Theory, 2nd ed. by Bethe and Morrison, Wiley: New York.
5.Nuclear Physics by D. C. Tayal.
1. Describe the Types of Nuclear Reaction with suitable examples.

2. Explain the Conservation Laws for Nuclear Reactions with examples.
3. Discriminate between different Charged Particle Induced Nuclear
Reactions along with suitable examples.
4. What do you understand by the Photo-disintegration process.
5. Derive an expression for the Q-value of the nuclear reaction.
6. Explain Nuclear Reaction Cross-Sections
7. Describe in detail the Partial Wave Analysis of Nuclear Reaction
Cross-sections.
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UNIT 11
FISSION AND FUSION
11.1 Introduction
11.2 Objectives
11.3 Compound Nucleus
11.3.1 Excitation energy of the Compound Nucleus
11.4 Direct Reactions
11.5 Theory of Stripping and Pick-up Reactions
11.6 Resonance Scattering and Reaction Cross-sections
11.6.1 Reaction Cross-Section
11.7 Continuum Theory of Nuclear Reactions
11.8 Optical Model Theory of Nuclear Reactions
11.9 Nuclear Fission
11.9.1 Discovery of Nuclear Fission
11.9.2 Energy Release in Fission
11.10 Nuclear Fusion
11.10.1 Source of Energy in Stars
11.11 Summary
11.12 References

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264
11.1 INTRODUCTION
. In this unit we shall study that there are many types of nuclear reactions but they all can
be broadly classified into two categories :
(i) Particle induced nuclear reactions.
(ii) Electromagnetic radiation induced nuclear reactions.
The particle induced reactions may further be divided into two more or less distinct
classes known as compound nucleus reactions and direct reactions.
. A unified theory of nuclear reactions simplified enough to be of practical utility is not

available and therefore it is necessary to use different models and approximations
according to the nature of the projectile particle and the target. We first consider particle
induced reactions. The target nucleus as seen from the projectile may be taken to be a
region with a potential and absorption coefficient. This potential is due to the single
nucleons composing the target nucleus. The projectile particles after hitting the target,
may be diffracted by the potential without any energy loss (elastic scattering). If the
energy of the projectile is enough, it may enter the nucleus, hit one target nucleon and lift
it to higher energy state or even to an unbounded state and still preserve enough energy to
leave the target nucleus. This type of interaction is termed as 'direct interaction' or a
direct reaction.
At lower energies, reactions may not involve a direct interaction of the type considered
above but may proceed via an unstable intermediate state called the compound nucleus
state. In this case the incident particle loses so much energy that it cannot escape the
struck nucleus. The compound nucleus thus formed has its mass number equal to the sum
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265

of the mass numbers of the target nucleus and that of the projectile. When nucleons are
the bombarding particles, the mass number of the compound nuclei us shall be 𝐴 + 1.
11.2 OBJECTIVES
• Explain Compound Nucleus

• Describe Direct Reactions
• Understand the Theory of Stripping and Pick-up Reactions
• Explain Resonance Scattering and Reaction Cross-sections
• Discuss the Continuum Theory of Nuclear Reactions
• Describe Optical Model Theory of Nuclear Reactions
• Explain Nuclear Fission
• Discuss the Nuclear Fusion
11.3 COMPOUND NUCLEUS
The compound nucleus being in a highly excited state, is unstable and gives rise to a
typical energy spectrum for the emitted particles having nearly a Maxwellian distribution
of velocities and a practically isotropic angular distribution. In contrast, the direct
processes show strong angular dependences and characteristic maxima of the cross-
section as a function of energy. Perhaps the simplest quantitative difference between
compound nucleus mechanism and the direct reaction mechanism is the time of
interaction. In direct processes the interaction time is of the order of the transit time of the
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266

incident particle over a nuclear diameter which is of the order of 10−22 sec. whereas in
compound nucleus formation mechanism, the interaction time is of the order of
10−14 sec . which is about 108 times greater than the direct reaction mechanism
interaction time. We first discuss the compound nucleus theory of nuclear reaction
mechanisms.
The compound nucleus theory of nuclear reactions was put forth by N. Bohr in 1936.
Prior to the publication of this classic paper, attempts were made to explain the variation
of neutron cross sections on the assumption that the incident projectile particle interacts
with a simple square well type potential. This problem is easily handled within scattering
theory in quantúm mechanics and calculations predict that an interaction with a square
potential well can neither change the energy of the particles nor can remove it from the
beam; it can only detlect it. In other words in an interaction with a simple square
potential-well, elastic scattering will be the dominant process. Calculations also predict
that scattering resonances should occur at wide energy intervals of 10 to 20MeV. But
Fermi and others discovered that slow neutron resonances in medium and heavry nuclei
occur as close as 1eV. Thus the simple square potential-well proves to be inadequate to
expla in them.
According to the Bohr's compound nucleus theory, the nuclear reaction is a two stage
process as outlined below :
(i) A low energy bombarding particle is absorbed by the target nucleus and both together
form a compound nucleus which is in a highly excited state, i.e. incident particle + target
nucle us → compound nucleus. The kinetic energy of the incident particle together with
its binding energy with the compound nucleus, represents the excitation energy of the
compound nucleus. After the incident particle has merged completely with the target
nucleus, its energy no longer remains concentrated on one particle but is rapidly shared
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267

by the collective motion of all the particles of the new system - the compound nucleus.
Thus each of the nucleons in the compound nucleus will have some additional energy but
none of them will have the likelihood of getting all the energy of the incoming particle or
even any large fraction of it. The sharing of the incoming energy is exactly like the
sharing of heat energy in a drop of water between all its constituent molecules in the form
of kinetic energy. Because of this thorough sharing of energy among all the nucleons, the
compound nucleus has no chances of immediate break up.
(ii) The second stage in the compound nucleus theory involves the break-up of the
compound nucleus us into a product nucleus and one or more emitted particles.
Compound nucleus → product nucleus + emitted particles.
The decay of the compound nucleus will occur only when sufficient energy gets
concentrated on some single nucleon or a group of nucleons such as the 𝛼-particles, for it
to break through the coulomb potential barrier and escape from the nucleus. As the life
time of the nucleus is large on a nuclear scale (about 108 times larger than the time taken
by the incident particle to traverse nuclear diameter), the intrinsically slow process of
electromagnetic radiation ( 𝛾 -emission) competes strongly with charged particle
emission. Since the excitation energy is thoroughly shared between all the nucleons, it
seems highly probable that then the mode of decay of the compound nucleus is
completely independent of the mode of its formation i.e. the break up does not take place
until the compound nucleus has completely forgotten its history of formation. In general,
the emitted particle is different from the one which entered the target nucleus to form the
compound nucleus and it will neither have the energy nor the direction of the incoming
particle. However, even if the emitted particle is the same as the incident projectile
particle ; e.g. a neutron escapes when a neutron enters, the process need not be an elastic
collision.
The assumption of the formation of the compound nucleus and its subsequent break-up is
in accordance with many facts of the nuclear reactions. The decay mode is independent
268

of the formation mode is demonstrated by the fact that in general when a target nucleus is
bombarded by means of a given type of projectile, several different nuclides are formed.
27
For example when 13 Al is bombarded by means of protons : nuclides like
24 17 28 24 27
12 Mg , 14 Si , 14 Si , 11 Na etc. are formed. The reaction involving 13 Al target and
proton as projectiles, on the basis of compound nucleus formation theory, may be written
as
Stage I (formation) 1 H1 + 13 Al
27
→ 14 Si28 (compound nucleus )
Stage II (decay) 14 Si28 → 12 Mg
24
+ 2 He4
We shall now derive and discuss the expression for the cross-section of a nuclear reaction
that proceeds by way of compound nucleus formation hypothesis. Since accroding to the
compound nucleus hypothesis as mentioned above, the formation and decay modes of the
compound nucleus are independent of each other and therefore to calculate the cross-
section for a nuclear reaction, it is necessary to determine the cross-sections for the two
processes 𝑖.e. the formation and decay of the compound nucleus.
When the incident particle bas a low energy, the energy level separation is greater than
the level width of the excited levels, and therefore the decay takes place from well
defined states of the compound nucleus. However at high projectile energies the excited
levels may overlap and the life time of the compound nucleus may be comparable with
the transit time of the projectile through the nucleus. In this case a departure from
compound nucleus concept is expected. Here we shall assume the validity of the
compound nucleus hypothesis i.e. consider two stage reactions. On the basis of this
hypothesis, the cross-section for the nuclear reaction 𝑋(𝑎, 𝑏)𝑌 through the formation of
the compound nucleus 𝐶 i.e.
𝑎+𝑋 →𝐶 →𝑌+𝑏
may be written as
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269

𝜎(𝑎, 𝑏) = 𝜎𝐶 (𝑎)𝐺𝐶 (𝑏) ………..(1)
where 𝜎𝐶 (𝑎) is the cross-section for the formation of the compound nucleus 𝐶 by the
incident particle ' 𝑎 ' with the target nucleus 𝑋 and 𝐺𝐶 (𝑏) is the relative probability that
the compound nucleus 𝐶 , once formed decays with the emission of the particle ' 𝑏 '
leaving a residual nucleus 𝑌. Probability 𝐺𝐶 (𝑏) is also referred to as the branching ratio
of the reaction into the emission of ' 𝑏 ' and it is a pure number. It is implied that the
particle ' 𝑎 ' and the target nucleus are in a given total angular momentum state specified
by the symbol ' 𝑎 ' and that the emission of the particle ' 𝑏 ' leaves the residual nucleus in
a given state. During the reaction, total angular momentum and parity were conserved.
The specified entrance and exit states were called channels by Weisskopf. In this
terminology, the reaction is initiated through the channel ' 𝑎 ' and the compound nucleus
decays through the channel ' 𝑏 '.
Consider now another case in which the compound nucleus 𝐶 decays through the
emission of another particle 𝑧 leaving a restdual nucleus 𝑍 so that
and
𝑎+𝑋 →𝐶 →𝑍+𝑧
…………(2)
𝜎(𝑎, 𝑧) = 𝜎𝐶 (𝑎)𝐺𝐶 (𝑧)
According to Bohr's assumption, the disintegration of the compound nucleus into

different channels depends only upon the excitation energy 𝐸𝐶 , the total angular
momentum 𝐽𝐶 and the partio 𝜋𝐶 , of the compound nucleus. Therefore for the same
excitation energy, angular momentum and parity, we assume that the cross-section for the
formation of the compound nucleus with particle 𝑎, viz. 𝜎𝐶 (𝑎) is the same. Consider now
the reaction
and then
𝑏+𝑌 →𝐶 → 𝑍+𝑧
…………(3)
𝜎(𝑏, 𝑧) = 𝜎𝐶 (𝑏)𝐺𝐶 ⋅ (𝑧)
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270

where for the same compound nucleus 𝐶. we have assumed the same branching ratio
𝐺𝐶 (𝑧).
Now, we have 𝜏 = ħ/Γ, where Γ is the total level width. If the partial level widths for the
decay modes 𝑎, 𝑏, 𝑧 are Γ𝑎 , Γ𝑏 , Γ𝑧 , then 𝐺𝐶 (𝑎)𝐺𝐶 (𝑏) may be writien, as
Γ𝑎 Γ𝑏
𝐺𝐶 (𝑎) = and 𝐺𝐶 (𝑏) = ………..(4)
Γ Γ
condition
Now since 𝐺𝐶 (𝑎) or 𝐺𝐶 (𝑏) is a probability, it must satisfy the nomalisation
Σ𝐺𝑐 (𝑎) = 1 and also ΣΓ𝑎 = Γ ………(5)
The mean life times for the modes 𝑎, 𝑏, … are given by [see equation
ℏ ℏ ℏ
𝜏𝑎 = , 𝜏𝑏 = , 𝜏 𝑧 = … … … … . (6)
Γ𝑎 Γ𝑏 Γ𝑧
1 Γ Γ𝑎 + Γ𝑏 + ⋯ 1 1
Hence = = = + +⋯ … … … (7)
𝜏 ℏ ℏ 𝜏𝑎 𝜏𝑏
Now, we know that formation and decay processes of the compound nucleus are
independent of each other, we find out a relationship between 𝑇𝑎 and 𝜎𝐶 (𝑎) making use
of the reciprocity theorem* which state that
𝑘𝑎2 𝜎(𝑎, 𝑏) = 𝑘𝑏2 𝜎(𝑏, 𝑎) ……….(8)
where 𝑘 = 1/ℷ, is the wave number
( ℷ is to be read as 'lambda cross)
Making use of equations (1) and (4). this may be written as
Γ𝑏 Γ𝑎
𝑘𝑎2 𝜎𝐶 (𝑎) = 𝑘𝑏2 𝜎𝐶 (𝑏) … … … … . . (9)
Γ Γ
𝑘𝑎2 𝜎𝐶 (𝑎) 𝜎𝐶 (𝑏) 𝜎𝐶 (𝑧)
= 𝑘𝑏2 = 𝑘𝑧2 = ⋯ = 𝑈 … … … … … (10)
Γ𝑎 Γ𝑏 Γ𝑧
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where 𝑈 is a function only of the state 𝑖. 𝑒. excitation energy, angular momentum and
parity, of the compound nucleus and does not depend upon the decay channel. From
equations (4) and (10), the relative probability of decay of the compound nucleus through
the channel, ' 𝑎 ' is given by,
Γ𝑎 2 𝜎 (𝑎)
𝑘𝑎 2 𝜎 (𝑎)
𝑘𝑎
𝐶 𝐶
𝐺𝐶 (𝑎) = = =Γ
Γ Γ𝑈 𝑎 𝑈+Γ𝑏 𝑈+Γ𝑧 𝑈…
2
𝑘𝑎 𝜎𝐶 (𝑎)
…………(11)
= 2 𝜎 (𝑎)+𝑘 2 𝜎 (𝑏)+𝑘 2 𝜎 (𝑧)+⋯ )
(𝑘𝑎 𝐶 𝑏 𝐶 𝑧 𝐶
Of course a number of experimental evidences are available which support Bohr's

hypothesis of compound nucleus formation and break up but the most interesting series
of experiments to test it was made by Ghoshal. His experiments were designed to study
64∗
the decay of the compound nucleus 30 Zn which could either be produced by the
60 63
bombardment of 28 Ni by means of 𝛼-particles or by bombarding 29 Cu by means of
protons,
4 60
2 He + 28 Ni → 30 Zn64∗
1 63 …………(12)
1 H + 29 Cu → 30 Zn64∗
64∗
In order to produce the same state of excitation of the compound nucleus 30 Zn , the
kinetic energy of the 𝛼-particles must exceed that of the protons by ≃ 7MeV i.e.
Thus
𝛼 𝑝
𝐸𝑘𝑖𝑛 = 𝐸𝑘𝑖𝑛 + 7MeV
60 2 63 )𝑐 2
…….(13)
(𝑀𝛼 + 𝑀𝑁𝑖 )𝑐 + 7MeV = (𝑀𝑃 + 𝑀𝐶𝑢
The reactions studied were the following since, after the formation of the compound
nucleus 30 Zn64∗, it may decay in any of the following ways :
6.3
30 Zn + 0 𝑛1
() 2 He4 + 28 Ni
60
→ 30 Zn
64∗
⟶[ 29 Cu
62
+ 1 H1 + 0 𝑛1
62
30 Zn + 0 𝑛1 + 0 𝑛1
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272

6.3
30 Zn + 0 𝑛1
(b) 1 H1 + 29 Cu
63
→ 30 Zn
64∗
⟶[ 29 Cu
62
+ 1 H1 + 0 𝑛1
62
30 Zn + 0 𝑛1 + 0 𝑛1
The cross-section for these reactions with the aid of equations (1) and (4) may be written
as
Γ𝑏
𝜎(𝑎, 𝑏) = 𝜎𝐶 (𝑎) ………(14)
Γ
where 𝜎𝐶 (𝑎) is the cross-section for the formation of the-compound nucleus with particle
𝑎 and Γ𝑎 is the partial level width for the decay of compound nucleus through the
64∗
emission of the particle 𝑏. If the decay of the compound nucleus 30 Zn which is in a
given excited state, is independent of its mode of formation, the ratio of the cross-section
for the first three reactions given by (a)must be equal to the radio of the cross-section for
the last three reactions given by (b), provided the excitation energy of the compound
64∗ 60
nucleus 30 Zn , when it is formed through the bombardment of 28 Ni by means of 𝛼-
63
particles and when it is formed by the bombardment of 29 Cu by protons is the same i.e.
𝜎(𝛼, 𝑛): 𝜎(𝛼, 𝑝𝑛): 𝜎(𝛼, 2𝑛) = 𝜎(𝑝, 𝑛): 𝜎(𝑝, 𝑝𝑛): 𝜎(𝑝, 2𝑛)
The excitation energy was made the same by making the kinetic energy of the 𝛼-particles
to exceed that of protons by about 7MeV.
The above relation was actually verified by Ghoshal. The experimental results of Ghoshal
with proton cross section curves shifted by 7MeV to the right are shown in fig. (1). The
results were found to be accurate within 10%. A notable feature of these curves is that
the probability for the decay of the compound nucleus by the emission of protons and
neutrons (𝑝𝑛) is much greater than the probability for decay by the emission of two
neutrons (2𝑛).
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273

Another confirmation of the compound nuclear hypothesis. was done by R.R. Roy
through the study of the decay of the compound nucleus 9 F18∗ which could be formed in
either of the following two ways :
2
1H + 8 O16 → 9 F18∗ → 8 O17 + 1 H1
𝑎𝑛𝑑 2 He4 + 7 N14 → 9 F18∗ → 8 O17 + 1 H1
The compound nucleus 9 F18∗ formed by the bombardment of 8 O16 by deuterons was
known to decay by the emission of two group of protons for a specified
excitation energy. R.R. Roy was able to observe the same two groups of protons
excitation energy. R.R. Roy was able to observe the same two groups of protons
Fig. 1: . Cross-section curves (https://www.researchgate.net)
Similar confirmation was also made by John in heavy elements by comparing the
excitation function of (𝛼, 𝑥𝑛) reaction in 82 Pb206 and (𝑝, 𝑥𝑛) reaction in 83 Bi
209
The
210∗
compound nucleus formed was 84 Po . According to Bohr's hypethesis we should
have
𝜎(𝑝, 2𝑛): 𝜎(𝑝, 3𝑛): 𝜎(𝑝, 4𝑛) … = 𝜎(𝛼, 2𝑛); 𝜎(𝛼, 3𝑛): 𝜎(𝛼, 4𝑛) …
These results were confirmed experimentally by John.

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11.3.1 Excitation Energy Of The Compound Nucleus

The relativistic relationship between the total energy 𝐸, momentum 𝑝, and rest mass 𝑚0
is as follows :
𝐸 2 = 𝑝2 𝑐 2 + 𝑚02 𝑐 4
………..(15)
∴ 𝑚0 𝑐 2 = [𝐸 2 − 𝑝2 𝑐 2 ]1/2
Now let,
∗
and 𝑀𝑁𝐶 = the rest mass of the compound nucleus in excited state
𝑀𝐶𝑁 = the rest mass of the compound nucleus in ground state
Energy of the excited state = Kinetic encrgy of the incident particle
+ [mass of the target nucleus + mass of the incident particle ]𝑐 2
And excitation energy, 𝐸exc = (𝐸cxcited state − 𝐸ground state )
∗
= (𝑀𝐶𝑁 − 𝑀𝐶𝑁 )𝑐 2 ………….(16)
Rest mass energy equivalent of the compound nucleus in the excited state in accordance
with equation (15) is
2 2 1/2
∗
𝑀𝐶𝑁 𝑐 2 = [{𝐾 𝑖 + (𝑀𝑖 + 𝑀𝑇 )𝑐 2 } − (𝑝𝑖 ) 𝑐 2 ]
2 2 1/2
∴ 𝐸𝑒𝑐𝑐 = [{𝐾 𝑖 + 𝑀𝑖 + 𝑀𝑇 )𝑐 2 } − (𝑝𝑖 ) 𝑐 2 ] − 𝑀𝐶𝑁 ⋅ 𝑐 2
here 𝑀𝑖 and 𝑀𝑇 refer to the masses of the incident particle and the target nucleus
respectively
2 2
∴ 𝐸𝑐𝑥𝑐 = [(𝐾 𝑖 ) + (𝑀𝑖 + 𝑀𝑇 ) 𝑐 4 + 2𝐾 𝑖 (𝑀𝑖 + 𝑀𝑇 )𝑐 2 ]
2 1/2
−(𝑝𝑖 ) 𝑐 2] − 𝑀𝐶𝑁 ⋅ 𝑐 2
2 2
= [(𝐾 𝑖 ) + 2𝐾 𝑖 𝑀𝑖 𝑐 2 + (𝑀𝑖 + 𝑀𝑇 ) 𝑐 4 + 2𝐾 𝑖 𝑀𝑇 ⋅ 𝑐 2
2 1/2
−(𝑝𝑖 ) 𝑐 2 ] − 𝑀𝐶𝑁 ⋅ 𝑐 2
2 1/2
= [(𝑀𝑖 + 𝑀𝑇 ) 𝑐 4 + 2𝐾 𝑖 𝑀𝑇 ⋅ 𝑐 2 ] − 𝑀𝐶𝑁 ⋅ 𝑐 2
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275

2 2
Since (𝐾 𝑖 ) + 2𝐾 𝑖 ⋅ 𝑀𝑖 𝑐 2 = (𝑝𝑖 ) ⋅ 𝑐 2
2𝑀𝑖 𝐾 𝑇
∴ 𝐸𝑒𝑥𝑐 = (𝑀𝑖 + 𝑀𝑇 )𝑐 2 [1 + ] − 𝑀𝐶𝑁 ⋅ 𝑐 2
(𝑀𝑖 + 𝑀𝑇 )2 𝑐 2
2𝐾 𝑖 𝑀𝑇
≈ (𝑀𝑖 + 𝑀𝑇 )𝑐 2 [1 + ] − 𝑀𝐶𝑁 ⋅ 𝑐 2
2(𝑀𝑖 + 𝑀𝑇 )2 ⋅ 𝑐 2
𝐾 𝑖 𝑀𝑇
≈ (𝑀𝑖 + 𝑀𝑇 )𝑐 2 + − 𝑀𝐶𝑁 ⋅ 𝑐 2
(𝑀𝑖 + 𝑀𝑇 )
𝑖 𝑇 2
𝐾𝑖
≈ (𝑀 + 𝑀 − 𝑀𝐶𝑁 )𝑐 + ) … … … … … (17)
𝑀𝑖
(1 + 𝑇
𝑀
This equation shows that the excitation energy of the compound nucleus is made up of
two parts
(i) Part I is independent of the kinetic energy of the incident particle.
(ii) Part II is proportional to the kinetic energy of the incident particle.
The first part as given above is just the 𝑄-value of the reaction,
viz, 𝑄 = (𝑀𝑖 + 𝑀𝑇 )𝑐 2 − (𝑀𝐶𝑁 )𝑐 2
where 𝑀𝐶𝑁 = (mass of the emitted particle + mass of the residual nucleus) and the
second part is the kinetic energy available in the centre of mass system.
Example:. Protons of energy 4.99MeV, are made to bombard 9 F19 as a result of which a
neutron is emitted according to the following reaction :
9 F19 + 1 H1 → 10 Ne
20∗
→ 10 Ne
19
+ 0 𝑛1
20∗
Calculate the excitation energy of the compound nucleus 10 Ne . By
Solution: Kinetic energy of the proton available in the C.M. system is given
𝐾𝑖 4.99MeV 4.99 × 19MeV

= = =
𝑀 𝑖 1 20
1+ 𝑇 1 + 19
𝑀
= 4.74MeV
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276

To calculate the first part
𝑀𝑖 = 1 H1 = 1 ⋅ 008146 a.m.u. 𝑀𝐶𝑁 = 10 Ne20 = 19 ⋅ 998772

𝑀𝑇 = 9 F19 = 18 ⋅ 004444 a.m.u.
𝑀𝑖 + 𝑀𝑇 = 20 ⋅ 012590 a.m.u..
∴ (𝑀𝑖 + 𝑀𝑇 ) = (20 ⋅ 012590 − 19 ⋅ 998772)
= 0.013818 a.m.u.
which is equivalent to (0.013818 × 931)MeV = 12.88MeV
∴ Excitation energy of the compound nucleus is given by
𝐸𝑒𝑥𝑐 = 4 ⋅ 74 + 12 ⋅ 88 = 17 ⋅ 62MeV
11.4 DIRECT REACTIONS

The compound nucleus process for the nuclear reactions is dominant only in the low
energy region i.e. for comparatively slow projectiles and as such is very important in
practical applications. However the information provided by it about the nucleus itself is
minimal. In fact it was responsible for a 'bum steer' which held back the development of
nuclear structure theory for more than a decade. However, back the development of
nuclear structure theory for more than a decade. However, as the projectile energy is
increased, direct reactions become increasingly and these reactions play a vital role in
revealing the structure of nuclei.
The term direct reaction is used for a wide variety of nuclear interactions such as :
(i) An incident nucleon enters a target nucleus, collides with it. During the collision it
loses some of its energy and then comes off (Inelastic collision).
(ii) An incident nucleon enters a target nucleus and in collision with it loses most of its
energy and then comes off (Inelastic collision).
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(iii) An incident nucleon comes close enough to the target nucleus to set it into vibration
or if the target nucleus is non-spherical, into rotation. The incident particle continues its
way though in a different direction with different energy (Inelastic collision).
(iv) A proton entering a target nucleus transfers its charge to a neutron in the nucleus
through meson exchange and then comes off as a neutron (Knock out process).
(v) An incident nucleon comes close enough to a target nuclets, picks ap a nucleon from
the nuclear surface and the two come off together as deteron. (Pick up reaction).
(vi) An incident deuteron comes close enough to a target nucleus for one of its two
constituent nucleons to settle in the nucleus
Thus direct reactions include inelastic nuclear collisions, stripping reactions and their
inverse i.e. the pick up reactions and knock out processes etc. All these processes have
one thing in common viz. the life time of the corresponding compound nucleus. Thus a
direct reaction is the one which proceeds without the intermediate step of formation of
the compound nucleus. Direct reactions are considered to be instantaneous processes and
take place only at the surface of the target nucleus and therefore direct reaction products
exhibit certain characteristics which are markedly different from those if the reaction
proceeds by way of compound nucleus formation. In these reactions the impinging
particle hits a specific nucleon in the target and the remainder of the nucleus simply acts
as a "spectator".
11.5 THEORY OF STRIPPING AND PICK-UP REACTIONS

The term 'stripping' is used for a type of direct reaction in which an incident compound
particle (an incident projectile having more than one nucleons) is split into two fragmets,
one of which is absorbed within the target nucleus while the other emerges at some angle
𝜃 with the incident direction. An example of such a reaction is provided by deuteron. The
most common reactions involving deuteron are the (𝑑, 𝑝) and the (𝑑, 𝑛) reactions.
Deuteron has a low binding energy of only 2.225MeV per nucleon and a comparatively
large neutron-proton separation of about 2 ⋅ 18 Fermi. This means that the neutron and
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the proton forming the deuteron are very loosely bound together. It is therefore quite
probable that when deuteron strikes against a target, one of its constituent nucleons
(either the proton or the neutron) may be captured by the target nucleus and the other
simply scattered which is pictorially depicted below :
Qualitatively, deuteron stripping mechanism may be understood as follows:
Fig.2: . Pictorial depiction of deuteron stripping.
Because of the finite size of the deuteron, it may happen that one of its constituent
nucleons (either the proton or the neutron) comes in contact with the target nuclear
surface before the other. Since nuclear interaction energies are much higher than the low
binding energy of only 2.225MeV per nucleon for deuteron, the nucleon arriving first at
the nuclear surface may be quickly separated from its partner and form a new residual
nucleus. If the second nucleon, hits the target nuclear surface an instant later, compound
nucleus formation takes place which decays at a later instant as described earlier.
However, if the seco place which decays at a later instant as described earlier. However,
if the second nucleon misses the target nuclear surface, the stripping process, which is a
direct process, takes place. The stripping process, is more probable at low energies of the
deuteron projectile since then the Coulomb repulsion between the target nucleus and the
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proton keeps the proton away from the target nucleus. The nucleon that misses the target,
proceeds with its original speed plus whatever is added to it during the breakup
(stripping) process.
The first experimental studies of the (𝑑, 𝑝) stripping reactions at low deuteron energies
revealed that the reaction cross-section was much higher than anticipated. The theory of
low energy deuteron stripping was first put forth by Oppenheimer and Phillips.
At very high deuteron projectile energies (> 100MeV), the stripping process has proved
to be of considerable practical use in producing nearly mono-energetic beams of fast
neutrons. For example, stripping of 190MeV deuterons from the Berkley Synchro-
cylotron, produced near mono-energetic beam of 90MeV neutrons confined to a narrow
cone in the forward direction.
The inverse of the stripping reaction is called a pick-up reaction such as the following
(𝑝, 𝑑) reaction :
27
13 Al + 1 H1 → 13 Al
26
+ 1H2
Other examples of pick-up reactions are (𝑝, 𝛼) and (𝑛, 𝑑) reactions. The (𝑝, 𝑑) pick-up
reaction takes place the following way :
As the incoming proton approaches the target nuclear surface, one of the surface neutrons
is plucked out from the nuclear surface by the proton-neutron (𝑝 − 𝑛) nuclear force and
the two (the incident proton and the plucked neutron) travel on as bound deuteron. In the
(𝑛, 𝑑) pick-up reaction, a proton is picked up from the nuclear surface and the incoming
neutron and this picked up proton travel on as deuteron. Similarly (𝑝, 𝛼) pick-up reaction
involves the pick-up of a triton out of the target nucleus.
Another reaction produced by deuteron is the so-called knock-out reaction in which

deuteron projectile while colliding with the target nucleus enters the target nuclear
surface but in the process, it may collide with a proton which is knocked out of the target
nucleus and moves away from it. Following are the examples of the knock-out reactions:
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2 27 28
1H + 13 Al → 13 Al + 1 H1
2 39 40
1H + 19 K → 19 K + 1 H1
From what has been said above, it is seen that the theory of stripping and pick-up
reactions is very much similar. A general theory of stripping and pick-up reactions was
put forward by S.T. Butler. For simplicity we have a semi-classical description of the
(𝑑, 𝑝) stripping reaction.
Let 𝐊 𝑑 ℏ be the momentum of deuteron projectile and 𝐊 𝑝 ℏ the montentim of the proton
just after stripping and 𝜃𝑝 be the angle between the incifent deuteron direction and the
direction of travel of the stripped
1
𝐊 𝑝 ℏ = 2 𝐊 𝑑 ℏ + 𝐊 0 ℏ ………..(18)
where 𝐊 𝑝 ℏ is the contribution to the momentum of the stripped proton from the internal
momentum of deuteron at the instant of stripping. The neutron from the deuteron
approaches the nuclear surface with a momentum 𝐊 𝑛 ℏ which is given by
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Fig.3:. Momentum relations in (𝑑, 𝑝) stripping reactions.
𝐊𝑛ℏ = 𝐊𝑑ℏ − 𝐊𝑝ℏ

1
= 𝐊𝑑 ℏ − 𝐊𝑑 ℏ − 𝐊0ℏ
2
1
= 𝐊 𝑑 ℏ − 𝐊 0 ℏ … … … … . (19)
2
In crude approximation 𝐾0 ℏ = |𝐊 0 ℏ| can be taken to be
𝐾0 ℏ = √(𝑀𝑛 𝐸𝐵 ) ………..(20)
where 𝑀𝑛 = neutron mass and 𝐸𝐵 = the binding energy of deuteron,
Now from fig. (3)
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2
2 1
(𝐾0 ℏ)2 = (𝐾𝑝 ℏ) + ( 𝐾𝑑 ℏ) − 𝐾𝑝 𝐾𝑑 ℏ2 cos 𝜃𝑝
2
1
𝑜𝑟 𝐾02 = 𝐾𝑝2 + 𝐾𝑑2 − 𝐾𝑝 𝐾𝑑 cos 𝜃𝑝
4
1
= 𝐾𝑝2 + 𝐾𝑑2 − 𝐾𝑝 𝐾𝑑 (1 − 2sin2 𝜃𝑝 /2)
4
2
1
= (𝐾𝑝 − 𝐾𝑑 ) + 2𝐾𝑝 𝐾𝑑 sin2 𝜃𝑝 /2)
2
or
2 1/2
1 2
𝐾0 = [(𝐾𝑝 − 2 𝐾𝑑 ) + 2𝐾𝑝 𝐾𝑑 sin 𝜃𝑝 /2] ……….(21)
and
2
(𝐾𝑛 ℏ)2 = (𝐾𝑝 ℏ) = (𝐾𝑑 ℏ)2 − 2𝐾𝑝 𝐾𝑑 ℏ2 cos 𝜃𝑝
𝐾𝑛2 = 𝐾𝑝2 + 𝐾𝑑2 − 2𝐾𝑝 𝐾𝑑 cos 𝜃𝑝
= 𝐾𝑝2 + 𝐾𝑑2 − 2𝐾𝑝 𝐾𝑑 (1 − 2sin2 𝜃𝑝 /2)
2
= (𝐾𝑝 − 𝐾𝑑 ) + 4𝐾𝑝 𝐾𝑑 sin2 𝜃𝑝 /2
2 1/2
∴ 𝐾𝑛 = [(𝐾𝑝 − 𝐾𝑑 ) + 4𝐾𝑝 𝐾𝑑 sin2 𝜃𝑝 /2] … … … … (22)
If the neutron has an impact parameter 𝑅 on the target nucleus, then the orbital angular
momentum of the captured neutron is given by
or
𝐾𝑛 ℏ𝑅 = √[𝑙(𝑙 + 1)]ℏ
…………..(23)
𝐾𝑛 ℏ𝑅 = √[𝑙(𝑙 + 1)] ≅ 𝑙
In order to satisfy this equation 𝐾𝑛 must have certain discrete values. For given values of
𝐾𝑝 and 𝐾𝑑 , this means that the possible values of 𝜃𝑝 are restricted by equations 21) and
(22). This is a semi-classicial argument and the uncertainly principle relaxes this rather
stringent condition. Nevertheless the emergent protons are expected to be peaked in a
direction 𝜃𝑝 . From equation (21) we find that for given values of 𝐾𝑝 and 𝐾𝑑 , 𝐾0 increases
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with 𝜃𝑝 . For 𝑆-wave neutrons (𝑙 = 0), we can take 𝜃𝑝 = 0 meaning thereby that the
protons are expected to be peaked in the forward direction.
11.6 RESONANACE SCATTERING AND REACTION CROSS-

SECTIONS .
Breit-Wigner Dispersion Formula For 𝒍 = 𝟎 neutrons.
When target nuclei are bombarded by means of projectiles (both charged and
uncharged), the phenomenon of resonance in unclear reaction and scattering cross-
sections is observed. The resonance is characterized by a sharp increase in the cross-
sections. We are familiar with the phenomenon of resonance in classical physics.
Resonance corresponds to driving a physical system capable of vibrations at one of its
natural frequencies and always corresponds to a sharp rise in the induced amplitude. In
quantum systems, resonance corresponds to driving a physical system having a large
number of energy levels, at a frequency appropriate to induce transitions from the
original state of the system represented by an energy 𝐸, to another state of energy 𝐸 ′ . The
resonance frequency here is given by ℎ𝑣 = 𝐸 ′ − 𝐸0 or ℏ𝜔 = 𝐸 ′ − 𝐸0 and the resonant
state or resonance refers to such an excited state.
In a quantised system, we have a ground state with a well defined energy say 𝐸0 and a
number of excited states each of which is unstable and as such does not possess a well
defined energy but has an energy spread characterised by a width t Γ and a central energy
𝐸1 = h𝜔1 . The resonant energy i.e., the energy corresponding to maximum scattering is
thus given by ℏ𝜔 = 𝐸1 − 𝐸0 .
If we have a large probability for the formation of the excited state which has a number
of decay modes then we may expect maxima in the cross-section for inelastic reactions at
the resonant energy or the elastic reactions. The criterion for the existence of resonance
then is sharp peaks in the cross-sections for both elastic and inelastic reactions.
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On entering the nucleus an individual nucleon may collide with nucleons in the bound
state (target nuclei) raising these nucleons to excited states and thus forming a many
nucleons excited state of the compound nucleus. For a discussion of the resonances
observed in scattering ad absorption cross-sections, we start with the assumption that the
nucleus has a well-defined surface of radius 𝑅. Outside this radius 𝑅, the incident particle
experiences no nuclear potential at all. Inside the nuclear surface, an incident particle is
subjected to a negative time-independent potential so that it moves with a kinetic energy
much larger than that it had outside the nuclear surface. As a consequence, its wave-
length changes but it still retains its original identity. We further assume that inside the
nuclear surface, the incident particle is subjected to strong interaction, so that it rapidly
shares its energy with other nucleons present within the nucleus.
The scattering coefficient 𝜂𝑙 which determines the absolute values of the scattering and
reaction cross-sections is itself determined by the logarithmic derivative of the modified
wave function at the nuclear surface. Of course evaluation of 𝜂𝑙 is a hopelessly difficult
task but even then under some extreme conditions 𝜂𝑙 is expected to behave in a simple
and predicatble manner. It is under this expectation that we have considered the nucleus
possessing a well-defined surface of radius 𝑅.
Here, we consider only the 𝑆 -wave neutrons viz, 𝑙 = 0 neutrons which avoids
complications through Coulomb or centrifugal barriers. We also refrain from
incorporating the introduction of a statistical spin factor. We shall here derive the Breit-
Wigner formula in the neighborhood of a single isolated resonance level. We define a
modified wave function as follows, by assuming that the wave function of the net system
comprising of a neutron and a scattering nucleus is determined only by the radial distance
𝑟 from the interaction centre. The radial wave function in terms of the outgoing and
incoming waves may be written as
𝑈(𝑟) = 𝑟𝜓(𝑟) = 𝑈𝑖𝑛 (𝑟) + 𝑈𝑜𝑢𝑡 (𝑟) ……..(24)
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Therefore for 𝑙 = 0 neutrons, we get the total wave function as : for 𝑟 ≥ 𝑅
1
𝜋 2 −𝑖(𝑘𝑟)
[𝜓(𝑟)]𝑙=0 = 𝑖[𝑒 − 𝜂0 𝑒 𝑖(𝑘𝑟) ]𝑌0,0 (𝜃) = [𝑈(𝑟)]𝑙=0 𝑌0,0 (𝜃) … (25)
𝑘𝑟
∴ The radial wave function [𝑈(𝑟)]𝑙=0 for 𝑙 = 0 neutrons is
𝜋 1/2
[𝑈(𝑟)]𝑙=0 = 𝑖[𝑒 −𝑖𝑘𝑟 − 𝜂0 𝑒 𝑖𝑘𝑟 ] ……..(26)
𝑘
Comparison of equations (24) and (26) shows that
𝜋 1/2
𝑈𝑖𝑛 (𝑟) = 𝑟𝜓𝑖𝑛 (𝑟) = 𝑖𝑒 −𝑖𝑘𝑟 ……….(27)
𝑘
𝜋 1/2
and 𝑈𝑜𝑢𝑡 = 𝑟𝜓𝑜𝑢𝑡 = − 𝑖𝜂0 𝑒 𝑖𝑘𝑟 ……..(28)
𝑘
We now require that at the nuclear surface (𝑟 = 𝑅) the radial wave functions
𝑈𝑖𝑛 (𝑟), 𝑈𝑜𝑢𝑡 (𝑟) and their first derivatives match each other i.e.,
𝑑𝑈𝑖𝑛 (𝑟) 𝑑𝑈𝑜𝑢𝑡 (𝑟)

[𝑈𝑖𝑛 (𝑟)]𝑟=𝑅 = [𝑈𝑜𝑢𝑡 (𝑟)]𝑟=𝑅 and [ ] =[ ] ……..(29)
𝑑𝑟 𝑟=𝑅 𝑑𝑟 𝑟=𝑅
This means that
1 𝑑𝑈𝑖𝑛 (𝑟) 1 𝑑𝑈𝑜𝑢𝑡 (𝑟)

[𝑈 ⋅ ] = [𝑈 ⋅ ] ……….(30)
𝑖𝑛 (𝑟) 𝑑𝑟 𝑟=𝑅 𝑜𝑢𝑡 (𝑟) 𝑑𝑟 𝑟=𝑅
Let us now define a dimensionless quantity ' 𝑓 ' as follows
𝑅𝑑𝑈
𝑓 ≡ Lim𝑟→𝑅 ( 𝑈𝑑𝑟 ) ………(31)
which is 𝑅 times the logarithmic derivative of radial wave function 𝑈 ≡ 𝑈(𝑟) evaluated
at the nuclear surface. From equation (26), we have
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𝑑𝑈 𝜋1/2
[ ] = 𝑖[(−𝑖𝑘)𝑒 −𝑖𝑘𝑟 − (𝑖𝑘)𝜂0 𝑒 𝑖𝑘𝑟 ]
𝑑𝑟 1 𝑘
𝑑𝑈
∴[ ] = 𝜋1/2 [𝑒 −𝑖𝑘𝑅 + 𝜂0 𝑒 𝑖𝑘𝑅 ]
𝑑𝑟 𝑟=𝑅
𝑅 𝑑𝑈 𝑘𝑅 𝜋1/2 [𝑒 −𝑖𝑘𝑅 + 𝜂0 𝑒 𝑖𝑘𝑅 ]
𝑓=[ ⋅ ] = ⋅
𝑈 𝑑𝑟 𝑟=𝑅 𝜋1/2 𝑖[𝑒 −𝑖𝑘𝑅 − 𝜂0 𝑒 𝑖𝑘𝑅 ]
𝑒 −𝑖𝑘𝑅 + 𝜂0 𝑒 𝑖𝑘𝑅
𝑓 = −𝑖𝑘𝑅 [ ] … … … … … (32)
𝑒 −𝑖𝑘𝑅 − 𝜂0 𝑒 𝑖𝑘𝑅
𝑓𝑒 −𝑖𝑘𝑅 − 𝑓𝜂0 𝑒 𝑖𝑘𝑅 = −𝑖𝑘𝑅𝑒 −𝑖𝑘𝑅 − 𝑖𝑘𝑅𝜂0 𝑒 𝑖𝑘𝑅
𝑓𝑒 −𝑖𝑘𝑅 + 𝑖𝑘𝑅𝑒 −𝑖𝑘𝑅 = 𝑓𝜂0 𝑒 𝑖𝑘𝑅 − 𝑖𝑘𝑅𝜂0 𝑒 𝑖𝑘𝑅
(𝑓 + 𝑖𝑘𝑅)𝑒 −𝑖𝑘𝑅 = 𝜂0 (𝑓 − 𝑖𝑘𝑅)𝑒 𝑖𝑘𝑅
[𝑓 + 𝑖𝑘𝑅]𝑒 −𝑖𝑘𝑅 [𝑓 + 𝑖𝑘𝑅] −2𝑖𝑘𝑅
𝜂0 = = 𝑒 … … … . . (33)
[𝑓 − 𝑖𝑘𝑅]𝑒 𝑖𝑘𝑅 [𝑓 − 𝑖𝑘𝑅]
We have the scattering crosssection for 𝑙 = 0 neutrons as,
𝜎𝑠𝑐,0 = 𝜋(ƛ)2 |1 − 𝜂0 |2
𝑓 + 𝑖𝑘𝑅 −2𝑖𝑘𝑅 2
= 𝜋(ƛ)2 |1 − 𝑒 |
𝑓 − 𝑖𝑘𝑅
2
2
(𝑓 − 𝑖𝑘𝑅)𝑒 2𝑖𝑘𝑅 − (𝑓 + 𝑖𝑘𝑅)
= 𝜋(ƛ) | |
(𝑓 − 𝑖𝑘𝑅)𝑒 2𝑖𝑘𝑅
2
2
𝑓𝑒 2𝑖𝑘𝑅 − 𝑖𝑘𝑅𝑒 2𝑖𝑘𝑅 − 𝑓 − 𝑖𝑘𝑅
= 𝜋(ƛ) | |
2
2
𝑓𝑒 2𝑖𝑘𝑅 − 𝑖𝑘𝑅𝑒 2𝑖𝑘𝑅 − 𝑓 − 2𝑖𝑘𝑅 + 𝑖𝑘𝑅
= 𝜋(ƛ) | |
2
2
(𝑓𝑒 2𝑖𝑘𝑅 − 1)𝑖𝑘𝑅(𝑒 2𝑖𝑘𝑅 − 1) − 2𝑖𝑘𝑅
= 𝜋(ƛ) | |
2
2
(𝑒 2𝑖𝑘𝑅 − 1)(𝑓 − 𝑖𝑘𝑅) − 2𝑖𝑘𝑅
= 𝜋(ƛ) | |
𝜋(ƛ)2 2𝑖𝑘𝑅 2𝑖𝑘𝑅 2
= |𝑒 − 1 − |
𝑒 ∣2𝑖𝑘𝑅 ∣ 𝑓 − 𝑖𝑘𝑅
2𝑖𝑘𝑅
= 𝜋(ƛ)2 |𝑒 2𝑖𝑘𝑅 − 1 − | … … … … . (34)
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287
This is so since 𝑒 |2𝑖𝑘𝑅| = |cos 2𝑘𝑅 + 𝑖sin 2𝑘𝑅|
= √ (|cos 2 2𝑘𝑅 + sin2 2𝑘𝑅|) = 1

2
∴ 𝜎𝑠𝑐,0 = 𝜋(𝜋)2 |𝐴𝑝𝑜𝑡 + 𝐴𝑟𝑒𝑠 | … … … … (35)
2𝑖𝑘𝑅
where 𝐴res = − 𝑓−𝑖𝑘𝑅 and 𝐴𝑝𝑜𝑡 = 𝑒 2𝑖𝑘𝑅 − 1 𝐴res is called internal or the resonance
scattering amplitude and 𝐴𝑝𝑜𝑡 is called the potential or hard sphere scattering amplitude.
This origin of the names comes from the consideration of nucleus as perfectly reflecting
sphere of radius 𝑅. This sphere forces the wave-function to vanish at 𝑟 = 𝑅. According to
𝑅𝑑𝑈
the definition of 𝑓 = Lim𝑟→𝑅 { 𝑈𝑑𝑟 }, this implies that 𝑓 becomes infinite at the nuclear
surface and in that case 𝐴res becomes zero for a hard sphere. Scattering then is
determined only by 𝐴𝑝𝑜𝑡 .
The separation of the scattering amplitude into an internal and external part depends upon
the assumption of the nucleus having a well defined surface of radius 𝑅 . Such a
separation has only theroretical meaning i.e. it can not be observed experimentally. The
only experimentally measurable quantity is 𝐴res + 𝐴𝑝𝑜𝑡 . In spite of these restrictions, the
separation of the scattering amplitude into an internal and external part is of prime
importance for the proper understanding of nuclear reaction theory.
Now if 𝑓 is real, we find from equation (33) that |𝜂0 | = 0 and then there is no reaction
(absorption). Therefore only pure elastic scattering would result. However, in general 𝑓
will be a complex quantity and therefore we can write it as
𝑓 = 𝑓Re + 𝑖𝑓Im …………(36)
where the symbols Re and Im stand for real and imaginary parts of 𝑓. In order that |𝜂0 | <
1, the imaginary part must be negative, as is evident from expression (11 ⋅ 140) giving
the value of 𝜂0 .
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Now
−2𝑖𝑘𝑅 −2𝑖𝑘𝑅
𝐴res = 𝑓−𝑖𝑘𝑅 = 𝑓 ……….(37)
𝑅𝑒 +𝑖𝑓Im −𝑖𝑘𝑅
by means of eqn. (36)
Showing that 𝐴res is maximum at 𝑓 = 0
and
𝐴𝑝𝑜𝑡 = 𝑒 2𝑖𝑘𝑅 − 1 = 𝑒 𝑖𝑘𝑅 (𝑒 𝑖𝑘𝑅 − 𝑒 −𝑖𝑘𝑅 )

= 2𝑖𝑒 𝑖𝑘𝑅 sin 𝑘𝑅 … … … (38)
Substituting for 𝐴res and 𝐴𝑝𝑜𝑡 in the above equation (34),we get,
2
2 𝑖𝑘𝑅
−2𝑖𝑘𝑅
𝜎𝑠𝑐,0 = 𝜋(ƛ) |2𝑖𝑒 sin 𝑘𝑅 + |
𝑓Re + 𝑖𝑓lm − 𝑖𝑘𝑅
2
−𝑘𝑅
= 4𝜋(ƛ)2 |𝑒 𝑖𝑘𝑅 sin 𝑘𝑅 + |
2
2 𝑖𝑘𝑅
𝑘𝑅
= 4𝜋(ƛ) |𝑒 sin 𝑘𝑅 + | … … … . . (39)
𝑖(𝑘𝑅 − 𝑓Im ) − 𝑓Re
Similarly, by the value of 𝜂0 from equations (33) with 𝑙 = 0, we obtain the reaction
cross-section,
2 (1
𝑓 + 𝑖𝑘𝑅 −2𝑖𝑘𝑅 2
𝜎𝑎,1 = 𝜋(ƛ) − |𝜂0 |− ) 2
= 𝜋(ƛ) [1 − | 𝑒 | ]
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289
𝑓Re + 𝑖𝑓Im + 𝑖𝑘𝑅 −2𝑖𝑘𝑅 2

= 𝜋(ƛ)2 [1 − | 𝑒 | ]
𝑓Re + 𝑖𝑓Im − 𝑖𝑘𝑅
2
2
𝑓Re + 𝑖(𝑘𝑅 + 𝑓Im )2
= 𝜋 (ƛ) ([1 − | | ]
𝑓Re2 − 𝑖(𝑘𝑅 − 𝑓Im )
𝑓Re2 + (𝑘𝑅 + 𝑓Im )2
= 𝜋(ƛ)2 [1 − ]
𝑓Re2 + (𝑘𝑅 − 𝑓Im )2
2
(𝑘𝑅 − 𝑓Im )2 − (𝑘𝑅 + 𝑓Im )2
= 𝜋(ƛ) [ ]
𝑓Re2 + (𝑘𝑅 − 𝑓Im )2
−4𝑓Im 𝑘𝑅
= 𝜋(ƛ)2 [ ] … … … … (40)
𝑓𝑅𝑒 + (𝑘𝑅 − 𝑓Im )2
We again find that for real 𝑓, i.e. 𝑓Im = 0, 𝜎𝑎,0 vanishes and the scattering cross-section
𝜎𝑠𝑐,0 takes on a form that may be easily interpreted if sin 𝑘𝑅 is very large or very small
𝑘𝑅
compared with |𝑖𝑘𝑅−𝑓 |. So,
Re
(i) If sin 𝑘𝑅 is very large i.e. 𝑘𝑅 is very large, then 𝐴𝑝𝑜𝑡 >> 𝐴res . It will correspond to a
situation far away from resonance, then from equations (11.141) and (11.145), 𝜎𝑠𝑐,0 is
given by
2
𝜎𝑠𝑐,0 = 𝜋(ℏ)2 |𝐴𝑝𝑜𝑡 |
2
= 𝜋(ℏ)2 |2𝑖𝑒 𝑖𝑘𝑅 sin 𝑘𝑅|
2
= 4(𝜋)2 |𝑒 𝑖𝑘𝑅 sin 𝑘𝑅|
2
= 4𝜋(𝜋)2 sin2 𝑘𝑅. Since 𝑒 |𝑖𝑘𝑅| = 1 … … … . . (41)
If 𝑘𝑅 << 1 viz. for low energy neutrons, we get
𝜎𝑠𝑐,0 ≃ 4𝜋(𝜋)2 ⋅ 𝑘 2 𝑅 2 ≃ 4𝜋𝑅 2 ………….(42)
which is four times the geometrical cross-section. This is the hard sphere scattering
formula and represents scattering from an impenetrable sphere of radius R.This holds
good for low energies. Since potential scattering is a function of nuclear radius 𝑅, it can
provide us information regarding the size of the nucleus.
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(ii) If 𝑠𝑖𝑛𝑘𝑅 is very small i. e kR is very small. This implies that 𝐴res ≫> 𝐴𝑝𝑜𝑡 . This
will correspond to a situation quite near to the resonance. In this case, the scattering
cross-section 𝜎𝑠𝑐,0 as given quite near to the resonance
2
−2𝑖𝑘𝑅
𝜎𝑠𝑐,0 = 𝜋(ℏ)2 |𝐴𝑟𝑒𝑠 |2 = 𝜋(ℏ)2 | |
𝑘𝑅
= 4𝜋(𝜋)2 | |
𝑖(𝑘𝑅 − 𝑓Im ) − 𝑓Re
4𝜋𝑅 2
= … … … . (43)
(𝑘𝑅 − 𝑓Im )2 + 𝑓Re 2
which is the resonance scattering formula. This will have maximum value when 𝑓𝑅𝑒 = 0,
so that
[𝜎𝑠𝑐,0 ]max = 4𝜋(ℏ)2
This shows that 𝜎𝑠𝑐 depends critically upon 𝑓Re . Assume now that for a certain value 𝐸0
of the energy of he incident particle, 𝑓𝑅𝑒 = 0, so that resonance occurs at an energy 𝐸0 .
To study the behaviour of the resonance amplitude 𝐴𝑟𝑒𝑠 and [𝜎𝑠𝑐 , 0]𝑟𝑒𝑠 , the scattering
cross-section for 𝑙 = 0 neutrons at an energy 𝐸 in the vicinity of resonance energy 𝐸0 , we
can expand 𝑓 in a Taylor series,
∂𝑓
𝑓 = 𝑓||𝐸=𝐸0 + ∂𝐸| (𝐸 − 𝐸0 ) + ⋯ ……..(44)
𝐸=𝐸0
and use the so called linear approximation method i.e. neglect all derivatives of orders
higher than the first while noting that 𝑓|𝐸=𝐸0 vanishes identicaly. If now we define a
width Γ𝑠𝑐,0 through the relation
∂𝑓 2𝑘𝑅
| = −Γ ……..(45)
∂𝐸 𝐸=𝐸0 𝑠𝑐,0
We can write
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2𝑘𝑅
𝑓 = −Γ (𝐸 − 𝐸0 ) ……….(46)
𝑠𝑐,0
and hence under the approximation; (for real f)
2𝑘𝑅
𝑓 = 𝑓𝑅𝑒 = − Γ (𝐸 − 𝐸0 ) while 𝑓Im = 0 …….(47)
𝑠𝑐,0
Substituting this value in the expression (43), we obtain
4𝜋(ƛ)2 ⋅ (𝑘𝑅)2
[𝜎𝑠𝑐,0 ]𝑟𝑒𝑠 =
4𝑘 2 𝑅 2 (𝐸 − 𝐸0 )2
(𝑘𝑅)2 + 2
Γ𝑠𝑐,0
2
4𝜋(ƛ)2 Γ𝑠𝑐,0
= 2
Γ𝑠𝑐,0 + 4(𝐸 − 𝐸0 )2
2
𝜋(ƛ)2 Γ𝑠𝑐,0
= 2 … … … . (48)
(Γ𝑠𝑐,0 /2) + (𝐸 − 𝐸0 )2
This is the famous Breit-Wigner Single-Level Formula For Scattering. The above
equation shows that the maximum possible cross-section is 4𝜋(ƛ)2 , which corresponds to
the case 𝐸 = 𝐸0 . In a region in the vicinity of resonance energy 𝐸0 , the scattering cross-
section shows the typical resonance shape common to many a physical phenomena,
notably the refractive index of a medium near an absorption line. The physical meaning
of Γ𝑠𝑐,0 is the full energy width of resonace peak at half its maximum value as shown in
the fig. (4).
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292
Fig.4: Schematic representation of the level width Γ𝑠𝑐,0 .
11.6.1 Reaction Cross-Section
For calculating this, we have to consider for the case where 𝑓 is complex, in order to
encompass the case where there are reactions besides clastic scattering. When reaction
takes place 𝑓lm is not zero. In this case we shall not be dealing with the ordinary
stationary states of the Schrodinger's equation. The square of the modulus of the wave-
function of the nucleus shows time dependence. In order to include such time
dependence, we formally introduce a complex value for energy.
Let
𝑖Γ𝑎,0
𝐸=𝜖− ……..(49)
2
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There will be a value of the complex variable near 𝐸0 for which 𝑓 will be zero. We call
𝑖Γ𝑎,0
this value 𝐸𝑎 − . We now desire to have 𝑓 in the vicinity of its zero value and for real
2
values of the energy. Expanding 𝑓 in a power series and applying the linear
approximation i.e. taking the first term only, we get, since
𝑖Γ𝑎,0
𝑓 (𝐸𝑎 − )=0
2
𝑖Γ𝑎,0 ∂𝑓 𝑖Γ𝑎,0
𝑓(𝐸) = (𝐸 − 𝐸𝑎 + )⋅ (𝐸𝑎 − )+⋯
2 ∂𝐸 2
∂𝑓 𝑖Γ𝑎,0
The imaginary part of ∂𝐸 (𝐸𝑎 − ) is small and so to a good approximation, we can
2
consider only its real part. Therefore our expansion gives
𝑖Γ𝑎,0 ∂𝑓
𝑓(𝐸) = (𝐸 − 𝐸𝑎 + )⋅ |
2 ∂𝐸 𝐸=𝐸0
Now from equation (45)
∂𝑓 2𝑘𝑅
| = −Γ
∂𝐸 𝐸=𝐸0 𝑠𝑐,0
2𝑘Γ𝑎,0 𝑅
∴ fRc = −(𝐸 − 𝐸𝑎 ) Γ𝑠𝑐,0
……..(50)
𝑘Γ𝑎,0
𝑎𝑛𝑑 flm = − Γ … … (51)
𝑠𝑐,0
Substituting these values in the expression (40) for the reaction cross-section, we obtain
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2
2
Γ𝑎,0
2 2
𝑘𝑅Γ𝑎,0 2
4𝑘 2 𝑅 2
𝜎𝑎,0 = 𝜋(ℏ) [4𝑘 𝑅 ⋅ ] / [(𝑘𝑅 + ) + (𝐸 − 𝐸𝑎 ) ]
Γ𝑠𝑐,0 Γ𝑠𝑐,0 Γ𝑠𝑐,0
4Γ𝑎,0 Γ𝑠𝑐,0
= 𝜋(ƛ)2 2
(Γ𝑠𝑐,0 + Γ𝑠𝑐,0 ) + 4(𝐸 − 𝐸𝑎 )2
2
Γ𝑎,0 , Γ𝑠𝑐,0
= 𝜋(ƛ) 2 … … … . . (52)
(Γ + Γ )
[ 𝑠𝑐,0 2 𝑎,0 ] + (𝐸 − 𝐸𝑎 )2
𝜋(ƛ)2 Γ𝑎,0 Γ𝑠𝑐,0

= (53)
Γ2
+ (𝐸 − 𝐸𝑎 )2
4
where 𝐸𝑎 ∼ 𝐸0 and Γ𝑎,0 + Γ𝑠𝑐,0 = total level width
Similarly from equation (43)

1
𝜎𝑠𝑐,0 = 4𝜋𝑅 2 𝑘𝑅Γ𝑎,0 4𝑘2 𝑅2
(𝑘𝑅+ )+(𝐸−𝐸𝑎 )2 2
Γ𝑠𝑐,0 Γ𝑠𝑐,0
Γ 2
2 𝑠𝑐,0
= 4𝜋(ƛ) 2
(Γ𝑠𝑐,0 +Γ𝑎,0 ) +4(𝐸−𝐸𝑎 )2
𝜋(ƛ)2 Γ𝑠𝑐,0
……………..(54)
= Γ𝑠𝑐,0 +Γ𝑎,0 2
( ) +(𝐸−𝐸𝑎 )2
2
𝜋(ƛ)2 Γ2𝑠𝑐,0
= Γ2
(𝐸−𝐸𝑎 )2
4
Equations (53) and (54) are the Breit-Wigner Fornula For Single Isolated Level For l =
0 Neutrons when 𝑓 is complex. Equation (53) shows that resonance reaction is always
accompanied by resonance scattering since 𝜎𝑎,0 is zero when 𝜎𝑠𝑐,0 zero.
In order to trace the resonance curve for scattering cross-section in detail, we shall have
to include 𝐴𝑝𝑜𝑡 also. Doing this, we get
2
Γ𝑠𝑐,0
𝜎𝑠𝑐,0 = 𝜋(ƛ)2 | + 2𝑒 𝑖𝑘𝑅 sin 𝑘𝑅|
𝑖
(𝐸 − 𝐸0 )
2 (Γ𝑠𝑐,0 + Γ𝑎,0 ) +
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If we assume that resonance shows scattering only, i.e., Γ𝑎,0 = 0, and also that the
neutron energy is low, i.e. 𝑘𝑅 << 1, then approximately
2
Γ𝑠𝑐,0
𝜎𝑠𝑐,0 = 𝜋(ƛ) |𝑖Γ𝑠𝑐,0 | …………(55)
+(𝐸−𝐸𝑎 )
2
As the energy 𝐸 approaches the resonance energy 𝐸𝑎 , (𝐸 − 𝐸𝑎 ) diminishes and becomes

exactly equal to zero at resonance, then
2
Γ𝑠𝑐,0
2
[𝜎𝑠𝑐,0 ]𝑟𝑒𝑠 = 𝜋(ƛ) | + 2𝑘𝑅|
𝑖Γ𝑠𝑐,0
= 4𝜋(ƛ)2 | − 𝑖| + 𝑘𝑅|2
= 4𝜋(ƛ)2 since 𝑘𝑅 is small ………….(56)
For values of 𝐸 smaller than 𝐸𝑎 , i.e. for the negative values of the term (𝐸 − 𝐸𝑎 ), the real
part of the first term in the expression for the cross-section (55) becomes negative and so
reduces the positive term 𝑘𝑅.
For values of 𝐸 > 𝐸𝑎 , (𝐸 − 𝐸𝑎 ) is positive, the first term decerases again and so also the
cross-section. A plot of 𝜎𝑠𝑐,0 for 𝑙 = 0 neutrons is shown in the fig. (11 ⋅ 10) below.
Fig.5:Elastic scattering cross-section for 𝑙 = 0 neutrons
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11.7 CONTINUM THEORY OF NUCLEAR REACTIONS

If we investigate the interaction of neutrons with nuclei at different energies, starting
from the lowest neutron energy, we find that at neutron energies below the lowest
excitation energy of the compound nucleus, the only decay channel which is open, is the
entrance channel. There can then be no energy loss and the only possible reaction is,
elastic scattering. Further because of low incident energy, only 𝑙 = 0 neutrons will be
affected by the nucleus. As the energy of the incident neutrons increases, the number of
available channels also increases. Let us consider the situation in which the number of
available channels is very large. For medium and heavy nuclei, this will be the case when
the energy of the incident neutrons is of the order of several MeV. The postulate, that a
large number of the channels are available, simplifies the problem in that the probability
of the incident particle leaving through the entrance channel becomes negligible. In this
case no resonances in cross-section will be observed because the individual levels
become broader and are very closely spaced. For this reason this theory is known as
continum theory of nuclear reactions. The continuum theory of nuclear cross-sections
treats the individual levels not separately but as an average over many resonances. We
now formulate the continuum theory for 𝑙 = 0 neutrons.
We have seen that scattering and absorption cross-sections could be completely

determined provided, the scattering coefficient 𝜂𝑙 , introduced earlier, is known. This in
turn requires the knowledge of 𝑓 - the logarithmic derivative of the modified radial wave
function 𝑈(𝑟). Hence the crux of the problem in nuclear reaction studies is the evaluation
of 𝜂 from some nuclear model.
Hence we start with the following model of the nucleus :
(i) The nucleus possesses a well defined surface of radius 𝑅. (ii) As the incident neutrons
enters the nuclear surface, it is subjected to a negative potential and hence it moves with a
much larger kinetic energy than what it had outside the nuclear surface.
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(iii) The incident neutron is subjected to very strong interactions within the nucleus so
that it rapidly shares its energy with other nucleons present within the
nucleus.
Under these assumptions, the incident neutron having once entered the nuclear surface,
has negligible probability of coming out of it. In other words, the amplitude
of the outgoing wave, for 𝑟 < 𝑅 would be of the outgoing wave, for 𝑟 < 𝑅, would be
negligibly small and hence the wave function inside the nuclear surface may be written as
𝑈0 = 𝑒 −𝑖𝑘0𝑟 for 𝑟 ≤ 𝑅 …………(57)
Now from equation (31), we have
𝑅 𝑑𝑈
𝑓 = Lim𝑟→𝑅 ( ) . Hence in this case,
𝑈 𝑑𝑟
𝑅
𝑓= (−𝑖𝑘0 )𝑒 −𝑖𝑘0 𝑅 = −𝑖𝜅0 ∧ … … … … (58)
𝑒 −𝑘0𝑅
Hence from equation (33),
𝑓 + 𝑖𝑘𝑅 −2𝑖𝑘𝑅 −𝑖𝑘0 𝑅 + 𝑖𝑘𝑅 −2𝑖𝑘𝑅

𝜂0 = ( )𝑒 =( )𝑒
𝑓 − 𝑖𝑘𝑅 −𝑖𝑘0 𝑅 − 𝑖𝑘𝑅
𝑘0 − 𝑘 −2𝑖𝑘𝑅
𝑜𝑟 𝜂0 = ( )𝑒
𝑘0 + 𝑘
𝑘0 −𝑘
and so, |𝜂0 | = ( )………..(59)
𝑘0 +𝑘
Hence the cross-section for the formation of the compound nucleus with 𝑙 = 0 neutrons,
is given by,
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298
𝜋 2
𝜋 𝑘0 − 𝑘 2
𝜎𝑎,0 = [1 − |𝜂0 | = {1 − ( ) }
𝑘2 𝑘2 𝑘0 + 𝑘
4𝜋𝑘0
= … … … … (60)
𝑘(𝑘0 + 𝑘)2
This expression may be considered to be the product of two terms (i) 𝜋/𝑘 2 and (ii)
4𝑘0 𝑘/(𝑘0 + 𝑘)2 of which the latter has the meaning of a 'transmission coefficient'. It has
values between 0 and 1 when 𝑘 << 𝑘0 and 𝑘 ≈ 𝑘0 respectively. 𝑘 and 𝑘0 are related as
under
ℏ2 𝑘02 ℏ2 𝑘 2
= + 𝑉0 ……………….(61)
2𝑀 2𝑀
where 𝑉0, represents the depth of the potential well.
According to the above formula, the cross-section becomes infinite for neutron energies
approaching zero. So this theory fails for this case as it is only an approximate theory. For
very small neutron energies 𝑘 << 𝑘0 , the cross-section for the formation of the
compound nucleus, as contained above, reduces to
4′ 𝜋𝑘0 4𝜋
𝜎𝑎,0 = = 𝑘𝑘 ……………..(62)
𝑘𝑘02 0
11.8 OPTICAL MODEL THEORY OF NUCLEAR REACTIONS

A high degree of precision in understanding nucleon-nucleon interaction is a prerequisite
for understanding the free nucleon-nuclei interaction. Even with such knowledge, it may
be extremely difficult to establish a theory of nucleon-nucleon or nucleus-nucleus
interactions due to pure mathematical challenges. Therefore, rather than focusing on the
forces that exist between specific single nucleons, we try to understand these interactions
by looking at the overall behaviour of the system, which is the nucleus. The optical
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model of the nucleus was created as a result of this. We already know that the simplest
sort of two-nucleon potential is a square-well, which only consists of a real part. In order
to fit in with the experimental data, the depth and width of the well had to be selected in
an appropriate manner. However this model predicts, (i) the values of the neutron
absorption cross-section that are far too low, (ii) that energy variation of the neutrons has
but a little effect on these cross-sections, (iii) widely spaced resonances, which are all
quite contrary to most reliable experimental evidences. This model therefore, was
replaced by another known as the compound nucleus model which envisages the nuclear
reactions to proceed by way of the formation of an intermediate state - the compound
nucleus, first put forward by Bohr. This strong absorption model is capable of explaining
successfully the existence of sharp, narrow resonances in the elastic scattering cross-
section of slow neutrons by atomic nuclei, but it fell short of providing a satisfactory
quantitative explanation of large variations (giant resonances) in the cross-sections with
energy as also quite marked forward peaking in elastic scattering for medium and fast
neutrons. These giant resonances were observed to have cross-sections with broad peaks
with a large width of the order of 1MeV, varying regularly with neutron energy and mass
number of target nuclei. This is in an obvious eontradiction with the compound nucleus
theory which ascribes sharp and less broader peaks in resonance cross-section. This then
led to the possibility of nuclear reactions proceeding by way of direct interaction
processes which do away with the intermediate stage of compound nucleus formation.
The direct interaction model gained importance chiefly because of its ability to explain
successfully deuteron stripping.
However the compound nucleus and direct interaction models are incomplete unless we
specify the interaction potential. These two models provide two extremes - compound
nucleus envisages strong absorption and the direct interaction no absorption. To account
for the experimental evidence of the presence of scattering along with absorption, a
compromise between the two extremes was needed. This led to a form of the interaction
potential that contained'both a real and an imaginary part - the imaginary part to provide
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an adequate explanation for the absorption of nucleons in nucleuar matter. So a semi-
transparent (cloudy crystal ball) model of the nucleus known as the optical model gained
prominence chiefly because of its ability to explain both the existence and location of
gaint resonances in neutron cross-sections, both at low and high neutron energies. The
model was put forth by Bethe, Feshbach, Porter and Weisskopf. In this model, the
nucleus is not taken to be completely black to the incident nucleons but rather translucent
to the particle waves 𝑖.e. the particle of a given kinetic energy may have a mean free path
in the nuclear matter of such a magnitude that they may and may not be absorbed in the
nucleus to form a compound nucleus. This fact may be introduced in mathematical
analysis by taking a complex potential in which both the real and imaginary parts are
energy-dependent. All that was needed, was to choose the complex potential parameters
so as to produce the required mean free path of the incident particle in the target nucleus
at various energies of the incident nucleon. A small value of the imaginary part means
long free path.
Tbe simplest form of the optical model complex potential is the simple square-well type
which may be written as :
−(𝑉0 + 𝑖𝑊) = −𝑉0 (1 − 𝑖𝜁) for 𝑟 ≤ 𝑅

𝑉(𝑟) = { ……….(63)
0 for 𝑟 > 𝑅
where 𝑅 = 𝑅0 𝐴1/3, the nuclear radius and both 𝑉0 , 𝑊 are positive and real numbers.
Impressive gross fits to the elastic scattering cross-sections of low energy neutrons (from
0.05 to 3MeV ) and target mass number upto 𝐴 = 238, were obtained by using this
square-well complex potential by Freshbach, Porter and Weisskopf with the following
values of the well-parameters :
𝑉0 = 42MeV, 𝑊 = 1.26MeV, 𝑅 = 1.45 A1/3 Fermi

𝑊
𝜁= = 0.03
𝑉0
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To gain insight into the inherent physical content of the complex potential, we write the
one-dimensional Schrodinger wave equation incorporating this potential into it.
2𝑀
∇2 𝜓 + (𝐸 + 𝑉0 + 𝑖𝑊)𝜓 = 0 ……..(64)
ℏ
Let us find out plane wave solutions of this equation of the form
𝜓 = 𝑒 ±𝑖𝑘𝑟 ……..(65)
wherein 𝑘 is now a complex wave number and plus sign betokens incoming waves. This
complex wave number 𝑘 is given by
1/2
2𝑀
𝑘 = { 2 (𝐸 + 𝑉0 + 𝑖𝑊)} … … … . . (66)
ℏ
(2𝑀)1/2 1/2
𝑖𝑊 1/2
= (𝐸 + 𝑉0 ) {1 + } … … … . (67)
ℏ 𝐸 + 𝑉0
If we assume that 𝑊 is small (long mean free path) compared with (𝐸 + 𝑉0 ), then the
above expression may be written as (expanding binomially and applying first order
approximation)
(2𝑀)1/2 𝑖𝑊
𝑘 = (𝐸 + 𝑉0 )1/2 { }
ℏ 2(𝐸 + 𝑉0 )
… … … . . (68)
(2𝑀)1/2 𝑖𝑊(2𝑀) 1/2 (𝐸
+ 𝑉0 )1/2
= (𝐸 + 𝑉0 )1/2 + (𝐸 + 𝑉0 )
ℏ 2ℏ
The complex wave number 𝑘 may be written as
𝑘 → 𝑘𝑟 + 𝑖𝑘𝑖 ………(69)
where 𝑘𝑟 stands for the real part of the wave number and 𝑘𝑖 for its imaginary part . Then
comparing (68) and (69), we can write for the real and imaginary part
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302

1/2
1 2𝑀
{ 2 (𝐸 + 𝑉0 )} … … … . (70)
𝑘𝑟 =
ƛ ℏ
1/2 (𝐸
𝑊((2𝑀) + 𝑉0 )1/2 𝑊
𝑘𝑖 = = 𝑘 … … . . (71)
2(𝐸 + 𝑉0 )ℏ 2(𝐸 + 𝑉0 ) 𝑟
𝑊
Now since (𝐸 + 𝑉0 ) > 𝑊 i.e. 𝐸+𝑉 < 1, therefore we conclude that within the potential-
0
well, real part of the potential (𝑘𝑟 ) is much larger than the imagincay part (𝑘𝑖 ). Equation
(70) gives the real part of the wave number (𝑘𝑟 ) within the potential well and outside the
2𝑀𝐸 1/2
well the wave number 𝑘0 = ( ) . We thus find that the real part of the wave number
ℏ2
𝑣𝑖𝑧, 𝑘𝑟 with in the well (i.e. within the nucleus) is larger than that outside the nucleus.
(𝑘𝑟 > 𝑘0 ). This means that the wave-length and consequently the velocity of the incident
particle within the nucleus is smaller than that outside the nucleus : The real part of the
refractive index of the nuclear matter in this model is
𝑐0 𝑘𝑟 𝐸+𝑉0 1/2
𝑛𝑟 = = =( ) ………(72)
𝑐𝑟 𝑘0 𝐸
If we take 𝐸 = 10MeV and 𝑉0 = 40MeV
𝑛𝑟 = √((10 + 40)/10) = √5 ≅ 2 ⋅ 24
which is much greater than 1.
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303

However the actual refractive index is given by
𝐸+𝑉0 +𝑖𝑊 1/2

𝑛=( ) ………(73)
𝐸
where 𝐸 = 10MeV, 𝑊 = 10MeV and 𝑉0 = 40MeV,
𝑘𝑟 = 1.55fm−1 and ∴ (Å)2 = 0.645fm
and
𝑘𝑟 𝑊 𝑘𝑟 × 10 𝑘𝑟
𝑘𝑖 = = = = 0 ⋅ 155fm−1
2(𝐸 + 𝑉0 ) 2(10 + 40) 10
Now, the plane wave solution for the outgoing wave is found to be
𝜓 = 𝑒 𝑖𝑘𝑥 = 𝑒 𝑖(𝑘𝑟 +𝑖𝑘𝑖 )𝑥 = 𝑒 𝑖𝑘𝑟 𝑥 𝑒 −𝑘𝑖 𝑥 ……….(74)
which obviously represents an exponentially decaying (attenuated) wave. It therefore

represents stream of particles, some of which are being absorbed. The wave number 𝑘𝑖
has the character of an absorption coefficient. The probability of compound nucleus
formation depends upon the square of the amplitude of the wave function.
An incident neutron of energy 𝐸 , has an energy 𝐸 + 𝑉0 within the well, therefore

velocity within the well is
1/2
2(𝐸 + 𝑉0 )
𝐶well ={ }
𝑀
ℏ
and mean life time 𝜏 = 2𝑊. Therefore the mean distance the neutron traverses within the
nucleus before it loses its energy in an inelastic collision i.e. Mean decay length = Λ is
1/2
2(𝐸 + 𝑉0 ) ℏ ℏ (𝐸 + 𝑉0 )
Λ = 𝐶well 𝜏 ={ } ⋅ =
𝑀 2𝑊 𝑊 [2𝑀(𝐸 + 𝑉0 )]1/2
1/2
𝐸 + 𝑉0 2𝑀
= since 𝑘𝑟 = { 2 (𝐸 + 𝑉0 )}
𝑊 ⋅ 𝐾𝑟 ℏ
𝐸 + 𝑉0
=( ) ƛ … … … . . (75)
𝑊
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304

with 𝐸 = 10MeV, 𝑉0 = 40MeV and 𝑊 = 10MeV, Λ = 3.226fm , which is not small
compared to nuclear dimensions. This lends credence to the hypothesis of compound
nucleus formation in which the particle is trapped within the nucleus for such a long time
that its initial energy is effectively shared with the entire ensemble of nucleons and it
loses its pristine identity.
One of the most unexpected features of the analysis is the large free path of neutrons in
nuclear matter. Strong interaction between nucleons should be revealed by a mean free
path much smaller than the nuclear radius. A larger free path corresponds to a small
value of the imaginary part of the potential, to account for strong absorption. Bethe
considered a larger imaginary term. At low energies (long mean free path), we may
expect that the attenuation of the incident beam is predominant at the nuclear surface.
This may be due to the fact that imaginary part of the potential near the nuclear surface
may be larger than its value deeper inside the nucleus. (The nucleus may be regarded as
a Fermi gas, the Pauli's exclusion principle then prevents nucleons being scattered into
energy levels already occupied and therefore nucleons only near the top of the Fermi
energy distribution, can interact with the incoming low energy neutrons). However, as
the energy of the incident neutrons is increased, this effect becomes less important and
more and more neutrons may be scattered as there is sufficient energy to lift them to
empty states above the top of the Fermi level. This we expect and do find that the
imaginary part of the potential increases with increase in energy i.e. the absorption of the
incident beam may take place uniformly throughout the whole volume of the nucleus.
One important aspect of the optical potential model is that it can never represent a black
nucleus, even if 𝑊 → ∞, because one cannot invent a potentialwell with boundary
condition such that all the incoming partial waves are completely absorbed. So it is a
grey sphere partly absorbing and partly transmitting.
Complex square well potential gives too large a cross-section for elastic scattering
process in the backward direction. A much better fit to the experimental cross-sections is
obtained by a modified form of the potential introduced by Woods and Saxon which
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replace the square-well potential by a smoothed out potential. This form of the potential
is
−1
𝑉(𝑟) = −(𝑉0 + 𝑖𝑊)[1 + 𝑒 (𝑟−𝑅)/𝑎 ] ……….(76)
A plot of the real part of this potential is shown in the fig.6 below.
Fig 6 : Woods and Saxon Potential diagram(http://inside.mines.edu)
R in this expression is the potential radius at half the well depth. Parameter ' 𝑎 ' is a
measure of the amount of tapering of the potential near the nuclear surface, a large ' 𝑎 '
meansing a diffuse nuclear surface. Typical values of ' 𝑎 ' range between 0.50 and
0.65fm.
The paramters 𝑊 and 𝑉0 are dependent upon the energy and type of the incident
particles.
Recently other forms of the optical model potentials have been introduced to take into
account the deformation of nuclei, the spin orbit coupling and also the energy region in
which experiments are being performed. With the development of giant electronic
computers, more and more sophisticated optical model calculations have been made with
more and more complex potentials. For charged particle interactions, coulomb potantial
has been introduced. Such a complex potantial used by Rosen, Berry, Goldhaber and
Auerbach which includes spin dependence also, is
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306

−1 𝑑 −1
𝑉(𝑟) = 𝑉couiomb − 𝑉[1 + 𝑒 (𝑟−𝑅)/𝑎 ] + 4𝑖𝑊𝑏 ( ) [1 + 𝑒 (𝑟−𝑅)/𝑏 ]
𝑑𝑟
1 𝑑𝑓(𝑟)
−𝑉𝑆 (ℏ𝜋 )2 ( ) | | 𝑙 ⋅ 𝑣⃗ … (76)
𝑟 𝑑𝑟
−1
where ƛ𝜋 is the pion compton wavelength and 𝑓(𝑟) = [1 + 𝑒 (𝑟−𝑅)/𝑎 ] .
The optical model, which is so successful in explaining the course of nuclear reactions,
has however a serious drawback in that, very little is known about the nuclear surface.
The imaginary part 𝑊 is expected to have a maximum value ther the nuclear surface
because it is in this region that there are maximum value near available for the scattered
nucleons.
In addition to the theory of neutron-nuclear interaction, the optical model has 𝛼-particles
and other heavier nuclei by including 𝑉coulomb term in the imaginary potential as above
11.9 NUCLEAR FISSION
11.9.1 Discovery of Nuclear Fission

Nuclear fission is a special type of nuclear reaction in which an excited compound
nucleus breaks up generally into two fragments of comparable mass numbers and atomic
numbers. Fission usually occurs amongst the isotopes of the heaviest elements known,
e.g., uranium, thorium etc.
Nuclear fission was discovered by the two German chemists Otto Hahn and F.
Strassmann in 1939. It happened to be one of the most important discoveries in nuclear
physics, since it paved the way for the utilization of the internal energy of the nucleus for
practical purposes.
We have seen earlier that Enrico Fermi and his associates in Rome investigated the
neutron capture (𝑛, 𝛾) reactions by various nuclei using neutrons, slowed down to very
low energies (1934). Along with other elements, when they bombarded uranium (𝑍 =
92), the last naturally occuring element in the periodic table by slow neutrons they found
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evidence for the production of a few 𝛽 − active isotopes of short half-lives. One of these
was the 23min239 U isotope, which by 𝛽 − emission should transform into the isotope
239
Np of the first transuranic element neptunium (𝑍 = 93). Though this transformation
239
was later actually observed from the radioactivity of Np, Fermi and his associates
failed to establish definitely that such transformation had actually taken place.
Subsequently, Hahn and Strassmann, along with Lise Meitner in Germany and
independently L. Curie-Joliot and L. Savich in France, tried to identify chemically the
radioactive products produced by the neutron bombardment on uranium. The former
group observed that one of the products was chemically similar to the element barium
(𝑍 = 56), while the latter group found evidence that one of the products was similar to
the element lanthanum (𝑍 = 57).
Since the mass number and atomic number of barium or lanthanum are much smaller
than those of uranium, it was not possible to explain the above observations on the basis
of the general ideas about nuclear reactions prevalent in those days. It was believed that
during a nuclear reaction, generally a nucleon (e.g., a proton or a neutron) or a group of
them comprising a light nucleus like a deuteron or an 𝛼-particle, was emitted. So the
residual nucleus produced in a reaction could differ from the target nucleus by only a
few units in 𝐴 and 𝑍. So it was at first thought that the bariumlike element, identified in
the experiments of Hahn and his associates, was in reality an element heavier than
barium and chemically similar to the latter. A glance at the periodic table shows that the
element radium with 𝑍 = 88 falls in the same vertical column as barium (Group II-B).
Hence that the chemically similar. An isotope of radium could conceivably be produced
from uranium bombarded with neutrons by the emission of two 𝛼-particles.
Soon afterwards, Hahn and Strassmann, by very careful chemical analysis, definitely
established that it was barium, and not radium, that was produced as one of the reaction-
products, when uranium was bombarded with slow neutrons. They also found lanthanum
and cerium (𝑍 = 58) amongst the reaction-products. Subsequently, Meitner and her
nephew Otto Frisch, working in Sweden (to which country they had escaped to save
themselves from Nazi persecution in Germany) provided the correct explanation of Hahn
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308

and Strassmann's results by suggesting that the uranium nucleus bombarded with
neutrons broke up into two large fragments. They gave the name nuclear fission to this
new phenomenon.Since the atomic number of barium is 56 , the other fragment produced
in the fission of uranium should have the atomic number (92 − 56) or 36 . It should thus
be the nucleus of an isotope of krypton, which was subsequently identified. The two
fragments produced in fission are known as fission fragments.
The nuclear fission discovered by Hahn and Strassmann can be symbolically written as
235 236 ∗
92 U + 10 𝑛 → 92 U → 56 Ba + 36 Kr
Because of the uncertainties in assigning the mass numbers to the fission fragments,
these are not shown in the above equation.
Since the fission fragments are heavy and carry positive charges, which are many times
the electronic charge (high 𝑍 ) they are expected to produce intense ionization in gases.
This was confirmed by Frisch, who observed large electrical pulses produced by them in
an ionization chamber connected to a linear amplifier.
11.9.2 Energy Release in Fission
Nuclear fission is a highly exoergic reaction. Various experiments have established that
the total kinetic energy of the two fission fragments is about 167MeV, which shows the
enormity of the energy release in the fission process, compared to the energy release in
an ordinary nuclear reactions. Besides, the fission fragments, some energy is also carried
by the 𝛾-rays and a few prompt neutrons emitted along with the fragments during fission.
To these must be added the energies of the 𝛽 − particles and the antineutrinos emitted by
the fission fragments, which are usually radioactive, as also the energies of the 𝛾-rays
associated with the 𝛽-disintegrations of the fragments. Thus the total energy evolved
during nuclear fission is higher than the value given above. In Table.1 there are shown
the distribution of the energies carried by the different components in nuclear fission.
Table.1
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309
Components Energy
(MeV)
Kinetic energy of the fission fragments Kinetic energy of the prompt

167
neutrons
Energy of the prompt 𝛾-rays 5
Energy of the 𝛽 − -particles emitted by the fission fragments 6
Energy of the antineutrinos emitted by the fission fragments 8
Energy of the 𝛾-rays emitted by the fission fragments 12
Total energy release in fission 604
The energy released during nuclear fission can be measured by bombarding a piece of
uranium with thermal neutrons, which is found to be heated due to the absorption of the
fission fragments and some of the other products. The heat thus generated can be
measured by calorimetric method, which gives a value of about 186MeV per uranium
nucleus fissioned. This falls short of the value given in the above table. This is due to the
fact that the antineutrinos and 𝛾-rays produced have very high penetrability and hence
escape from the uranium piece.
The enormous quantity of energy release in nuclear fission can be understood

qualitatively with the help of the binding fraction (𝑓𝐵 ) curve given in Ch. II (see Fig.
2.2). A heavy nucleus like uranium has a value of 𝑓𝐵 = 𝐵/𝐴 ≈ 7.6MeV per nucleon. The
fragments produced in its fission have mass numbers near the middle of the periodic
table and hence the values of 𝑓𝐵 for them are 8.5MeV per nucleon. Thus, during the
fission process, about 0.9MeV energy per nucleon is released, so that the total energy
release is around 238 × 0.9 = 212MeV.
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310

We can determine the energy release in fission quantitatively from the known atomic
masses of the nuclei involved. If we assume that three prompt neutrons are released
when the fission occurs, we can write in a typical case
235 236 ∗ 141 92

92 U + 10 𝑛 → 92 U → 56 Ba + 36 Kr + 3 10 𝑛
From the mass energy equivalence principle, we then get
𝑄 = 𝑀( 235 U) + 𝑀𝑛 − 𝑀( 141 Ba) − 𝑀( 92 Kr) − 3𝑀𝑛

= 235.04278 + 1.00866 − 140.9129 − 91.89719 − 3 × 1.00866
= 0.21537u = 200.6MeV
Energies of more or less the same order of magnitude are released in the fission of other
nuclei.
Because of this enormous energy release during fission, it is possible to obtain very large
quantity of energy by the nuclear fission of a small amount of uranium. For example if
235
1 g of U is completely fissioned, we can calculate the energy released from the above
235
estimate of the 𝑄 value: The number of atoms of U per kilogram is
6.025 × 1023 × 103

𝑛= = 2.564 × 1024
235
235
Hence the energy release per gram of U is
𝑛𝑄 2.564 × 1024
𝐸 = = × 200.6 × 1.6 × 10−13
103 103
= 8.229 × 1010
= 2.29 × 104 kWh
A thermal power generator having a capacity of 1MW (heat) would have to be run for
229 hours to generate this amount of energy. The mass of coal which must be burnt to
produce an equivalent amount of energy can be estimated as follows. Since the energy
release in the chemical process of burning of coal (C + O2 = CO2 ) is 4eV per atom of
carbon, the quantity of energy released when 1 kg of carbon is completely burnt is
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4 × 1.6 × 10−19 × 6.025 × 1023 × 103
𝜀 =
12
= 3.213 × 107 J
= 8.926kWh
So mass of carbon required is
2.29 × 104
𝑚= = 2.56 × 103 kg
8.926
The above estimates clearly demonstrate the advantage of using uranium for power
generation.
11.10 NUCLEAR FUSION

It is an accepted fact that the energy created during nuclear fission can be used
practically for both constructive and destructive purposes. Extensive attempts are
currently being conducted to put the energy created in another sort of nuclear reaction to
use for peaceful reasons. The fusion reaction is the name given to this process. As the
name suggests, two (or more) light nuclei combine to form a heavier nucleus in this sort
of reaction. Such reactions are typically exoergic for very light nuclei, which can be
deduced qualitatively from the binding fraction curve. For very light nuclei, the binding
fraction 𝑓𝐵 is a rapidly rising function of 𝐴 which means that a nucleus produced as a
result of the fusion of two lighter nuclei may have greater binding energy than the
combined binding. energies of the latter.
For an example, if we consider the fusion of two deuterons to produce an 𝛼-particle

according to the equation 2 H + 2 H → 4 He, we get the 𝑄 value of the reaction as
𝑄 = 𝐵𝛼 − 2𝐵𝑑 = 4𝑓𝐵𝛼 − 2 × (2𝑓𝐵𝑑 )

= 28.3 − 2 × 2.225 = 23.84MeV
Thus the energy released per nucleon in this reaction is 23.84/4 = 5.96MeV. This is
much larger than the energy released per nucleon in fission. The latter is about (200/
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238) or 0.84MeV. So mass for mass, the fusion reaction usually gives more energy than
the fission reaction.
It may be noted that the fusion reaction given in the above example does not actually
occur. Due to the huge quantity of energy release, the 4 He nucleus has so great an
excitation energy that it breaks up by the emission of a proton or a neutron, as soon as it
is formed, giving rise to the following reactions:
2
H + 2 H → 3 He + 1 𝑛 (𝑄 = 3.26MeV)
2
H + 2H → 3H + 1H (𝑄 = 4.04MeV)
Other examples of exoergic fusion reactions are
3
H + 2 H → 4 He + 1 𝑛 (𝑄 = 17.6MeV)
3
He + 2 H → 4 He + 1 H (𝑄 = 18.3MeV)
6
Li + 2 H → 24 He (𝑄 = 22.4MeV)
7
Li + 1 H → 4 He (𝑄 = 17.3MeV)
The energy released in the last four reactions is much greater than in the first two. The
energy released in the 𝑑 − 𝑡 reaction is 17.6MeV or 3.5MeV per nucleon.
The fusion reactions in the examples given above can occur only if the incident
projectile (usually the deuteron) has sufficiently high energy so that it can come into
close contact with the target nucleus by over-coming the electrostatic repulsion. For two
deuterons the electrostatic with the when in contact (𝑅𝑑 = 4.3 × 10−15 m) is
𝑒2
𝑉𝑏 = = 0.17MeV
8𝜋𝜀0 𝑅𝑑
Classically the fusion can take place only if the incident particle has energy at least equal
to 𝑉s. . However, quantum mechanically there may be tunnelling through the barrier, so
that the reaction may take place at much lower energy ( ∼ a few keV). Of course the
probability of penetration through the barrier will be higher at higher energies.
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313

Reactions using charged particles as projectiles are carried out in the laboratory, using
particle accelerators. However, there is another way in which the above fusion reactions
may be made to occur. The incident particles (𝑝, d etc.) can acquire fairly high energies
if the gases e.g., hydrogen or deuterium are heated to very high temperatures (several
hundred million degrees). According to the kinetic theory of gases, the mean thermal
energy of the gas molecules increases as the temperature increases. At the temperature
𝑇 K, the mean thermal enengy is of the order
𝐸 = 𝑘𝑇 = 0.87 × 10−4 𝑇eV
Here 𝑘 = 1.38 × 10−23 J/ degree is the Boltzmann constant. For 𝑇 ≈ 107 K, 𝐸 is of the
order of a keV,
If a gas is heated to such a high temperature, many changes will take place in the
structure of the molecules and atoms of the gas. At the temperature of a few thousand
degrees kelvin, the molecules of the gas dissociate and the bare atoms move about freely
at random. If the temperature is raised to 104 K or higher, then the electrons in the orbits
of the atoms begin to separate from the latter, thereby ionising them. The process, known
as thermal ionization is governed by Saha ionization equation deduced by the Indian
physicist M.N. Saha. shows that the energies of the atoms are then of the order of a few
electrons volts, i.e., comparable to the ionization energy of the atoms. Thus, at these
temperatures, the positively charged atomic ions and the negative electrons move about
freely. They constitute what is known as a plasma, which is regarded as the fourth state
of matter. The plasma as a whole is electrically neutral. So unless electric or magnetic
fields are applied, the only force acting on the plasma is the gravitational force.
As the temperature of the plasma increases all the electrons in the atoms get loose from
the latter and we get a mixture of bare nuclei and electrons. As the temperature further
rises, the mean thermal energy of the nuclei and the electrons increases. When the
temperature is 107 K or higher, this energy is of the order of kilo electron volt or higher.
The energy distribution of the plasma particles is similar to the Maxwellian distribution.
Hence the plasma contains some particles which have much higher energies than the
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314

mean thermal energy 𝑘𝑇 for a particular temperature 𝑇 . At higher temperatures, the
number of the high energy particles increases. Such higher energy particles (nuclei) have
finite probability of overcoming the mutual Coulomb repulsion between them which
makes possible for them to penetrate through the potential barrier and produce fusion
reaction. This is known as the thermo-nuclear reaction.
11.10.1 Source of Energy in Stars

It is known that the sun is an average sized star, emitting 4 × 1026 J energy per second.
Its mass is 2 × 1030 kg, which means that the mean energy production in the sun is
2 × 10−4 J/kg ⋅ s. Actually the rate of energy liberation in the central region, where the
energy is produced, must be larger than the above figure. The energy liberation in the
sun is going on at the above rate for at least 4 × 10∘ years.
Neither chemical reactions nor gravitational energy changes can account for the above
huge rate of energy release for such a long time. For example, if the sun consisted of
carbon and oxygen only and the solar energy were generated due to the burning of
carbon, then all of it would burn up in only 1500 years. That leaves nuclear energy as the
only possible source of solar (and stellar) energy. The sun is known to be mainly made
up of hydrogen and helium (90%) in about equal proportions. If by some suitable series
of nuclear reactions, four hydrogen nuclei combine to produce one helium nucleus, then
the energy release for each such fusion will be
𝐸 = 4𝑀𝐻 − 𝑀He = 4 × 1.007825 − 4.002603

= 0.028697u = 26.73MeV = 4.28 × 10−12 J
Since each kilogram of hydrogen contains about 6 × 1026 protons, the energy content of
such a source will be about 2.4 × 1015 J per kg, which could liberate energy at the rate
of 2 × 10−4 J/kg.s for 1012 y. So the nuclear reactions leading to the fusion of four
hydrogen nuclei to produce one helium
R. Atkinson and F. Houtermans (1928) were the first to suggest that the successive
capture of four protons by some light nuclei to produce an 𝛼-particle might be the
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315

processes which would release energy at reasonable rates for the sun to continue burning
for such a long time. They had suggested the idea of cyclic nuclear reactions, the
importance of which was proved later.
Two such thermonuclear reaction cycles have been suggested for the production of a
helium nucleus by the fusion of four protons.
(A) Proton-proton cycle:
1
1 H + 1 H → 2 H + 𝛽 + + 𝑣 + 0.42MeV
1
2 H + 2 H → 3 He + 𝛾 + 5.5MeV
3
3 He + 3 He → 4 He + 21 H + 12.8MeV
Two reactions, each of (1) and (2), must occur for each reaction (3) to take place. When
these are written out in detail and all the reactions are added, we get
21 H + 21 H + 21 H + 22 H + 3 He + 3 He → 22 H + 2𝛽 + + 2𝑣 + 23 He + 2𝛾 + 4 He +
21 H + 24.64MeV So the net result is
41 H → 4 He + 2𝛽 + + 2𝑣 + 2𝛾 + 24.64MeV
Since the neutrinos escape, we are left with about 24.3MeV energy after the fusion of
four protons to produce one helium nucleus. Thus the result is the gradual depletion of
the protons and the building up of the helium concentration.
The mean reaction rates for the different thermonuclear reactions can be theoretically
calculated. For the three reactions considered above, the mean reaction times under the
conditions prevailing at the centre of the sun (Density 𝜌 = 105 kg/m3 ; 𝑇 = 2 × 107 K;
Hydrogen concentration = 0.35 ) are:
1.5 × 1010 y for reaction (1); 6𝑠 for reaction (2); 9 × 105 y for reaction (3)
Since the mean time for the 𝑝 − 𝑝 reaction is much longer than the other two, we must
suppose that the sequence of reactions to be in equilibrium. The first reaction has never
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been observed in the laboratory. It is actually the 𝛽 + transformation of the excited 2 He∗
produced as the compound nucleus in the reaction. The above estimated life-times tell us
that the concentration of hydrogen in the sun will fall to 1/𝑒 times its present value in a
time about 5 times the present age of the sun (5 × 109 y)
If the 𝛽 + transformation in reaction (1) is allowed, then this reaction would be of primary
importance in the sun and more so in less massive stars.
(B) Carbon cycle:
This cycle was proposed by H.A. Bethe (1939) and comprises of the following reactions:
12
1 C + 1H → 13
N + 𝛾 + 1.95MeV
13 13
2 N→ C + 𝛽 + + 𝜈 + 2.22MeV
13
3 C + 1H → 14
N + 𝛾 + 7.54MeV 4. 14
N + 1H → 15
O + 𝛾 + 7.35MeV
15 15
4 O→ N + 𝛽 + + 𝑣 + 2.7MeV
15
5 N + 1H → 12
C + 4 He + 4.96MeV
The net result is
12
C + 41 H → 12
C + 4 He + 2𝛽 + + 2𝑣 + 𝛾 + 26.72MeV
Notice that the net result is the fusion of four protons to produce one 4 He a presence of
12
C, which must be present, but is not destroyed in produce one 4 He nucleus in the
Since about 1.7MeV energy is carried away by the neutrinos which escape, the it acts
like a catalyst. 25.02MeV. The reaction period of the cycle is determined essentially by
the reaction release is reactions (1) and (4) which are 2.5 × 106 y and 3 × 108 y
respectively by the reaction periods of the
The present evident are 2.5 × 106 y and 3 × 108 y respectively.
316
317

main part of the energy production is due to the with masses between 0.4 to 2.5 solar
masses, the massive stars, the situation is reversed. Since according ty ascle rather than
the 𝑝 − 𝑝 cycle. For less solar masses form the bulk of the stellar population in our
galaxy, we can say stars of 0.4 or lower cycle probably goes on in some very bright
stars, the in our galaxy, we can say that whereas carbon the 𝑝 − 𝑝 cycle.
Thermo-nuclear fusion opens up immense possibilities for the peaceful utilization of

nuclear energy, in which the sea water may serve as the main and an almost
inexhaustible source of fuel.
11.11 SUMMARY
• Explain Compound Nucleus
• Describe Direct Reactions
• Understand the Theory of Stripping and Pick-up Reactions
• Explain Resonance Scattering and Reaction Cross-sections
• Discuss the Continuum Theory of Nuclear Reactions
• Describe Optical Model Theory of Nuclear Reactions
• Explain Nuclear Fission
• Discuss the Nuclear Fusion
11.12 REFERENCES
1.Nuclear Physics by Irving Kaplan, Narosa Publishing House
2.The Atomic Nucleus by R D. Evans, McGraw-Hill Publications.
3.Elements of Nuclear Physics by M.L.Pandya,R.P.S.Yadav
4.Nuclear Physics An Introduction by S.B.Patel
6.https://www.researchgate.net/publication/45868627_Nuclear_Reactions
317
318

7.http://inside.mines.edu/~kleach/PHGN422/lectures/Lecture10.pdf

1. “The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of
Physics, Berlin: Springer Ver.
3. Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum Press,
New York.

1.Explain the theory of Compound Nucleus in detail
2.Describe direct reactions with some suitable examples.
3.Explain about the resonance scattering and reaction cross-sections
4.Discuss the continuum theory of nuclear reactions in detail.
5.Describe optical model theory of nuclear reactions .
6.Explain the Nuclear Fission and Nuclear Fusion processes.
318

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MSCPH 511

M. Sc. III SEMESTER

Prof. P. D. Pant Prof. S.R. Jha,

Department of Physics (School of Sciences)

Unit Writing and Editing

Course Title and Code : Nuclear Physics (MSCPH511)

Unit Block and Unit Title Page

1 Nuclear Properties 1-29

2 Nuclear Binding 30-55

3 Nuclear forces I 56-82

4 Nuclear Forces II 83-101

5 Nuclear Models 102-129

7 Alpha Decay 158-173

8 Beta Decay 174-193

9 Gamma Decay 194-215

BLOCK 3: Nuclear Reactions

10 Nuclear Reactions 216-262

11 Fission and Fusion 263-318

NUCLEAR PHYSICS MSCPH511

Structure of the Unit

NUCLEAR PHYSICS MSCPH511

NUCLEAR PHYSICS MSCPH511

1.3 INTRODUCTION TO NUCLEAR TERMINOLOGY

NUCLEAR PHYSICS MSCPH511

observed spin of the deuterium nucleus is 1.

NUCLEAR PHYSICS MSCPH511

1.4 DISCOVERY OF NEUTRON

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𝑍𝑘𝑒 2 79(1.44 𝑀𝑒𝑉. 𝑓𝑚)

So, from above result there are no electrons in the nucleus.

Fig. 01: Curie and Joliot experiment

NUCLEAR PHYSICS MSCPH511

Figure 02: Discovery of neutron

Solving for the mass energy of the neutron gives

NUCLEAR PHYSICS MSCPH511

1.5 RUTHERFORD SCATTERING AND NUCLEAR SIZE

NUCLEAR PHYSICS MSCPH511

Figure 03: Rutherford’s gold foil experiment

1.5.1 Nuclear Radius

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Figure 04: Deflection of an alpha particle by a gold nucleus

Fig.05: Particle trajectory with impact parameter b and small deflection θ

NUCLEAR PHYSICS MSCPH511

From fig.05 the deflection θ is

Where m denotes the mass of the α-particle.

NUCLEAR PHYSICS MSCPH511

If we put other values Z= 79 for gold, mass of α particle 𝑚 = 6 × 10−27 𝑘𝑔

𝑣 ~ 107 m/s for speed of α particle

𝑒 = 1.6 × 10−19 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

We obtain the radius has order of 10-14m.

NUCLEAR PHYSICS MSCPH511

1.5.2 Measurement of Nuclear Radius

Incident beams of high energy Thin target Beam collector

electrons from accelerator

Detector for scattered electrons

Fig.06: Experimental Setup for nuclear radius measurement

NUCLEAR PHYSICS MSCPH511

measures how rapidly 𝜌 falls to zero at the nuclear surface.

NUCLEAR PHYSICS MSCPH511