Nuclei

the nuclear mass M is found to be always less than this. For example, let
us consider 168 O ; a nucleus which has 8 neutrons and 8 protons. We
have
Mass of 8 neutrons = 8 × 1.00866 u
Mass of 8 protons = 8 × 1.00727 u
Mass of 8 electrons = 8 × 0.00055 u
Therefore the expected mass of 168 O nucleus
= 8 × 2.01593 u = 16.12744 u.
The atomic mass of 168 O found from mass spectroscopy experiments
is seen to be 15.99493 u. Substracting the mass of 8 electrons (8 × 0.00055 u)
from this, we get the experimental mass of 168 O nucleus to be 15.99053 u.
Thus, we find that the mass of the 168 O nucleus is less than the total
mass of its constituents by 0.13691u. The difference in mass of a nucleus
and its constituents, ΔM, is called the mass defect, and is given by
ΔM = [ Zm p + ( A − Z ) m n ] − M (13.7)
What is the meaning of the mass defect? It is here that Einstein’s
equivalence of mass and energy plays a role. Since the mass of the oxygen
nucleus is less that the sum of the masses of its constituents (8 protons
and 8 neutrons, in the unbound state), the equivalent energy of the oxygen
nucleus is less than that of the sum of the equivalent energies of its
constituents. If one wants to break the oxygen nucleus into 8 protons
and 8 neutrons, this extra energy ΔM c2, has to supplied. This energy
required Eb is related to the mass defect by
Eb = Δ M c 2 (13.8)

Example 13.3 Find the energy equivalent of one atomic mass unit,
first in Joules and then in MeV. Using this, express the mass defect
of 168 O in MeV/c 2.
Solution
1u = 1.6605 × 10–27 kg
To convert it into energy units, we multiply it by c 2 and find that
energy equivalent = 1.6605 × 10–27 × (2.9979 × 108)2 kg m2/s2
= 1.4924 × 10–10 J
1.4924 ×10 −10
= eV
1.602 × 10 −19
= 0.9315 × 109 eV
= 931.5 MeV
EXAMPLE 13.3

or, 1u = 931.5 MeV/c 2

For 168 O , ΔM = 0.13691 u = 0.13691×931.5 MeV/c 2
= 127.5 MeV/c 2
16
The energy needed to separate 8 O into its constituents is thus
2
127.5 MeV/c .

If a certain number of neutrons and protons are brought together to

form a nucleus of a certain charge and mass, an energy Eb will be released 443

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 Physics
in the process. The energy Eb is called the binding energy of the nucleus.
If we separate a nucleus into its nucleons, we would have to supply a
total energy equal to E b, to those particles. Although we cannot tear
apart a nucleus in this way, the nuclear binding energy is still a convenient
measure of how well a nucleus is held together. A more useful measure
of the binding between the constituents of the nucleus is the binding
energy per nucleon, E bn, which is the ratio of the binding energy Eb of a
nucleus to the number of the nucleons, A, in that nucleus:
E bn = E b / A (13.9)
We can think of binding energy per nucleon as the average energy
per nucleon needed to separate a nucleus into its individual nucleons.
Figure 13.1 is a plot of the
binding energy per nucleon E bn
versus the mass number A for a
large number of nuclei. We notice
the following main features of
the plot:
(i) the binding energy per
nucleon, E bn, is practically
constant, i.e. practically
independent of the atomic
number for nuclei of middle
mass number ( 30 < A < 170).
The curve has a maximum of
about 8.75 MeV for A = 56
and has a value of 7.6 MeV
FIGURE 13.1 The binding energy per nucleon for A = 238.
as a function of mass number.
(ii) E bn is lower for both light
nuclei (A<30) and heavy
nuclei (A>170).
We can draw some conclusions from these two observations:
(i) The force is attractive and sufficiently strong to produce a binding
energy of a few MeV per nucleon.
(ii) The constancy of the binding energy in the range 30 < A < 170 is a
consequence of the fact that the nuclear force is short-ranged. Consider
a particular nucleon inside a sufficiently large nucleus. It will be under
the influence of only some of its neighbours, which come within the
range of the nuclear force. If any other nucleon is at a distance more
than the range of the nuclear force from the particular nucleon it will
have no influence on the binding energy of the nucleon under
consideration. If a nucleon can have a maximum of p neighbours
within the range of nuclear force, its binding energy would be
proportional to p. Let the binding energy of the nucleus be pk, where
k is a constant having the dimensions of energy. If we increase A by
adding nucleons they will not change the binding energy of a nucleon
inside. Since most of the nucleons in a large nucleus reside inside it
and not on the surface, the change in binding energy per nucleon
would be small. The binding energy per nucleon is a constant and is
444 approximately equal to pk. The property that a given nucleon

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 Nuclei
influences only nucleons close to it is also referred to as saturation
property of the nuclear force.
(iii) A very heavy nucleus, say A = 240, has lower binding energy per
nucleon compared to that of a nucleus with A = 120. Thus if a
nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more
tightly bound. This implies energy would be released in the process.
It has very important implications for energy production through
fission, to be discussed later in Section 13.7.1.
(iv) Consider two very light nuclei (A ≤ 10) joining to form a heavier
nucleus. The binding energy per nucleon of the fused heavier nuclei
is more than the binding energy per nucleon of the lighter nuclei.
This means that the final system is more tightly bound than the initial
system. Again energy would be released in such a process of
fusion. This is the energy source of sun, to be discussed later in
Section 13.7.3.

13.5 NUCLEAR FORCE

The force that determines the motion of atomic electrons is the familiar
Coulomb force. In Section 13.4, we have seen that for average mass
nuclei the binding energy per nucleon is approximately 8 MeV, which is
much larger than the binding energy in atoms. Therefore, to bind a
nucleus together there must be a strong attractive force of a totally
different kind. It must be strong enough to overcome the repulsion
between the (positively charged) protons and to bind both protons and
neutrons into the tiny nuclear volume. We have already seen
that the constancy of binding energy per nucleon can be
understood in terms of its short-range. Many features of the
nuclear binding force are summarised below. These are
obtained from a variety of experiments carried out during 1930
to 1950.
(i) The nuclear force is much stronger than the Coulomb force
acting between charges or the gravitational forces between
masses. The nuclear binding force has to dominate over
the Coulomb repulsive force between protons inside the
nucleus. This happens only because the nuclear force is
much stronger than the coulomb force. The gravitational
force is much weaker than even Coulomb force.
FIGURE 13.2 Potential energy
(ii) The nuclear force between two nucleons falls rapidly to
of a pair of nucleons as a
zero as their distance is more than a few femtometres. This function of their separation.
leads to saturation of forces in a medium or a large-sized For a separation greater
nucleus, which is the reason for the constancy of the than r0, the force is attractive
binding energy per nucleon. and for separations less
A rough plot of the potential energy between two nucleons than r0, the force is
as a function of distance is shown in the Fig. 13.2. The strongly repulsive.
potential energy is a minimum at a distance r0 of about
0.8 fm. This means that the force is attractive for distances larger
than 0.8 fm and repulsive if they are separated by distances less
than 0.8 fm. 445

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