PHYSICS – II (PH - 1002) 3 credits (2-1-0)

Content
Statistical Mechanics: Statistical distributions, Maxwell-Boltzmann statistics, molecular energies in ideal gas,
quantum statistics: Bose-Einstein and Fermi-Dirac statistics, Rayleigh-Jeans formula, Planck's law of radiation, specific
heats of solids, Dulong- Petit’s law, Einstein’s formula and Debye theory.
Solid State Physics : Crystalline and amorphous solids, crystal structure, Bravais lattice, packing fraction, atomic
radius, reciprocal lattice, Bragg’s diffraction point defect, dislocations, ionic crystals, co-valent crystals, Van der Waals
bond, metallic bond, band theory of solids, classification of solid on band theory, semi-conductor, energy bands,
superconductivity, high temperature superconductor, bound electron pairs.
Nuclear Physics: Nuclear structure, atomic masses, nuclear properties, stable nuclei, binding energy, radioactive
decay, half life, radio-active series, alpha, beta and gamma decay, liquid drop model and shell model, Meson theory
of nuclear forces, concept of magic number, nuclear cross-section, nuclear reactions, nuclear fissions, nuclear
reactors, fusion in stars, fusion reactors: energy source for future.
Particle Physics: Interactions and Particles , Leptons , Hadrons , Elementary Particle Quantum Numbers
 Solid State Physics : Crystalline and amorphous solids, crystal structure, .
Bravais lattice, packing fraction, atomic radius, reciprocal lattice, Bragg’s
diffraction point defect, dislocations, ionic crystals, co-valent crystals, Van der
Waals bond, metallic bond, band theory of solids, classification of solid on band
theory, semi-conductor, energy bands, superconductivity, high temperature
superconductor, bound electron pairs.
Nuclear Physics: Nuclear structure, atomic masses, nuclear properties, .
stable nuclei, binding energy, radioactive decay, half life, radio-active series,
alpha, beta and gamma decay, liquid drop model and shell model, Meson
theory of nuclear forces, concept of magic number, nuclear cross-section,
nuclear reactions, nuclear fissions, nuclear reactors, fusion in stars, fusion
reactors: energy source for future.
Total .

References:
• AG; A. Ghatak, Optics, Tata-McGraw Hill, 2004.
•AB; A. Beiser, Concept of Modern Physics, Tata-McGraw Hill, 2005.
•CK; C. Kittel, Introduction to Solid State Physics, Wiley & Sons, 2004.
 Microscopic and Macroscopic systems
Macroscopic systems:
Large compare to atomic dimensions
Consists of many atoms and molecules
Solids , liquids, gases, biological organisms
Microscopic systems:
Small scale of the order of atomic dimensions (10-10 m)
All the matter consists of molecules, molecules built up of atoms,
atoms consists of nucleus and electron and so on.
Any microscopic system consists of many atoms.
For microscopic systems gravitational forces are negligible as
compare to other fundamental forces.

A typical macroscopic system contains ~ 1025 interacting atoms. Our

aim is to understand and predict the properties of such systems on
the basics of a minimum number of fundamental concepts.
Consider a gas of identical simple atoms (e.g. helium atoms). We
know all the basic forces among the He atoms.

But this microscopic information is not sufficient to know at what

temperature it will liquefy. (Problem)

Understanding of macroscopic systems requires new concepts:

(i) Fundamental laws of microscopic physics
(ii) Essential characteristic and regularities exhibited by such
system
(iii) Simple methods to predict qualitatively the properties of the
system
 Statistical Mechanics
Describes the physics of many particles.

We give up trying to keep track of individual particles. If we can’t

solve Schrödinger’s equation in closed form for helium (4 particles)
what hope do we have of solving it for the gas molecules in this
room (1023 particles).

Statistical mechanics handles many particles by calculating the

most probable behavior of the system as a whole, rather than by
being concerned with the behavior of individual particles.

It tells about the probability. (e.g. a particle has certain amount of
energy at certain moment)
Applicable to both classical (molecules in a gas) and Quantum
systems (photons & free electrons in metal )

It is not associated with

Motion of single particle.
Interactions of individual particles.
Life history of one particle in the system

Statistical Statistical consideration

Mechanics + Laws of Mechanics
 Statistical Mechanics cont….

In statistical mechanics, we assume that the more ways there are to

arrange the particles to give a particular distribution of energies, the
more probable is that distribution. (Seems reasonable?)

6 units of energy, 3 particles to give it to

321 411

312 141
213 114
231 3 ways

123
more likely
132
6 ways
 Statistical Mechanics cont….

(repeating) In statistical mechanics, we assume that the more ways

there are to arrange the particles to give a particular distribution of
energies, the more probable is that distribution. (Seems reasonable.)

We begin with an assumption that we believe describes nature.

We see if the consequences of the assumption correspond in any

way with reality.

It is not “bad” to begin with an assumption, as long as we realize

what we have done, and discard (or modify) the assumption when
it fails to describe things we measure and observe.
 Statistical Distribution
What kind of particles are?
Particles can interact with each other and with the walls of the container.
Interaction with the walls of the container is not so violently that temperature changes.
Total number of particles=N Number of particles of energy ε
Total energy of the system=E
Temperature of the system=T n(ε)=g(ε)f(ε) (1.a)

g(ε)=number of states of energy ε

The function f(ε) for quantum =Statistical weight corresponding to energy ε
statistics depends on whether or
not the particles obey the Pauli f(ε)=distribution function
exclusion principle. =Average number of particle in each state of energy ε
=Probability of occupancy of each state of energy ε

If energy is continuous

n(ε)=g(ε)f(ε)d ε (1.b)

g(ε) dε =number of states with energies between ε and ε+d ε

 Maxwell-Boltzmann Statistics
We take another step back in time from quantum mechanics (1930’s)
to statistical mechanics (late 1800’s).

Classical particles which are identical but far enough apart to be

distinguishable obey Maxwell-Boltzmann statistics.

classical  “slow,” wave functions don’t overlap

distinguishable  you would know if two particles changed places

(you could put your finger on one and follow it as it moves about)

Two particles can be considered distinguishable if their separation is

large compared to their de Broglie wavelength.

Example: ideal gas molecules.

Boltzmann discovered statistical mechanics and was a pioneer
of quantum mechanics.

His work contained elements of relativity and quantum

mechanics, including discrete atomic energy levels.

“In his statistical interpretation of the second law of

thermodynamics he introduced the theory of probability into
a fundamental law of physics and thus broke with the classical
prejudice, that fundamental laws have to be strictly
deterministic.”

“With Boltzmann's pioneering work the probabilistic

interpretation of quantum mechanics had already a
precedent.”

Boltzmann constantly battled for acceptance of his work. He

also struggled with depression and poor health. He
committed suicide in 1906. Most of us believe
thermodynamics was the cause.
Paul Eherenfest, who wrote Boltzmann’s eulogy, carried on
(among other things) the development of statistical
thermodynamics for nearly three decades.

Ehrenfest was filled with self-doubt and deeply troubled

by the disagreements between his friends (Bohr, Einstein,
etc.) which arose during the development of quantum
mechanics.

Ehrenfest shot himself in 1933.

US physicist Percy Bridgmann (the man on the

right, winner of the 1946 Nobel Prize) took up
the banner of thermodynamics, and studied
the physics of matter under high pressure.

Bridgman committed suicide in 1961.

 Problems

1. A system has three energy levels of 0, 100 kB, and 200 k B

with degeneracies of 1, 3, and 5 respectively. Calculate
the relative population of each level and average energy at
temperature of 100 K.

2. The average Kinetic energy (=3kBT/2) of hydrogen atom in a

stellar gas is 1 eV. What is the ratio of the number of
atoms in the second excited state (n=3) to the number in
the ground state (n=1)? The energy levels in the hydrogen
atoms are εn=-α/n2 where α=13.6 eV and the degeneracy
of the nth level is 2n2.
 Maxwell-Boltzmann Distribution Function
The average number of particles in a state of energy ε is a system of particles at absolute
temperature T is


f MB ( )  Ae kT (2)

A= Constant
=Depends on number of particle in the system
=It has the same role as that of normalization constant in case of wave function
k=Boltzmann’s constant
=1.381X10-23 J/K=8.617X10-3 eV/K

Combining Eq. (1.a) and (2), we get n   g   f MB  



n( )  Ag ( )e kT (3)

Let’s apply MB statistics to find the distribution of energies among the molecules of an ideal gas
 The Kinetic Molecular Model for Gases ( Postulates )
• Gas consists of large number of small individual particles
with negligible size
• Particles in constant random motion and collisions
• No forces exerted among each other
• Kinetic energy directly proportional to temperature in
Kelvin

3
KE  RT (4)
2

Let’s use the Maxwell-Boltzmann distribution to learn about the

energies and speeds of molecules in an ideal gas.
 Molecular energies in an Ideal gas
N is very large → We can consider continuous distribution of energies instead of discrete.

Now Eq. (3) can be rewritten as


n( )d  Ag ( )e d kT (5)

where
n(ε)dε = Number of molecules with energies between ε and ε+d ε
g(ε)dε = Number of states with energies between ε and ε+d ε
Evaluation of g(ε)dε

We know that energy and momentum are related with each other.

p2
  p  2m  px2  p y2  pz2 (6)
2m

So let’s evaluate g(p)dp [number of states with momentum whose magnitudes are in
between p and p+dp]
 MB Distribution Cont…

Because each p corresponds to a single ε,

g ( )d  g ( p)dp  Bp dp 2 (7)

Now,
1  12
p  2m  p  2m  dp  2m  d
2

so that
1
p dp  
2 2
d ,

1
g ( )d   2
d (8)

and


n( )d  C  e kT
d (9)

The constant C contains B and all the other proportionality constants lumped together.
 MB Distribution Cont…


n( )d  C  e kT
d
To find the constant C, we evaluate
 

N   n( )d  C   e kT
d
0 0
where N is the total number of particles in the system. Standard integral

C 3

x
N  (kT ) 2 n 1  x
2 e dx  (n)
0
2N 
so that n( )d  3
e kT
d
(kT ) 2

This is the number of molecules having energy between ε and ε+dε in a sample containing N
molecules at temperature T.
 MB Distribution Cont…

Wikipedia says: “The Maxwell-Boltzmann distribution is an important relationship that finds

many applications in physics and chemistry.”

“It forms the basis of the kinetic theory of gases, which accurately explains many
fundamental gas properties, including pressure and diffusion.”

“The Maxwell-Boltzmann distribution also finds important applications in electron transport

and other phenomena.”

Here’s a plot of the distribution:

Notice how “no” molecules have E=0, few

molecules have high energy (a few kT or
greater), and there is no maximum of
molecular energy.

k has units of [energy]/[temperature] so kT has units of energy.

 MB Distribution Cont…

Here’s how the distribution changes with temperature (each vertical grid line corresponds to
1 kT).

Notice how the distribution for higher temperature

is skewed towards higher energies (makes sense!)
but all three curves have the same total area (also
makes sense).

Notice how the probability of a particle having

energy greater than 3kT (in this example) increases
as T increases.

Continuing with the physics, the total energy of the system is

Standard integral
 

2N
x
 n 1  x
e dx  (n)
3
E   n( )d  3  e 2 kT
d
0 (kT ) 2
0
0
 MB Distribution Cont…

Evaluation of the integral gives 3NkT

E
2
3
So, the average energy per molecule is
  kT
2
exactly the result you get from elementary kinetic theory of gases.
Things to note about our ideal gas energy:

• The energy is independent of the molecular mass.

• Which gas molecules will move faster at a given temperature: lighter or

heavier ones? Why?

• at room temperature is about 40 meV, or (1/25) eV. This is not a large

amount of energy.

• kT/2 of energy "goes with" each degree of freedom.

 MB Distribution Cont…

Because ε = mv2/2, we can also calculate the number of molecules having speeds between v
and v + dv.
3
2Nm 2  mv 2
The result is n(v)dv  3
2
ve 2 kT
dv
(kT ) 2

Here’s a plot (number having a “We” (Beiser) call this n(v). The
given speed vs. speed): hyperphysics web page calls it f(v).

“Looks like” n( ) plot—nothing at

speed=0, long exponential tail.
Mean speed
3
2
 m   kT 

vnv dv
1 2
 v  4   dx
x
v  2  xe
N0
 2kT   m  0
3
 mv 2
 m  2  3
  4 
2
 ve  m   kT 
3 2 kT
dv 2
 2kT   4   2  (2)
 2kT   m 
0
3
 mv 2
 m  2 
 4   v e
3 2 kT
dv 8kT
 2kT  0 
m

mv 2 kT
put  x  vdv  dx
2kT m
8kT
as v  0, x  0 v
as v  , x   m
Most probable speed
3
mv 2
 m  2 
n(v)dv  4N   ve
2 2 kT
dv
 2kT 
dn(v)
for most probable speed 0
dv
 mv 2 
d 2  2 kT
v e 0
dv  
mv 2 mv 2
 
2
ve 2 kT
m
 2v   e 2 kT
2v  0
2kT
mv 2
  2 m 
2ve 2 kT
  v  1 0
 2kT 
2kT
vMP 
m
Mean square speed

1 2
v   v n(v)dv
2

N0
3
 mv 2
 m  2 
 4   v e
4 2 kT
dv
 2kT  0
3
 3
 m  kT  2kT  2  x
3
2
 4  
 2kT 
0 m  m  x e dx
4 kT  5 
  
 m 2 vrms 
3kT
3kT m

m
 MB Distribution Cont…

The speed of a molecule having the average energy comes from solving

mv2 3
  kT
2 2

for v. The result is

3kT
vrms  v 
2

vrms is the speed of a molecule having the average Energy .

It is an rms speed because we took the square root of the square of an average quantity.
 MB Distribution Cont…

The average speed can be calculated from



 vn(v)dv
v 0


 n(v)dv
0
The result is
8kT
v
m

Comparing this with vrms , we find that

vrms  1.09v

Because the velocity distribution curve is skewed towards high energies, this result makes
sense (why?).
 MB Distribution Cont…

You can also set dn(v) / dv = 0 to find the most probable speed. The result is

2kT
vp 
m
The subscript “p” means “most probable.”

Summarizing the different velocity results:

3kT
vrms  v 2 
m

8kT vrms  1.09v

v
m
2kT
vp 
m
 MB Distribution Cont…

Plot of velocity distribution again:

n(v)

This plot comes from the hyperphysics web site. The R’s and M’s in the equations are a result
of a different scaling than we used. See here for how it works (not testable material).
 Bosons & Fermions
Electrons and other particles with half-integral spin (1/2, 3/2, 5/2, etc.) are fermions and
obey the Pauli exclusion principle.

The wave function of a system of fermions is antisymmetric because it changes sign upon
the exchange of any pair of fermions. We will find that fermions follow Fermi-Dirac
statistics.

Recall also that photons and other particles with integral spin (0, 1, 2, etc.) are bosons
and are not subject to the Pauli exclusion principle.

The wave function of a system of bosons is symmetric because it sign remains

unchanged upon the exchange of any pair of bosons. We will find that bosons follow Bose-
Einstein statistics.
 Symmetric & anti-symmetric wave function
Consider two identical particles (1 and 2) which may exist in two different states (a and b).

I = a (1) b (2) II = a (2) b (1)

If the particles are indistinguishable, then we cannot tell whether the system is in state  I
or  II, and, because both states are equally likely, we write the system wave function as a
linear combination of  I and  II.

If the particles are bosons, the system wave function is symmetric:

1
B = a (1) b (2) + a (2) b (1)  = S
2
If the particles are fermions, the wave function is antisymmetric:

1
F = a (1) b (2) -  a (2) b (1)  =  A
2

What happens if we try to put both particles 1 and 2 in the same state????
 Classical Particles
If the particles are distinguishable, we can simply write

M = a (1) a (2) .
The subscript M indicates Maxwell-Boltzmann statistics,* because the particles are
distinguishable.

The probability density for distinguishable particles is

M*M =  a (1)*  a (2)*   a (1)  a (2)  .

*“Huh? I thought you said Maxwell-Boltzmann statistics is for classical (not quantum)
particles. How come the wave functions?”

Be quiet! Actually, recall that you can always use QM. You usually don’t use QM unless you
have to. In this case, it is useful for comparing MB results with quantum statistics.
For bosons B = 2 a (1) a (2) .
with a probability density

B *B = 2  a (1)*  a (2)*   a (1)  a (2) 

B *B = 2 M*M .
In other words, if the particles are bosons, they are twice as likely to be in the same state as
distinguishable particles!

On the other hand, if the particles are fermions,

1
F = a (1) a (2) - a (2) a (1)  ,
2
with a probability density
F * F = 0 .
If the particles are fermions, it is impossible for both particles to be found in the same state.

In general, the presence of a boson in a particular quantum state increases the probability
that other bosons will be found in the same state…

…but the presence of a fermion in a particular quantum state prevents other fermions from
being in that state.

We are now (almost) ready to write down the distribution functions for bosons and fermions.
Remember, the distribution function gives the probability that a state of energy  is occupied
by a particle.

For bosons, we use a function called the Bose-Einstein (BE) distribution function.

In 1924, Indian physicist S. N. Bose submitted a paper

using statistical mechanics and the idea of light quanta
to explain Planck’s radiation law. The paper was
rejected. Einstein translated the paper, got it accepted,
and extended Bose’s ideas from photons to other
particles. Bose’s paper removed any final objections to
the photon theory of light. Bose

For fermions, we use a function called the Fermi-Dirac (FD) distribution function.
In 1926, Enrico Fermi and Paul Dirac independently realized that
the Pauli exclusion principle leads to a different kind of statistics
for fermions (including electrons). Dirac shared the 1933 Nobel
prize with Schrödinger.

Dirac was a brilliant mathematician. Sometimes it seems like

all the great mathematicians of the 1930’s worked on
Dirac
problems involving quantum mechanics.

Dirac was famous for saying exactly what he meant, and no more (typical
mathematician?). Once when someone, making polite conversation at dinner,
commented that it was windy, Dirac excused himself, left the table and went to
the door, looked out, returned to the table and replied that indeed it was windy.*

When Dirac won the 1933 Nobel prize, he decided to turn it down because he
hated publicity. When it was pointed out he would receive far more publicity by
turning it down, he changed his mind.

*http://www-groups.dcs.st-and.ac.uk/history/Mathematicians/Dirac.html
For bosons, the distribution function is
the difference
1
fBE (ε) =  ε / kT
.
e e -1
For fermions, the distribution function is

1
fFD (ε) =  ε / kT
.
e e +1

These were derived in an appendix in the previous edition of Beiser, but are just given as
truth here.

Remember, bosons are particles with integral spins. There is no limit on the number of bosons which
may occupy any particular state (typically the ground state). Examples are photons in a cavity, phonons,
and liquid 4He.

Also remember, fermions are particles with half integral spin, with only one particle per state n, ℓ, mℓ,
ms. The +1 in the denominator of f(ε) means that the f(ε) is never greater than 1. Examples are free
electrons in metals and nuclei in collapsed stars.
 The Fermi-Dirac Distribution Function
1
fFD (ε) =
e eε / kT +1
The Fermi energy εF is the energy at which fFD = 1/2. From the above equation we see that ε F
= -kT, and we can write fFD as

1
fFD (ε) =  ε - ε F  / kT
e +1
On the next slide is a plot of the Fermi-Dirac distribution function at T=0, 150, 300, and
1000K.
 T=0
T = 150
T = 300

T = 1000

F = 3 eV
 MB, BE and FD Distribution Function:
Comparison
Bosons “like” to be in the
same energy state, so you
can cram many of them
in together.

Fermions don’t “like” to

be in the same energy
state, so the probability is
the least.
 BE Condensatation
When bosons get very cold, they all fall into the same (lowest) energy state. Their wave
packets merge and form a single wave packet. If the bosons are atoms, the individual atoms
lose their identity and form a “super atom.”

Cornell and Wieman of U. Colorado and Ketterle won the 2001 Nobel prize for making a
Bose-Einstein condensate out of 2000 rubidium atoms.

They had to cool the atoms to 0.000001 K to do this.

Formation of the BEC…

The BEC is a new state of matter.

BEC : ordinary matter :: laser : light bulb light.

Practical applications of BEC: none yet. (Who knows…)

Comparison between Distributions
Bose Fermi
Boltzmann
Einstein Dirac
1 1 1
nk  nk  nk 
        
exp   exp    1 exp    1
 k BT   k BT   k BT 
indistinguishable indistinguishable indistinguishable
Z=(Z1)N/N! integer spin 0,1,2 … half-integer spin 1/2,3/2,5/2 …
nK<<1

spin doesn’t matter bosons fermions

localized particles wave functions overlap wave functions overlap

 don’t overlap total  symmetric total  anti-symmetric

gas molecules photons free electrons in metals

at low densities 4He atoms electrons in white dwarfs

“unlimited” number of unlimited number of never more than 1

particles per state particles per state particle per state
nK<<1