Prakash Gupta Statistical Physics
1 Classical Statistical Mechanics:
Statistical Mechanics deal with systems containing of many particles and methods employed are to get
a collective or macroscopic property of the system without taking into account the individual motion
of the particles.
It is not practically possible to determine the property of each particle individually, a statistical approach
is made using the concept of probability of distribution. This statistical approach helps in determining
the bulk or macroscopic property of the system as a whole.
The idea of probability does not imply that the particles move in a random way without obeying the
laws.
Phase Space
From the statistical point of view, a mono-atomic gas constitutes the simplest system. The state of the
gas is completely known if the position and momentum of each molecule of the gas is specified, i.e., we
must specify six quantities x, y, z, px, py, pzfor each of the molecules.
In a purely mathematical concept, we may imagine six dimensional space in which the state of a point
will be described a set of six co-ordinates x, y, z, px, py, pz. This six-dimensional space is called phase
space (µ-space).
The 6N dimensional space where we specify co-ordinates and momenta of N molecules is called τ-space.
Volume of Phase Space
In three dimensional space, we consider an element of volume dxdydz. Similarly, we consider three
dimesional volume element dpxdpydpzin this momentum space.
thus, dxdydzdpxdpydpz= elementary volume of phase space.
The element of volume in this space is termed as a cell. According to Hisenberg’s uncertainty principle,
dxdpxh, dydpyh&dzdpzh
so, volume of each state is 'h3
Thus, in phase space, the co-ordinates of a particle can be specified only to the extent that the particle
under consideration has the position and momentum lying within an element of phase space of volume
h3.
Microstates and Macrostates
The specification of a molecule in a cell is known as microstate. If the individual molecules remains
always in the same region of cell, there is no change in microstate.
The specification of number of molecules in a cell is known as macrostate. if the number of molecules
always remains same in the confined region of space (cell) whichever may be the molecule, the macrostate
remains same. If the initial no of molecules in the cell is not same with final number of molecules, there
will be change in macrostate.
cell C
cell B
cell A
a, b, c
p,q
f
The specification of molecule a in cell A, p in cell B and f
in cell C are their respective microstates.The total number
of molecules in cell A is 3,in cell B is 2 and in cell C is 1
refers to macrostate of cell B and C respectively.
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If we exchange the molecules in cell A and B
in such a way that total number of molecules
before and after exchange remains same, there
will be change in microstate but no change in
macrostate.
If there is difference in total number of
molecules in cell A and B before and after
exchange, there will be a change in both mi-
crostate and macrostate.
cell B
cell A
a, b, c
p,q
a, b, q
p, c
2 Ensemble:
A system is defined as a collection of a number of particles. An ensemble is defined as a collection of a
large number of macroscopically identical but essentially independent systems.
By the term macroscopically, it mean that each of the systems constituting an ensemble satifies the
same macroscopic conditions e.g., volume, energy, pressure,total number of particles, etc.
By the term independent systems, it mean that the systems constituting an ensemble are mutually
non-interacting.
In an ensemble the systems play the same role as the non-interacting molecules do in a gas.
3 Kinds of Ensemble:
Microcanonical Ensemble
It is the collection of a large number of essen-
tially independent systems having the same en-
ergy E, volume V, and the number of particle
N.
We consider identical particles. The individual
systems of a microcanonical ensemble are sep-
arated by rigid, imperable and well insulated
walls such that the values of E, V and N for a
particular system are not affected by the pres-
ence of other systems.
E, V, N
E, V, N
E, V, N
E, V, N
E, V, N
E, V, N
E, V, NE, V, N E, V, N
Canonical Ensemble
It is the collection of a large number of es-
sentially independent systems having the same
temperature T, volume V, and the same num-
ber of particles N.
The equality of temperature of all systems can
be achieved by bringing each in thermal contact
with large heat reservoir at constant tempera-
ture T or bringing all the systems in thermal
contact with one another. The individual sys-
tems of a canonical ensemble are separated by
rigid, impermeable but conducting walls.
T, V, N
T, V, N
T, V, N
T, V, N
T, V, N
T, V, N
T, V, NT, V, N T, V, N
Constraints and Accessible States
The condition imposed in the selection of system, ensemble, etc are called constraints. In others words,
the restrictions employed in the selection of parameters is called constraints. For example, the total
number of particles be N is constant or the movement of balls in unidimensional abacus in movement
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Grand-canonical Ensemble
It is the collection of a large number of essentially
independent systems having the same temperature
T, volume V, and chemical potential µ.
The individual systems of a grand canonical en-
semble are separated by rigid, permeable and con-
ducting walls. As the separating walls are con-
ducting and permeable, the exchange of heat en-
ergy as well as that of particles between the sys-
tems takes place in such a way that all the systems
arrive at common temperature T and chemical po-
tential µ.
T, V, µ
T, V, µ
T, V, µ
T, V, µ
T, V, µ
T, V, µ
T, V, µT, V, µT, V, µ
of ball.
Accessible states are the states consistent with the given constraints of the system.
Thermodynamical Probability
The thermodynamic probability of a particular macrostate is defined as the number of microstates
corresponding to that macrostate. In general, this is a very large number and is represented by Ω.
Consider two cells i and j in phase space and four phase points or molecules a, b, c, and d. If Ni
and Njare the number of phase points or molecules in the cells i and j respectively, then the possible
macrostates are represented as
Ni43210
Nj01234
It shows that total number of macrostates are 5.
In general, for each of these five macrostates, there corresponds different number of microstates.
For example, the number of microstates corresponding to the macrostate Ni= 3 and Nj= 1 is shown
below:
abc
d
abd
c
acd
b
bcd
a
cell i
cell j
Thus, for a macrostate Ni= 3 and Nj= 1, there are 4 possible microstates.
Therefore, the thermodynamic probability for the macrostate Ni= 3 and Nj= 1 is 4,
i.e., = 4
For N phase points, total number of permutations is N!.
If n1is the number of phase points in cell 1, n2in the cell 2 etc, then there are n1! permutations in cell
1, n2! permutations in cell 2 etc.
Then, thermodynamical probability for such a system is
= N!
n1!n2!...ni!...
4 Fundamental Postulates of Statistical Mechanics:
For the application of Statistical Mechanics to gases, we make the following hypothesis:
1. Any gas may be considered to be composed of molecules that are in motion and behave like very
small elastic spheres.
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2. All the cells in the phase space are of equal size.
3. All accessible microstates corresponding to possible macrostates are equally probable.
4. The equilibrium state of a gas corresponds to the macrostate of maximum probability.
5. The total number of molecules is constant
i.e., N=n1+n2+n3+..... +ni+..... = constant
6. The total energy of the system is constant
i.e., E=1n1+2n2+3n3+..... +ini+..... = constant
5 Division of Phase Space into cells:
Consider a 2fdimensional phase space defined by position coordinates q1, q2, q3, ......qi....qfand mo-
mentum coordinates p1, p2, p3, .....pi.......pf.
An element of phase volume may be represented by,
= (dq1.dq2.dq3......dqi....dqf)(dp1.dp2.dp3.....dpi......dpf)
=
f
Y
i=1
dqidpi
where Π represents the product of the terms.
The dimensions of this volume element are
(length ×momentum)f= (joule - sec)f
Let us divide any finite volume of phase space into a large number of cells. Let the size of each cell be
hf.
Here h is an arbitrary constant and has the dimensions of joule - sec.
i.e., h=δqiδpi
Thus, the number of microstates or phase cells in this volume element is
hf=(dq1.dq2.dq3......dqi....dqf)(dp1.dp2.dp3.....dpi......dpf)
hf
6 Entropy and Probability:
The equilibrium state of the system is the state of maximum probability, i.e., the probability of the
system in equilibrium state is maximum. But from the thermodynamically point of view the equilibrium
state of a system is the state of maximum entropy. If the system is not in equilibrium, then changes
take place within the system until the equilibrium state or the state of maximum entropy is reached.
Thus in equilibrium state both the entropy and thermodynamical probability have the maximum values
which led Boltzmann to expect some co-relation between them.
i.e., S=f(Ω)
Where Sis entropy and is the thermodynamical probability of the state.
According to thermodynamics, entropy Sof a system is related with temperature by the relation
1
T=S
E (1)
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According to Statistical mechanics, we have
1
T=kβ =k
E log (2)
From equation (1) and (2),
S
E =k
E log
Integrating, we get
S=kloge
This is the required relation between entropy and probability and is called Boltzmann’s Statistical
definition for entropy.
7 Boltzmann’s Canonical Distribution Law:
Let n1, n2, n3, .....ni..... be the number of gas molecules in cell 1, cell 2, cell 3, ... cell i, .... in the
equilibrium state. The gas molecules are moving continously, and the ni’s will change continously. But,
according to fundamental hypothesis of statistical mechanics,
i) The total number of molecules is constant
i.e., N=n1+n2+n3+..... +ni+..... = constant
or, δN =δn1+δn2+δn3+.... +δni+..... = 0 (1)
ii) The total energy of the system is constant
i.e., E=1n1+2n2+3n3+..... +ini+..... = constant
or, δE =1δn1+2δn2+3δn3+.... +iδni+..... = 0 (2)
iii) If the system is in equilibrium, the therodynamical probability is maximum, i.e., δ = 0.
since, is maximum, log is maximum
δlog = 0.(3)
We have,
= N!
n1!n2!...ni!...
or, log = log N!(log n1! + log n2! + .. log ni! + ...)
or, log = log N!X
i
log ni!
or, δlog = δ"log N!X
i
log ni!#
Using Striling Theorem, log x! = xlog xx, if x is large
Then,
δlog = δ"Nlog NNX
i
(nilog nini)#
or, 0 = 0 X
i
(δ(nilog ni)δni)
or, 0 = 0 X
i
(log niδni+niδlog niδni)
or, 0 = 0 X
i
(log niδni+ni
1
ni
δniδni)
or, X
i
log niδni= 0
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log n1δn1+ log n2δn2+.. log niδni+... = 0 (4)
eqn (4), (2) and (1) are independent of one another and they must satisfy at the same time.
Now, multiplying each equation by arbitrary constant and adding them, we get
(log n1δn1+log n2δn2+.. log niδni+...)+α(δn1+δn2+....+δni+.....)+β(1δn1+2δn2+....+iδni+.....) = 0
Collecting the coefficient of δn1, δn2, ....δni....., we have
(log n1+α+β1)δn1+ (log n2+α+β2)δn2+.. + (log ni+α+βi)δni+.. = 0 (5)
Eqn (5) is the combination of independent equations where δn1, δn2, ....δni..... cannot be zero.
So, we assign
(log ni+α+βi) = 0
or, log ni=(α+βi)
or, ni=e(α+βi)
or, ni=eαeβi
ni=Aeβiwhere, A=eα= constant
This gives the no. of molecules in each cell as a function of energy associated with each particle in that
cell and is called Boltmann’s Canonical Distribution law.
8 Maxwell’s Distribution Law of velocities:
Let us consider an ideal gas in a vessel of volume V. If the gas is in equilibrium, then according to
Maxwell’s Boltzmann canonical distribution law, the number of molecules in a cell of energy iwill be
ni=Aeβi(1)
The no. of molecules having energy iand having position coordinate between
x and x + dx
y and y + dy
z and z + dz
and velocity components between
vxand vx+dvx
vyand vy+dvy
vzand vz+dvz
is given by
nidxdydzdvxdvydvz=Aeβidxdydzdvxdvydvz
But
i= energy of particle = 1
2mv2=1
2m(v2
x+v2
y+v2
z)
Thus, we have
nidxdydzdvxdvydvz=Aeβm
2(v2
x+v2
y+v2
z)dxdydzdvxdvydvz(2)
Integrating equation (2) over all available volume and all ranges of velocities, we get total number of
molecules, N
i.e.,
N=ZZZZZZAeβm
2(v2
x+v2
y+v2
z))dxdydzdvxdvydvz
=AV ZZZeβm
2(v2
x+v2
y+v2
z))dvxdvydvzV=ZZZdxdydz = vol. of vessel with gas
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N=AV Z
eβm
2v2
xdvxZ
eβm
2v2
ydvyZ
eβm
2v2
zdvz(3)
or, N=AV 2π
1/22π
1/22π
1/2Z
eax2dx =π
a1/2
or, N=AV 2π
3/2
or, A=N
V
2π3/2
Since, β=1
kBT, where, kB= Boltzmann constant and T= Absolute temperature
or, A=N
Vm
2πkBT3/2
(4)
using eqn(4), eqn(2) can be written as
nidxdydzdvxdvydvz=N
Vm
2πkBT3/2
em
2kBT(v2
x+v2
y+v2
z)dxdydzdvxdvydvz(5)
Now,
The no. of molecules having velocity co-ordinate in the range
vxand vx+dvx
vyand vy+dvy
vzand vz+dvz
irrespective of position of co-ordinates is given by integrating equ(5) with respect to x,y,z.
nidvxdvydvz=N
Vm
2πkBT3/2
em
2kBT(v2
x+v2
y+v2
z)dvxdvydvzV
=Nm
2πkBT3/2
em
2kBT(v2
x+v2
y+v2
z)dvxdvydvz(6)
And
no. of molecules having velocity component in the range vxand vx+dvxirrespective of vy, vz, x, y, z is
given by integrating eqn(6), with respect to vyand vz.
nidvx=Nm
2πkBT3/2Z Z em
2kBT(v2
x+v2
y+v2
z)dvxdvydvz
=Nm
2πkBT3/2
em
2kBTv2
xdvxZem
2kBTv2
ydvyZem
2kBTv2
zdvz
=Nm
2πkBT3/2
em
2kBTv2
xdvx2πkBT
m1/22πkBT
m1/2
nidvx=Nm
2πkBT1/2
em
2kBTv2
xdvx(7)
The probability that a molecule will have x-component of velocity in the range vxand vx+dvxis given
by
P(vx)dvx=nidvx
N
or, P(vx)dvx=m
2πkBT1/2
em
2kBTv2
xdvx(8)
Eqn (7) and (8) gives Maxwell’s distribution law of velocities.
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9 Law of Equipartition of energy:
”The total energy available to dynamic system consisting ofa large no. of particles in thermal equilib-
rium is equally divided among its all the degree of freedom and the energy, associated with each degree
of freedom is 1
2kBT”.
According to kinetic theory of gas, the mean KE of a molecule at absolute temp T is given by
1
2mc2=3
2kBT(1)
where, kB= Boltzmann constant and c= rms speed
But, c2=u2+v2+w2
As x, y and zare all equivalent, u2=v2=w2
1
2mu2=1
2mv2=1
2mw2
so,
1
2mc2=1
2m(u2+v2+w2) = 3 1
2mu2=3
2kBT
or, 1
2mu2=1
2kBTsimilarly,1
2mv2=1
2mw2=1
2kBT
Thus, the average energy associated with each degree of freedom = 1
2kBT.
10 Maxwell’s Boltmann Statistics:
Let us consider a system of N particles n1, n2, ..., ni, .... consisting of energy ε1, ε2, ..., εi, ... The no. of
particles can exchange the state so that the particles in each state be the same. The condidition of
Maxwell’s Boltzmann Statistics are as follows:
1) Any no of particles (n = 0, 1, 2, 3 ...) can be accomodated in a state.
2) The particles are considered to be distinguishable.
3) The sum of particles in each state is the total number of particles. (Conservaton of mass).
i.e., N=n1+n2+n3+..... +ni+..... = constant
or, δN =δn1+δn2+δn3+.... +δni+..... = 0 (1)
4) The sum of energy of each particle in the state is the total energy (Conservation of energy)
i.e., E=ε1n1+ε2n2+ε3n3+..... +εini+..... = constant
or, δE =ε1δn1+ε2δn2+ε3δn3+.... +εiδni+..... = 0 (2)
If giis the probability of locating a particle in certian energy state εi, then probability of locating ni
particles is gni
i.
The probability of given microstate is
= N!
n1!n2!...ni!...gn1
1gn2
2.....gni
i.... (3)
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Taking log, we get
log = log N! + log gn1
1
n1!+ log gn2
2
n2!+..... + log gni
i
ni!+....
or, log = log N! + X
i
log gni
i
ni!
or, log = log N! + X
i
(nilog gilog ni!)
Using Stirling approximation,[log n! = nlog nn]
log = Nlog NN+X
i
(nilog ginilog ni+ni)
δlog = δ"Nlog NN+X
i
(nilog ginilog ni+ni)#
= 0 + X
i
δ(nilog ginilog ni+ni)
= 0 + X
i
(log giδnilog niδniniδlog ni+δni)
= 0 + X
i
(log giδnilog niδnini
1
ni
δni+δni)
= 0 + X
i
(log gilog ni)δni
For maximum, δlog = 0
X
i
(log nilog gi)δni= 0
or, X
i
(log ni
gi
)δni= 0 (4)
Multiplying eqn(1) by αand eqn(2) by βand adding to eqn(4), we get
X
i
(log ni
gi
)δni+αX
i
δni+βX
i
εiδni= 0
X
ilog ni
gi
+α+βεiδni= 0
For each independent δni, we have
log ni
gi
+α+βεi= 0
log ni
gi
=(α+βεi)
ni
gi
=e(α+βεi)
ni=gi
eα+βεi
This is Maxwell’s Boltzmann Distribution Law.
11 Bose Einstein’s Statistics:
In classical statistics, the particles are distinguishable from one another. If two particles interchange
their positions or energy states, a new complexion would occur. But in Bose-Einstein’s Statistics, we
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start with identical particles which are indistinguishable. Thus the interchange of two particles between
two energy states will not give rise to a new complexion or macrostate.
The condition of Bose-Einstein’s Statistics are:
1) The particles are indistinguishable.
2) The sum of particles in each state is the total number of particles. (Conservaton of mass).
i.e., N=n1+n2+n3+..... +ni+..... = constant
or, δN =δn1+δn2+δn3+.... +δni+..... = 0 (1)
3) The sum of energy of each particle in the state is the total energy (Conservation of energy)
i.e., E=ε1n1+ε2n2+ε3n3+..... +εini+..... = constant
or, δE =ε1δn1+ε2δn2+ε3δn3+.... +εiδni+..... = 0 (2)
Let us now consider a box divided into gisections by (gi1) partitions and niindistinguishable particles
to be distributed among these sections.
The permutation of niparticles and (gi1) partitions simultaneously is given by (ni+gi1)!. Since,
both the groups are internally indistinguishable, the actual no. of ways in which niparticles are to be
distributed in gisublevels of ith quantum state is.
(ni+gi1)!
ni!(gi1)
Thus, the thermodynamic probability of the microstate is
= (n1+g11)!
n1!(g11)! .(n2+g21)!
ni!(g21)! .....(ni+gi1)!
ni!(gi1)! ....
Ω=Π(ni+gi1)!
ni!(gi1)! (3)
As niand giare large numbers, 1 is neglected,
= Π(ni+gi)!
ni!gi!(4)
Taking log, we get
log = X
i
log (ni+gi)!
ni!gi!
or, log = X
i
[log(ni+gi)! log ni!log gi!]
Using Stirling approximation,[log n! = nlog nn]
log = X
i
[(ni+gi) log(ni+gi)(ni+gi)nilog ni+nigilog gi+gi]
or, log = X
i
[(ni+gi) log(ni+gi)nilog nigilog gi]
or, δlog = X
i
δ[(ni+gi) log(ni+gi)nilog nigilog gi]
or, δlog = X
i
[log(ni+gi)log ni]δni
or, δlog = X
i
log ni
(ni+gi)δni
For most probable distribution, δlog = 0
X
i
log ni
(ni+gi)δni= 0 (5)
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Multiplying eqn(1) by αand eqn(2) by βand adding to eqn(5), we get
X
i
(log ni
(ni+gi))δni+αX
i
δni+βX
i
εiδni= 0
X
ilog ni
(ni+gi)+α+βεiδni= 0
For each independent δni, we have
log ni
(ni+gi)+α+βεi= 0
log ni
(ni+gi)=(α+βεi)
ni
(ni+gi)=e(α+βεi)
(ni+gi)
ni
=eα+βεi
gi
ni
=eα+βεi1
ni=gi
eα+βεi1
This is Bose Einstein’s Distribution Law.
12 Fermi-Dirac Statistics:
This statistics is obeyed by indistinguishable particles of half-integral spin that have antisymmetric
wave functions and obey Pauli’s exclusion principle.
This statistics is obeyed by indistinguishable particles of half-spin that have antisymmtric wave functions
and obey Pauli’s exclusion principle.
1) The particles are indistinguishable.
2) The particles obey Pauli’s exclusion principle.
3) The sum of particles in each state is the total number of particles. (Conservaton of mass).
i.e., N=n1+n2+n3+..... +ni+..... = constant
or, δN =δn1+δn2+δn3+.... +δni+..... = 0 (1)
3) The sum of energy of all the particles in the different quantum state is the total energy (Conservation
of energy)
i.e., E=ε1n1+ε2n2+ε3n3+..... +εini+..... = constant
or, δE =ε1δn1+ε2δn2+ε3δn3+.... +εiδni+..... = 0 (2)
Let nibe the number of particle for energy level εi. Each particle can be placed in any one of the
available gistates. Thus the total number of different ways of arranging niparticles among the gi
states with energy level εiis
gi!
(gini)!
If the particles are taken to be indistinguishable. Then, the no. of distinguishable arrangement of ni
particles in gistate is
gi!
ni!(gini)!
Thus, the thermodynamic probability of the microstate is
= g1!
n1!(g1n1)!.g2!
n2!(g2n2)!..... gi!
ni!(gini)!....
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= Π gi!
ni!(gini)! (3)
Taking log, we get
log = X
i
log gi!
ni!(gini)!
or, log = X
i
[log gi!log ni!log(gini)!]
Using Stirling approximation,[log n! = nlog nn]
log = X
i
[gilog giginilog ni+ni(gini) log(gini)+(gini)]
or, log = X
i
[gilog ginilog ni(gini) log(gini)]
or, δlog = X
i
δ[gilog ginilog ni(gini) log(gini)]
or, δlog = X
i
[log(gini)log ni]δni
or, δlog = X
i
log ni
(gini)δni
For most probable distribution, δlog = 0
X
i
log ni
(gini)δni= 0 (4)
Multiplying eqn(1) by αand eqn(2) by βand adding to eqn(4), we get
X
i
(log ni
(gini))δni+αX
i
δni+βX
i
εiδni= 0
X
ilog ni
(gini)+α+βεiδni= 0
For each independent δni, we have
log ni
(gini)+α+βεi= 0
log ni
(gini)=(α+βεi)
ni
(gini)=e(α+βεi)
(gini)
ni
=eα+βεi
gi
ni
=eα+βεi+ 1
ni=gi
eα+βεi+ 1
This is Fermi-Dirac Distribution Law.
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13 Comparison of Three Statistics:
The expression for the most probable distribution in the three statistics are:
Maxwell-Boltzmann Bose-Einstein Fermi-Dirac
ni=gi
eα+βεi
ni=gi
eα+βεi1ni=gi
eα+βεi+1
gi
ni=e(α+βεi)
gi
ni=eα+βεi1gi
ni=eα+βεi+ 1
if gi
niis very large in comparison to unity,
gi
nigi
ni
+ 1 gi
ni1 (1)
i.e., for large value of gi
niBose-Einstein and Fermi-Dirac distribution approaches the Maxwell Boltzmann
Distribution.
This is the normal case of gases when the temperature is not two low and pressure is not too high. Hence
classical statistics is capable of describing their behaviour quite adequately. The classical statistics fails
only at certain extreme conditions. The examples of these conditions are radiation, electron gas in
metals.
14 Electron gas in metal:
Electrons in a metal belong to a most characteristics system of fermions because electrons obey the
exclusion principle. For electrons in a metal, the energy levels are ground in bands. Practically at
all temperature, the lower energy bands are filled with electrons. The upper level energy bands are
partially filled with electrons. The distribution of electrons is to be considered only in the upperbands
called conduction bands. The zero energy level is taken at the lowest level of conduction band. It is
assumed that the electrons have free movements within the conductor, provided the energy associated
with the electrons is of the order of upper energy bands.
From Fermi-Dirac Distribution law,
ni=gi
eα+βεi+ 1 (1)
where niis the no. of particles in ith cell and giis the degeneracy factor for ith cell.
As the energy of electron in the conduction band is continous the degeneracy factor giis replaced by
g(E)dE and niby n(E)dE.
n(E)dE =g(E)dE
eαeεi/kBT+ 1 (2)
Here n(E) refers to number of electrons, g(E) refers to number of phase cells.
But g(E)dE in terms of momentum is
g(p)dp =8πv
h3p2dp =8πv
h3p.pdP
Prakash Gupta Statistical Physics
But,
E=1
2mv2andp=mv
or, p= (2mE)1/2
or, p2= 2mE
or, 2pdp = 2mdE
or, pdp =mdE
g(E)dE =8πv
h3(2mE)1/2mdE
=82πv
h3m3/2E1/2mdE
n(E)dE =82πvm3/2E1/2mdE
h3eεi/kBT+ 1
This is exact form of Fermi Dirac law of energy distribution electrons. As the temperature T = 0K,
no. of electron is equal to the total no. of energy states occupied by the electron from zero to EF.
n=ZEF
o
g(E)dE =82πv
h3m3/2ZEF
o
E1/2dE
or, n=82πv
h3m3/22
3E3/2EF
0
or, n= 162πv m3/2
3h3E3/2
F
or, E3/2
F=3nh3
162πvm3/2
, EF=h2
2m3n
8πv 2/3
15 Fermi level and Fermi Energy:
As the value of T decreases, the energy of electrons goes on decreasesing. But at 0K, the energy of
elctron is not zero. Moreover according to Pauli-exclusion principle, only one electron can occupy one
energy level. Therefore, there will be only one lectron in the groung state having zero energy. If n is
the total number of electron at 0K, then total no of energy states which will be occupied will also be
n, each state with one electron. Let EFbe the energy of the highest state occupied by the electrons at
0K, then all the energy states having energy values greater then EFwill be empty and all the energy
Prakash Gupta Statistical Physics
states having energy values lesser than EFwill be fully occupied, each with on electron.
Thus, the energy value upto which all the energy states are completely filled by electrons at absolute
zero (i.e. OK) of temperature and above which all the energy states are completely empty is known as
Fermi energy. It is denoted by EF.
Thus, the occupation index f(E) = n(E)
g(E)= 0forE > Effor E > EF
and, the occupation index f(E) = n(E)
g(E)= 1forE > Effor E > EF
From thses two conditions, we can prove that
α=EF
kBT
The physical interpretation of the Fermi energy at absolute zero is that it represents highest energy
level at T = 0K in which all the quantum states are occupied.