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2021 – PG- TRB - PHYSICS - UNIT – 4 – PART – 1

UNIT –IV STATISTICAL MECHANICS – Material (1/2)

1. Maxwell Boltzmann Statistics-

2. Maxwell’s Distribution Of Velocities

3. Expression For Mean, RMS,MPV, Velocities

4. BE Statistics- Distribution Function

5. Photon Gas

6. Black Body Radiation

7. FD Statistics - Distribution Function

8. Electron Gas

9. Pauli Paramagnetism

10. Thermionic Emission

11. Elementary Idea Of Phase Transition

12. Properties Of Liquid Helium

13. Phase Space

14. Liouvillie’s Theorem Statistical Equation

15. Micro, Canonical Ensembles

16. Equation Of State

17. Thermodynamic Functions Of An Ideal Gas

18. Equi Partition Of Energy

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118MAXWELL BOLTZMANN STATISTICS: ( CLASSICAL STATISTICS )

MAXWELL BOLTZ MANN STATISTICS:( CLASSICAL STATISTICS )

STATISTICS

Classical Statistics Quantum Statistics

(Maxwell Boltzmann Statistics) 1.Bose Einstein Statistics

2. Fermi-Dirac Statistics

1. This is based on classical theory where particle has only particle character.

2. This is applicable to distinguishable particles.

3. No restriction on the number of particles in each state

The probability of distribution W = N! ∏1

k (gi)¿

¿! [ ∏❑ represents multiplication ]

The NUMBER OF WAYS W = (g1) n1

Which is not true about Maxwell Boltzmann Statistics

a. It is based on classical theory

b. This is applicable to distinguishable particles.

c. No restriction on the number of particles in each state

If priori probability is G and thermodynamic probability is P, total probability W of the

distribution is

W = G- P

The priori probability for the distribution of N particles among gi states such that n1

particles in the first state g1, n2 particles in the second state g2 etc is given by

G = (g1) n1 (g2)n2

Thermodynamic probability P is given by . N !

n 1!× n2 !× ….

Expression for most probable distribution of particles among various energy levels:

ni = gi

eα+β ∈ i

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NOT FOR EXAMINATION:

MAXWELL BOLTZ MANN STATISTICS: ( CLASSICAL STATISTICS )

1. This is based on classical theory only particle character is considered.

2. This is applicable to distinguishable particles..

Distribution function

The priori probability that ‘ ni ‘ particles fall simultaneously in the ‘ gi ‘ states is ( gi) ni

. The priori probability that ‘ n2 ‘ particles fall simultaneously in the ‘ g2 ‘ states is ( gi) n2

.The priori probability for the distribution of ‘ N’ particles among ‘ gi’ states such that ‘ n1’

particles in the state’ g1’ , ‘n2’ particles in state’ g2’ etc is given by

G = (g1) n1 (g2)n2 (g3)n3

The number of ways in which ‘n1’ particles out of ‘N’ particles may fall in the first state is

N cn 1 = N !

n1!×(N−n1)!

The number of ways in which ’ n2’ particles out of remaining’ (N - n1) ‘ particles may fall in

the second state is

(N- n1) cn 2 = (N−n1)!

n 2!×(N−n1−n2)! and so on.

Therefore the total number of ways in which ‘n1’ particles fall in state ‘1’ , ‘n2 ‘ particles in

state ‘ 2’... is

P = N !n 1!×(N−n1) !

× (N−n1)!n2 !×(N−n 1−n2) ! ....

= N !

n1!× n2 !× ….

This is known as thermodynamic probability.

Total probability ‘W’ of the distribution is the product of priori probability G and the

thermodynamic probability P

W = G×P

= (g1) n1 (g2)n2 (g3)n3× N !

n1!× n2 !× ….

= N! ∏1

k (gi)¿

¿! [ ∏❑ represents multiplication ]

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Consider a system of ‘N’ similar but distinguishable molecules of a gas.

The number of ways of arranging the particles in ‘g1 ‘ state is given by

W = N! ∏1

k (gi)¿

¿! ------------------------1

Taking ‘ln’ on both sides

ln W = ln N! + ∑ ¿ lngi - ∑ lnni! [ when we take’ log’ ∏❑ becomes ∑❑ ]

By Striling,s formula ln x ! = x ln x –x

Applying this formula we get

ln W = N lnN – N + ∑ ¿ lngi -∑ (¿ ln ¿−¿)

= N lnN – N + ∑ ¿ lngi -∑ (¿ ln ¿) + ni

= N lnN – N+∑ ¿ lngi -∑ (¿ ln ¿) + ∑ ni

Differentiating with respect to ‘n1’,

d lnWdni = ∑ 0+0+¿ (0 )+¿¿ ln gi (1) – [ ni (

1ni

) + lnni (1) ] +1

= ∑ ln g i−1−ln ni+1

= ∑ ln g i−ln ni

d ln W = ∑ ln ( gi¿ ) dni

The condition for most probable distribution is d ln W = 0

Therefore the above equation becomes ∑ ln( gi¿ ) dni = 0 -------------------------2

If we confine our investigation to a closed system of particles the following conditions are

satisfied.

1.The total number of particles is constant i.e ∑ ni = constant

∴∑ d ni = 0 [differentiating] -----3

2. The total energy of the system is constant. i.e ∑∈i ni = constant

∴ ∑∈i d ni = 0 [differentiating] ------4

Multiply equation 3 by ∝ and equation 4 by β and add the result,

α∑ d ni +β∑ ∈i d ni} = 0 --------------------5

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According to the method of Langrage’s undetermined multiplier, equations 2 and 5 are equal

∑ ln ( gi¿ )dni = α∑ d ni +β∑ ∈i d ni

Dividing by dni, we get ln ( gi¿ )= ∝ + β ∈ i

Taking exponential gi¿

= eα+β∈ i

gi

eα+β ∈ i = ni

Rearranging we get

nii = gi

eα+β ∈ i

This is the expression for most probable distribution.

Problem: Find the number of ways of arranging 2 distinguishable particles in two states Solution:g = 2, n=2,

The number of ways W = (g) n

Number of ways = 22

= 4Let the particles be A,B The possibilities are

Number of ways = 4

Problem: Find the number of ways of arranging 2 distinguishable particles in three states Solutiong = 3, n=2,

Number of ways W =32

= 9

Let the particles be A,BThe possibilities are

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AB AB A B B A

AB AB AB

A B A B A B

B A B A B A


Number of ways = 9Problem: Find the number of ways of arranging 3 distinguishable particles in two states

Solution

g = 2, n=3,

Number of ways W = 23

= 8

Let the particles be A,B,CThe possibilities are

Number of ways = 8

Problem: Calculate the number of ways of distributing two particles among four energy levels when the particles are MaxwellonsSolutionNumber of ways for Maxwellons = gn = 42

= 16 Problem: Calculate the number of ways of distributing three particles among four energy

levels when the particles are Maxwellons

Solution

Number of ways for Maxwellons = 43

= 64

218MAXWELL’S DISTRIBUTION OF MOLECULAR VELOCITIES :

The fraction of molecules with velocity between c and c+dc is

ρ(c) dc = 4π ( M2 πRT

)32 c2 exp ( −M c2

2 RT ) dc

Important features:

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ABC ABC

BC AAC BAB C

A BCB ACC AB


1. The fraction of molecules having velocity greater than zero increases with increase in

velocity, reaches a maximum and then falls off towards zero at higher velocities.

2. The fraction of molecules with too low or too high velocities is very small

3. There is certain velocity for which the fraction of molecules is maximum. This is called

most probable velocity.

4. Most probable velocity increases with rise in temperature.

5. exp ( −M c2

2 RT ) is called Boltzmann factor.

6. Boltzmann factor. Increases with increase in temperature.

MAXWELL’S DISTRIBUTION OF MOLECULAR VELOCITIES

Maxwell and Boltzmann derived an expression from statistical considerations.

DERIVATION:

Number of molecules in a cell of energy E is n = A e−βϵ

Energy of particle ϵ = 12 mv2

Total Number of molecules N = ∫−∞

∞

e−βϵ dv

N = ∫−∞

∞

e−β 1

2 mv2

dv

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= √ 2πmβ

β = 1

kT

Number of molecules having energy ϵ and having position co ordinates between ‘x’ and x+dx’

is

N = √ 2 πkTm

Number of molecules having energy ϵ and having position co ordinates between ‘x’ and x+dx’

y’ and y+dy’ and z’ andz+dz’ is

N= √ 2 πkTm

× √ 2 πkTm

× √ 2 πkTm

N = ( m2 πkT

)32

A= N ( m2 πkT

)32

Number of molecules in a cell of energy E is

n = A e−βϵ

= A e−β 1

2 mv2

= A e−mv2

2 kT

= N ( m2 πkT

)32 e

−mv2

2kT

N∝ v2

n = ( m2 πkT

)32 e

−mv2

2 kT v2

fraction of molecules between the two spheres of radii ‘p’ and p+dp” is

= 43 π ( p+dp)3 -

43 π ( p)3

= 4 π p2

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= 4 π ( m2πkT

)32 e

−−m v2

2 kT v2

318EXPRESSION FOR AVERAGE, MPV, RMS VELOCITIES-

1. Average velocity : Cav = √ 8 RTπM

2. Most probable velocity : Cp = √ 2 RTM

3. Root Mean Square velocity(RMS) : C RMS = √ 3 RTM

The ratio between RMS , Cav, and MPV is 1.00 : 0.92:0.82

1. MEAN (Average ) VELOCITY ( <c> :

It is given by the arithmetic mean of different velocities possessed by the molecules at

the given temperature. If c1,c2,c3 ... cn are the velocities of the gas molecules and n is the total

number then the average velocity is given by

<c> = c1+c 2+c3+ ....+cn

n

Average velocity CAV

= √ 8 RTπM

DERIVATION:

p(c) dc = 4π( m2πRT

)32 e

−mc2

2 RT c2 dc

Average velocity = 4π ( m2 πRT

)32 ∫

0

∞

ce−mc2

2 RT c2 dc

= 4π ( m2πRT

)32 ∫

0

∞

e−m c2

2 RT c3 dc

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= 4π ( m2πRT

)32 [

1

2( m2 RT

)2 ∫

0

∞

x2 n+1 e−a x2

dx =

n !2 an+1

= 4π m

2πRT × ( m2 πRT

)12 ×

2(RT )2

m2

= 4 ( m2 πRT

)12 ×

RTm

= √2√2√2 π

× 2 ( mRT

)12 ×

RTm

= √ 2π

× 2 ( RTm

)12

= √ 8RTπm

2. ROOT MEAN SQUARE VELOCITY( RMS VELOCITY):

It is defined as the square root of the mean of the squares of different velocities

possessed by the molecules of the gas at the given temperature. If c1,c2,c3 ... cn are the

velocities of the gas molecules and n is the total number then

(< c2 > ) 1/2 = (c12+c 22+c 3 2+ .....

n) ½

CRMS

= √ 3 RTM

DERIVATION:

CRMS

= √∫0

∞

c2 p(c)dc

∫0

∞

c2 p(c ) dc = ∫0

∞

c2 4 π ( m2πRT

)32 e

−mc2

2 RT c2 dc

= 4 π ( m2 πRT

)32 ∫

0

∞

c4 e−mc2

2 RT dc

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= 4 π ( m2 πRT

)32 ×

1.322+1 √ π

( m2 RT )

4+1 ∫0

∞

x2 n e−a x2

dx = 1.3 ….. 2n−1

2n+1

√ πa2 n+1

= 4 π ( m2 πRT

)32 ×

38 ( 2RT

m)

52 √π

=4π m

2πRT × ( m2 πRT

)12 ×

38 ( 2 RT

m)

52 √π

= m

2 RT × ( m2 πRT

)12 ×

32 ( 2 RT

m)

52 √π

= m

2 RT × ( m2 RT

)12 ×

32 ( 2RT

m)

52

= 32 ×

2 RTm

= 3 RT

m

CRMS

= √ 3 RTm

3. MOST PROBABLE VELOCITY( CP):

It is defined as the velocity possessed by maximum number of molecules of a gas

at a given temperature.

Most probable velocity CMPV

= √ 2 RTM

DERIVATION:

At the most probable state, the first derivative is equal to zero

c = 4π( m2 πRT

)32 e

−mc2

2 RT c2

differentiating with respect to ‘c’

dc = 4π( m2 πRT

)32 [e

−mc2

2 RT (2c) + c2 ¿ ) ]

equating to zero we get

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4π( m2 πRT

)32 [e

−mc2

2RT (2c) + c2 ¿ ) ] = 0

Dividing by 4π( m2 πRT

)32

¿ (2c) + c2 ¿ ) ] = 0

Dividing by e−mc2

2 RT (2c)

1 - m c2

2 RT = 0

m c2

2RT = 1

c2 = 2RT

m

c =√ 2 RTm

Problem: Calculate the RMS velocity of a gas whose molecular weight is 83.1 at 270 K

Solution:

T = 1000

M = 83.1

R = 8.314

CRMS

= √ 3 RTM

= √ 3 × 8.314 ×27083.1

= √3 ×27

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= 9 m/s

Problem: Calculate the RMS velocity of CO2 at 440K

Solution:

T = 440 K

M = 12 + 2(16)

= 44

R = 8.314

CRMS

= √ 3 RTM

= √ 3 × 8.314 × 44044

= √3×83

= √249 m/s

Relation between average velocity, RMS velocity and most probable velocity

Average velocity CAV

= √ 8 RTπM

CRMS

= √ 3 RTM

C AV

CRMS = √ 8 RT

πM× √ M

3 RT

= √ 83 π

= 0.9213

Average velocity = 0.9213 × RMS velocity

Most probable velocity CMPV

= √ 2 RTM

CRMS

= √ 3 RTM

CMPV

CRMS = √ 2 RT

M× √ M

3 RT

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= √ 23

= 0.8165

Most probable velocity = 0.8165 × RMS velocity

Problem: Calculate the Most probable velocity of N2 at 7 oC

Solution:

T = 7 +273 K

= 280 K

M = 28

R = 8.314

CMPV

= √ 2 RTM

= √ 2 × 8.314 ×28028

= √2× 83.14

= √166.28 m/sec

Problem: You are given the following group of particles with speed

Ni 2 4 4 7 3

vi 1 2 3 4 5

Find the average velocity and most probable velocity

Solution:

average velocity = ∑ nivi

∑ ¿

= 2+8+12+28+15

20

= 6520

= 3. 25 m/sec

most probable velocity = velocity possessed by maximum number of molecules

= 4 m/sec ( velocity corresponding to n= 7)


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Ni 2 4 7 3

vi 1 2 3 4

Find the rms velocity and most probable velocity

Solution:

Mean square velocity = ∑ ¿vi2

∑ ¿

= 2+4 (4 )+7 (9)+3(16)

10

= 2+16+63+48

10

=12.9 m/sec

most probable velocity = velocity possessed by maximum number of molecules

= 3 m/sec ( velocity corresponding to n= 7)

Problem: At what temperature will the average speed of molecule of hydrogen gas be double

the average speed of oxygen at 100 K

Solution:

Cav = √ 8 kTπm

CH 2 = √ 8 kT 1

π (2)

CO2 = √ 8 k T 2

π (32)

CH 2

CO2 = √ 32 T 1

2 T 2

2CO2

CO 2 = √ 16 T 1

T 2

2 = 4 √ T 1

T 2

squaring on both sides

4= 16 T1

T2

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T 2 = 4 (100)1

= 400 K

Problem: At what temperature will the average speed of molecule of hydrogen gas will be

same as that of nitrogen molecules at 280 K

Solution:

Cav = √ 8 kTπm

√ 8 kT 1

π (2) = √ 8 k T 2

π (28)

√ T 1

(2) = √ T 2

(28)

Squaring on both sides

T 1

(2) =

T2

(28)

T 1

(2) =

280(28)

T 1 = 20

4

18 . BOSE EINSTEIN STATISTICS(QUANTUM STATISTICS)

This statistics deals with

1. Particles which are indistinguishable.

2. Any number of particles may occupy a given energy level.[ Paulis principle not obeyed]

3. This statistics is obeyed by particles having integral spin such as hydrogen ( H2),

deuterium (D2), nitrogen ( N2), helium-4 ( 4 He).

4. Particles obeying BE statistics are called bosons.

5. The total number of ways in which ‘n’ particles are to be distributed among ‘g’ levels

W = ∏1

g [ni+gi−1]!(gi−1)!× ni!

6. Expression for most probable distribution of ‘ni’ particles in ‘gi’ energy levels is

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ni = gi

eα+β ∈ i−1

NOT FOR EXAM:

This statistics deals with

1. Particles which are indistinguishable.

2. Any number of particles may occupy a given energy level.

3. particles having integral spin

This statistics is obeyed by particles having integral spin such as hydrogen ( H2), deuterium

(D2), nitrogen ( N2), helium-4 ( 4 He). Particles obeying BE statistics are called bosons.

Distribution function:

Consider a system of ‘N’ indistinguishable particles such that ni particles are in ‘gi’ states.

The number of ways in which one state is selected out of ‘gi’ state = (gi ) !.

The total number of permutations of ‘ni’ particles and ( gi – 1 ) partitions is = ( ni + gi – 1) !

The priori probability = gi ×( ni + gi – 1) !

Since the states are indistinguishable, interchanging of partitions does not alter an

arrangement . Hence the above expression must be divided by’ (gi ) !.

And particles are also indistinguishable and interchanging of particles also does not

alter an arrangement. Hence the above expression must also be divided by ‘ni !’

The total number of ways in which ‘n’ particles are to be distributed among g levels

= gi ×[n+g−1]!

g !× n!

= gi×[n+g−1]!

gi ×(gi−1)!×n!

= [n+g−1]!

(gi−1)!× n!

Similar expressions can be found out for various other quantum states.

Considering all the available quantum groups, ‘n1’ particles with energy ‘E1,’, ‘n2 ‘ particles with

energy’ E2 ‘ and so on, The total number of distinguishable arrangements is given by

W = [n1+g1−1]!

(g1i−1) !×n1 ! × [n2+g 2−1]!

(g2−1)!× n 2! × [n 3+g 3−1]!

(g 3−1) !×n3 !

= π [¿+gi−1]!

(g i−1)!×∋! -------------------------------------1

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Consider a system of ‘N’ similar but indistinguishable molecules of a gas. The number of

ways of arranging the particles in ‘gi’ state is given by

W = π [¿+gi−1]!

(g i−1)!×∋!

As ‘ni’ and ‘gi’ are large numbers ,we may neglect one in the above expression.

Therfore the above equation becomes

W = π [¿+gi ]!

(g i)!×∋! ------------1

Taking logarithm on both sides ln W = ∑ ln[ni+gi] ! - ∑ ln(gi ) !- ∑ lnni!

using Stirlings formula [ ln x ! = x ln x – x ]

ln W = ∑❑ { [ni+gi] ln [ni+gi] - [ni+gi} - {gi ln (gi ) - gi } – { ni ln ni - ni }

= ∑❑ { [ni+gi] ln [ni+gi] – ni- gi - gi ln (gi) + gi – ni ln ni + ni

= ∑❑ [ni+gi] ln [ni+gi] - gi ln (gi ) - ni ln ni

Differentiating with respect to ni,

d lnWd∋¿¿ = ∑❑ { ( ni +gi) (

1¿+gi ) + ln [ni+gi](1) } – {0} –{ ni ( 1

¿ ) + ln ni(1)}

= ∑❑ 1 + ln [ni+gi] – 1 - ln ni

= ∑ ln [¿+gi ]– ln∋¿¿

= ∑ ln ([¿+gi ]

¿ )

d ln W = ∑ ln ([¿+gi ]

¿ )dn i


∴∑ ln([¿+gi ]

¿ )dn = 0 ----------2

If we confine our investigation to a closed system of particles then

1.The total number of particles is constant ∑ ni = constant

∴∑ dni = 0 -----------------3

2. The total energy of the system is constant. ∑∈i ni = constant

∴ ∑∈i d ni = 0 -----------------4

, Multiply equation 3 by ∝ and equation 4 by β and add the result,

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α∑ d ni +β∑ ∈i d ni} = 0 --------------------5


∑ ln ([¿+gi ]

¿ )dni = { α∑ d ni +β∑ d (∈ini¿)¿}

Dividing by dni, we get ln ( [¿+gi ]¿ )= ∝ + β ∈ i

Taking exponential [¿+gi ]¿ = eα+ β∈ ie

1+ gi¿

= eα+ β∈ i

gi¿

= eα+ β∈ i - 1

∴ ni = gi

eα+β ∈ i−1

Problem: Calculate the number of ways of distributing two particles among four energy levels

when the particles are Bosons

Solution

n= 2, g=4

Number of ways for Bosons = [n+g−1]!

(g−1) !×n !

= [2+4−1]![ 4−1 ] !∗2 !

= 10.

Problem: Calculate the number of ways of distributing three particles among four energy

levels when the particles are Bosons

Solution

n= 3, g=4


(g−1) !×n !

= [3+4−1]![ 4−1 ] !∗3 !

= 6 !

3!3 !

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= 6 ×5 × 4×3 !3 × 2×1 ×3 !

= 20.

Problem . Calculate the number of ways of distributing two particles among four energy levels

when the particles are 1. Bosons 2. Maxwellons

Solution

n= 2, g=4


(g−1) !×n !

= [2+4−1] ![ 4−1 ] !∗2 !

= 10.

Number of ways for Maxwellons = N ! gi¿

¿ !

= 2!. 42

2!

= 16

5

18 APPLICATION OF BE STATISTICS TO PHOTON GAS

PLANCK’S RADIATION LAW ( Application of BE statistics to photon gas )

Planck's law describes the spectral density of electromagnetic radiation emitted by a black

body in thermal equilibrium at a given temperature T.

Eϒ dγ = 8 πh γ3

c3 d γ

(e hγKT

−1)

This equation is known as Planck’ energy distribution law in terms of 𝛄.

Energy density in terms of λ isE λ d λ = 8πh( c

ƛ)

3

c3

1

e h γKT

−1 × - ( cλ2 ) dƛ

= - 8πhC

λ5 1

e hCƛ KT

−1 dƛ

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This equation is known as Planck’ energy distribution law in terms of λ,

DERIVATION:

An assembly of bosons ( indistinguishable particles with zero or integral spin ) is termed

as photon gas.

Consider an enclosure containing electromagnetic radiation. If the enclosure is

maintained at temperature T , it will emit and reabsorb photons. After certain time there will be

a thermodynamic equilibrium.. This electromagnetic radiation within the enclosure is called

black body radiation.

Energy of one photon = hγ

If there are ‘n’ photons , total energy = n hγ

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Iv ‘V represents the volume of the container then Energy density = n h γV

Energy density in the given frequency interval between γ and γ + d γ , in volume V is given

by Eϒ dγ = h γ nV -----------------------------1

h - planc constant, γ – frequency, V – total volume, n- number of photons

According to Bose – Einstein distribution law,

n = g

eα+β ∈−1 ------------------------------2

where g is the number of degeneracies

Photons are indistinguishable from one another. Every process of such emission

creates photons and every process of absorption results in the annihilation of photons. In this

condition their number in the system is not constant. ∑ dni = 0 is invalid .Therefore the

undetermined multiplier α = 0 , put β = 1

KT

n = g

e∈

KT −1 ---------------------------------3

To find the number of degeneracies( g)

number of eigen states (g)= Degeneracy × V shell

V element -------------------4

where V shell – volume of shell, V element - volume of element

= ( 2S +1) × V shell

V element [ Degeneracy = 2S +1]

= ( 2 (12 ) +1) ×

V shell

V element [For photon s = ½ ]

g = ( 2) × V shell

V element ----------------5

According to quantum idea , the element of volume in the momentum space

V element = h3

V

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The volume of the shell having their momenta between p and p+dp

= 43 π ( p + dp ) 3 -

43 π ( p ) 3

= 43 π [ p 3 + 3 p2 dp + 3 p (dp) 2 + (dp ) 3 ] -

43 π p 3

= 43 π [ p 3 + 3 p2 dp + 3 p (dp) 2 + (dp ) 3 - p 3 ]

= 43 π [ 3 p2 dp + 3 p (dp) 2 + (dp) 3 ]

= 43 π [ 3 p2 dp ] [ neglecting the higher terms of dp]

V shell = 4 π p2 dp ---------------------------6

For a photon, p = hλ ---------------------7

= hγc ------------------8

Differentiating

∴ dp = hdγc -----------------------9

Substituting in equation 5 we get

number of eigen states (g) = 2 × 4 π ( hγ

c)

2

h3

V

×hdγc [above equation]

g = 8π V γ 2

c3 d γ ------------------------10

Substituting in 3 we get

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n =

8 π V γ 2c3

e h γKT

−1 d γ

Substituting the value of ‘n’ in 1 we get

Eϒ dγ = hγV ×

8 π V γ 2c3

e h γKT

−1 d γ

= 8 πh γ3

c3 d γ

(e hγKT

−1)

This equation is known as Planck’ energy distribution law in terms of 𝛄.

In terms of λ,

put γ = cƛ

∴ dγ = - ( cλ2 ) dƛ

Energy density in terms of λ isE λ d λ = 8πh( c

ƛ)

3

c3

1

e h γKT

−1 × - ( cλ2 ) dƛ

= - 8πhC

λ5 1

e hCƛ KT

−1 dƛ

This equation is known as Planck’ energy distribution law in terms of λ,

.Rayleigh –Jeans law: When hγ >>> KT

Eϒ dγ = 8 πh γ3

c3 d γ

(1+ h γKT

+…… ..−1) [ ex = 1 + x + …..]

= 8πh γ3

c3 d γ

( h γKT

)

= 8πγKT

c3 dγ of 54


Wiien’s law When hγ <<< KT

Eϒ dγ = 8πh γ3

c3 d γ

e hCƛ KT

This is wien’s law

6

18 BLACK BODY RADIATION

Any object, which absorbs the radiations of all wavelength , incident on it is called black body and when such a body is placed inside an isothermal enclosure at high temperature , it emits radiations of all wavelengths.

This heat radiation emitted by a black body is known as black body radiation.

Diagram

Incident radiation

Black body radiations

Energy distribution:

1. The energy is not uniformly distributed

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2. At a given temperature , the intensity of radiations increases with increase in wavelength to a maximum and then decreases.

3. With increase in temperature , the wavelength at which maximum emission of energy decreases..

Laws of black body radiation:

1. Planc’s quantum theory of black body radiation:

A black body can not absorb or emit energy in a continuous manner. It can absorb or

emit energy in the multiples of small units called quanta.

The energy density in the wavelength range λ and λ + dλ is given by

E= 8 πhc

λ5 (¿ ehϑKT −1 )¿

For short wavelengths:

λ is very low and ϑ = cλ is high e

hϑKT is also very high when compared to 1 . Therfore the

above equation becomes

E= 8πhc

λ5 ehϑKT

Which is Wein’s law

For longt wavelengths:

λ is veryhigh and ϑ = cλ is low e

hϑKT = 1+

hϑKT . Therefore the above equation becomes

E= 8 πhc

λ5(1+ hϑKT

−1) =

8 πhc

λ5( hϑKT )

= 8 πhcKT

λ5 hϑ

= 8 πcKT

λ5 ϑ

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∴ E = 8πKT

λ4 [ϑ = cλ ]

Which is Rayleigh – Jeans law

Thus this law holds good both for long wavelengths and short wavelengths.

718 . FERMI-DIRAC STATISTICS

This statistics is obeyed by

1. Indistinguishable particles

2. Obey pauli’s exclusion principle.

3. This statistics is obeyed by particles having half integral spin such as electron, proton .

and the particles obeying FD statistics are called Fermions

4. The number of ways in which gi quantum states will each be occupied by one particle is

a. W = ∏1

g gi!( g i−ni ) !×n i!

5. Expression for most probable distribution of ‘ni’ particles in ‘gi’ energy levels is

ni = gi

eα+β ∈ i+1

Fermions

The atoms having odd number of particles are called fermions. They obey Fermi-

Dirac statistics Example: 1 H2( proton = 1,electron = 1 neutron =1 ) 2 He + ( 2+1+ 2)

NOT FOR EXAM:

Distribution function

Consider a system of ‘n’ indistinguishable particles. Let gi- denotes the degeneracy.

The possible number of distribution is given by ‘gi !’

Due to pauli’s principle, no cell can occupy more than one particle. Therefore among

‘gi’ cells only’ ni’ cells are occupied by one particle each and the remaining ‘( gi - ni )’ cells

are empty.

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Since the partitions are indistinguishable, interchanging of partitions does not alter an

arrangement . Hence the above expression must be divided by (gi -ni) !.

Since particles are also indistinguishable and interchanging of particles also does not

alter an arrangement. Hence the above expression must also be divided by ‘ni !’

Therefore the number of ways in which ‘ gi ‘quantum states will each be occupied by one

particle is given by

W = ∏ gi!( gi−¿ )!×∋!


Consider a system of ‘N’ similar but indistinguishable molecules of a gas. The number

of ways of arranging the particles in ‘gi’ state is given by

W = ∏ gi!( gi−¿ )!×∋!

Taking ‘ln ‘ on both sides

ln W = ∑ lngi ! - ∑ ln(gi-ni ) !- ∑ lnni!

using Stirlings formula [ ln x ! = x ln x – x ]

ln W = gi ln gi – gi - { gi-ni ) ln (gi-ni ) - (gi-ni ) } - {ni ln ni - ni }

= gi ln gi – gi - (gi-ni) ln (gi-ni) + gi - ni - ni ln ni + ni

= gi ln gi – gi - (gi-ni) ln (gi-ni) + gi - ni - ni ln ni + ni

= gi ln gi - (gi-ni ) ln gi-ni ) - ni ln ni

Differentiating with respect to’ ni’

d lnWd∋¿¿ = 0 – [ (gi – ni) (

1(gi−¿)) (-1) + ln [gi-ni]( 0 -1) ] – [ ni ( 1

¿) + ln ni(1)]

= +1 + ln [gi-ni] – 1 - ln ni

= ∑ ln([ gi−¿]

¿ )

d ln W =∑ ln([ gi−¿]

¿ )dn i


Therefore we get ∑ ln ([ gi−¿]

¿ )dn = 0 --------2

If we confine our investigation to a closed system of particles the following conditions are

satisfied.

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1.The total number of particles is constant∑ ni = constant

∴∑ dni = 0 ----------------------3

2. The total energy of the system is constant. ∑∈i ni = constant

∑∈i d ni = 0 ------------4

, Multiply equation 3 by ∝ and equation 4 by β and add the result,

α∑ d ni +β∑ ∈i d ni} = 0 ---------------------5


∑ ln ([ gi−¿]

¿ )dn i = α∑ d ni +β∑ ∈i dni

Dividing by dni, we get ln ([gi−¿]

¿ )= ∝ + β ∈ i

Taking exponential [ gi−¿]¿ = eα+ β∈ ie

gi¿

- 1 = eα+ β∈ i

gi¿

= eα+ β∈ i + 1

Taking reciprocal ni

g i =

1eα+βE+1

∴ ni = gi

eα+β ∈ i+1

1.Calculate the number of ways of distributing two particles among four energy levels when the

particles are 1. . Fermions 2. Bosons

Number of ways for Fermions = [ gi ]!

(gi−¿)!∗¿!

= [4 ]!

(4−2) !∗2!

= 6.

Let the particles be a,a

1 2 3 4 5 6

a a a

a a a

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a a a

a a a

Number of ways for Bosons = [¿+gi−1]!(gi−1)!∗¿ !

= [2+4−1] ![ 4−1 ] !∗2 !

= 10.

2.Calculate the number of ways of distributing three particles among four energy levels when

the particles are 1. Fermions 2. Bosons

Number of ways for Fermions = [ gi ]!

(gi−¿)!∗¿!

= [4 ] !

(4−3)!∗3 !

= 4× 3!1×3 !

= 4

Let the particles be a,a,a

1 2 3 4

a a a

a a a

a a a

a a a


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2 3 4 5 6 7 8 9 10

a a a aa

a a a aa

a a a aa

a a a aa


= [3+4−1]![ 4−1 ] !∗3 !

= 6 !

3!3 !

= 6 ×5 × 4×3 !3 × 2×1 ×3 !

= 20.

Let the particles be a,a,a

1 2 3 4 5 6 7 8 9 10 11 12 1

3

14 15 16 17 18 1

9

20

aa

a

aa aa aa a a a a a a

aaa a aa aa aa a a a a a

aa

a

a a aa aa aa a a a a

aaa a a a aa aa aa a a a

Problem: Calculate the number of ways of distributing two particles among four energy levels

when the particles are Fermions

Solution

Number of ways for Fermions = g!

(g−n)!∗n !

= [4 ]!

(4−2) !∗2!

= 6.Problem: Calculate the number of ways of distributing three particles among four energy

levels when the particles are Fermions

Solution

Number of ways for Fermions = [ g]!

(g−n)!∗n !

= [4 ]!

(4−3) !∗3 !

= 4× 3!1×3 !

= 4

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Problem Calculate the number of ways of distributing three particles among four energy levels when the particles are 1. Fermions 2. Bosons 3.. Maxwellons Solution

ni = 3gi = 4

Number of ways for Fermions = [ g]!

(g−n)!∗n !

= [4 ]!

(4−3) !∗3 !

= 4 × 3!1 ×3 !

= 4


= [3+4−1]![ 4−1 ] !∗3 !

= 6 !

3!3 !

= 6×5× 4×3 !3× 2×1 ×3 !

= 20.

Number of ways for Maxwellons = N! gi¿

¿ !

= 3 !. 43

3!

= 64

COMPARISON OF MB , BE and FD STATIISTICS

S.NO

MAXWELL –BOLTZMANN BOSE EINSTEIN FERMI-DIRAC

1 Classical statistics ( particle aspect of particle)

Quantum statistics( wave aspect of particle)

Quantum statistics ( wave aspect of particle)

2Particles are distinguishable

Particles are indistinguishable Particles are indistinguishable

3 No restriction for the number of particles in a given state

No restriction for the number of particles in a given state

Only two particles are permitted in a given state.

4 Number of distinguishable Number of distinguishable Number of distinguishable

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ways is N! ∏1

i (gi)¿

¿! ways is ∏1

g [ni+gi−1]!(gi−1)!× ni!

ways is = ∏1

g gi!( g i−ni ) !×n i!

5Applicable to ideal gas Applicable to Photons, bosons Applicable to Electrons,

fermions

6 most probable distribution is

gi

eα+β ∈ i

most probable distribution is gi

eα+β ∈ i−1

most probable distribution is gi

eα+β ∈ i+1

8. No question of spin of particles arises.

Particle posses zero or integral spin

Particle posses half integral spin

818APPLICATION OF FD STATISTICS TO ELECTRON GAS IN METALS

A metal is composed of system of fixed positive nuclei and a number of mobile

electrons. These mobile electrons are assumed to move freely in the metal like the particles of a

gas and constitute a perfect gas known as electron gas.

Since these electrons are assumed to obey pauli’s exclusion principle, they should obey FD

statistics. According to FD distribution law, the most probable distribution is given by,

ni = gi

eα+βε+1

from this we can derive the expression for Fermi energy which is

εF = h2

2m [

3n4 π V gs

¿2/3

for electrons s = ½ , therefore gs = 2s +1 = 2

therefore

εF = h2

2m [

3n4 π V .2

¿2/3

= 0.625 × 10 -17 J

Thus specific heat of FD gas is proportional to the absolute temperature and vanishes at

absolute zero.

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4.Calculate the Fermi energy for copper whose density is 8.84 × 10 3 Kg /m3 .Atomic weight of

copper is 93.5 g/mole.

Fermi energy E F = h28m

(3 N )2/3( πV )2/3

h = 6.626 × 10 -34 Js, m = 9.1 × 10 -31 Kg, N = 6.023 × 10 23

V = atomic weight/ density

= 0.0935/8.84 × 10 3

= 10.58 × 10 -6

E F = (6.626 ×10−34)2

8× 9.1× 10−31 ×

[3 ×6.023 ×10 23]2/3[3.14 ×10.58 ×10−6]2/3

= 8.72 × 10 -19 J

A metal is composed of system of fixed positive nuclei and a number of mobile

electrons. These mobile electrons are assumed to move freely in the metal like the particles of a

gas . This is known as electron gas. Since these electrons are assumed to obey pauli’s exclusion

principle, they should obey FD statistics.

Their energy at 0 K is called Fermi energy ( EF) which is given by

EF = h2

2m ×( 3 N

8 π V)

23

Where h – planc constant, m- mass of electron, N – number of electrons V- volume of the electron gas DERIVATION:

Total number of electrons N = ∫0

∞

n dn ----------------------1

According to FD distribution law, the most probable distribution is given by,

n = g

eα+βε+1 --------------------2

When T = 0 K , E= Fermi Energy EF eβε = e−∞

= 0

Therefore the above equation becomes

n = g -------------------3

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To find the number of degeneracies( g)

number of eigen states (g)= Degeneracy × V shell

V element -------------------4

where V shell – volume of shell, V element - volume of element

= ( 2S +1) × V shell

V element [ Degeneracy = 2S +1]

= ( 2 (12 ) +1) ×

V shell

V element [For electron s = ½ ]

g = ( 2) × V shell

V element ----------------5

According to quantum idea , the element of volume in the momentum space = h3

V

The volume of the shell having their momenta between p and p+dp

= 43 π ( p + dp ) 3 -

43 π ( p ) 3

= 43 π [ p 3 + 3 p2 dp + 3 p (dp) 2 + (dp ) 3 ] -

43 π p 3

= 43 π [ p 3 + 3 p2 dp + 3 p (dp) 2 + (dp ) 3 - p 3 ]

= 43 π [ 3 p2 dp + 3 p (dp) 2 + (dp) 3 ]

= 43 π [ 3 p2 dp ] [ neglecting the higher terms of dp]

V shell = 4 π p2 dp --------------------------------6

The kinetic energy E of the electron is given by

½ mv2 = E ----------------------7

m2v2 = 2mE [ multiplying by ‘m’ on both sides]

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p 2 = 2mE [ p = mv]---------------8

∴ p = √2 mE

Differentiating dp = √2 m × 1

2√ E ------------9 [ d( √x =

12√ x

)

Substituting in 6 we get

V shell = 4 π ( 2mE )×√2m × 12√E

= 4√2 π m32 E

12

g = ( 2) × 4 √2π m

32 E

12

h3

V

= 8√2 πV

h3 × m32 E

12

Total number of electrons N = ∫0

E 8√2 πVh3 m

32 E

12 dE

= 8√2 πV

h3 m32 ∫

0

∞

E12 dE

= 8√2 πV

h3 m32

E32

32

[ ∫0

∞

x1 /2 dx=¿¿ x

32

32

]

= 8√2 πV

h3 m32 ×2

3 × E32

N = 8 π V3 h3 × 2

32 ×m

32 × EF

32

∴ EF

32 = 3N h3

8π V ×

1

( (2 m)32 )

Taking ( 23 ) power on both sides

EF = ( 3 N8 π V

)23 × h2

2m

This the expression for Fermi energy.

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918PAULI PARAMAGNETISM

1018APPLICATION OF FD STATISTICS TO THERMIONIC EMISSION:

When metal is heated, there occurs an emission of electrons. This effect is known as

thermionic emission. The electrons thus emitted are called thermions.

According to Fermi- Dirac statistics, an electron at absolute zero possesses a maximum

energy equal to Fermi energy. EF Free electron theory assumes that potential within the metal

is constant (W). This is the minimum energy of the electron for its emission from the metallic

surface. Therefore the minimum energy to be supplied to the electron for its emission is W-EF

and is defined as the work function of the metal.

When the metal is heated, the free electrons gain the thermal energy , and when the

thermal energy is more than W-EF , the electrons are emitted from the metallic surface. The

thermal energy in excess of W-EF will be converted in to kinetic energy of the electrons. On

the other hand , on the classical picture, the amount of energy supplied to emit an electron from

the metal is W., a value much greater than indicated by experiments.

Free electron model of a metal according to Fermi-Dirac statistics

W-EF

W

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1118ELEMENTARY IDEA OF PHASE TRANSITION

1218 PROPERTIES OF LIQUID HELIUM[ APPLICATION OF BE STATISTICS TO

LIQUID HELIUM]

BOSE – EINSTEIN CONDENSATE(BEC)

BEC is a state of matter in which separate atoms cooled to near absolute zero coalese in to

single entity , which can be described by a wave function.

When the temperature of BE gas is lowered below the critical temperature( T0), the number

of particles in the ground state rapidly increases. This rapid increase in the population of

ground state below the critical temperature , for BE gas is known as Bose – Einstein

condensation.

In the case of Helium in liquid state, the critical temperature is found to be 2.19 K.

Liquid helium above 2.19 K is called He I and below this is called HeII . this 2.19 is called

--- point

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1. Helium occurs in two stable isotopic forms

2. He4 has a spin of zero . therefore it is a boson

3. He3 has nuclear spin of ½ . therefore it is a fermion

4. The lamda point occurs at 2.19 K

5. liquid helium above lamda line is called helium –I and below is called He-II

6. The viscosity of He-II is zero and hence it flows through capillary tubes

7. It has zero entropy

Which is not true:?

a. He4 is a boson

b. He3 is a fermion

c. In BEC, large particles occupy the highest energy state

d. In BEC separate atoms cooled to near absolute zero coalese in to single entity

Properties

He4 is called super fluid liquid

It has zero viscosity

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1318 PHASE SPACE:

In classical mechanics, the state of the system of a particle at any given time is specified by,

three position components x ,y, z and three momentum components px.py and pz. This six –

dimensional space for a single particle is termed as phase space.or μ- space. Each component is

called phase point

X ( x,y,z, px,py,pz)

1418LIOUVILLE'S THEOREM

Part – 1

Conservation of density

states that the density of particles in phase space is a constant

dρdt = 0

Conservation of extension

states that the volume of system in phase space is a constant

dvdt = 0

1518 ENSEMBLES [Micro, Canonical Ensembles]

ENSEMBLES

An ensemble is defined as a collection of a large number of microscopically identical ,but independent systems.

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1.Micro Canonical Ensemble:

It is a collection of large number of independent systems having the same energy ,volume and number of identical particles. The individual systems are separated by rigid, impermeable and well insulated walls

The values of E,V and N for a particular system are not affected by then presence of other systems.

2.Canonical Ensemble:

It is a collection of large number of independent systems having the same temperature ,volume and number of identical particles The individual systems are separated by rigid, impermeable but conducting walls

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impermeable but insulated wallsE,V,N E,V,N

E,V,N

E,V,N

E,V,N

E,V,N E,V,N

E,V,N

E,V,N


The values of V and N for a particular system are not affected by other systems.

All the systems will arrive at a common temperature T.

3. Grand Canonical Ensemble:

It is a collection of large number of independent systems having the same

temperature ,volume and chemical potential. The individual systems are separated by

rigid, permeable but conducting walls

The values of V for a particular system is not affected by then presence of other systems.

The systems will arrive at a common temperature T.

The systems arrive same chemical potential.

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permeable but conducting walls

An ensemble is defined as a collection of a large number of microscopically

identical ,but independent systems.

By the term macroscopically identical, we mean that each of the systems constituting an

ensemble should have same volume, energy, pressure and total number of particles.

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By the term independent systems we mean that the system constituting are mutually

non-interacting. The following are the types of ensembles.

1618EQUATION OF STATE

1718 THERMODYNAMIC FUNCTIONS OF AN IDEAL GAS

1818 EQUI PARTITION OF ENERGY

Standard definite integrals

∫0

∞

e−a x2

dx = 12 √ πa

∫0

∞

x e−a x2

dx = 12 a

∫0

∞

x2 e−a x2

dx = 14 √ πa3

∫0

∞

x3 e−a x2

dx = 1

2a2

∫0

∞

x4 e−a x2

dx = 38 √ πa5

∫0

∞

x5 e−a x2

dx = 1a3

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UNIT –IV STATISTICAL MECHANICS – TEST

This paper contains 20 questions CODE Code: 4433221111223344

1.The fraction of molecules with velocity between c and c+dc is

c. ρ(c) dc = 4π ( M2 πT

)32 c2 exp ( −M c2

2 RT ) dc b. ρ(c) dc = 4π ( M

2 πR)

32 c2 exp ( −M c2

2 RT ) dc

c. ρ(c) dc = 4π ( M2 πR

)32 c2 exp ( −M c2

2 RT ) dc d. ρ(c) dc = 4π ( M

2 πRT)

32 c2 exp ( −M c2

2 RT )

dc

2. The fraction of molecules having velocity greater than zero

a. increases with increase in velocity,

b. decreases with increase in velocity,

c. decreases with increase in velocity reaches a minimum and then increases

d. increases with increase in velocity, reaches a maximum and then falls off .

3. The fraction of molecules with too low or too high velocities is

a. zero b. very high

c. very small d. infinite

4. The velocity possessed by maximum number of molecules is called

a. average velocity b. root mean square velocity

c. most probable velocity d. mean square velocity

5. Most probable velocity

a. decreases with rise in temperature. b. increases with rise in temperature

c. remains constant d. first increases and then decreases

6. The factor exp ( −M c2

2 RT ) is called

a. frequency factor b. Boltzmann factor.

c. kinetic factor d. velocity factor

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7.Expression for average velocity is

a.√ 8 RTπM

b. √ 2 RTM

c. √ 3 RTM

d. √ 8RTM

8.Expression for most probable velocity is

a.√ 2 RTM

b. √ 3 RTM

c. √ 8 RTπM

d. √ 8RTM

9.Expression for RMS velocity is

a. √ 3 RTM

b. √ 2 RTM

c. √ 8 RTπM

d. √ 8RTM

10.The RMS velocity of a gas whose molecular weight is 83.1 at 270 K is

a. 9 m/s b. 19 m/s

c. 29 m/s d. 35 m/s

11. The RMS velocity of CO2 at 440K is

a.√265 m/s b. √249 m/s

c. 29 m/s d. 35 m/s

12 The Most probable velocity of N2 at 7 oC

a.√265 m/s b. √166.3 m/s

c. 29 m/s d. 35 m/s

13. The six –dimensional space for a single particle is termed as

a.. critical space b. vortex region

c. μ- space d. hysteresis space

14. Phase point is represented by

a.X ( x,y,z, px,) b. X ( x,y, px,py)

c.X ( x,y,z, px,py,pz) d.X ( x, px,)

15.ASSERTION(A) : At lower temperatures the speeds of the molecules are lower .

REASON(R): At lower temperatures, the molecules have less energy

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a Both A and R are true but R is not correct explanation of A

b.. A is true but R is false

c. Both A and B are false

.d. Both A and R are true and R is the correct explanation of A

16. Which is true ?

a. As the temperature of the molecules increases, the distribution flattens out.

b. heavier molecules move more slowly than lighter molecules.

c. lighter molecules will have a speed distribution that is more spread out.

d.All

17. Which is true ?

1. The fraction of molecules having too low or too high velocities is very low.

2. There is a certain velocity for which the fraction of molecules is maximum. This is

called the most probable velocity.

3. The rise in temperature increases the fraction of molecules having high velocities

a.. I only b. I and II only

c. II and III d. I, II and III

18. The square root of the mean of the squares of different velocities possessed by the

molecules of the gas at the given temperature is called

a. most probable speed b. average speed

c. critical speed d. root-mean-square speed

19. Pauli’s principle is followed in

a. Maxwell statistics b. Bose- Einstein statistics

c. Fermi- Dirac statistics d. classical statistics

20. Most probable distribution of electron is

a. g i

eα+β ∈ i b. gi

eα+β ∈ i−1

c. gi

eα+β ∈ i+1 d.

gi

eβ∈ i−1

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21. In Micro canonical ensemble the constant quantities are

a. E,V,N b. T,V,N

c. T,V,μ d. P,V,N

22. In Canonical ensemble which is/are true ?

I. systems having the same temperature ,volume and number of identical particles

II. The individual systems are separated by rigid, impermeable but conducting walls

III The values of V and N of a particular system are not affected by other systems.

IV. Heat can not be exchanged between the systems

a. I only b. I and II only

c. I,II and III only d. II and III only

23. In grand canonical ensemble which is/ are not true ?

I. independent systems having the same temperature ,volume and chemical potential.

II. The individual systems are separated by rigid, permeable but conducting walls

III. Heat can not be exchanged between the systems

I. systems have same chemical potential.

a. III only b. I and II only

c. I only d. II and III only

24. The state of the system of a particle in phase space at any given time is specified by

a. three position components and three momentum components

b. one position component and three momentum components .

b. two position components and three momentum components .

c. . one position component and one momentum component

25... Energy density in the given frequency interval between γ and γ + d γ , in volume V is

given by

a. Eϒ dγ = 8 πh γ3

c3 d γ

(e h γKT

−1) b. Eϒ dγ = 8h γ 3

c3 d γ

(e hγKT

−1)

c. Eϒ dγ = 8πhc3

d γ

(e hγKT

−1) d. Eϒ dγ = 8 π γ3

c3 d γ

(e h γKT

−1)

26 Planc’s radiation formula is

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a. Eϒ dγ = 8 πh γ3

c3 d γ

(e h γKT

−1) b. Eϒ dγ = 8h γ 3

c3 d γ

(e hγKT

−1)

c. Eϒ dγ = 8πhc3

d γ

(e hγKT

−1) d. Eϒ dγ = 8 π γ3

c3 d γ

(e h γKT

−1)

27. Planc’s energy distribution in terms of wavelength is

a. Eϒ dγ = 8 πhc3

d γ

(e h γKT

−1) b. E λ d λ = - 8 πhC

λ5 1

e hCƛ KT

−1 dƛ

c.Eλ d λ = - 8 πhC

λ5 1

e hCƛ KT

−1 dƛ d. Eϒ dγ = 8πhc3

d γ

(e hγKT

−1)

28. Wien’s law is

a. Eϒ dγ = 8πKT

c3 dγ b. Eϒ dγ = 8 πh γ3

c3 d γ

e hCƛ KT

c. Eϒ dγ = 8πh γ3

c3 d γ

e hCƛ KT

d. Eϒ dγ = 8πh γ3

c3

29. Independent systems having the same energy ,volume and number of identical particles

is called

a.Canonical ensemble b. Grand canonical ensemble

c.Microcanonical ensemble d. none of the above

30. If the individual systems are separated by rigid, impermeable and well insulated walls

the ensemble is called

a.Canonical ensemble b. Grand canonical ensemble

c. Micro canonical ensemble d. none of the above

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Platform : Https://Www.Csaslearningcenter.Com/ Online Tests/M.Sc/ Physical

PHYSICAL - TEST-5 -STATISTICAL THERMODYNAMICS- 1

This paper contains -- questions CODE : 43211234

1. Striling’s formula is a. ln x ! = x ln x+ x b. ln x ! = ln x –xc. ln x ! = x ln x –2x d. ln x ! = x ln x –x

2. Calculate the value of ‘ln 100!’ a. 606.5 b. 256.8c. 360.6 d. 678.9

3. Calculate the value of ’ln N!’ if N = 6.023 ×1023 given ln (6.023 ×1023) = 54.76a. 8.65×1025 b. 3.23 ×1025

c.5.66×1025 d. 6.78×1025

4. Which is not true about Maxwell Boltzmann Statistics

II. This is based on classical theory

III. According to this particle has wave character.

IV. . This is applicable to distinguishable particles.

V. Total probability W of the distribution is W = N! ∏1

i (gi)¿

¿ !

a. II only b. III only

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https://www.csaslearningcenter.com/


c. III and II only a. IV and II only

5. The number of ways of arranging 2 distinguishable particles in two states is

a. 4 b. 2c. 3 d 20

6. The number of ways of arranging 2 distinguishable particles in three states isa. 2 b. 9c. 8 d 20

7..The number of ways of arranging 3 distinguishable particles in two states is a. 2 b. 4c. 8 d 18

8 The number of ways of distributing two Maxwellons among four energy levels is

a. 10 b.64

c. 14 d. 16

9 The number of ways of distributing three Maxwellons among four energy levels is

a. 10 b.6

c. 14 d. 64

10. Which is not true about Bose Einstein statistics

I. This is based on classical theory

II. According to this ,particle has wave character.

III. This statistics deals with Particles which are indistinguishable.

IV. Any number of particles may occupy a given energy level


c. I only a. IV and II only

11. Which is not true about Bose Einstein statistics

I. This statistics is obeyed by particles having integral spin

II. Any number of particles may occupy a given energy level

III, The number of ways in which n particles are distributed among g levels is

∏1

g [¿+gi−1]!(gi+1)!×∋!

IV. Expression for most probable distribution of ‘ni’ particles in ‘gi’ energy levels is gi

eα+β ∈ i−1


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c. I only a. IV and II only

12. The number of ways of arranging 2 bosons in four states is

a. 10 b. 2c. 3 d 20

13. The number of ways of arranging 3 bosons in four states isa. 20 b. 9c. 8 d 2

14 . Calculate the number of ways of distributing two particles among four energy levels when

the particles are 1. Bosons 2. Maxwellons

a. 2,4 b. 10,16c. 8,16 d 18,10

15. Which is not true about Fermi -Dirac statistics

I. This statistics is obeyed by particles having integral spin

II. Any number of particles may occupy a given energy level

III, The number of ways in which n particles are distributed among g levels is

∏1

g [¿+gi−1]!(gi+1)!×∋!

IV. Obey pauli’s exclusion principle


c. I , II and III only a. IV and II only

16. The number of ways of arranging 2 fermions in four states is

a. 10 b. 2c. 3 d 6

17. The number of ways of arranging 3 fermions in four states isa. 20 b. 9c. 8 d 4

18 . Calculate the number of ways of distributing three particles among four energy levels

when the particles are 1. Fermions 2. Bosons

a. 2,4 b. 8,16 c. 4,20 d 18,10

NO CODE


Ni 2 4 4 7 3

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vi 1 2 3 4 5

Find the average velocity and most probable velocity

a. 3.25 m/sec , 4 m/sec b. 8 m/sec,16 m/secc. 4 m/sec,20 m/sec d 18 m/sec,10 m/sec


Ni 2 4 7 3

vi 1 2 3 4

Find the rms velocity and most probable velocity

a. 12.9 m/sec , 3 m/sec b. 8 m/sec,16 m/secc. 4 m/sec,20 m/sec d 18 m/sec,10 m/sec

Problem: At what temperature will the average speed of molecule of hydrogen gas be double

the average speed of oxygen at 100 K

a. 400K b. 900Kc. 800K d 530K

Problem: At what temperature will the average speed of molecule of hydrogen gas will be

same as that of nitrogen molecules at 280 K

a. 20K b. 900Kc. 800K d 530K

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