Statistical Mechanics

Statistical Distribution

This determines the most probable way in which a certain total amount of energy ‘E’ is distributed among the ‘N’ members of a system of particles in thermal equilibrium at absolute temperature, T.

Thus Statistical Mechanics reflects overall behavior of system of many particles.

Suppose n(є) is the no. of particles having energy, є

then )()()( εεε fgn = where g(є) = no. of states of energy є or statistical weight

corresponding to energy є

f(є) = distribution function

= average no. of particles in each state of energy є

= probability of occupancy of each state of energy є

Statistical Distribution are of three kinds

1. Identical particles that are sufficiently far apart to be distinguishable.

Example: molecules of a gas.

Negligible overlapping of ψ

Maxwell-Boltzmann Statistics

2. Indistinguishable identical particles of ‘0’ or integral spin.

Example: Bosons (don’t obey the exclusion principle)

Overlapping of ψ

Bose-Einstein Statistics

3. Indistinguishable identical particles with odd-half integral spin (1/2, 3/2, 5/2 ..).

Example: Fermions (obey the exclusion principle)

Fermi-Dirac Statistics

Maxwell-Boltzmann Statistics

According to this law number of identical and distinguishable particles in a system at temperature, T having energy є is

n(є) = (No. of states of energy є).(average no. of particles in a

state of energy є) kTAegn

εεε −= ).()(

Here A is a constant and

This law cannot explain the behavior of photons (Black body radiation) or of electrons in metals (specific heat, conductivity)

kTAef BM

εε −=)(..

(i)

Equation (i) represents the Maxwell-Boltzmann Distribution Law

(Classical Approach)

Applications of M.B. Statistics

(i) Molecular energies in ideal gas

No. of molecules having energies between є and є + dє is given by

)}(}{)({)( εεεεε fdgdn =

No. of states in momentum space having momentum between p and p + dp is proportional to the volume element i.e.

The momentum of a molecule having energy є is given by

εεε εdeAgn kT

−= )()( (i)

εmp 2=ε

εm

mddp

2=⇒

zyx dpdpdp∫ ∫ ∫α

dppg )(

Each momentum corresponds to a single є

33

3

4)(

3

4pdpp ππ −+

dpp24πα

dpp24πα

α

dpBpdppg 2)( =

dpBpdg 2)( =εε

B is a constant

εεεεεm

mdmBdg

2.2.)( =

No. of molecules with energy between є and є + dє

Put in (i)

εεεε dBmdg 2/32)( =

kTeBAmdnεεεε −= 2/32)(

(ii)kTeCdnεεεε −=)(

Total No. of molecules is N

∫∫∞∞

−==00

)( εεεε εdeCdnN kT

Put value of C in equation (ii)

This equation gives the number of molecules with energies between є and є + dє in a sample of an ideal gas that contains N molecules and whose absolute temperature is T. It is called molecular energy distribution equation.

23

)(

2

kT

NC

ππ=

∫∞

− =0 2

121

aadxex ax π Using identity

εεπ

πεε εde

kT

Ndn kT

−=23

)(

2)(

Maxwell-Boltzmann energy distribution curve for the molecules of an ideal gas.

0 kT 2kT 3kT

n(є)

Total energy of the system

Average Molecular Energy

∫∞

=0

)( εεε dnE

∫∞

−=0

23

23

)(

2 εεπ

π εde

kT

NE kT

= kTkT

kT

NE π

ππ 2)(

4

3

)(

223

∫∞

− =0

24

323

aadxex ax π

Using

Average energy per molecule

NkTE2

3=

N

E==−ε

This is the average molecular energy and is independent of the molecular mass

kT2

3=−

ε

At Room Temperature its value is 0.04 eV.

Equipartition of Energy: Equipartition Theorem

Average Kinetic Energy associated with each degree of freedom of a molecule is

Put this in molecular energy distribution equation

ε⇒..EK

kT2

1

Distribution of molecular speed

2

2

1mv=

mvdvd =⇒ ε

It is called molecular speed distribution formula.

It is the square root of the average squared molecular speed

dvevkT

mNdvvn kT

mv22

23

2

24)(

−

=

ππ

−

= 2vvrms

kTvm2

3

2

1 2 ==−−

ε

m

kTvrms

3=

RMS Speed

Using

Average Speed

∫∞−

=0

)(1

dvvvnN

v

dvevkT

mkT

mv22

23

0

3

24

−

∫∞

=

ππ

∫∞

− =0

23

2

12

adxex ax

=

−

2

22

2

4.

24

23

m

Tk

kT

mv

ππ

m

kT

π8=

Most Probable Speed

m

kTvp

2=⇒

kTmv

evkT

mNvn 2

223

2

24)(

−

=

ππ

[ ] 02

4)(

22

23

2 =

= −

kTmv

evdv

d

kT

mN

dv

vdn

ππ

02

423

≠

kT

mN

ππ

[ ] 022

2 =−kT

mv

evdv

d

(i) Speed distribution is not symmetrical

M.B. Speed Distribution Curve

n(v)

vmost probable = vp

−vvrms

v

rmsp vvv <<−

(ii)

m

kT

m

kT

m

kT 382 <<π

Variation in molecular speed with ’T’ and ‘ molecular mass’

This curve suggests that most probable speed α T

Also the most probable speed α 1/m

Quantum Statistics

1. ‘0’ or integral spin

(Indistinguishable identical particles)

Bosons Fermions

1. Odd half integral spin

2. Do not obey exclusion

principle

2. Obey exclusion principle

3. Symmetric wave function 3. Anti-symmetric wave function

4. Any number of bosons

can exist in the same

quantum state of the

system

4. Only one fermion

can exist in a particular

quantum state of the

system

Consider a system of two particles, 1 and 2, one of which is in state ‘a’ and the other in state ‘b’.

)2()1(' ba ψψψ = When particles are distinguishable then

)1()2(" ba ψψψ = When particles are indistinguishable then

[ ])1()2()2()1(2

1babaB ψψψψψ += For Bosons,

symmetric

For Fermions, [ ])1()2()2()1(2

1babaF ψψψψψ −=

anti-symmetric

ψF becomes zero when ‘a’ is replaced with ‘b’ i.e.

Thus two Fermions can’t exist in same state (Obey Exclusion Principle)

[ ] 0)1()2()2()1(2

1 =−= aaaaF ψψψψψ

Bose-Einstein Statistics

Any number of particles can exist in one quantum state.

Distribution function can be given by

1

1)(.. −

=kTee

f EB εαε

where α may be a function of temperature, T.

For Photon gas 10 =⇒= αα e

-1 in denominator indicates multiple occupancy of an energy state by Bosons

Bose-Einstein Condensation

If the temperature of any gas is reduced, the wave packets grow larger as the atoms lose momentum according to uncertainty principle.

When the gas becomes very cold, the dimensions of the wave packets exceeds the average atomic spacing resulting into overlapping of the wave packets. If the atoms are bosons, eventually, all the atoms fall into the lowest possible energy state resulting into a single wave packet. This is called Bose-Einstein condensation.

The atoms in such a Bose-Einstein condensate are barely moving, are indistinguishable, and form one entity – a superatom.

Fermi-Dirac Statistics

Obey Pauli’s exclusion Principle

Distribution function can be given by

1

1)(.. +

=kTee

f DF εαε

f (є) can never exceed 1, whatever be the value of α, є and T. So only one particle can exist in one quantum state.

α is given by

kTFεα −=

Fε where is the Fermi Energy

1

1)( )(..

+= −

kTF

ef DF εεε

then

−∞=− kTF )( εε

Case I: T = 0

+∞=− kTF )( εε For є < єF

For є > єF

For є < єF

11

1)(.. =

+= ∞−e

f DF ε

For є > єF

01

1)(.. =

+= ∞+e

f DF ε

Thus at T = 0, For є < єF all energy states from є = 0 to єF are occupied as 1)(.. =εDFf

Thus at T = 0, all energy states for which є > єF are vacant.

0)(.. =εDFf

Case II: T > 0

Some of the filled states just below єF becomes vacant while some just above єF become occupied.

2

1

1

1)(

0.. =+

=e

f DF ε

Case III:

At є = єF

Probability of finding a fermion (i.e. electron in metal) having energy equal to fermi energy (єF) is ½ at any temperature.

For all T

0.5

Є (eV) 0

1

єF 1

)(.. εDFf0 K

10000 K

1000 K

Electromagnetic Spectrum

λ (µm)

1000 100 10 1 0.1 0.01

ultravioletvisiblelightinfraredmicrowaves x-rays

HighEnergy

LowEnergy

Black Body Radiation

All objects radiate electromagnetic energy continuously regardless of their temperatures.

Though which frequencies predominate depends on the temperature.

At room temperature most of the radiation is in infrared part of the spectrum and hence is invisible.

The ability of a body to radiate is closely related to its ability to absorb radiation.

A body at a constant temperature is in thermal equilibrium with its surroundings and must absorb energy from them at the same rate as it emits energy.

A perfectly black body is the one which absorbs completely all the radiation, of whatever wavelength, incident on it.

Since it neither transmits any radiation, it appears black whatever the color of the incident radiation may be.

There is no surface available in practice which will absorb all the radiation falling on it.

The cavity walls are constantly emitting and absorbing radiation, and this radiation is known as black body radiation.

(i) The total energy emitted per second per unit area (radiancy E or area under curve) increases rapidly with increasing temperature.

(ii) At a particular temperature, the spectral radiancy is maximum at a particular frequency.

Characteristics of Black Body Radiation

(iii) The frequency (or wavelength, λm) for maximum spectral radiancy decreases in direct proportion to the increase in temperature. This is called “Wien’s displacement law”

=×Tmλ constant (2.898 x 10-3 m.K)

Planck’s Radiation Law

Planck assumed that the atoms of the walls of cavity radiator behave as oscillators with energy

nhvn =ε ....3,2,1,0=n The average energy of an oscillator is

N

εε =−

N is total no. of Oscillators

1−=

−

kThv

e

hvε (i)

Thus the energy density (uv) of radiation in the frequency range v to v + dv is

−×= επ

3

28

c

dvvdvuv

−=

1

83

2

kThv

e

hv

c

dvvdvuv

π

1

83

3

−=

kThv

e

dv

c

hvdvuv

π

In terms of wavelength

1

85 −

=kT

hc

e

dhcdu

λ

λλπλλ

This is Planck’s Radiation formula in terms of frequency.

This is Planck’s Radiation formula in terms of wavelength.

When λ is large then

1

85 −

=kT

hc

e

dhcdu

λ

λλπλλ

Case I : (Rayleigh-Jeans Law)

This is Rayleigh-Jeans Law for longer λ’s.

kT

hce kThc

λλ +≈1

λ

λλ

πλλ d

kThchc

du

−+

=11

8

5

λλπλλ dkT

du4

8=

When λ is very small then

1

85 −

=kT

hc

e

dhcdu

λ

λλπλλ

Case II : (Wein’s Law)

This is Wein’s Law for small λ’s.

1>>kThc

e λ

λλπλ λ

λ dehc

du kThc−=

5

8

λλ

λ λλ de

Adu kT

B

5= This is another form of

Wein’s Law..

Here A and B are constants.

Stefan’s Law

The spectral radiancy Ev is related to the energy density uv by

1

83

3

−=

kThv

e

dv

c

hvdvuv

π Planck’s Radiation formula in terms of frequency is

vv uc

E4

=

1

2 3

2 −=

kThv

e

dvv

c

hdvEv

π

1

2

4 2

3

−=

kThv

e

dv

c

hvdvu

cv

π

xkT

hv =

∫∫∞∞

−==

0

3

20 1

2kT

hv

e

dvv

c

hdvEE v

π

Let xh

kTv =⇒ and dx

h

kTdv =

∫∞

−=

0

3

23

44

1

2xe

dxx

ch

TkE

π

151

4

0

3 π=−∫

∞

xe

dxx But

Let

423

45

15

2T

ch

kE

π=

Here

σπ =23

45

15

2

ch

k (Stefan’s constant)

4TE σ= This is Stefan’s Law.

23

45

15

2

ch

kπσ = = 5.67 x 10-5 erg/(cm2.sec.K4)

= 5.67 x 10-8 W/(m2.K4)

Wein’s Displacement Law

Using

Planck’s Radiation formula in terms of frequency is

0)( =

λλ

d

ud

1

185 −

=kT

hc

e

hcu

λλπ

λ

for maxλλ =

=maxλkThc

Constant

=×Tmλ constant (2.898 x 10-3 m.K)

Peaks in Black Body radiation shifts to shorter wavelength with increase in temperature.

Specific Heat of Solids

Each atom in the solid should have total energy = 3kT (3 degrees of freedom).

Atoms in solid behave as oscillators.

RT

EC

Vv 3=

∂∂=

In case of solids total average energy per atom per degree of freedom is kT (0.5kT from K.E. and 0.5kT from P.E.).

For one mole of solid, total energy, E = 3NokT (classically)

E = 3RT

Here No is the Avogadro number.

Specific heat at constant volume is

~ 6 kcal/kmol.K

This is Dulong & Petit’s Law.

This means that atomic specific heat (at constant volume) for all solids is approx 6 kcal/kmol.K and is independent of T.

Dulong and Petit’s Law fails for light elements such as B, Be and C.

Also it is not applicable at low temperatures for all solids.

According to it the motion of the atoms in a solid is oscillatory.

Einstein’s theory of Specific Heat of Solids

The average energy per oscillator atom) is

1−=

−

kThv

e

hvε

The total energy for one mole of solid in three degrees of freedom.

13

−=

kThv

e

hvNE o

The molar specific heat of solid isV

v T

EC

∂∂=

This is Einstein’s specific heat formula.

22 )1(3

−

=

kThv

kThv

e

e

kT

hvhvNC ov

Case I:

2

2

)1(3

−

=

kThv

kThv

e

e

kT

hvkNo

2

2

)1(3

−

=

kThv

kThv

e

e

kT

hvRCv

At high T, kT >> hv then

kT

hve kThv +≈1

2

2

11

13

−+

+

≈

kThv

kThv

kT

hvRCv

+≈

kT

hvRCv 13

RCv 3≈ )( hvkT >>∴

This is in agreement with Dulong & Petit’s Law at high T.

This implies that as

Case II:

kThv

ekT

hvRCv

−

≈

2

3

At low T, hv >> kT then 1>>kThv

e

i.e. is in agreement with experimental results at low T.

0,0 →→ vCT

Typically, one metal atom gives one electron.

Free electrons in a metal

One mole atoms gives one mole of free electrons, (No)

RTkTNE oe 2

3

2

3 ==

Then total specific heat in metals at high T should be

( ) RT

EC

V

eev 2

3=

∂∂=

If each free electron can behave like molecules of an ideal gas. Then

Average K.E. for 1 mole of electron gas,

Molar specific heat of electron gas,

RRRCv 2

9

2

33 =+=

But experimentally at high T

Free electrons don’t contribute in specific heat,

Why?

The no. of electrons having energy є is

Electrons are fermions and have upper limit on the occupancy of the quantum state.

By definition highest state of energy to be filled by a free electron at T = 0 is obtained at є = єF

∫=F

dgNε

εε0

)(

RCv 3≈⇒

where εεπεε dh

Vmdg

3

23

28)( =

Here V is the volume of the metal

where N/V is the density of free electrons.

∫=F

dh

VmN

ε

εεπ

03

23

28

No. of electrons in the electron gas having energy between є and є + dє is

23

23

33

216Fh

Vm επ=

23

8

3

2

2

=⇒V

N

m

hF π

ε (i)

Electron energy distribution

εεεεε dfgdn )()()( =

This is electron energy distribution formula, according to which distribution of electrons can be found at different temperatures.

εεεπεε εε deh

Vmdn kTF 1

28)( /)(3

23

23

+= −

using (i)

εεεεε εε de

Ndn

kTF 12

3)(

/)(

23

+

= −

−

When a metal is heated then only those electrons which are near the top of the fermi level (kT of the Fermi energy) are excited to higher vacant energy states.

The electrons in lower energy states cannot absorb more energy because the states above them are already filled.

kT = 0.0025 eV at 300K

kT = 0.043 eV at 500K

This is why the free electrons contribution in specific heat is negligible even at high T.

Total energy at 0K is

Average electron energy at 0K

∫=F

dnEo

ε

εεε0

)(

Then average electron energy

0/)( == −∞− ee kTFεε

Since at 0K all electrons have energy less than or equal to єF

εεεε

dN

EF

Fo ∫−

=0

23

23

2

3F

N ε5

3=

Fo

o N

E εε5

3==

−

at 0K.