6. fermi-dirac statistics 1

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1 Fermi-Dirac distribution and the Fermi-level Density of states tells us how many states exist at a given energy E. The Fermi function f(E) specifies how many of the existing states at the energy E will be filled with electrons. The function f(E) specifies, under equilibrium conditions, the probability that an available state at an energy E will be occupied by an electron. It is a probability distribution function. E F = Fermi energy or Fermi level k = Boltzmann constant = 1.38 10 23 J/K = 8.6 10 5 eV/K T = absolute temperature in K

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Page 1: 6. Fermi-Dirac Statistics 1

1

Fermi-Dirac distribution and the Fermi-level

Density of states tells us how many states exist at a given energy E. The Fermi function f(E) specifies how many of the existing states at the energy E will be filled with electrons. The function f(E) specifies, under equilibrium conditions, the probability that an available state at an energy E will be occupied by an electron. It is a probability distribution function.

EF = Fermi energy or Fermi level

k = Boltzmann constant = 1.38 1023 J/K = 8.6 105 eV/K

T = absolute temperature in K

Page 2: 6. Fermi-Dirac Statistics 1

2

Fermi-Dirac distribution: Consider T 0 K

For E > EF :

For E < EF :

0)(exp1

1)( F

EEf

1)(exp1

1)( F

EEf

E

EF

0 1 f(E)

Page 3: 6. Fermi-Dirac Statistics 1

3

If E = EF then f(EF) = ½

If then

Thus the following approximation is valid:

i.e., most states at energies 3kT above EF are empty.

If then

Thus the following approximation is valid:

So, 1f(E) = Probability that a state is empty, decays to zero.

So, most states will be filled.

kT (at 300 K) = 0.025eV, Eg(Si) = 1.1eV, so 3kT is very small in comparison.

kTEE 3F 1exp F

kT

EE

kT

EEEf

)(exp)( F

kTEE 3F 1exp F

kT

EE

kT

EEEf Fexp1)(

Fermi-Dirac distribution: Consider T > 0 K

Page 4: 6. Fermi-Dirac Statistics 1

4

Temperature dependence of Fermi-Dirac distribution

Page 5: 6. Fermi-Dirac Statistics 1

5

Page 6: 6. Fermi-Dirac Statistics 1

6

Equilibrium distribution of carriers

Distribution of carriers = DOS probability of occupancy = g(E) f(E)

(where DOS = Density of states)

Total number of electrons in CB (conduction band) =

top

Cd)()(C0

E

EEEfEgn

Total number of holes in VB (valence band) =

V

Bottomd)(1)(V0

E

EEEfEgp

Page 7: 6. Fermi-Dirac Statistics 1

Properties of a Fermion gas

Page 8: 6. Fermi-Dirac Statistics 1

The internal energy of a gas of N fermions

Page 9: 6. Fermi-Dirac Statistics 1

Integration by parts (I)

In calculus, integration by parts is a rule that transforms the integral of products of functions into other, hopefully simpler, integrals. The rule arises from the product rule of differentiation.

If u = f(x), v = g(x), and the differentials du = f '(x) dx and dv = g'(x) dx; then in its simplest form the product rule is

duvuvdvu

)(')()()('))'()(( xgxfxgxfxgxf

Page 10: 6. Fermi-Dirac Statistics 1

Integration by parts (II)

In the traditional calculus curriculum, this rule is often stated using indefinite integrals in the form

As a simple example, consider

Since ln x simplifies to 1/x when differentiated, we make this part of ƒ; since 1/x2 simplifies to −1/x when integrated, we make this part of g. The formula now yields

dxxgxfxgxfdxxgxf )()(')()()(')(

dxx

x 2

ln

dxxxx

xdx

x

x)/1)(/1(

lnln2

Page 11: 6. Fermi-Dirac Statistics 1

At T = 0, U = (3/5)NεF , this energy is large because all the electrons must occupy the lowest energy states up to the Fermi level.

The average energy of a free electron in silver at T = 0 is

The mean kinetic energy of an electron, even at absolute zero, is two orders of magnitude greater than the mean kinetic energy of an ordinary gas molecule at room temperature.

Page 12: 6. Fermi-Dirac Statistics 1

Heat capacity

The electronic heat capacity Ce can be found by taking the derivative of Equation (19.18):

For temperatures that are small compared with the Fermi temperature, we can neglect the second term in the expansion compared with the first and obtain

Page 13: 6. Fermi-Dirac Statistics 1

• Thus the electronic specific heat capacity is 2.2 x 10-2 R. This small value explains why metals have a specific heat capacity of about 3R, the same as for other solids.

• It was originally believed that their free electrons should contribute an additional (3/2) R associated with their three translational degrees of freedom. Our last calculation shows that the contribution is negligible.

•The energy of the electrons changes only slightly with temperature (dU/dT is small) because only those electrons near the Fermi level can increase their energies as the temperature is raised, and there are precious few of them.

Page 14: 6. Fermi-Dirac Statistics 1

At very low temperatures the picture is different. From the Debye theory, Cv is proportional to T3 and so the heat capacity of a metal takes the form Cv = AT + BT3,

where the first term is the electronic contribution and the second is associated with the crystal lattice.

At sufficiently low temperatures, the AT term can dominate, as the sketch of Figure

19.9 indicates.

Figure 19.9 Sketch of the heat capacity of a metal as a function of temperature showing the electronic and lattice contributions.

Page 15: 6. Fermi-Dirac Statistics 1

S = 0 at T = 0, as it must be. The Helmholtz function F = U -TS is

The fermion gas pressure is found from

Page 16: 6. Fermi-Dirac Statistics 1

For silver we find that

N/V = 5.9 x1028 m-3 and TF = 65,000K .

Thus

P = 2/5 *5.9*1028 *(1.38*10-23) (6.5*104)

= 2.1*1010 Pa = 2.1*105 atm.

Given this tremendous pressure, we can appreciate the role of the surface potential barrier in keeping the electrons from evaporating from the metal.

Page 17: 6. Fermi-Dirac Statistics 1

19.5 Applications to White Dwarf Stars

The temperature inside the core of a typical star is at the order of 107 K.

The atoms are completely ionized at such a high T, which creates a hugh electron gas

The loss of gravitational energy balances with an increase in the kinetic energy of the electrons and ions, which prevent the collapse of star!

Page 18: 6. Fermi-Dirac Statistics 1

Example: The pressure of the electron gas in Sirius B can be calculated with the formula

Using the following numbers

Mass M = 2.09 × 1030 kg

Radius R= 5.57 × 106 m

Volume V= 7.23 × 1020 m3

Page 19: 6. Fermi-Dirac Statistics 1

Assuming that nuclear fusion has ceased after all the core hydrogen has been converted to helium!

The number nucleons =

Since the ratio of nucleons and electrons is 2:1

there are electrons

Page 20: 6. Fermi-Dirac Statistics 1

Therefore, T(=107 K) is much smaller then TF .

i.e. is a valid assumption !

Thus: P can be calculated as

Page 21: 6. Fermi-Dirac Statistics 1

A white dwarf is stable when its total energy is minimum

For

Since

can be expressed as

Where

Page 22: 6. Fermi-Dirac Statistics 1

For gravitational energy of a solid

With

In summary

To find the minimum U with respect to R

Page 23: 6. Fermi-Dirac Statistics 1

19.7 a) Calculate Fermi energy for Aluminum assuming three electrons per Aluminum atom.

eV

V

NelectronsforDensity

molekilo

atoms

kilomoleKg

mKg

V

N

F 8.111.26.58

108.13

1011.92

1063.6

108.1

3#

1099.51002.627

1069.2

32

29

31

234

29

28263

3

Page 24: 6. Fermi-Dirac Statistics 1

19.7b) Show that the aluminum at T=100 K, μ differs from εF by less than 0.01%. (The density of aluminum is 2.69 x 103 kg m-3 and its atomic weight is 27.)

%01.0

1038.41

1062.88.11

1000

121

121

50

2

15

2

0

22

0

thanless

eVKeVK

T

T

F

Page 25: 6. Fermi-Dirac Statistics 1

19.7c) Calculate the electronic contribution to the specific heat capacity of aluminum at room temperature and compare it to 3R.

Using the following equation

FFe

kTNk

T

TNkC

22

2

Page 26: 6. Fermi-Dirac Statistics 1

19.13. Consider the collapse of the sun into a white dwarf. For the sun, M= 2 x 1030 kg, R = 7 x 108 m, V= 1.4 x 1027 m3.

Calculate the Fermi energy of the Sun’s electrons.

eV

vm

vm

V

N

m

h

V

N

nucleousofelectronsof

electronsofNo

F

eF

eF

eF

6.20

10248.633.0

104.1

1023.7

26.1

205.133.0

8

3

2

104.1

102205.1

2

1#

10205.11066.1

102.

5

32

27

27

322

27

57

5727

30

Page 27: 6. Fermi-Dirac Statistics 1

(b) What is the Fermi temperature?

(c) What is the average speed of the electrons in the fermion gas (see problem 19-4). Compare your answer with the speed of light.

keVk

eV

kT F

F

5

151039.2

1062.8

6.20