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 1023 J/K
= 8.6  105 eV/K
T = absolute temperature in K
1
Fermi-Dirac distribution: Consider T  0 K
For E > EF :
f ( E  EF ) 
1
 0
1  exp ()
For E < EF :
f ( E  EF ) 
1
 1
1  exp ()
E
EF
0
1
f(E)
2
Fermi-Dirac distribution: Consider T > 0 K
If E = EF then f(EF) = ½
If
E  EF  3kT
then
 E  EF 
exp 
  1
 kT 
Thus the following approximation is valid:
  ( E  EF ) 
f ( E )  exp 

kT


i.e., most states at energies 3kT above EF are empty.
If
E  EF  3kT
then
 E  EF 
exp 
  1
 kT 
 E  EF 
f ( E )  1  exp 

 kT 
So, 1f(E) = Probability that a state is empty, decays to zero.
Thus the following approximation is valid:
So, most states will be filled.
kT (at 300 K) = 0.025eV, Eg(Si) = 1.1eV, so 3kT is very small
in comparison.
3
Temperature dependence of Fermi-Dirac distribution
4
5
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) =
n0  
E top
EC
g C ( E ) f ( E ) dE
Total number of holes in VB (valence band) =
p0  
EV
EBottom
g V ( E ) 1  f ( E )  dE
6
Properties of a Fermion gas
The internal energy of a gas of N fermions
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.
(
f
( x ) g ( x ))'

f
' ( x) g ( x)

f
( x) g ' ( x)
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
 u dv  uv   v du
Integration by parts (II)
In the traditional calculus curriculum, this rule is often stated using indefinite integrals in
the form

f
( x ) g '( x ) dx

f
( x) g ( x)



l
n
x
f
' ( x ) g ( x ) dx
x
2
dx
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
ln x
ln x
 x 2 dx   x   (1 / x)(1 / x)dx
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.
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
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.
• Thus
• 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.
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.
S = 0 at T = 0, as it must be. The Helmholtz function F = U -TS is
The fermion gas pressure is found from
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.
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!
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
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
Therefore, T(=107 K) is much smaller then TF .
i.e.
is a valid assumption !
Thus: P can be calculated as
A white dwarf is stable when its total energy is minimum
For
Since
can be expressed as
Where
For gravitational energy of a solid
With
In summary
To find the minimum U with respect to R
19.7 a) Calculate Fermi energy for Aluminum assuming three
electrons per Aluminum atom.
2.69 103 Kg
N

V
27 Kg
m 3  6.02 10 26
kilomole
atoms
 5.99 10 28
kilo  mole
N
V
 1.8 10 29
# Density for electrons  3 
F
6.63 10 

 3 1.8 10

31 
2  9.1110 
8
34 2
29



2
3
 5.6  2.1  11.8eV
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 m3 and its atomic weight is 27.)
 2  T
  0  1  
 12  TF



2





 2 
1000 K


   0  1 
11.8eV
 12 


 8.62 10 5 eVK 1
  0  1  4.38  10 5

less than 0.01%







2





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
T
Ce  Nk 
2  TF

 2
 kT 
 
Nk  
 2
 F 
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.
2  10 30
No. of electrons 
 1.205  10 57
 27
1.66  10
1
# of electrons  of nucleous
2
N 1.205 2  10 57

V
1.4  10 27
h 2  3N 
F 


2me  8V 
2
3
 1.205 7.23  10 27 

 F  0.33me v  

27 
1
.
26
1
.
4

10


 F  0.33me v  6.248  10 5
 F  20.6eV
2
3
(b) What is the Fermi temperature?
F
20.6eV
5
TF 

 2.39  10 k
5
1
k 8.62  10 eVk
(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.