18.38
19.2 The calculation of μ(T)
• To obtain these curves, we must determine
μ(T) . The calculation is considerably more
complicated than it was for T = 0. We have
• the number of particles at T = 0 is the same as
at T ≠ 0,
• At high temperatures the fermion gas approximates
the classical ideal gas. In the classical limit,
• The spin degeneracy factor is 2 for fermions.
• For T >> TF ,μ/kT takes on a large negative value and exp(-μ/kT) >>1.
• As an example, consider a kilomole of 3He gas atoms (which are fermions)
at STP. The Fermi temperature is 0.069 K, so that T /TF = 3900. Using eqns
(9.16) and (9.17), we find thatμ/kT = -12.7 and exp(-μ/kT) = 3.3x105. The
average occupancy of single particle states is indeed very small, as in the
case of an ideal dilute gas obeying the M-B distribution.
19.2) consider a kilomole of 3He gas
atoms under STP conditions
(a) What is the Fermi temperature of the
gas?
h  N 
TF 


2mk  1.504v 
2
2
3
6.626 10 
 6.023  10 



 27
 23 
2  4.98  10  1.38  10  1.504  22.4 
 0.069 K  0.07 K
Fermi Temperature  TF  0.07 K
34 2
26
2
3
(b) Calculate


and exp    
kT
kT 

?
kT
we knowthat
  2mkT  3 2 V 

  ln 2

2
kT
h
N

 

3
 

 27
 23
 2
2


4
.
98

10

1
.
38

10

273
22
.
4


  ln 2
2
26

34
 6.023 10 
 
6.626 10

 




kT
 12.7
exp      ?
kT 


kT
 12.7

e
kT
 3.28 105

f  a single
(c) Find the average occupancy
of

particle state that has energy of

3 kT
2
Energy :   3 kT
2
Chemical Potential :   12.7 kT
Average Occupancy : f    ?
We knowthat
1
f       
KT
e
1
1
 3 12.7 kT
e
2
kT
1
1
1.469  10 6
f    6.8  10 7

Average
Occupancy  f    6.8  10 7
19.3) For a system of noninteracting electrons,
show that the probability of finding an
electron in a state with energy ∆ above the
chemical potential μ is the same as the
probability of finding an electron absent from
the with energy ∆ below μ at any given
temperature T
19.4 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)   f ' ( x) g ( x)dx
• As a simple example, consider
ln x
 x 2 dx
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 found 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 m-3 and
its atomic weight is 27.)
  2  T 2 
   0  1    
 12  TF  


 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.
(a) Calculate the Fermi energy of the Sun’s
electrons.
2  10
No. of electrons 
30
1.66  10
 27
 1.205  10 57
1
of nucleous
2
N 1.205 2  10 57

V
1.4  10 27
# of electrons 
h2
F 
2me
 3N 


 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.