19.2) consider a kilomole of 3 He gas atoms under STP conditions
No. of 3 He atoms at STP = N = 1 kilomole = 6.023  10 26
Volume at STP =V = 22.4 dm 3
Mass of 3 He atom = m =
3
 27

4
.
98

10
6.023  10 26
a) What is the Fermi temperature of the gas?
Fermi Temperature = TF  ?
We know that
h2  N 
TF 


2mk  1.504v 
2
3
6.626  10 
 6.023  10 26 



2  4.98  10  27  1.38  10  23  1.504  22.4 
 0.069 K  0.07 K
Fermi Temperature  TF  0.07 K
 34 2
b) Calculate 
kT
and exp    
kT 

2
3

?
kT
we knowthat
  2mkT  3 2 V 

  ln 2

2
kT
 N 
  h
3
 

 27
 23
 2
2


4
.
98

10

1
.
38

10

273
22.4




  ln 2
26 
34 2

 
6
.
023

10
6.626  10

 




kT
 12.7
exp      e
kT 


e

kT

kT
?
 3.28  10 5
 
c) Find the average occupancy f   of a single particle state that has energy of 3 kT
2
Energy :   3 kT
2
Chemical Potential :   12.7kT
Average Occupancy : f    ?
We knowthat
f   
   
e
1
KT
1
1
 3
12.7 kT
2
kT
e
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 f  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
N  
1
    
g   e kT  1
where     
1
f    
e kT  1
for the probability of absent at     
1
1
1  f       
 1  
e kT  1
e kT  1
f   
1  f   
e
kT

1
1


1 1 1
1  e kT

e kT
therefore they are equal to each other
e


kT
19.4a) Verify that the average energy per fermion is

a direct calculation of U 0
Use equation 19.18
35 
F
at absolute zero by making
N
3
U  N F for T  0
5
U
3
  T o    F
N
5
b) Similarly, prove that the average speed of a fermion gas particle at T=0 is
where the Fermi velocity
VF
v  f v   g v   dv
0
N

v
is defined by
 
2
 F  3 5 mvF
3 4 
VF
,
f v  
N v 

g v 
1
1 2

 mv   
2

kT
e
1
g    d  g v   dv
8 2V 3 2  1 2  2 1
g v   dv 
m   mv   mv  dv
h3
2
 2
1
1
1 2 2 1
 mv   mv  dv
8 2V 3 2   2
 2
v
m 
1

3
 mv  

0
h
2

kT



e
1
2
8 2V
v
h3
3
1 2 3
  m
0
2
v3
1 2
 mv  
2
kT







dv
e
1
need to use int egration in part !
19.6) At very low temperature, the electronic specific heat capacity of a metal is
approximately ce  AT where A can be determined by experiment. For gold,
the constant A is found to be A  0.73J kilomole K . Compare this to the
value obtained from Equation (19.20). (The atomic weight of gold is 197 and its
3
3
density is18.9  10 kgm )
1
1
From eqn (19.20), one identifies
 2 Nk

2
TF

1
1
6.63  1034   3  18.9  103

with TF   F  

 6.02  1026 

5
1
31
k
8.62  10 eVk
2  9.11 10  8  197

TF  62028k
2
A
 2 6.02  1026  1.381 1023 Jk 1

2
62028k
1
 0.66 J  kilomole K 2
19.7 a) Calculate
F
for aluminum assuming three electrons per aluminum atom.
3 Kg
N 2.69 10
m 3  6.02 10 26 atoms  5.99 10 28

V
kilo  mole
27 Kg
kilomole
N
# Density for electrons  3 
V
 1.8 10 29
F
6.63 10 

 3 1.8 10 29 


2  9.1110 31 
8

34 2
b) Show that the aluminum at T=100 K, μ differs from
aluminum is
2.69 10 kgm
3
3
2
3
 5.6  2.1  11.8eV
 F by less than 0.01%. (The density of
and its atomic weight is 27.)
2
3
 2
   0  1 
 12
T

 TF



2





 2 
1000k

   0  1 
11.8eV
 12 


 8.62  10 5 eVk 1

  0  1  4.38  10 5








2






less than 0.01%
c) Calculate the electronic contribution to the specific heat capacity of aluminum at room
temperature and compare it to 3R
 T  2
 kT 
Nk   
Nk  
2
 TF  2
F 
 8.617  10 5 eVk 1  2.43  10 3 


  3  n  R  
2


1
 297 J  kilomole  k
Ce 

19.8) In sodium there are approximately 2.6  10 conduction electrons per cubic meter,
which behave as a free electron gas. Give an approximate value for the electronic specific
heat of sodium at room temperature. (The atomic weight of sodium is 23.) In the series
expansion for c e , show that at this temperature the cubic term is negligible compared
with the linear term.
26
N
 2.6  1028
V
6.63  10 
28
 3  2.6  10 
F 


2  9.11 1031 
8

 5.6eV  0.579
 3.243eV
 34 2

2
3
 k  298  
Ce  Nk 
R  0.008

2  3.243  2
 325J kilomole1k 1
2
19.10) calculate the isothermal compressibility of the fermion gas consisting of the free electrons
in silver. Compare your answer with the experimental value for silver of
0.99 1011 Pa 1
1  2v 

V  2 p T
   
from eq 19.24
P
2 NkTF
5 V
2
h2  N  3  1 
TF 

  
2mk  1.504   V 
2
2
3
eq 19.8
2
h2  N  3  1 
P  Nk 

  
5
2mk  1.504   V 
2
Nh 2  N  3 1
P

  5
5m  1.504  V 3
5
3
2
V
5
3
Nh 2  N  3 1


 
5m  1.504 
p
3
 Nh 2  N  3  5 1
V 

   35
5

m
1
.
504

 
p

2
3
5
dv  Nh 2  N  3 
 3  58


    P
dp  5m  1.504  
 5
2
3
5
1  Nh 2  N  3 
 3  58
   

    P
v  5m  1.504  
 5
3
this equation leads to   P with P  2.1  1010 Pa from textbook
5
3
3
  P  2.1 1010 Pa   1.26  1011 Pa 1
5
5
2
19.12) For the white dwarf star Sirius B, do the computation leading to Rmin  7  10 6 m
Use Equation (19.28) and (19.30) with the appropriate parameter value.
Rmin  7  10 6 m
We knowthat
2a
19.32
b
calculatin g b
Rmin 
3
b  GM 2 19.30
5
G  Gravitational energy  6.673  10 11
M  Mass of the sun  2.09  10 30 kg

3
b  6.673  10 11 2.09  10 30
5
b  1.75  10 50

2
calculatin g a
2
3h 2  9  3 5 3
a

 N
10me  32 2 
calculatin g N


5 1.82  10 22 7.23  10 20
 6.10  10 56
2
5.33  10 14
so value of a is
N


2
2
3 6.626  10 34  9  3
a
6.10  10 56
31 
2 
10  9.1  10
 32 
a  1.45  10 37  0.0933  4.388  10 94
a  5.92  10 56
put values of a and b in 19.32
Rmin


2 5.92  10 56
1.75  10 50
 6.77  10 6  7  10 6
Rmin 


5
3
19.13) consider the collapse of the sun into a white dwarf. For the sun, M= 2  10
R= 7  10
8
30
kg
m , V=1.4 1027 m3
a) Calculate the Fermi energy of the sun’s electrons
No. of electrons 
2  10 30
 1.205  10 57
1.66  10  27
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


5
 F  0.33me v  6.248  10
2
3
 F  20.6eV
b) What is the Fermi Temperature?
TF 
F
k

20.6eV
 2.39  105 k
5
1
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.
N
1.205 2  1057
31
Density  9.11 10   9.11 10 
V
1.4  1027
 0.39 kg 3
m
31