1) The total mean energy (E) of an ideal gas of N electrons in a volume V at absolute zero is calculated. E is expressed in terms of the Fermi energy and is shown to be proportional to N2/3 when volume is fixed, rather than directly proportional to N. 2) This is because, although there are no interactions between particles, the Fermi energy (and hence accessible energy states) depends on the particle density N/V. As more particles are added at fixed volume, higher energy states must be occupied. 3) E is an extensive quantity, scaling with the number of particles N. But at fixed volume, E is not directly proportional to N as the density of

1. 1
R eif 9.22) A m etal has n conduction electrons per unit volum e each electron havin g spin 2
and
an associated m agnetic m om ent  m . T he m etal is at T = 0 K and is placed in a sm all exte rnal
0
m agnetic field H . T he total energy of the conduction electrons in presence of a m agnetic
field H m ust then be as sm all as possible . U se this fact to find out an explicit expression for
the param agentic susceptibility du e to the spin m agnetic m om entof this con duction electrons.
A 9.22
W hen the field H is turned on,
they energy levels of the parallel spins shift by -  m H
w hile the levels of the antiparallel spin s shift by   m H
T he electrons fill up these states to th e Ferm i energy  as show n in the figure.
T he m agnetic m om ent is determ ined by the num ber of electrons
w hich spill over from antiparallel state s to parallel states
to m inim ize the energy. T his is (  m H )  (  0 ) w here  (  0 ) is the density of states
at the f erm i energy  0 ,C onsequently the m agnetization is
M  2  m  (  0 ) N /V
2
3 N
B ut sin ce  (  0 )  w e hav e
4 0
3 m N
2
M  H
2 0 V
and
 M M 3 n m N
    w here n 
B H 2 0 V

3. R eif 9.16) A n ideal F erm i gas is at rest at absolute zero and has a F erm i energy  .
T he rest m ass of each particle is m . If  denotes the velocity of a m olecule find out  x and  x
2
T he Ferm i-D irac function is
1
F    
1
kT
e
W hen T is absolute zero tem perature   
1
 F    F v 
2
1
B ecause w e can w rite  as a function of v h int  
2
mv
2

v F v  d v
3
x

v x  
 F v  d
3
v

B ecause v x
is equals in any direction for the gas

v F v  d v  0
3
then x

v x 0
1

2 2
W e consider v x
v because the m ation of the electron gas are
3
v f
v F v  d v
2 3
x
v f

2
equals in any direction : v x vf
 F v  d v
3
v f
4
the volum e   v  d v  4  v dv
3 3 2
3
vf vf
4 4
v v
2 2 4
v dv dv 5
3 3 1vf 1
v    
2 0 0 2
x vf vf 3
vf
5v 5
4 v 4 v
2 2 f
dv dv
0 0
1
f  ,f  
2
mvf
2
2 2 
v f  v 
2 2
x
m 5 m

4. R eif 9.17) C onsider an ideal gas of N ele ctrons in a volum e V at absolute zero
(a) C alculate the total m ean energy E of this gas.
(b) E xpress E in term s of the Ferm i energ y .
(c) S how that E is proberly a n ex tensiv e quantity , but that f or a f ix ed v olum e V , E
is not proportional to the num ber of par ticles N of particles in the container . H ow
do you account for this last result desp ite the fact that there is no in ter action potential
betw een the particles?
a) A t zero tem perature, as in the previou s steps w e have all the states up to the Ferm i energy
filled w ith one electron (there is a m ultiplicity of 2 due to t he electron's spin in the density of sta tes).

E    ( )d 
0
1 (4  k )dk
k 2
N=2
  
3
0
8
;w here the f actor 2 in Introduced sin ce electrons hav e tw o spin states
1 1
2m   2m  
2
1  2 m  2 21
 ;k    , dk   2   d 
2
w e hav e k 2 2
    2  
1
2
v v 2 dk 2m 2
 (k )  .4  k dk  .k dk ;  
d
3 2 2
 2  2 
1 3
2
2m 
2
1 1
v (2 m ) v (2 m )
p ( )   
2 2
2 2
 2 3
2
2  2 
T he density of states is given by:-
1 3
3/2  3
v (2 m ) 2v  2m  2 2
N =  (  )d   2
4
2

3   d   3 2   2  
 
0
3
3
v  2m  2 2
 N  2  2 

3   
hence the average e nergy w ill be:
3/2 3
  3/2 5
v (2 m ) 2 v (2 m )
   (  )d   d  
2
E   2 3 2 3
0 2  0 5 
3/2 5
v (2 m )

2
E  2 3
5 

5. b Ferm i E nergy is given by
3
3
v  2m  2 2
N  2  2 

3   
2
1 
3 3
  3
 3
v  2m  2  v  2m  2 
2       
 3 N   2    3 N   2  
2 2
   
2
2
  2 N 
3
   3 
2m  V 
3

2
 d

5

2
2
2 N 
2
E 2
3 3
 0
1
 5
5
 3
5
   3 
N 
2
2
3
 2 10 m  V 
 d
0
c)W e can substitut e the previous expressions for E and th en w e get the follow ing expression:
2
2
E 3 2 N 3
  3 
N 10 m  V 
2
2

3 2 N 3
E   3  N
10 m  V 
w hich m akes E an extensive quantity
, w hen w e rescale N   N , V   V , E   E .
W hen volum e is fixed though, it m eans that the Ferm i E nergy w ill have som e sc aling w ith the
particle num ber, because particles w ill start filling up higher energy levels, and one can see th at
the Ferm i E nergy is an intensive quan tity from the expression for µ .