Show that the average energy of an electron in a metal is 3 5 at T = 0 K.
Answers
Answered by
0
Answer:
The average energy of the conduction electrons is given by
E
avg
=
n
1
∫
0
∞
EN(E)P(E)dE
where n is the number of free electrons per unit volume, N(E) is the density of states, and P(E) is the occupation probability. The density of states is proportional to E
1/2
, so we may
write N(E)=CE
1/2
, where C is a constant of proportionality. The occupation probability
is one for energies below the Fermi energy and zero for energies above. Thus,
E
avg
n
C
∫
0
E
F
E
3/2
dE=
5n
2C
E
F
5/2
Now
n=∫
0
∞
N(E)P(E)dE=C∫
0
E
F
E
1/2
dE=
3
2C
E
F
3/2
We substitute this expression into the formula for the average energy and obtain
E
avg
=(
5
2C
)E
F
5/2
(
2CE
F
3/2
3
)=
5
3
E
F
Similar questions