39.
(a) The number of electrons in the valence band is
N
ev
= N
v
P (E
v
) =
N
v
e
(E
v
−E
F
)/kT
+ 1
.
Since there are a total of N
v
states in the valence band, the number of holes in the valence band is
N
hv
=
N
v
− N
ev
= N
v
1
−
1
e
(E
v
−E
F
)/kT
+ 1
=
N
v
e
−(E
v
−E
F
)/kT
+ 1
.
Now, the number of electrons in the conduction band is
N
ec
= N
c
P (E
c
) =
N
c
e
(E
c
−E
F
)/kT
+ 1
,
Hence, from N
ev
= N
hc
, we g et
N
v
e
−(E
v
−E
F
)/kT
+ 1
=
N
c
e
(E
c
−E
F
)/kT
+ 1
.
(b) In this case, e
(E
c
−E
F
)/kT
1 and e
−(E
v
−E
F
)/kT
1. Thus, from the result of part (a),
N
c
e
(E
c
−E
F
)/kT
≈
N
v
e
−(E
v
−E
F
)/kT
,
or e
(E
v
−E
c
+2E
F
)/kT
≈ N
v
/N
c
. We solve for E
F
:
E
F
≈
1
2
(E
c
+ E
v
) +
1
2
kT ln
N
v
N
c
.