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NN
P
E
N
e
cc
c
EE k
T
c
ec
F
==
+
−
bg
1
,
Hence, from
N
ev
=
N
hc
, we get
N
e
N
e
v
T
c
T
vc
−−
−
+
=
+
FF
11
.
(b) In this case,
F
()
1
c
T
e
−
>>
and
e
T
v
−
−
F
1
. Thus, from the result of part (a),
,
EE
cF
vF
cv
kT
kT
ee
−
≈
or
(
)
2
vc F
EE E k
T
eN
N
−+
≈
. We solve for
E
F
:
ln
.
22
c
v
Fv
c
N
E
k
T
N
⎛⎞
≈+
+
⎜⎟
⎝⎠
32. (a) The number of electrons in the valence band is
P
E
N
e
vv
v
T
v
ev
F
+
−
1
.
Since there are a total of
N
v
states in the valence band, the number of holes in the valence
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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