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31. (a) From Eq. 4129, we know that
NN e
EkT
21
=
−∆
. We solve for
∆
E
:
()
5
1
2
4
1 0.25
ln
8.62 10
eV K
2.7K ln
0.25
2.56 10 eV
0.26
eV.
N
Ek
T
N
m
−
−
−
§·
∆=
=
×
¨¸
©¹
=×
≈
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This note was uploaded on 01/23/2011 for the course ＰＨＹ 214 taught by Professor Timothybolton during the Fall '10 term at Kansas State University.
 Fall '10
 TimothyBolton

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