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SAMPLE PROBLEM (Class 24)
Ans: As the hint says:
E
n
EN
E dE
av
=
∫
1
0
(
)
.
The limits on the integral are not given in the hint, but the integral should extend over the
entire range of electron energies; at
T
= 0, that is from
E
= 0 to
E
=
E
F
. Then, the limits and
the expression for
N
0
(
E
) can be substituted, and
E
n
EN
E dE
n
E
m
h
E
dE
n
m
h
E
dE
av
E
E
E
F
F
F
=
∫
=
∫
=
∫
1
1
8
2
1 8
2
0
0
3 2
3
12
0
3 2
3
3 2
0
(
)
/
/
/
/
π
This is an easy integral to carry out, but the constants are messy. This can be simplified if
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Unformatted text preview: n is replaced by the expression for n given on page 1046, just above equation 429. E n m h E dE m h E m h E dE E E dE E E E E E E av E F E F E av F E F F F F F F F = = ∫ = ∫ ∫ = = = 1 8 2 1 8 2 2 3 8 2 3 2 3 2 2 5 3 2 2 5 3 5 3 2 3 3 2 3 2 3 3 2 3 2 3 3 2 3 2 3 2 3 2 5 2 3 2 5 2 / / / / / / / / / / / /...
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This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler

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