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27. (a) Setting
E
=
E
F
(see Eq. 429), Eq. 425 becomes
N
(
E
F
)=
8
πm
√
2
m
h
3
µ
3
16
π
√
2
¶
1
/
3
h
√
m
n
1
/
3
.
Noting that 16
√
2=2
4
2
1
/
2
=2
9
/
2
so that the cube root of this is 2
3
/
2
=2
√
2, we are able to
simplify the above expression and obtain
N
(
E
F
)=
4
m
h
2
3
√
3
π
2
n
which is equivalent to the result shown in the problem statement. Since the desired numerical
answer uses eV units, we multiply numerator and denominator of our result by
c
2
and make use of
the
mc
2
value for an electron in Table 383 as well as the
hc
value found in problem 3 of Chapter 39:
N
(
E
F
)=
µ
4
mc
2
(
hc
)
2
3
√
3
π
2
¶
n
1
/
3
=
µ
4(511
×
10
3
eV)
(1240 eV
·
nm)
2
3
√
3
π
2
¶
n
1
/
3
=
(
4
.
11 nm
−
2
·
eV
−
1
)
n
1
/
3
which is equivalent to the value indicated in the problem statement.
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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