32. From the expression for
n
(
K
) given we may write
n
(
K
)
∝
K
1/2
e
– K
/
kT
. Thus, with
k
= 8.62
×
10
– 5
eV/K = 8.62
×
10
– 8
keV/K,
we have
avg
1/2
()
/
87
avg
avg
( )
5.00keV
5.00keV 1.94keV
exp
(
)
1.94keV
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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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