Km
v
m
kT
m
kT
vp
p
,
==
F
H
G
I
K
J
=
1
2
1
2
2
2
2
which is twice as large as that found in part (a). Thus, at
T
= 1.5
×
10
7
K we have
K
v,p
=
1.3 keV, which is indicated in Fig. 4310 by a single vertical line.
54. (a) Rather than use
P
(
v
) as it is written in Eq. 1927, we use the more convenient
nK
expression given in Problem 4332. The
n
(
K
) expression can be derived from Eq. 1927,
but we do not show that derivation here. To find the most probable energy, we take the
derivative of
n
(
K
) and set the result equal to zero:
dn K
dK
n
kT
K
K
kT
e
KK
KkT
p
p
()
.
,
//
/
/
=
−
=
=−
F
H
G
I
K
J
=
113
1
2
0
32
12
which gives
Kk
T
p
=
1
2
. Specifically, for
T
= 1.5
×
10
7
K we find
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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, Energy

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