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42. (a) We are given the energy release per fusion (calculated in §437:
Q
= 26.7 MeV =
4.28
×
10
– 12
J) and that four protons are consumed in each fusion event. To find how
many sets of four protons are in the sample, we adapt Eq. 4221:
()
23
26
sam
4A
1000g
6.02 10
mol
1.5 10
.
44
1
.
0
g
m
o
l
p
H
M
NN
M
⎛⎞
==
×=
×
⎜⎟
⎝⎠
Multiplying this by
Q
gives the total energy released: 6.4
×
10
14
J. It is not required that
the answer be in SI units; we could have used MeV throughout (in which case the answer
is 4.0
×
10
27
MeV).
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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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