This preview shows page 1. Sign up to view the full content.
51. (a)
Q
=(5
m
2
H
−
m
3
He
−
m
4
He
−
m
1
H
−
2
m
n
)
c
2
=[5(2
.
014102 u)
−
3
.
016029 u
−
4
.
002603 u
−
1
.
007825 u
−
2(1
.
008665 u)](931
.
5MeV
/
u) = 24
.
9MeV.
(b) Assuming 30
.
0% of the deuterium undergoes fusion, the total energy released is
E
=
NQ
=
µ
0
.
300
M
5
m
2
H
¶
Q.
Thus, the rating is
This is the end of the preview. Sign up
to
access the rest of the document.
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, Energy

Click to edit the document details