462
8 Nuclear Physics – II
8.3.10 Nuclear Reactions (General)
8.45 The given decay is
13 N →
13 C + β
+
+ ν + 1.2 MeV
(1)
M N − M C − 2m e = 1.2 MeV
(2)
where masses are atomic and c = 1
For the reaction
p +
13 C →
13 N + n
Q = M p + M C − M N − M n
where the masses are nuclear.
Add and subtract 7m e to get Q in atomic masses
Q = M H + M C − M N − M n = −[(M n − M H ) + (M N − M C )]
Q = −[0.78 + 1.2 + 2m e ] = −3 MeV
where we have used (2) and the mass difference M n − M H = 0.78 MeV and
m e = 0.51 MeV
E threshold = |Q|
1 +
m p
m c
= 3 ×
1 +
1
13
= 3.23 MeV
8.46 Given reaction is
p +
7 Li →
7 Be + n − 1.62 MeV
(1)
M Li + M P − M Be − M n = −1.62 MeV
(2)
where the masses are nuclear.
It follows that
M Be − M Li = 1.62 + 938.23 − 939.52 = 0.33 MeV
Add and subtract 4m e to get the Q mass difference of Be and Li
M Be − M Li − m e = 0.33
or M Be − M Li = 0.33 + m e = 0.84 MeV
where m e = 0.51 MeV. Thus the total energy released in the electron capture
e
−
+
7 Be →
7 Li + ν is 0.84 MeV
This energy is shared between the neutrino and the recoil nucleus. Energy
and momentum conservation give
E N + E υ = 0.84 MeV
(1)
P
2
υ = E
2
υ = P
2
N = 2 M N E N
(2)
Using M N ∼ = 7 amu = 6520.5 MeV, (1) and (2) can be solved to obtain E N =
7.5 keV and E υ ∼ = 0.84 MeV.
8.47 The Q-value is −1.37 MeV. Minimum energy required is
E a = |Q|
1 +
m a
m x
= 1.37
1 +
1
24
= 1.427 MeV