8.3 Solutions
455
8.20 Three types of decays are possible
64
29 Cu →
64
30 Zn + β
−
+ ¯
ν e
(1)
64
29 Cu →
64
28 Ni + β
+
+ ν e
(2)
64
29 Cu + e
−
→
64
28 Ni + ν e
(3)
The energy released for the three processes are as follows:
Q 1 = m cu − m zn = (63.929759 − 63.929145) × 931.5 = 0.572 MeV
Q 2 = m cu − m Ni − 2m e = (63.929759 − 63.927959) × 931.5 − 2 × 0.511
= 0.655 MeV
Q 3 = m cu − m Ni = (63.929759 − 63.927959) × 931.5 = 1.677 MeV
We calculate the recoil energy in the process (3) and show that it is
negligible.
T Ni + T ν = Q = 1.677 (energy conservation)
(4)
P Ni = P ν
or P
2
Ni = 2M Ni T Ni = P
2
ν = T
2
ν
(5)
Solve (4) and (5) to obtain T Ni = 6 eV
8.21 Q = (M Al − M Si ) × 931.5
= E max + E γ
= (27.981908 − 27.976929) × 931.5
= 4.638 MeV
= E max + E γ
Therefore, E γ = Q − E max = 4.638 − 2.865
= 1.773 MeV
Fig. 8.10 β
− decay of
28
13 Al
followed by γ decay
455
8.20 Three types of decays are possible
64
29 Cu →
64
30 Zn + β
−
+ ¯
ν e
(1)
64
29 Cu →
64
28 Ni + β
+
+ ν e
(2)
64
29 Cu + e
−
→
64
28 Ni + ν e
(3)
The energy released for the three processes are as follows:
Q 1 = m cu − m zn = (63.929759 − 63.929145) × 931.5 = 0.572 MeV
Q 2 = m cu − m Ni − 2m e = (63.929759 − 63.927959) × 931.5 − 2 × 0.511
= 0.655 MeV
Q 3 = m cu − m Ni = (63.929759 − 63.927959) × 931.5 = 1.677 MeV
We calculate the recoil energy in the process (3) and show that it is
negligible.
T Ni + T ν = Q = 1.677 (energy conservation)
(4)
P Ni = P ν
or P
2
Ni = 2M Ni T Ni = P
2
ν = T
2
ν
(5)
Solve (4) and (5) to obtain T Ni = 6 eV
8.21 Q = (M Al − M Si ) × 931.5
= E max + E γ
= (27.981908 − 27.976929) × 931.5
= 4.638 MeV
= E max + E γ
Therefore, E γ = Q − E max = 4.638 − 2.865
= 1.773 MeV
Fig. 8.10 β
− decay of
28
13 Al
followed by γ decay