466
8 Nuclear Physics – II
8.58 E (He) + E γ = 5.3 MeV (energy conservation)
(1)
P He =
2M (He) E (He) = P γ = E γ /c (momentum conservation)
(2)
Or E γ =
2M (He) c 2 E (He)
(3)
Use M (He) c
2
= 3, 728 MeV and solve (1) and (3) to find
E (He) = 3.775 × 10
−3 MeV
= 3.78 keV
Heat energy =
3
2
kT =
3
2
× 1.38 × 10
−23
× 1.7 × 10
7
= 3.5 × 10
−16 J = 2.19 × 10
−3 MeV
Equating this to the electrostatic energy
(1.44 × 1 × 1)/r = 2.19 × 10
−3
we get r = 640 fm
8.59 d +
16 O → n +
17 F − 1.631
d +
16 O → p +
17 O + 1.918
It follows that
m n + m F − 1.631 = m p + m O + 1.918
or m F − m O = 3.549 − (m n − m p )
= 3.549 − (m n − m H + m e )
= 3.549 − (0.782 + 0.511)
= 2.256 MeV
Therefore
17 F is heavier than
17 O. Actually
17 F decays to
17 O by β
+ emission.
17 F →
17 O + β
+
+ ν
Q = E β (max) + 2m e = E max + 1.022
E max = Q − 1.022 = 2.256 − 1.022 = 1.234 MeV
8.60 In the reaction X (a, b)Y at 90
◦ ,
Q = E b
1 +
mb
m y
− E a
1 −
ma
m y
= E p
1 +
1
30
− 7.68
1 −
4
30
For E p = 8.63 MeV, Q 0 = 2.262
E p = 6.41 MeV, Q 1 = −0.032
E p = 5.15 MeV, Q 2 = −1.334
E p = 3.98 MeV, Q 3 = −2.543
The energy levels are determined by
E 0 = Q 0 − Q 0 = 0 (ground state)
E 1 = Q 0 − Q 1 = 2.294
E 2 = Q 0 − Q 2 = 3.596
E 3 = Q 0 − Q 3 = 4.805
The energy levels of
30 Si are shown in Fig. 8.11.