1000 Solved Problems in Modern Physics

466 8 Nuclear Physics – II

8.58 E(He)+Eγ= 5 .3 MeV (energy conservation) (1)

PHe=

√

2 M(He)E(He)=Pγ=Eγ/c (momentum conservation) (2)

OrEγ=

√

2 M(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 × 107

= 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 getr=640 fm

8.59 d+^16 O→n+^17 F− 1. 631
d+^16 O→p+^17 O+ 1. 918
It follows that
mn+mF− 1. 631 =mp+mO+ 1. 918
or mF−mO= 3. 549 −(mn−mp)
= 3. 549 −(mn−mH+me)
= 3. 549 −(0. 782 + 0 .511)
= 2 .256 MeV
Therefore^17 F is heavier than^17 O. Actually^17 F decays to^17 Obyβ+emission.

(^17) F→ (^17) O+β++ν
Q=Eβ(max)+ 2 me=Emax+ 1. 022
Emax=Q− 1. 022 = 2. 256 − 1. 022 = 1 .234 MeV
8.60 In the reactionX(a,b)Yat 90◦,
Q=Eb

(

1 +mmby

)

−Ea

(

1 −mmay

)

=Ep

(

1 + 301

)

− 7. 68

(

1 − 304

)

ForEp= 8 .63 MeV, Q 0 = 2. 262

Ep= 6 .41 MeV, Q 1 =− 0. 032

Ep= 5 .15 MeV, Q 2 =− 1. 334

Ep= 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.