1000 Solved Problems in Modern Physics

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

++νe (2)

64

29 Cu+e

−→ 64

28 Ni+νe (3)

The energy released for the three processes are as follows:

Q 1 =mcu−mzn=(63. 929759 − 63 .929145)× 931. 5 = 0 .572 MeV

Q 2 =mcu−mNi− 2 me=(63. 929759 − 63 .927959)× 931. 5 − 2 × 0. 511

= 0 .655 MeV

Q 3 =mcu−mNi=(63. 929759 − 63 .927959)× 931. 5 = 1 .677 MeV

We calculate the recoil energy in the process (3) and show that it is

negligible.

TNi+Tν=Q= 1 .677 (energy conservation) (4)

PNi=Pν

orPNi^2 = 2 MNiTNi=Pν^2 =Tν^2 (5)

Solve (4) and (5) to obtainTNi=6eV

8.21 Q=(MAl−MSi)×^931.^5

=Emax+Eγ

=(27. 981908 − 27 .976929)× 931. 5

= 4 .638 MeV

=Emax+Eγ

Therefore,Eγ=Q−Emax= 4. 638 − 2. 865

= 1 .773 MeV

Fig. 8.10β−decay of^2813 Al
followed byγdecay