Relativistic analysis of the beta decay of bismuth 210

83Bi210 ^ 84Po210 + e- + v e- end-point energy = 1.16 MeV

A feature of beta decay is that there are three particles present after the decay: the emitted electron, the recoil nucleus and a neutrino (a 'three-body final state'). As a result, the energy of any of these decay products is not unique. We consider here the special case in which the neutrino takes a negligible part of the energy available. The process can then be considered as a simple projectile-recoil 'two-body final state', with the electron at its maximum possible energy (the 'end-point energy' of the beta ray spectrum).

The electron and recoil nucleus go off with equal and opposite momenta:

Rest energy of electron = m0 = 0.511 MeV/c2

Total energy of electron = rest energy + KE = 0.5511 + 1.16

Gamma factor for electron = total energy =

rest energy m0

0.511

(Momentum of electron)2 = E2 - m2 = 1.6712 - 0.5112

Momentum of electron = 1.591 MeV/c

Speed of electron = c^ 1 - -1 c = (1 - 0.30582) c = 0.952 c

= 1.591 MeV/c (checks out!) = Momentum of 84Po210 recoil

Rest energy of 84Po210 recoil = m0 = 209.9829 u

Total energy of 84Po210 =V m2 + p 2 = 1.956 x 105 MeV/c2

Gamma factor for 84Po210 = 1.0000

P 1 591

Speed of 84Po210 recoil = — = — _ = 8.1 x 10-6 c m0 1.956 x 105

Ratio of rest masses of 'gun' to 'bullet' = 3.8 x 105 Ratio of speeds of 'bullet' to 'gun' = 1.19 x 105

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