Beta Decay — Question 2

Question 2

A beta-minus endpoint follows from the neutral atomic mass difference

Neutral parent and daughter atomic masses differ by $M_P-M_D=.001500\,\mathrm u$ in a beta-minus transition to the daughter ground state. Use $uc^2=931.5\,\mathrm{MeV}$ and ignore atomic rearrangement and nuclear recoil.

1. Derive the beta-minus Q value from nuclear masses and show how electron rest masses cancel.

2. Find the maximum electron kinetic energy and the neutrino energy when the electron has kinetic energy .400 MeV.

Question 2 - Solution

A beta-minus endpoint follows from the neutral atomic mass difference

(a) Method and calculation

$$Q=[m_P-m_D-m_e]c^2,\quad m_P\simeq M_P-Zm_e,\quad m_D\simeq M_D-(Z+1)m_e.$$$$\boxed{Q=(M_P-M_D)c^2=.001500(931.5)=1.39725\,\mathrm{MeV}}.$$

(b) Interpretation and further calculation

$$\boxed{T_{e,\max}\simeq Q=1.39725\,\mathrm{MeV}}.$$$$\boxed{E_{\bar\nu}\simeq Q-T_e=.99725\,\mathrm{MeV}}.$$

Physical check

Q denotes available kinetic energy, excluding electron rest energy. Most decays share Q between the electron and antineutrino rather than placing all of it in the electron.