Beta Decay — Question 4

Question 4

A free neutron can beta decay while a free proton cannot

Use rest energies $m_nc^2=939.565\,\mathrm{MeV}$, $m_pc^2=938.272\,\mathrm{MeV}$ and $m_ec^2=.511\,\mathrm{MeV}$. Neutrino masses are neglected.

1. Find Q for free-neutron beta-minus decay and for hypothetical free-proton beta-plus decay.

2. Find the minimum proton-to-neutron change in nuclear binding energy that would permit beta-plus conversion inside a nucleus, neglecting recoil and electronic binding.

Question 4 - Solution

A free neutron can beta decay while a free proton cannot

(a) Method and calculation

$$\boxed{Q_{n\to pe^-\bar\nu}=939.565-938.272-.511=.782\,\mathrm{MeV}}.$$$$\boxed{Q_{p\to ne^+\nu}=938.272-939.565-.511=-1.804\,\mathrm{MeV}}.$$

(b) Interpretation and further calculation

$$Q_{\mathrm{nucleus}}=Q_{\mathrm{free}}+B_D-B_P.$$$$\boxed{B_D-B_P>1.804\,\mathrm{MeV}}\quad\text{is required for positive }Q.$$

Physical check

Nuclear binding changes the energy balance of a bound nucleon conversion. The proton need not be able to undergo the corresponding decay in isolation.