Question

For the ${\beta }^{+}$ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K shell, is captured by the nucleus and a neutrino is emitted).
$e^{+}+{}_Z^{A}X\to {}_{Z-1}^{A}Y+v$
Show that if ${\beta }^{+}$ emission is energetically allowed, electron capture is necessarily allowed but not vice versa.

Answer 1

Consider the competing processes:
${}_Z^{A}W\to {}_{Z-1}^{A}Y+e^{+}+v_e+Q_1$ (positron cature)
$e^{-}+{}_Z^{A}W\to {}_{Z-1}^{A}Y+v_e+Q_2$ (electron capture)

$Q_1=\left[m_N{(}_Z^{A}X\right)-m_N{(}_{Z-1}^{A}Y)-m_e]c^2$
$=\left[m_N{(}_Z^{A}X\right)-Z{m}_e-m{(}_{Z-1}^{A}Y)-(Z-1)m_e-m_e]c^2$
$=\left[m_N{(}_Z^{A}X\right)-m_N\left(A_{Z-1}Y\right)-2m_e]c^2$

$Q_2=\left[m_N{(}_Z^{A}X\right)+m_e-m_N{(}_{Z-1}^{A}Y\left)\right]c^2=\left[m{(}_Z^{A}X\right)-m{(}_{Z-1}^{A}Y\left)\right]c^2$
This means $Q_1>0$ implies $Q_2>0$, but $Q_2>0$ does not necessarily mean $Q_1>0$.