For the (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 $β^{+ +}$ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa.

Solution

Consider the two competing reaction processes:

{}_{Z}^{A}X \to {}_{Z - 1}^{A}Y + e^{+} + v_{e} + Q_{1}

(positron emission)

e^{-} + {}_{Z}^{A}X \to {}_{Z - 1}^{A}Y + v_{e} + Q_{2} (electron capture)

Q-value for the first reaction process is,

Q_{1} = [m_{N} ({}_{Z}^{A}X) - m_{N} ({}_{Z - 1}^{A}Y) - m_{e}] c^{2} = [m ({}_{Z}^{A}X) - {Zm}_{e} - m ({}_{Z - 1}^{A}Y) + (Z - 1) m_{e} - m_{e}] c^{2} = [m ({}_{Z}^{A}X) - m ({}_{Z - 1}^{A}Y) - 2 m_{e}] c^{2}

Q-value for the second reaction is given by,

Q_{2} = [m_{N} ({}_{Z}^{A}X) + m_{e} - m_{N} ({}_{Z - 1}^{A}Y)] c^{2}

= [m ({}_{Z}^{A}X) - m ({}_{Z - 1}^{A}Y)] c^{2}

where,
m_N is the mass of nucleus and,
m denotes the mass of atom.

Therefore, if positron emission is allowed energetically then, electron capture is allowed.

That is, if Q₁ > 0 implies Q₂ > 0 but,
Q₂ > 0 does not necessarily mean Q₁ > 0.

Hence the result.

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