For the

Solution:

Let the amount of energy released during the electron capture process be Q1. The nuclear reaction can be written as:

$e^{+}+{ }_{Z}^{A} X \rightarrow{ }_{Z-1}^{A} Y+v+Q_{1}$     ...(1)

Let the amount of energy released during the positron capture process be Q2. The nuclear reaction can be written as:

${ }_{Z}^{A} X \rightarrow{ }_{Z-1}^{A} Y+e^{+}+v+Q_{2}$      ...(2)

$m_{N}\left({ }_{Z}^{A} X\right)=$ Nuclear mass of ${ }_{Z}^{A} X$

$m_{N}\left({ }_{Z-1}^{A} Y\right)=$ Nuclear mass of ${ }_{Z-1}{ }^{A} Y$

$m\left({ }_{z}^{A} X\right)=$ Atomic mass of ${ }_{z}^{A} X$

$m\left({ }_{z-1}^{A} Y\right)=$ Atomic mass of ${ }_{z-1}^{A} Y$

me = Mass of an electron

c = Speed of light

Q-value of the electron capture reaction is given as:

$Q_{1}=\left[m_{N}\left({ }_{z}^{A} X\right)+m_{e}-m_{N}\left({ }_{z-1}^{A} Y\right)\right] c^{2}$

$=\left[m\left({ }_{z}^{A} X\right)-Z m_{e}+m_{e}-m\left({ }_{z-1}^{A} Y\right)+(Z-1) m_{e}\right] c^{2}$

$=\left[m\left({ }_{Z}^{A} X\right)-m\left({ }_{Z-1}^{A} Y\right)\right] c^{2}$     ...(3)

Q-value of the positron capture reaction is given as:

$Q_{2}=\left[m_{N}\left({ }_{z}^{A} X\right)-m_{N}\left({z_{-1}^{A}} Y\right)-m_{e}\right] c^{2}$

$=\left[m\left({ }_{z}^{A} X\right)-Z m_{e}-m\left({ }_{z-1}^{A} Y\right)+(Z-1) m_{e}-m_{e}\right] c^{2}$

$=\left[m\left({ }_{z}^{A} X\right)-m\left({ }_{z-1}{ }^{A} Y\right)-2 m_{e}\right] c^{2}$      ...(4)

It can be inferred that if Q2 > 0, then Q1 > 0; Also, if Q1> 0, it does not necessarily mean that Q2 > 0.

In other words, this means that if $\beta^{+}$emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically-allowed nuclear reaction.