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A proton, within a nucleus decays into a neutron during a process called $β^{+}$ decay which occurs is some unstable nuclei. This process is represented by the equation
$_{Z}^{A} X \to_{Z -}_{1}^{A} Y + e^{+} + V_{e}$
Let $m (_{Z}^{A} X)$ represent the mass of an atom of $_{Z}^{A} X,$ and c represent the velocity of light in vaccum. Assume that $v_{e}$ is massless. The Q value of the process is given by

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

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

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

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

answer is B.

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Detailed Solution

$Q = c^{2} [m_{n u m} (_{Z}^{A} X) - m_{n u m} (Z -_{1}^{A} Y) - m_{e +}]$

By definition.

Putting,

$m_{n u c} (_{Z}^{A} X) = m (_{Z}^{A} X) - Z m_{e +}$

and similarly for $Z -_{1}^{A} Y,$

we get the final result.

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

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