Question: Q. 3. $M_{x}$ and $M_{y}$ denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The $Q$-value for a $\beta$-decay is $Q_{1}$ and that for a $\boldsymbol{\beta}^{+}$decay is $Q_{2}$. If $m_{e}$ denotes the mass of an electron, then which of the following statements is correct?
(a) $Q_{1}=\left(M_{x}-M_{y}\right) c^{2}$ and $Q_{2}=\left(M_{x}-M_{y}-2 m_{e}\right) c^{2}$
(b) $Q_{1}=\left(M_{x}-M_{y}\right) c^{2}$ and $Q_{2}=\left(M_{x}-M_{y}\right) c^{2}$
(c) $Q_{1}=\left(M_{x}-M_{y}-2 m_{e}\right) c^{2}$ and $Q_{2}=\left(M_{x}-M_{y}+2 m_{e}\right) c^{2}$
(d) $Q_{1}=\left(M_{x}-M_{y}+2 m_{e}\right) c^{2}$ and $Q_{2}=\left(M_{x}-M_{y}+2 m_{e}\right) c^{2}$
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Solution:
Ans. Correct option : (a)
Explanation: Let the parent nuclei ${ }_{z}^{A} X$ is radioactive atom and decay $\beta^{-}$as :
$$ \begin{aligned} { }{\mathrm{Z}} X^{A} & \rightarrow{ }{z+1} Y^{A}+{ }{-1} e^{0}+\bar{v}+Q{1} \ Q_{1} & =\left[m_{n}\left({ }{z} X^{A}\right)-m{n}\left({ }{z+1} Y^{A}\right)-m{e}\right] c^{2} \ & =\left[m_{n}\left({ }{z} X^{A}\right)+m{\varepsilon} Z-m_{n}\left({ }{z+1} Y^{A}\right)-(Z+1) M{e}\right] c^{2} \ & =\left[m\left({ }{z} X^{A}\right)-m\left({ }{z+1} Y^{A}\right)\right] c^{2} \end{aligned} $$
Let the nucleus $={ }_{z} X^{A}$ radiate $\beta$ decay
$$ \begin{aligned} & { }{z} x^{A} \rightarrow{z-1} Y^{A}+{1} e^{0}+v+Q{2} \ & Q_{2}=\left[m_{n}\left({ }{z} X^{A}\right)-m\left({ }{z-1} Y^{A}\right)-2 m_{e}\right] c^{2} \ & =\left[m_{n}\left({ }{z} X^{A}\right)+m{e} Z-m_{n}\left({ }{z-1} Y^{A}\right)-(Z-1) m{e}-2 m_{e}\right] c^{2} \ & =\left[m\left({ }{z} X^{A}\right)-m{e}\left({ }{z-1} Y^{A}\right)-2 m{e}\right] c^{2} \ & =\left(M_{x}-M_{y}-2 m_{e}\right) c^{2} \end{aligned} $$
