Q. In $\beta^{+}$ decay process, the following changes take place inside the nucleus
A) $\begin{matrix}
A \\
Z
\end{matrix}X \rightarrow \begin{matrix}
A\ \ \ \ \\
Z_{- 1}
\end{matrix}Y + e^{+} + \gamma$
B) $\begin{matrix}
A \\
Z
\end{matrix}X \rightarrow \begin{matrix}
A\ \ \ \ \ \ \\
{Z +}_{1}
\end{matrix}Y + e^{-} + \overline{r}$
C) $\begin{matrix}
A \\
Z
\end{matrix}X \rightarrow_{Z}^{A}Y + e^{-} + \gamma$
D) $\begin{matrix}
A \\
Z
\end{matrix}X \rightarrow_{Z}^{A}Y + e^{-} + \overline{\gamma}$
Explanation
(a) Nuclides for which $\frac{N}{Z}$ is too small for stability can emit a positron, the electron’s antiparticle which is identical to the electron, but with positive charge. The basic process called $\beta^{+}$ decay.
$_{Z}^{A}X\ \rightarrow_{Z - 1}^{A}Y + \ e^{+} + \gamma\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\gamma =$neutrino$)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.