Question

If a nucleus $\begin{array}{c}A\\ Z\end{array}X$ emits an $\alpha -$ particle and a $\beta -$ particle, then the daughter nucleus will have the total number of neutrons as $A-Z-n=y.$ The value of $n$ is (integer only)

Answer 1

$\begin{array}{c}A\\ Z\end{array}X\to {}_{(Z-2+1)}{X}^{(A-4)}$

$\because A-Z=y$

For the daughter nuclei,

$\Rightarrow y=(A-4)-(Z-2+1)$

$\Rightarrow y=A-Z-4+1$

$\Rightarrow y=A-Z-3$

$\therefore n=3$