Question

The number of $\beta$-particles emitted during the transformation of ${}_y^{x}A$ to ${}_n^{m}B$

• A. $\frac{x-m}{4}$
• B. $n+\frac{x-m}{2}+y$
• C. $n+\frac{x-m}{2}-y$
• D. $2y-n+x-m$

Answer 1

Answer: (C)

For the transformation ${}_y^{x}A$ to ${}_n^{m}B$, the complete nuclear reaction is
${}_y^{x}A⟶{}_n^{m}B+p_{2}H{e}^4+q_{-1}{\beta }^0$
[Let $p$ and $q$ are the number of $\alpha$ and $\beta$ particles emitted respectively.]
On comparing atomic mass, we get
$x=m+4p+0$
$p=\frac{x-m}{4}$
On comparing atomic number, we get
$y=n+2p-q$
$y=n+\frac{2(x-m)}{4}-q$
$=n+\frac{x-m}{2}-q$
$\therefore q=n+\frac{x-m}{2}-y$