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
${ }_{y}^{x} A \longrightarrow { }_{n}^{m} B +p_{2} He ^{4}+q_{-1} \beta^{0}$
[Let $p$ and $q$ are the number of $\alpha$ and $\beta$ particles emitted respectively.]
On comparing atomic mass, we get
$x=m+4 p+0$
$p=\frac{x-m}{4}$
On comparing atomic number, we get
$y =n+2 p-q$
$y =n+\frac{2(x-m)}{4}-q$
$=n+\frac{x-m}{2}-q$
$\therefore q =n+\frac{x-m}{2}-y$