Question

What is the respective number of $\alpha$ and $\beta$ particles emitted in the following radioactive decay ?
${ }_{90} X ^{200} \longrightarrow { }_{80} Y^{168}$

• A. 6 and 8
• B. 8 and 8
• C. 6 and 6
• D. 8 and 6

Answer 1

Answer: (D)

$n_{\alpha}=\frac{A-A'}{4}$
$=\frac{200-168}{4}$
$=8$
$n_{\beta} =2 n_{\alpha}-Z+Z'$
$ =2 \times 8-90+80=6$