Question

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

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

Answer 1

The correct option is D 8 and 6

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