Q. What is the respective number of $\alpha$ and $\beta$ particles emitted in the following radioactive decay $\ _{90}X^{200} \rightarrow \ _{80}Y^{168}$
A) 6 and 8
B) 8 and 8
C) 6 and 6
D) 8 and 6
Explanation
(d) $n_{\alpha} = \frac{A - A'}{4} = \frac{200 - 168}{4} = 8$
$n_{\beta} = 2n_{a} - Z + Z^{'} = 2 \times 8 - 90 + 80 = 6$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.