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

A, A' - Atomic number of reactant and product respectively
Z, Z' - Mass number of reactant and product respectively
In Alpha decay - loose two protons and two neutrons.
In Beta decay - loose one neutron and gain one proton.
$n_{\alpha }=\frac{A-A^{\prime }}{4}=\frac{200-168}{4}=8$
$n_{\beta }=2n_{\alpha }-Z+Z^{\prime }=2\times 8-90+80=6$