Question

What is the respective number of $\alpha$ and $\beta$ particles emitted in the following radioactive decay?

${}^{200}{X}_{90}{\to }^{168}{Y}_{80}$

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

Answer 1

The correct option is D

8 and 6

Balance the nucleon numbers and nuclear charge separately to find out the number of $\alpha$ and $\beta$ particles emitted.

200 = 168 + n $\times$ 4 or 32 = n $\times$ 4 or n = 8

and 90 = 80 + 2 $\times$ 8 - m or m = 80 - 90 + 2 $\times$ 8 = 6

$\therefore$ 8 $\alpha$ and 6 $\beta$ particles are emitted.