Question

The end product of the decay of ${}_{90}^{232}\text{Th}$ is ${}_{82}^{208}\text{Pb}$. The number of alpha and beta particles emitted are

• A. $3,3$
• B. $6,4$
• C. $6,0$
• D. $4,6$

Answer 1

The correct option is B $6,4$

We know that due to one $\alpha$-particle emission, mass number of the nucleus decreases by 4 and atomic number decreases by 2 and due to one $\beta$-particle emission, mass number of the nucleus remains unchanged and atomic number increases by 1.

In the given reaction:

Change in mass number $=208-232=-24$ (only due to $\alpha -$emission)

Hence, number of $\alpha -$ particles emitted $=24/4=6$

Now, due to emission of 6 $\alpha -$ particles, atomic number will be $=90-6\times 2=78$, but the final atomic number is 82 therefore difference in atomic number $=82-78=4$.

As emission of one $\beta -$ particle increases the atomic number by one, hence atomic number will be increased by 6 (upto 82), by the emission of 4 $\beta -$ particles.