Question

In a radioactive decay chain, the initial nucleus is ${}_{90}^{232}\text{Th}$. At the end there are $6\alpha -$particles and $4\beta -$particles which are emitted. If the end nucleus is ${}_Z^A\text{X},A$ and $Z$ are given by:

• A. $\text{A = 200 ; Z = 81}$
• B. $\text{A = 208 : Z = 82}$
• C. $\text{A = 208; Z = 80}$
• D. $\text{A = 202 ; Z = 80}$

Answer 1

The correct option is B $\text{A = 208 : Z = 82}$

When one $\alpha -$particle is emitted, then the mass number $\left(A\right)$ of daughter nuclei decreases by $4$ and the atomic number decreases by $2$.

${}_{90}^{232}\text{Th}\to {}_{78}^{208}\text{Y}+6\left({}_2^4\text{He}\right)$

When one $\beta -$particle is emitted, then the mass number $\left(A\right)$ of daughter nuclei increases by $1$ and the atomic number remains the same.

${}_{78}^{208}\text{Y}\to {}_{82}^{208}\text{X}+4\beta$

Therefore, for the end nucleus, $\text{A = 208 : Z = 82}$

Hence, option $\left(C\right)$ is correct.