Q 12-13-128JEE MainJEE Main 2019 (12 Jan, Shift 2)Easy
In a radioactive decay chain, the initial nucleus is ${}^{232}_{90}\text{Th}$. At the end, there are $6$ $\alpha$-particles and $4$ $\beta$-particles which are emitted. If the end nucleus is ${}^{A}_{Z}\text{X}$, $A$ and $Z$ are given by:
Answer: (A) $A = 208$; $Z = 82$
Each $\alpha$ lowers $A$ by 4 and $Z$ by 2; each $\beta^-$ raises $Z$ by 1.
$$A = 232 - 6\times4 = 208,\qquad Z = 90 - 6\times2 + 4 = 82$$
Solution by Sreeraj P, M.Sc Physics