Q 12-13-023NEETJEE MainMedium
$^{238}_{92}$U decays to $^{206}_{82}$Pb through a series of alpha and beta-minus decays. The numbers of alpha and beta particles emitted are
Answer: (B) $8$ and $6$
Concept. Only $\alpha$ decays change $A$ (by $4$ each); then $\beta^-$ decays fix $Z$ (each raises it by $1$).
Step 1. Alpha particles. $\dfrac{238 - 206}{4} = 8$. These lower $Z$ by $16$: $92 - 16 = 76$.
Step 2. Beta particles. To reach $Z = 82$: $82 - 76 = 6$ $\beta^-$ decays.
Answer. $8$ and $6$ (Option B).
Solution by Sreeraj P, M.Sc Physics