Q 12-13-006NEETNEET 2021Top questionMedium
A radioactive nucleus $^A_ZX$ undergoes spontaneous decay in the sequence
$$^A_ZX \rightarrow {}_{Z-1}B \rightarrow {}_{Z-3}C \rightarrow {}_{Z-2}D$$
where Z is the atomic number of element X. The possible decay particles in the sequence are :
Answer: (D) $\beta^+,\ \alpha,\ \beta^-$
$Z \to Z - 1$: the atomic number drops by 1, which is $\beta^+$ emission.
$Z - 1 \to Z - 3$: the atomic number drops by 2, which is $\alpha$ emission.
$Z - 3 \to Z - 2$: the atomic number rises by 1, which is $\beta^-$ emission.
Sequence: $\beta^+$, $\alpha$, $\beta^-$.
Solution by Sreeraj P, M.Sc Physics