Q 12-13-002NEETNEET 2024Top questionMedium
$$^{290}_{82}X \xrightarrow{\ \alpha\ } Y \xrightarrow{\ e^+\ } Z \xrightarrow{\ \beta^-\ } P \xrightarrow{\ e^-\ } Q$$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are :
Answer: (B) $286,\ 81$
Concept. Track $(A, Z)$ through each decay: $\alpha$: $A - 4$, $Z - 2$. $e^+$ ($\beta^+$): $Z - 1$. $\beta^-$ (same as $e^-$): $Z + 1$. Beta decays leave $A$ unchanged.
Step 1. Follow the chain.
$$X(290, 82) \xrightarrow{\alpha} Y(286, 80) \xrightarrow{e^+} Z(286, 79) \xrightarrow{\beta^-} P(286, 80) \xrightarrow{e^-} Q(286, 81)$$
Answer. $286,\ 81$ (Option B).
Solution by Sreeraj P, M.Sc Physics