MHTCET 23 Apr 2025 Shift 2 Paper

Concept:Radioactive emissions change the mass number $A$ and atomic number $Z$ of a nucleus in specific ways depending on the particle emitted.Explanation:Each $\alpha$-particle is a helium nucleus ${}_{2}^{4}\mathrm{He}$, so it decreases mass number $A$ by 4 and atomic number $Z$ by 2.For two $\alpha$-emissions: decrease in $A$ is $2 \times 4 = 8$, and decrease in $Z$ is $2 \times 2 = 4$.After two $\alpha$-emissions: $A = 242 - 8 = 234$ and $Z = 92 - 4 = 88$.The intermediate nucleus becomes ${}_{88}^{234}$.One electron emission (a $\beta^{-}$ particle) converts a neutron into a proton.So atomic number $Z$ increases by 1, while mass number $A$ remains unchanged.After this emission: $Z = 88 + 1 = 89$ and $A = 234$.Each positron emission (a $\beta^{+}$ particle) converts a proton into a neutron, so $Z$ decreases by 1.Two positron emissions decrease $Z$ by 2.After these emissions: $Z = 89 - 2 = 87$ and $A = 234$.Hence, the product nucleus is ${}_{87}^{234}\mathrm{Y}$, giving $P = 87$.Answer:$P = 87$, so the correct option is A.