Q 12-13-051JEE MainJEE Main 2024 (8 Apr, Shift 2)Easy
In a hypothetical fission reaction
$$^{236}_{92}X \to {}^{141}_{56}Y + {}^{92}_{36}Z + 3R$$
the identity of the emitted particles ($R$) is:
Answer: (B) Neutron
Mass number: $236 = 141 + 92 + 3A \Rightarrow A = 1$.
Charge number: $92 = 56 + 36 + 3Z \Rightarrow Z = 0$.
A particle with $A = 1$ and $Z = 0$ is a neutron.
Solution by Sreeraj P, M.Sc Physics