In the decay chain given below the sequential order in which the radioactive radiation are emitted will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \beta,\ \alpha,\ \gamma \)
Compare the mass number A and the atomic number Z at each step. An alpha decay lowers A by 4 and Z by 2, a beta-minus decay leaves A unchanged and raises Z by 1, and a gamma emission changes neither A nor Z (an excited nucleus, marked by *, drops to its ground state).
From X to Y: A stays the same and Z rises by 1.
\( {}^{A}_{Z}\mathrm{X} \rightarrow {}^{A}_{Z+1}\mathrm{Y} \) → β emission
From Y to P*: A falls by 4 and Z falls by 2, from \( Z+1 \) to \( Z-1 \).
\( {}^{A}_{Z+1}\mathrm{Y} \rightarrow {}^{A-4}_{Z-1}\mathrm{P}^{*} \) → α emission
From P* to P: A and Z are unchanged; only the excitation is lost.
\( {}^{A-4}_{Z-1}\mathrm{P}^{*} \rightarrow {}^{A-4}_{Z-1}\mathrm{P} \) → γ emission
The order of emission is therefore \( \beta,\ \alpha,\ \gamma \).