A radioactive nucleus AZX undergoes spontaneous decay in the sequence AZX → Z-1B → Z-3C → Z-2D where Z is the atomic number of element. The possible decay particles in the sequence are (a) α, β⁺, β (b) β⁺, α, β (c) β, α, β⁺ (d) α, β, β⁺

Answer: (b). Z → Z-1: β⁺ emission. Z-1 → Z-3: Z falls by 2, so α emission. Z-3 → Z-2: Z rises by 1, so β⁻ emission. So the answer is (b) β⁺, α, β.

A radioactive nucleus AZX undergoes spontaneous decay in the sequence… | MHT CET 2026 (15 May, Shift 2)

Q1MHT CET 2026 · 15 MAY · SHIFT 2Medium

Radioactive Decays: Alpha, Beta, Gamma

A radioactive nucleus ${}^{A}_{Z}X$ undergoes spontaneous decay in the sequence

${}^{A}_{Z}X \rightarrow {}^{\square}_{Z-1}B \rightarrow {}^{\square}_{Z-3}C \rightarrow {}^{\square}_{Z-2}D$ where $Z$ is the atomic number of element. The possible decay particles in the sequence are

Show answer and solution

Answer: (b)

$Z \to Z-1$ : $\beta^+$ emission.
$Z-1 \to Z-3$ : $Z$ falls by 2, so $\alpha$ emission.
$Z-3 \to Z-2$ : $Z$ rises by 1, so $\beta^-$ emission.
So the answer is (b) $\beta^{+}, \alpha, \beta$ .

Report a mistake

Radioactive Decays: Alpha, Beta, Gamma question from MHT CET 2026 (15 May, Shift 2)

More questions on Radioactive Decays: Alpha, Beta, Gamma