A nucleus with Z = 92 emits the following in a sequence: α, β⁻, β⁻, α, α, α, α, α, β⁻, β⁻, α, β⁺, β⁺, α Then Z of the resulting nucleus is (a) 76 (b) 78 (c) 82 (d) 74

Answer: (b). Count the emissions: 8 α, 4 β⁻ and 2 β⁺. Each α: Δ Z = -2; each β⁻: +1; each β⁺: -1. Z = 92 - 8(2) + 4(1) - 2(1) = 78 So the answer is (b) 78.

A nucleus with Z = 92 emits the following in a sequence: α, β⁻, β⁻… | JEE Main 2003 (AIEEE)

Q38JEE Main 2003 · AIEEEEasy

Radioactive Decays: Alpha, Beta, Gamma

A nucleus with $Z = 92$ emits the following in a sequence:

$\alpha, \beta^{-}, \beta^{-}, \alpha, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}, \alpha$

Then $Z$ of the resulting nucleus is

Show answer and solution

Answer: (b)

Count the emissions: $8\ \alpha$ , $4\ \beta^-$ and $2\ \beta^+$ .
Each $\alpha$ : $\Delta Z = -2$ ; each $\beta^-$ : $+1$ ; each $\beta^+$ : $-1$ .
$Z = 92 - 8(2) + 4(1) - 2(1) = 78$
So the answer is (b) $78$ .

Report a mistake

Radioactive Decays: Alpha, Beta, Gamma question from JEE Main 2003 (AIEEE)

More questions on Radioactive Decays: Alpha, Beta, Gamma