₈₈Ra²²⁶ is converted into ₈₂Pb²⁰⁶ by emission of alpha ( α) and beta ( β) particles. The number of alpha and beta particles emitted are respectively (a) 5, 4 (b) 4, 5 (c) 6, 4 (d) 4, 6

Answer: (a). Only α changes A: nα = (226 - 206)/(4) = 5 After 5α: Z = 88 - 10 = 78. Each β⁻ raises Z by 1: nβ = 82 - 78 = 4 So the answer is (a) 5, 4.

₈₈Ra²²⁶ is converted into ₈₂Pb²⁰⁶ by emission of alpha ( α) and beta… | MHT CET 2025 (22 Apr, Shift 1)

Q5MHT CET 2025 · 22 APR · SHIFT 1Easy

Radioactive Decays: Alpha, Beta, Gamma

${}_{88}\text{Ra}^{226}$ is converted into ${}_{82}\text{Pb}^{206}$ by emission of alpha ( $\alpha$ ) and beta ( $\beta$ ) particles. The number of alpha and beta particles emitted are respectively

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Answer: (a)

Only $\alpha$ changes $A$ : $n_\alpha = \dfrac{226 - 206}{4} = 5$
After $5\alpha$ : $Z = 88 - 10 = 78$ .
Each $\beta^-$ raises $Z$ by 1: $n_\beta = 82 - 78 = 4$
So the answer is (a) 5, 4.

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