How many alpha and beta particles are emitted | vedclass

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Question

(A) Let $n$ be the number of alpha particles and $m$ be the number of beta particles emitted.The decay process is: ${}_{92}U^{238} \rightarrow {}_{82}

Vedclass · Accepted Answer

$8$ alpha particles and $6$ beta particles

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(A) Let $n$ be the number of alpha particles and $m$ be the number of beta particles emitted.The decay process is: ${}_{92}U^{238} \rightarrow {}_{82}Pb^{206} + n({}_{2}He^{4}) + m({}_{-1}e^{0})$.Equating the mass numbers: $238 = 206 + 4n \implies 4n = 32 \implies n = 8$.Equating the atomic numbers: $92 = 82 + 2n - m$.Substituting $n = 8$: $92 = 82 + 2(8) - m \implies 92 = 82 + 16 - m \implies 92 = 98 - m \implies m = 6$.Thus,$8$ alpha particles and $6$ beta particles are emitted.

How many alpha and beta particles are emitted when Uranium ${}_{92}U^{238}$ decays to lead ${}_{82}Pb^{206}$?

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