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\(_{92}^{238}U\) is known to undergo radioactive decay to form \(_{82}^{206}Pb\) by emitting alpha and beta particles. A rock initially contained 68 × 10–6 g of \(_{92}^{238}U\) . If the number of alpha particles that it would emit during its radioactive decay of \(_{92}^{238}U\) to \(_{82}^{206}Pb\) in three half-lives is Z × 1018 , then what is the value of Z ?

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1 Answer

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by (40.2k points)

In decay from U238 to Pb206, each U238 atom decays and produces 8 \(\alpha\)-particles and hence, total number of \(\alpha\)-particles emitted out

= 1.204 x 1018

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