JEE Advanced 2020 · Paper 1 · Chemistry Q17

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ChemistryNumerical+4Question 35 of 54

$\ce{^{238}_{92}U}$ is known to undergo radioactive decay to form $\ce{^{206}_{82}Pb}$ by emitting alpha and beta particles. A rock initially contained $68\times10^{-6}$ g of $\ce{^{238}_{92}U}$ . If the number of alpha particles that it would emit during its radioactive decay of $\ce{^{238}_{92}U}$ to $\ce{^{206}_{82}Pb}$ in three half-lives is $Z\times10^{18}$ , then what is the value of $Z$ ?

Your answer

Answer and worked solution

Official answer: 1.19

α particles per atom. The mass number falls by $238-206=32$ , so each $\ce{^{238}U}$ atom emits $32/4=8$ α particles on its way to $\ce{^{206}Pb}$ .

Atoms decayed.
$N_0=\frac{68\times10^{-6}}{238}\times6.022\times10^{23}=1.721\times10^{17}.$
In three half-lives $1-\frac18=\frac78$ of them decay: $1.506\times10^{17}$ atoms.

α count.
$8\times1.506\times10^{17}=1.20\times10^{18}\implies Z=1.20.$
Answer: 1.20 (key: 1.19 to 1.21)

Topics:Radioactive decay kinetics Mole concept

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JEE Advanced 2020 · Paper 1, Chemistry question 17