First slide

Kinetics of nuclear reactions or nuclear chemistry

Question

${}_{92}^{238}U$ is known to undergo radioactive decay to form ${}_{82}^{206}{Pb}$ by emitting alpha and beta particles. A rock initially contained 68 × 10^–6g of ${}_{92}^{238}{Pb}$ . 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 ?

Difficult

Solution

$Initial moles of U^{238} = \frac{68 \times 10^{- 6}}{238} = x$
$Moles of U^{238} decayed in three half-lives = \frac{7}{8} x$
In decay from $U^{238}$ to ${Pb}^{206}$ , each U²³⁸ atom decays and produces 8 $α$ -particles and hence, total number of $α$ -particles emitted out
$\begin{array}{l} = (\frac{7}{8} x) \times 8 \times N_{A} \\ = 7 \times \frac{68 \times 10^{- 6}}{238} \times 6.022 \times 10^{23} \\ = 1.204 \times 10^{18} \end{array}$

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