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 $68\times 10^{-6}g$ of ${}_{92}{}^{238}U$. If the number of alpha particles that itwould emit during its radioactive decay of ${}_{92}^{238}U$ to ${}_{82}{}^{206}Pb$ in three half-lives is $Z\times 10^{18}$, then what is the value of $Z$ ?

Answer 1

Answer: 1.2

${}_{92}^{238}U\to {}_{82}^{206}Pb$
Number of a-particles emitted per nucleii of
${}_{92}^{238}U=\frac{238-206}{4}=8$
Moles of ${}_{92}^{238}U$ present initially $=\frac{68\times 10^{-6}}{238}$
After three half lives, moles of ${}_{92}^{238}U$ decayed $=\frac{7}{8}\times \frac{68\times 10^{-6}}{238}$
$\therefore$ Number of alpha -particles emitted $=8\times \frac{7}{8}\times \frac{68\times 10^{-6}}{238}\times 6.02\times 10^{23}$
$=12.04\times 10^{17}$
$=1.2\times 10^{18}$
$\therefore Z=1.2$