Question

${}_{92}^{238}\text{U}$ is known to undergo radioactive decay to form ${}_{82}^{206}Pb$ by emitting alpha and beta particles. A rock initially contained $68\times {10}^{-6}\text{g}$ of ${}_{92}^{238}U.$ If the number of alpha particles that it would emit during its radioactive decay of ${}_{92}^{238}U{\to }_{82}^{206}\text{Pb}$ in three half-lives is $Z\times {10}^{18}$, then what is the value of $Z$?

Answer 1

${}_{92}{U}^{238}{\to }_{82}P{b}^{206}+8_{2}H{e}^4+6_{-1}{\beta }^0\left(antineutrino\right)92U238\to 82Pb206+82He4+6-1\beta 0\left(antineutrino\right)$

Initial mole of $U^{238}=\left[\frac{68\times {10}^{-6}}{238}\right]=x$

Mole of $U^{238}$ decayed in three half-lives $=7x/8$

In decay from $U^{238}$ to $P{b}^{206}$, each $U^{238}$ atom decays and produces $8\alpha$ particles and hence, total number of $\alpha$ particles emitted out $=\left(\frac{7x}{8}\right)\times 8\times N_A$

$=\frac{7}{8}\times \frac{68\times {10}^{-6}}{238}\times 8\times 6.022\times {10}^{23}=1.2046\times {10}^{18}$
$Z=1.21$