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

${ }_{92}^{238} U \rightarrow{ }_{82}^{206} Pb$
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 $=8 \times \frac{7}{8} \times \frac{68 \times 10^{-6}}{238} \times 6.02 \times 10^{23}$
$=12.04 \times 10^{17}$
$=1.2 \times 10^{18}$
$\therefore Z=1.2$