Question

A $1.5\times 10^{9}$ years old rock contains ${ }^{238} \mathrm{U}$ which disintegrates to form ${ }^{206} \mathrm{~Pb}$. Assume that there was no lead in the rock initially and it is the only stable product formed by the decay. Calculate the ratio of the number of nuclei of lead to that of uranium. [Given, half-life of $\_{}^{238}U$ = $4.5\times 10^{9}$ years, $2^{1/3}\approx 1.26$ ]

Answer 1

Let $N_{0}$ be the initial number of uranium nuclei. After time $t$ , let $N_{U}$ be the number of uranium nuclei and $N_{Pb}$ be the number of lead nuclei.
The number of half-lives passed is
$n=\frac{t}{T_{half}}=\frac{1 . 5 \times 10^{9}}{4 . 5 \times 10^{9}}=\frac{1}{3}$
$N_{U}=N_{0}\left(\frac{1}{2}\right)^{n}=N_{0}\left(\frac{1}{2}\right)^{1/3}$
$N_{Pb}=N_{0}-N_{0}\left(\frac{1}{2}\right)^{1/3}$
$\frac{N_{Pb}}{N_{U}}=2^{1/3}-1=0.26$