Question

A rock is $12.5\times {10}^{9}years$ old. The rock contains ${}^{238}U$ which disintegrates to form ${}^{206}U$. Assume that there was no ${}^{206}Pb$ in the rock initially and it is the only stable product formed by the decay. The ratio of number of nuclei of ${}^{238}U$ to that of ${}^{206}Pb$ in the rock is $\frac{1}{2^{1/x}-1}$. Find $x$ (Approximately). Half-life of ${}^{238}U$ is $4.5\times {10}^{9}years.(2^{1/3}=1.259)$

Answer 1

${}^{238}U$ atoms decay to form ${}^{206}Pb$ atoms. Let $N_0$ be an initial number of U atoms. After time t, Let $N_U$ be the number of U atoms left. Then

$\frac{N_U}{N_0}={\left(\frac{1}{2}\right)}^n$, where n is number of half-lives

$n=\frac{t}{T}=\frac{12.5\times {10}^{9}years}{4.5\times {10}^{9}years}-\frac{1}{3}$

$\therefore N_U={\left(\frac{1}{2}\right)}^{1/3}{N}_0$ (i)

Number of Pb atoms, $N_{Pb}=N_0-N_U$

Hence, required ratio is: $R=\frac{N_U}{N_{Pb}}=\frac{1}{2^{1/3}-1}=\frac{1}{1.259-1}=\frac{1}{0.259}=3.86$