Question

${}^{238}U$ decays with a half-life of $4.5\times 10^9$ years, the decay series eventually ending at ${}^{206}Pb$, which is stable. A rock sample analysis shows that the ratio of the number of atoms of ${}^{206}Pb$ to ${}^{238}U$ is $0.0058$. Assuming that all the ${}^{206}Pb$ is produced by the decay of ${}^{238}U$ and that all other half-lives on the chain are negligible, the age of the rock sample is $(\ln 1.0058=5.78\times 10^{-3}$ )

• A. $38\times 10^8$ years
• B. $38\times 10^6$ years
• C. $19\times 10^8$ years
• D. $19\times 10^6$ years

Answer 1

Answer: (B)

After a time $t$, a sample of ${}^{238}U$ originally consisting of $N$ atoms will have decayed to $N{e}^{-\lambda t}$. The number of ${}^{206}Pb$ atoms,
$N_{Pb}=N\left(1-e^{-\lambda t}\right)$
$\therefore \frac{N_{Pb}}{N_U}=N\frac{\left(1-e^{-\lambda t}\right)}{N{e}^{-\lambda t}}=0.0058$
$e^{\lambda t}-1=0.0058$
$\Rightarrow e^{\lambda t}=1.0058$
$\therefore t=\frac{1}{\lambda }\ln (1.0058)=\frac{\left(4.5\times 10^9\text{ years }\right)}{\ln 2}\ln (1.0058)$
$=0.0376\times 10^9$ years $=38\times 10^6$ years