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 $\left(\ln 1.0058=5.78 \times 10^{-3}\right.$ )

• 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