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 $(ln1.0058=5.78\times {10}^{-3})$

• 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

The correct option is B $38\times {10}^{6}years$

$N=N_o{e}^{-\lambda t}$
Given: $\frac{x}{1-x}=0.0058$ $\Rightarrow x=0.005766$
$1-0.005766=e^{-\frac{ln2}{4.5\times {10}^9}t}$

$\Rightarrow t=3.75\times {10}^7$ years