Question

${}^{238}\text{U}$ decays to ${}^{206}\text{Pb}$ with a half-life of $4.47\times {10}^9$. This happens in a number of steps. Can you justify a single half-life for this chain of process ? A sample of rock is found to contain 2.00 mg of ${}^{238}\text{U}$ and 0.600 mg of ${}^{206}\text{Pb}$. Assuming that all the lead has come from uranium, find the life of the rock.

Answer 1

Half life period can be a single for all the process. It is the time taken for $\frac{1}{2}$ of the uranium to convert to lead.

No. of atoms of

${\text{U}}^{238}=\frac{6\times {10}^{28}\times 2\times {10}^{-3}}{238}$

=$\frac{12}{238}\times {10}^{20}$

=$0.05042\times {10}^{20}$

Initially total no. of uranium atoms

$=(\frac{12}{238}+\frac{3.6}{206})\times {10}^{20}$

= $0.06789$

$\text{N}=N_0{e}^{-\lambda t}$

$\Rightarrow N=e^{\frac{-0.098}{t_{\frac{1}{2}}}}$

$\Rightarrow 0.05042=0.06789=e^{\frac{-0.691}{4.47\times {10}^9}}$

$\Rightarrow \text{log}\left(\frac{0.05042}{0.06789}\right)=\frac{-0.6931}{4.47\times {10}^9}$

$t=1.92\times {10}^9\text{years}$