Question

In an ore containing uranium, the ratio of ${}^{238}U$ to ${}^{206}Pb$ nuclei is $3$. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of ${}^{238}U$. Take the half-life of ${}^{238}U$ to be $4.5\times {10}^{9}years$.

Answer 1

Number of atoms of ${}^{238}U$ initially / Number of atoms of ${}^{238}U$ finally = $\frac{4}{3}=\frac{a}{(a-x)}$

[∵ Initially one part lead is present with three parts Uranium]

∴ $t=\frac{2.303}{\lambda log\frac{\alpha }{(\alpha –x)}}=\frac{2.303\times 4.5\times {10}^9}{0.693log4/3}$

=$1.868\times {10}^9$ years.