Assume that all the { }^{206} Pb found in a given sample of uranium ore resulted from decay of { }^{238} U and that the ratio of ^{206} Pb / ^{238} U is 0.60. How old is the ore?
Strategy Let N_{0} be the original number of { }^{238} U nuclei that existed. The { }^{238} U nuclei eventually decay to { }^{206} Pb, and the longest time in the radioactive decay chain { }^{238} U \rightarrow{ }^{206} Pb is the half-life of { }^{238} U , t_{1 / 2}=4.47 \times 10^{9} y. The numbers of nuclei for { }^{238} U \text { and }{ }^{206} Pb are then
\begin{aligned}N\left({ }^{238} U \right) &=N_{0} e^{-\lambda t} \\N\left({ }^{206} Pb \right) &=N_{0}-N\left({ }^{238} U \right)=N_{0}\left(1-e^{-\lambda t}\right)\end{aligned}The abundance ratio is
R^{\prime}=\frac{N\left({ }^{206} Pb \right)}{N\left({ }^{238} U \right)}=\frac{1-e^{-\lambda t}}{e^{-\lambda t}}=e^{\lambda t}-1 (12.50)
We can solve Equation (12.50) for t, because we know experimentally the ratio R^{\prime} \text { and the decay constant } \lambda \text { for }{ }^{238} U.
