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## Q. 12.18

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 $R^{\prime} \text { and the decay constant } \lambda \text { for }{ }^{238} U$.

## Verified Solution

The result for t from Equation (12.50) is

\begin{aligned}t &=\frac{1}{\lambda} \ln \left(R^{\prime}+1\right)=\frac{t_{1 / 2}}{\ln (2)} \ln \left(R^{\prime}+1\right) \\&=\frac{4.47 \times 10^{9} y }{\ln (2)} \ln (1.60)=3.0 \times 10^{9} y\end{aligned}

The sample of uranium ore is about 3 billion years old.