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Chapter 12

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.

Step-by-Step

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.