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## Q. 12.19

A bone suspected to have originated during the period of the Roman emperors was found in Great Britain. Accelerator techniques gave its ${ }^{14} C /{ }^{12} C \text { ratio as } 1.10 \times 10^{-12}$. Is the bone old enough to have Roman origins?

Strategy Remember that the initial ratio of ${ }^{14} C /{ }^{12} C$ at the time of death was $R_{0}=1.2 \times 10^{-12}$. We use the radioactive decay law to determine the time t that it will take for the ratio to decrease to $1.10 \times 10^{-12}$.

## Verified Solution

The number of ${ }^{14} C \text { atoms decays as } e^{-\lambda t}$.

$N\left({ }^{14} C \right)=N_{0} e^{-\lambda t}$

The ratio of ions is given by

$R=\frac{N\left({ }^{14} C \right)}{N\left({ }^{12} C \right)}=\frac{N_{0}\left({ }^{14} C \right) e^{-\lambda t}}{N\left({ }^{12} C \right)}=R_{0} e^{-\lambda t}$

where $R_{0}$ is the original ratio. We can solve this equation for t.

\begin{aligned}e^{-\lambda t} &=\frac{R}{R_{0}} \\t &=\frac{-\ln \left(R / R_{0}\right)}{\lambda}=-t_{1 / 2} \frac{\ln \left(R / R_{0}\right)}{\ln (2)} \\&=-(5730 y )\left(\frac{-0.087}{0.693}\right)=720 y\end{aligned}

where we have inserted the known values of $t_{1 / 2}, R_{0}$, and R to find the age of the bone. The bone does not date from the Roman Empire, but from the medieval period.