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## Q. 12.11

A sample of ${ }^{18} F$ is used internally as a medical diagnostic tool by observing this isotope’s positron decay $\left(t_{1 / 2}=110 min \right)$. How much time does it take for 99% of the ${ }^{18} F$ to decay?

Strategy We use the radioactive decay law, Equation (12.24), to determine the time needed.

$N(t)=N_{0} e^{-\lambda t}$ (12.24)

$N=N_{0} e^{-\lambda t}=N_{0} e^{-\ln (2) t / t_{1 / 2}}$

## Verified Solution

If we want 99% of the initial sample to decay, then only 1% will be left, and $N / N_{0}=0.01$. We then have

$\frac{N}{N_{0}}=0.01=e^{-\ln (2) t / t_{1 / 2}}$

If we take the natural logarithm, we have

\begin{aligned}\ln (0.01) &=-\ln (2)\left(\frac{t}{t_{1 / 2}}\right) \\t &=-\left[\frac{\ln (0.01)}{\ln (2)}\right] t_{1 / 2}=-\left(\frac{-4.61}{0.693}\right)(110 min ) \\&=731 \min =12.2 h\end{aligned}