Answer
$11\text{ years}$.
Work Step by Step
$P\left( t \right)={{P}_{0}}{{e}^{-kt}},k>0$.
The exponential decay rate of krypton-85 is $6.3\%$ per year. Therefore, $k=0.063$.
Now, the half-life is the amount of time necessary for half of the quantity to decay. So, if the initial quantity at time 0 is ${{P}_{0}}$, then half the quantity is $\frac{1}{2}{{P}_{0}}$.
Substitute $k=0.063$ and $P\left( t \right)=\frac{1}{2}{{P}_{0}}$ in $P\left( t \right)={{P}_{0}}{{e}^{-kt}}$ and solve for $t$ as follows:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{-kt}} \\
& \frac{1}{2}{{P}_{0}}={{P}_{0}}{{e}^{-0.063t}} \\
\end{align}$
Divide by ${{P}_{0}}$ on both sides:
$\begin{align}
& \frac{\frac{1}{2}{{P}_{0}}}{{{P}_{0}}}=\frac{{{P}_{0}}{{e}^{-0.063t}}}{{{P}_{0}}} \\
& \frac{1}{2}={{e}^{-0.063t}} \\
\end{align}$
Take the natural logarithm on both the sides:
$\ln \frac{1}{2}=\ln {{e}^{-0.063t}}$
Use the formula $\ln {{m}^{n}}=n\ln m$ in $\ln {{e}^{-0.063t}}$:
$\begin{align}
& \ln \frac{1}{2}=\ln {{e}^{-0.063t}} \\
& =-0.063t\ln e
\end{align}$
$\begin{align}
& \ln \frac{1}{2}=-0.063t\ln e \\
& -0.6931\approx -0.063t \\
\end{align}$
$\begin{align}
& \frac{-0.6931}{-0.063}\approx \frac{-0.063t}{-0.063} \\
& 11\approx t
\end{align}$
