Answer
it will take $386.75yr$
Work Step by Step
Given Half life of $^{90}_{38}Sr $ is equal to $T_{1/2}=29.1yr$
Since $T_{1/2}=\frac{0.693}{\lambda}$
So Deacy constant $\lambda=\frac{0.693}{T_{1/2}}$
$\lambda=\frac{0.693}{29.1yr}$
Suppose it takes $T$ time to disappear 99.99% of $^{90}_{38}Sr $release in a nuclear reactor accident.
so After time $T$ only $0.01$% of $^{90}_{38}Sr$ will be present.
Suppose at the time of nuclear accident number of $^{90}_{38}Sr$ nuclei present is $N_{0}$
After time $T$ it will decrease to $N= 0.01$% $\times N_{0}$=$10^{-4}N_{0}$
so applying decay equation for time $T$
$N=10^{-4}N_{0}=N_{0}e^{-\lambda T}$
or $10^{-4}=e^{-\lambda T}$
taking natural logarithm both side on base 10
$ln 10^{-4}$=$lne^{-\lambda T}$
from logarithm base conversion formula $Log_{b} x =\frac{ Log_{a} x}{Log_{a} b}$
$\frac{log_{10}10^{-4}}{log_{10}e}=-\lambda T$
$-4=-\lambda T\times log_{10}e$
$T=\frac{4}{\lambda\times log_{10}e}$
putting the value of $\lambda=\frac{0.693}{29.1yr}$
$ log_{10}e=0.43429$
$T=\frac{4}{\frac{0.693}{29.1yr}\times 0.43429}$
$T=\frac{4\times29.1yr}{0.300966}$
$T=386.75yr$
