Answer
Percentage remaining when $t=1000\,y$= 97.2 %
Percentage remaining when $t=25000\,y$= 48.7 %
Percentage remaining when $t=100000\,y$= 5.61 %
Work Step by Step
Decay constant $\lambda=2.88\times10^{-5}\,y^{-1}$
$t=1000\,y$
Recall that $\ln(\frac{A}{A_{0}})=-\lambda t$ where $A_{0}$ is the amount of sample at the beginning and $A$ is the amount sample after time $t$.
$\implies \ln(\frac{A}{A_{0}})=-(2.88\times10^{-5}\,y^{-1})(1000\,y)=-0.0288$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-0.0288}=0.972$
Percentage of the sample remaining=$0.972\times100\%=97.2\%$
When $t=25000\,y$,
$ \ln(\frac{A}{A_{0}})=-(2.88\times10^{-5}\,y^{-1})(25000\,y)=-0.0288$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-0.72}=0.487$
Percentage of the sample remaining=$0.487\times100\%=48.7\%$
When $t=100,000\,y$ ,
$ \ln(\frac{A}{A_{0}})=-(2.88\times10^{-5}\,y^{-1})(100,000\,y)=-2.88$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-2.88}=0.0561$
Percentage of the sample remaining=$0.0561\times100\%=5.61\%$
