Chemistry - Chemical Kinetics Question with Solution | TestHub

Given decay constant $\lambda =10^{-1}{\text{ year}}^{-1}=0.1{\text{ year}}^{-1}$.

Time $t=13.86\text{ years}$.

The half-life of the radioactive substance is given by $T_{1/2}=\frac{\ln 2}{\lambda }=\frac{0.693}{0.1}=6.93\text{ years}$.

The number of half-lives elapsed in time $t$ is $n=\frac{t}{T_{1/2}}=\frac{13.86}{6.93}=2$.

The amount of substance remaining after $n$ half-lives is $N=\frac{N_0}{2^n}=\frac{N_0}{2^2}=\frac{N_0}{4}=0.25N_0$.

The percentage of substance remaining is $25\%$.

The percentage of substance that has undergone decay is $100\%-25\%=75\%$.

Answer: $75$