Question

A radioactive sample contains two radionuclids $A$ and $B$ having decay constant $\lambda h ^{-1}$ and $2 \lambda h ^{-1}$. Initially, $25 \%$ of total decay comes from $A$. How long (in h) will it take before $75 \%$ of total decay comes from $A$. (Take $\lambda=\ln 3$ )

Answer 1

Initially, let the number of molecules of $A$ be $N_{A}$ and that of $B$ be $N_{B}$.
$\frac{\lambda N_{A}}{\lambda N_{A}+2 \lambda N_{B}}=\frac{25}{100}$
$\therefore 100 \lambda N_{A}=25 \lambda N_{A}+50 \lambda N_{B}$
$\Rightarrow N_{A}=\frac{2}{3} N_{B}$
Now, $N_{A}'=N_{A} e^{-\lambda t}=\frac{2}{3} N_{B} e^{-\lambda t}$
$N_{B}'=N_{B} e^{-2 \lambda t}$
$\therefore \frac{\frac{2}{3} \lambda N_{B} e^{-\lambda t}}{\frac{2}{3} \lambda N_{B} e^{-\lambda t}+2 \lambda N_{B} e^{-2 \lambda t}}=\frac{75}{100}=\frac{3}{4}$
$\Rightarrow \frac{1}{1+3 e^{-\lambda t}}=\frac{3}{4}$
$\Rightarrow 1=9 e^{-\lambda t}$
$\Rightarrow e^{\lambda t}=9$
So, $t=\frac{\ln (9)}{\lambda} h =\frac{2 \ln (3)}{\ln (3)} h =2 h$