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

Answer: 2

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, $N_A^{\prime }=N_A{e}^{-\lambda t}=\frac{2}{3}{N}_B{e}^{-\lambda t}$
$N_B^{\prime }=N_B{e}^{-2\lambda t}$
$\therefore \frac{\frac{2}{3}\lambda N_B{e}^{-\lambda t}}{\frac{2}{3}\lambda N_B{e}^{-\lambda t}+2\lambda N_B{e}^{-2\lambda t}}=\frac{75}{100}=\frac{3}{4}$
$\Rightarrow \frac{1}{1+3e^{-\lambda t}}=\frac{3}{4}$
$\Rightarrow 1=9e^{-\lambda t}$
$\Rightarrow e^{\lambda t}=9$
So, $t=\frac{\ln (9)}{\lambda }h=\frac{2\ln (3)}{\ln (3)}h=2h$