A radioactive sample contains two radio nuclides A and B having decay constant  $\lambda h{r}^{-1}$ and $2\lambda h{r}^{-1}.$  Initially 25% of total decay comes from A. How long (in hr) will it take before 75% of total decay comes from A.  $[\text{Take,\hspace{0.17em}}\lambda =\ln 3]$

Initially let no. of molecules of  $A=N_A$ and that of  $B=N_B$

$$\begin{gathered} \frac{\lambda N_A}{\lambda N_A+2\lambda N_B}=\frac{25}{100} \\ \therefore 100\lambda N_A=25\lambda N_A+50\lambda N_B\Rightarrow N_A=\frac{2}{3}{N}_B \end{gathered}$$

Now,  $N_A^{\text{'}}=N_A{e}^{-\lambda t}=\frac{2}{3}{N}_B{e}^{-\lambda t}$

$$\begin{gathered} N_B^{\text{'}}=N_B{e}^{-2\lambda t} \\ \therefore \frac{\frac{2}{3}\lambda N_B{e}^{-\lambda t}}{\frac{2}{3}\lambda N_B{e}^{-\lambda t}+2\lambda N_B{e}^{-2\lambda t}}=\frac{75}{100}=\frac{3}{4} \\ \Rightarrow \frac{1}{1+3e^{-\lambda t}}=\frac{3}{4}\Rightarrow 1=9e^{-\lambda t}\Rightarrow e^{\lambda t}=9 \\ So, t=\frac{\ln \left(9\right)}{\lambda }hr=\frac{2\ln \left(3\right)}{\ln \left(3\right)}hr=2hr \end{gathered}$$