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

Initially let no. of molecules of A = N_A and that of B = N_B
$\frac{{\mathrm{\lambda N}}_{\mathrm{A}}}{{\mathrm{\lambda N}}_{\mathrm{A}}+2{\mathrm{\lambda N}}_{\mathrm{B}}}=\frac{25}{100}\therefore 100{\mathrm{\lambda N}}_{\mathrm{A}}=25{\mathrm{\lambda N}}_{\mathrm{A}}+50{\mathrm{\lambda N}}_{\mathrm{B}}\Rightarrow {\mathrm{N}}_{\mathrm{A}}=\frac{2}{3}{\mathrm{N}}_{\mathrm{B}}$
Now, ${\mathrm{N}}_{\mathrm{A}}^{\mathrm{\prime }}={\mathrm{N}}_{\mathrm{A}}{\mathrm{e}}^{-\mathrm{\lambda t}}=\frac{2}{3}{\mathrm{N}}_{\mathrm{B}}{\mathrm{e}}^{-\mathrm{\lambda t}}$

and ${\mathrm{N}}_{\mathrm{B}}^{\mathrm{\prime }}={\mathrm{N}}_{\mathrm{B}}{\mathrm{e}}^{-2\mathrm{\lambda t}}\hspace{0.25em}$

So,
$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\frac{\frac{2}{3}{\mathrm{\lambda N}}_{\mathrm{B}}{\mathrm{e}}^{-\mathrm{\lambda t}}}{\frac{2}{3}{\mathrm{\lambda N}}_{\mathrm{B}}{\mathrm{e}}^{-\mathrm{\lambda t}}+2{\mathrm{\lambda N}}_{\mathrm{B}}{\mathrm{e}}^{-2\mathrm{\lambda t}}}=\frac{75}{100}=\frac{3}{4}\\ \Rightarrow \frac{1}{1+3{\mathrm{e}}^{-\mathrm{\lambda t}}}=\frac{3}{4}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$

$\Rightarrow e^{\mathrm{\lambda t}}=9\text{ So, }\mathrm{t}=\frac{\ln \left(9\right)}{\mathrm{\lambda }}\mathrm{hr}=\frac{2\ln \left(3\right)}{\ln \left(3\right)}\mathrm{hr}=2\mathrm{hr}$