Physics - Modern Physics Question with Solution | TestHub

Hints : For $20\%$ decay $\frac{80{\mathrm{N}}_0}{100}={\mathrm{N}}_0{e}^{-\lambda t_1}$ For $80\%$ decay $\frac{20N_0}{100}=N_0{e}^{-\lambda t_2}$ On dividing $\begin{array}{rl}& 4=e^{\lambda \left(t_2-t\right)}\\ & 2\ln 2=\frac{\ln 2}{t_{1/2}}\left(t_2-t_1\right)\\ & \Rightarrow t_2-t_1=2\times 20=40\min \end{array}$