Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 minutes.

$$\begin{gathered} In time t_1, 33\% decays , means 67\% is present no of nucleus=\frac{2N_0}{3} \\ In time t_2, 67\% decays , means 33\% is present no of nucleus=\frac{N_0}{3} \end{gathered}$$

$\begin{array}{l}\begin{array}{l}{T}_{1/2}=20\min \Rightarrow \frac{\ln 2}{\lambda }=20\min \Rightarrow \lambda =\frac{\ln 2}{20(min)}\end{array}\\ t=\frac{1}{\lambda }\ln \frac{N_o}{N}\Rightarrow t_1=\frac{1}{\lambda }\ln \frac{N_o}{\frac{2N_0}{3}}=\frac{1}{\lambda }\ln \frac{3}{2}\\ \mathrm{\&} t_2=\frac{1}{\lambda }\ln \frac{N_o}{\frac{N_0}{3}}=\frac{1}{\lambda }\ln 3\\ \\ Thus , t_2-t_1=\frac{1}{\lambda }\left(\ln 3-\ln \frac{3}{2}\right)=\frac{1}{\lambda }\ln 2=20\min \\ \end{array}$