The half life of a radioactive substance is 20 minutes. The approximate time interval $(t_2-t_1)$in min, between the time $t_2$ when 2/3 of it has decayed and time $t_1$ and 1/3 of it and decayed is

Given half life T = 20min, and in time $t_{2,}$ the material decays by $\frac{2}{3}$ thus the material left is $\frac{1}{3}$of its original.

From the equation, $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{\frac{t}{T}}$

we get, $\frac{N}{N_0}=\frac{1}{3}={\left(\frac{1}{2}\right)}^{\frac{t_2}{20}}\_\_\_\_\_\_\_\_\_\left(1\right)$

Similarly, $\frac{1}{3}$material is decayed in time $t_1$

$\frac{N}{N_0}=\frac{2}{3}={\left(\frac{1}{2}\right)}^{\frac{t_1}{20}}\_\_\_\_\_\_\_\_\_\left(2\right)$

dividing equation 1 by equation 2,

$\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{2}{3}}}={\left(\frac{1}{2}\right)}^1={\left(\frac{1}{2}\right)}^{\frac{t_2-t_1}{20}}$

$\frac{t_2-t_1}{20}=1$

$t_2-t_1=20min$

Hence the correct answer is 20.