The half-life of a substance is 20 minutes. What is the time interval between 33% decay and 67% decay? 

Given, half-life, $T=$20 min. We know that 

 $\frac{N}{N_0}= {\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$T$}\right.}$

For 33% decay, $\frac{N}{N_0}=\frac{100-33}{100}=\frac{67}{100}$

If ${t_1}_{ }$is the time for 33% decay, then 

 $\frac{67}{100}={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t_1$}\!\left/ \!\raisebox{-1ex}{$T$}\right.}$                       (i) 

For 67% decay, $\frac{N}{N_0}=\frac{100-67}{100}=\frac{33}{100}$

If $t_2$ is the time for 67% decay, then 

$\frac{33}{100}={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t_2$}\!\left/ \!\raisebox{-1ex}{$T$}\right.}$                   (ii)

Dividing (ii) by (i) , we have 

 $\frac{33}{67}={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{${\left(t_2-t_1\right)}_{ }$}\!\left/ \!\raisebox{-1ex}{$T$}\right.}$          (iii) 

Now $\frac{33}{67}\simeq \frac{1}{2}$. Hence Eq.(iii) reduces to 

 ${\left(\frac{1}{2}\right)}^1={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$\left(t_2-t_1\right)$}\!\left/ \!\raisebox{-1ex}{$T$}\right.}$

which gives $1=\frac{t_2-t_1}{T}$ or, $t_2-t_1=T=20 min$