Question

Half life of a radioactive substance is $20$ minutes. The time between $20$ % and $80$% decay will be:

• A. $40$ minutes
• B. $20$ minutes
• C. $25$ minutes
• D. $30$ minutes

Answer 1

The correct option is A $40$ minutes

Let the initial activity of the radioactive substance be $R_o$ and activity be $R$ after time $t$.

Given : $T_{1/2}=20$ min

Using $\lambda t=ln(\frac{R_o}{R})$ where $\lambda =\frac{0.693}{T_{1/2}}$

$\therefore$ $\frac{0.693}{20}t=ln(\frac{R_o}{R})$

OR $\frac{0.693}{20}(t_2-t_1)=ln(\frac{R_1}{R_2})$ .........(1)

Let time be $t_1$ for $20$ % decay i.e $R_1=R_o-0.2R_o=0.8R_o$

Also the time be $t_2$ for $80$ % decay i.e $R_2=R_o-0.8R_o=0.2R_o$

$\therefore$ $\frac{0.693}{20}(t_2-t_1)=ln(\frac{0.8}{0.2})$

$⟹$ $(t_2-t_1)=\frac{20}{0.693}\times ln4=40$ min $(ln4=1.386)$