Question

Half life of a substance is $20$ minutes. What is the time between $33\%$ decay and $67\%$ decay?

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

Answer 1

Answer: (B)

20 minutes

Answer 2

Let N be the number of nuclei at the beginning. Number of undecayed nuclei after 33% decay = 0.67 N Number of undecayed nuclei after 67% decay = 0.33 N Also 0.33 $N_{0}\approx\frac{0.67N_{0}}{2}$ And in one half life the number of undecayed nuclei becomes half. Exact calculation : Let number of nucler at the beginning = N Let the time required for 33% decay = t Then 0.67 N = N $e^{-\lambda t_{1}}$ $\Rightarrow e^{-\lambda t_{1}} =0.67$ ...................(1) Time required for 67% decay = t $\therefore e^{-\lambda t_{1}}$ = 0.33 ...................(2) [Since after 33% decay, 67% will remain and after 67% decay, 33% will remain]. $\therefore\left(2\right)\div\left(1\right) \Rightarrow e^{-\lambda\left(t_{2-t_1}\right)}=\frac{0.33}{0.67}=\frac{1}{2}$ $\therefore-\lambda\left(t_{2}-t_{1}\right)=in\left(\frac{1}{2}\right)$ \therefore-\lambda\left(t_{2}-t_{1}\right)=\frac{in\left(\frac{1}{2}\right)}{\lambda}=-\frac{in\left(\frac{1}{2}\right)\times T_{??}{0.693} $=- \frac{in\left(\frac{1}{2}\right)\times20}{-in\left(\frac{1}{2}\right)}=20$ minutes