Half-life of a radioactive substance is 20 m | vedclass

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(B) Given,half-life $T_{1/2} = 20 \ minutes$.We use the radioactive decay law: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{1/2}}$.For $20\%$ dec

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Given,half-life $T_{1/2} = 20 \ minutes$.We use the radioactive decay law: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{1/2}}$.For $20\%$ decay,the remaining amount is $N_1 = 80\%$ of $N_0$,so $\frac{N_1}{N_0} = 0.8 = \left( \frac{1}{2} \right)^{t_1/20}$ ... $(i)$.For $80\%$ decay,the remaining amount is $N_2 = 20\%$ of $N_0$,so $\frac{N_2}{N_0} = 0.2 = \left( \frac{1}{2} \right)^{t_2/20}$ ... $(ii)$.Dividing equation $(i)$ by $(ii)$:$\frac{0.8}{0.2} = \frac{(1/2)^{t_1/20}}{(1/2)^{t_2/20}} = \left( \frac{1}{2} \right)^{(t_1-t_2)/20} = \left( \frac{1}{2} \right)^{-(t_2-t_1)/20} = 2^{(t_2-t_1)/20}$.$4 = 2^{(t_2-t_1)/20}$.Since $4 = 2^2$,we have $2 = \frac{t_2-t_1}{20}$.Therefore,$t_2 - t_1 = 40 \ minutes$.

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

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