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(D) The number of undecayed atoms $N$ after time $t$ is given by $N = N_0 \left(\frac{1}{2}ight)^{\frac{t}{T_{1/2}}}$,where $T_{1/2} = 18 	ext{ min

Vedclass · Accepted Answer

$36$

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(D) The number of undecayed atoms $N$ after time $t$ is given by $N = N_0 \left(\frac{1}{2}ight)^{\frac{t}{T_{1/2}}}$,where $T_{1/2} = 18 	ext{ min}$.For $20 \%$ decay,the remaining amount is $N_1 = N_0 - 0.20 N_0 = 0.8 N_0$.Thus,$0.8 N_0 = N_0 \left(\frac{1}{2}ight)^{\frac{t_1}{18}} \implies \left(\frac{1}{2}ight)^{\frac{t_1}{18}} = 0.8 \quad (i)$.For $80 \%$ decay,the remaining amount is $N_2 = N_0 - 0.80 N_0 = 0.2 N_0$.Thus,$0.2 N_0 = N_0 \left(\frac{1}{2}ight)^{\frac{t_2}{18}} \implies \left(\frac{1}{2}ight)^{\frac{t_2}{18}} = 0.2 \quad (ii)$.Dividing equation $(i)$ by $(ii)$:$\frac{(1/2)^{t_1/18}}{(1/2)^{t_2/18}} = \frac{0.8}{0.2} = 4$.$\left(\frac{1}{2}ight)^{\frac{t_1-t_2}{18}} = 4 = 2^2 = \left(\frac{1}{2}ight)^{-2}$.Equating the exponents: $\frac{t_1-t_2}{18} = -2 \implies t_2 - t_1 = 36 	ext{ min}$.

The half-life of a radioactive substance is $18 \text{ minutes}$. The time interval between its $20 \%$ decay and $80 \%$ decay in minutes is

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