The half-life of a radioactive substance is 3 | vedclass

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(C) Let the initial amount of the radioactive substance be $N_i = 100 \%$. At $40 \%$ decay, the remaining amount is $N_1 = 100 \% - 40 \% = 60 \%$. A

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$60 	ext{ minutes}$

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(C) Let the initial amount of the radioactive substance be $N_i = 100 \%$. At $40 \%$ decay, the remaining amount is $N_1 = 100 \% - 40 \% = 60 \%$. At $85 \%$ decay, the remaining amount is $N_2 = 100 \% - 85 \% = 15 \%$. We know that the amount remaining after time $t$ is given by $N(t) = N_i \left( \frac{1}{2} ight)^{t/T_{1/2}}$, where $T_{1/2} = 30 	ext{ minutes}$. For the interval between $N_1$ and $N_2$, the ratio of remaining nuclei is $\frac{N_2}{N_1} = \frac{15 \%}{60 \%} = \frac{1}{4}$. Since $\frac{1}{4} = \left( \frac{1}{2} ight)^2$, the time taken corresponds to two half-lives. Therefore, the time taken is $t = 2 	imes T_{1/2} = 2 	imes 30 	ext{ minutes} = 60 	ext{ minutes}$.

The half-life of a radioactive substance is $30 \text{ minutes}$. The time taken between $40 \%$ decay and $85 \%$ decay of the same radioactive substance is

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