The half-life of a radioactive substance is 2 | vedclass

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(B) The decay constant $\lambda$ is given by $\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{20} \approx 0.03465 \ min^{-1}$.The time $t$ required fo

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$20$

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(B) The decay constant $\lambda$ is given by $\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{20} \approx 0.03465 \ min^{-1}$.The time $t$ required for a substance to decay is given by $t = \frac{1}{\lambda} \ln \left( \frac{N_0}{N} ight)$.For $33\%$ disintegration,the remaining amount $N_1 = N_0 - 0.33N_0 = 0.67N_0$. The time $t_1$ is:$t_1 = \frac{1}{0.03465} \ln \left( \frac{1}{0.67} ight) \approx 28.95 	imes 0.4005 \approx 11.6 \ min$.For $67\%$ disintegration,the remaining amount $N_2 = N_0 - 0.67N_0 = 0.33N_0$. The time $t_2$ is:$t_2 = \frac{1}{0.03465} \ln \left( \frac{1}{0.33} ight) \approx 28.95 	imes 1.1087 \approx 32.1 \ min$.The time difference is $\Delta t = t_2 - t_1 = 32.1 - 11.6 = 20.5 \ min$.Rounding to the nearest integer,the difference is approximately $20 \ min$.

The half-life of a radioactive substance is $20$ minutes. The difference between the points of time when it is $33\%$ disintegrated and $67\%$ disintegrated is approximately ......... $min$.

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