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

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

Vedclass · Accepted Answer

$20$

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(B) The decay constant $\lambda$ is given by $\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{20} \, min^{-1}$.The time $t$ required for a substance to decay to a fraction $N/N_0$ is given by $t = \frac{1}{\lambda} \ln(\frac{N_0}{N})$.For $33 \%$ disintegration,the remaining amount is $N_1 = 67 \%$ of $N_0$,so $t_1 = \frac{1}{\lambda} \ln(\frac{100}{67}) = \frac{20}{0.693} \ln(1.4925) \approx 28.85 	imes 0.4005 \approx 11.56 \, min$.For $67 \%$ disintegration,the remaining amount is $N_2 = 33 \%$ of $N_0$,so $t_2 = \frac{1}{\lambda} \ln(\frac{100}{33}) = \frac{20}{0.693} \ln(3.0303) \approx 28.85 	imes 1.1087 \approx 31.99 \, min$.The time difference is $\Delta t = t_2 - t_1 = 31.99 - 11.56 = 20.43 \, min$.Rounding to the nearest integer,the difference is approximately $20 \, min$.

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

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