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

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

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$20$

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(B) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{20} = 0.03465 \ min^{-1}$.The time $t$ required for a substance to decay is given by $t = \frac{2.303}{\lambda} \log_{10} \left( \frac{N_0}{N} ight)$.For $33\%$ disintegration,the remaining amount $N = N_0 - 0.33N_0 = 0.67N_0$. Thus,$t_1 = \frac{2.303}{0.03465} \log_{10} \left( \frac{100}{67} ight) \approx 66.46 	imes 0.1739 \approx 11.56 \ min$.For $67\%$ disintegration,the remaining amount $N = N_0 - 0.67N_0 = 0.33N_0$. Thus,$t_2 = \frac{2.303}{0.03465} \log_{10} \left( \frac{100}{33} ight) \approx 66.46 	imes 0.4815 \approx 32.00 \ min$.The time difference is $\Delta t = t_2 - t_1 = 32.00 - 11.56 = 20.44 \ min \approx 20 \ min$.

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

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