Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 minutes.
Solution
${T_{1/2}} = 20 \Rightarrow {{\ln 2} \over \lambda } = 20$ min<br><br>$\Rightarrow \lambda = {{\ln 2} \over {20(\min )}}$<br><br>$\because$ ${N_t} = {N_0}{e^{ - \lambda t}}$<br><br>$${{{N_t}} \over {{N_0}}} = {e^{ - \lambda {t_1}}} \Rightarrow 0.67 = {e^{ - \lambda {t_1}}}$$<br><br>$\Rightarrow \ln (0.67) = - \lambda {t_1}$<br><br>$\Rightarrow \ln \left( {{{100} \over {67}}} \right) = \lambda {t_1}$
<br><br>$$ \Rightarrow {t_1} = {{\ln \left( {{{100} \over {67}}} \right) \times 20(\min )} \over {(\ln 2)}}$$<br><br>Similarly, $${t_2} = {{\ln \left( {{{100} \over {34}}} \right) \times 20(\min )} \over {(\ln 2)}}$$<br><br>${t_2} - {t_1} = 19.57$ min $\approx 20$ min.
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
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