The half-life of a radioactive element is 20 | vedclass

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(B) The decay law is given by $N(t) = N_0 e^{-\lambda t}$,where $N(t)$ is the number of undecayed nuclei at time $t$.After $33\%$ decay,the number of

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$20$

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(B) The decay law is given by $N(t) = N_0 e^{-\lambda t}$,where $N(t)$ is the number of undecayed nuclei at time $t$.After $33\%$ decay,the number of undecayed nuclei is $N_1 = N_0(1 - 0.33) = 0.67 N_0$.After $67\%$ decay,the number of undecayed nuclei is $N_2 = N_0(1 - 0.67) = 0.33 N_0$.Note that $0.33 N_0 \approx \frac{0.67 N_0}{2}$.Since the number of undecayed nuclei becomes half in one half-life period $(T_{1/2} = 20 \, minutes)$,the time interval required for the population to reduce from $0.67 N_0$ to $0.33 N_0$ is exactly one half-life.Therefore,the time interval is $20 \, minutes$.

The half-life of a radioactive element is $20 \, minutes$. What is the time interval (in $minutes$) between $33\%$ decay and $67\%$ decay?

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