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(C) Let $N_0$ be the initial number of nuclei.$1$. Time $t_1$ when a quarter of the nuclei have decayed:Remaining nuclei $N_1 = N_0 - \frac{1}{4}N_0 =

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$\frac{\ln (3/2)}{\lambda}$

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(C) Let $N_0$ be the initial number of nuclei.$1$. Time $t_1$ when a quarter of the nuclei have decayed:Remaining nuclei $N_1 = N_0 - \frac{1}{4}N_0 = \frac{3}{4}N_0$.Using the decay law $N = N_0 e^{-\lambda t}$,we have $\frac{3}{4}N_0 = N_0 e^{-\lambda t_1}$,which gives $\ln(3/4) = -\lambda t_1$,or $t_1 = \frac{\ln(4/3)}{\lambda}$.$2$. Time $t_2$ when half of the nuclei have decayed:Remaining nuclei $N_2 = N_0 - \frac{1}{2}N_0 = \frac{1}{2}N_0$.Using the decay law,$\frac{1}{2}N_0 = N_0 e^{-\lambda t_2}$,which gives $\ln(1/2) = -\lambda t_2$,or $t_2 = \frac{\ln 2}{\lambda}$.$3$. The time gap $\Delta t = t_2 - t_1$:$\Delta t = \frac{\ln 2}{\lambda} - \frac{\ln(4/3)}{\lambda} = \frac{1}{\lambda} [\ln 2 - (\ln 4 - \ln 3)] = \frac{1}{\lambda} [\ln 2 - 2\ln 2 + \ln 3] = \frac{\ln(3/2)}{\lambda}$.

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as: (where $\lambda$ is the decay constant)

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