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(C) Let $N_0$ be the initial quantity kept the first time. Then the quantity kept the second time is $2N_0$.Let $t_1$ be the time elapsed for the firs

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$\frac{T}{\ln 2} \ln \frac{A_2}{2A_1}$

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(C) Let $N_0$ be the initial quantity kept the first time. Then the quantity kept the second time is $2N_0$.Let $t_1$ be the time elapsed for the first sample and $t_2$ be the time elapsed for the second sample. The age difference is $	au = t_1 - t_2$.The activity is given by $A = \lambda N = \lambda N_{initial} e^{-\lambda t}$.For the first sample: $A_1 = \lambda N_0 e^{-\lambda t_1}$.For the second sample: $A_2 = \lambda (2N_0) e^{-\lambda t_2}$.Dividing the two equations: $\frac{A_1}{A_2} = \frac{\lambda N_0 e^{-\lambda t_1}}{2 \lambda N_0 e^{-\lambda t_2}} = \frac{1}{2} e^{-\lambda (t_1 - t_2)} = \frac{1}{2} e^{-\lambda 	au}$.Rearranging for $	au$: $e^{\lambda 	au} = \frac{A_2}{2A_1}$.Taking the natural logarithm on both sides: $\lambda 	au = \ln \left( \frac{A_2}{2A_1} ight)$.Since $\lambda = \frac{\ln 2}{T}$,we have $	au = \frac{T}{\ln 2} \ln \left( \frac{A_2}{2A_1} ight)$.

$A$ radioactive material of half-life $T$ was kept in a nuclear reactor at two different instants. The quantity kept the second time was twice that kept the first time. If their present activities are $A_1$ (first) and $A_2$ (second) respectively,then their age difference is equal to:

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