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(C) Let $N_1$ and $N_2$ be the initial quantities produced at times $t_1$ and $t_2$ respectively.Given $N_2 = 2N_1$. Let $N_1 = N_0$,so $N_2 = 2N_0$.T

Vedclass · Accepted Answer

$\frac{T}{\ln 2} \left| \ln \frac{A_2}{2A_1} \right|$

Vedclass · Answer

(C) Let $N_1$ and $N_2$ be the initial quantities produced at times $t_1$ and $t_2$ respectively.Given $N_2 = 2N_1$. Let $N_1 = N_0$,so $N_2 = 2N_0$.The activity $A$ is given by $A = \lambda N$,where $\lambda = \frac{\ln 2}{T}$.The activity at time $t$ is $A = A_0 e^{-\lambda t} = \lambda N_0 e^{-\lambda t}$.For the first sample: $A_1 = \lambda N_0 e^{-\lambda t_1} \implies e^{-\lambda t_1} = \frac{A_1}{\lambda N_0} \implies t_1 = \frac{1}{\lambda} \ln \left( \frac{\lambda N_0}{A_1} \right)$.For the second sample: $A_2 = \lambda (2N_0) e^{-\lambda t_2} \implies e^{-\lambda t_2} = \frac{A_2}{2\lambda N_0} \implies t_2 = \frac{1}{\lambda} \ln \left( \frac{2\lambda N_0}{A_2} \right)$.The age difference is $|t_2 - t_1| = \left| \frac{1}{\lambda} \ln \left( \frac{2\lambda N_0}{A_2} \right) - \frac{1}{\lambda} \ln \left( \frac{\lambda N_0}{A_1} \right) \right|$.$|t_2 - t_1| = \frac{1}{\lambda} \left| \ln \left( \frac{2\lambda N_0 / A_2}{\lambda N_0 / A_1} \right) \right| = \frac{1}{\lambda} \left| \ln \left( \frac{2A_1}{A_2} \right) \right|$.Since $\frac{1}{\lambda} = \frac{T}{\ln 2}$,we have $|t_2 - t_1| = \frac{T}{\ln 2} \left| \ln \left( \frac{2A_1}{A_2} \right) \right| = \frac{T}{\ln 2} \left| \ln \left( \frac{A_2}{2A_1} \right) \right|$.

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

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