Two radioactive samples A and B have half-liv | vedclass

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(A) Let the initial activities be $R_{A,0}$ and $R_{B,0}$. Given $R_{B,0} = 2 R_{A,0}$.The activity at time $t$ is given by $R(t) = R_0 e^{-\lambda t}

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$\frac{T_1 T_2}{T_1-T_2} \ln(2)$

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(A) Let the initial activities be $R_{A,0}$ and $R_{B,0}$. Given $R_{B,0} = 2 R_{A,0}$.The activity at time $t$ is given by $R(t) = R_0 e^{-\lambda t}$,where $\lambda = \frac{\ln(2)}{T}$.Equating the activities at time $t$: $R_{A,0} e^{-\lambda_1 t} = R_{B,0} e^{-\lambda_2 t}$.Substituting $R_{B,0} = 2 R_{A,0}$: $R_{A,0} e^{-\lambda_1 t} = 2 R_{A,0} e^{-\lambda_2 t}$.$e^{(\lambda_2 - \lambda_1)t} = 2$.Taking the natural logarithm on both sides: $(\lambda_2 - \lambda_1)t = \ln(2)$.Since $\lambda = \frac{\ln(2)}{T}$,we have $(\frac{\ln(2)}{T_2} - \frac{\ln(2)}{T_1})t = \ln(2)$.Dividing by $\ln(2)$: $(\frac{1}{T_2} - \frac{1}{T_1})t = 1$.$(\frac{T_1 - T_2}{T_1 T_2})t = 1$.Therefore,$t = \frac{T_1 T_2}{T_1 - T_2}$.

Two radioactive samples $A$ and $B$ have half-lives $T_1$ and $T_2$ $(T_1 > T_2)$ respectively. At $t=0$,the activity of $B$ was twice the activity of $A$. Their activity will become equal after a time:

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