A radio nuclide A_1 with decay constant lambd | vedclass

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(A) Let $N_1$ and $N_2$ be the number of nuclei of $A_1$ and $A_2$ at time $t$. Given $\frac{dN_1}{dt} = -\lambda_1 N_1$,which integrates to $N_1(t) =

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$\frac{\ln(\lambda_2/\lambda_1)}{\lambda_2 - \lambda_1}$

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(A) Let $N_1$ and $N_2$ be the number of nuclei of $A_1$ and $A_2$ at time $t$. Given $\frac{dN_1}{dt} = -\lambda_1 N_1$,which integrates to $N_1(t) = N_{10} e^{-\lambda_1 t}$. The rate of change of $N_2$ is $\frac{dN_2}{dt} = \lambda_1 N_1 - \lambda_2 N_2$. Substituting $N_1$,we get $\frac{dN_2}{dt} + \lambda_2 N_2 = \lambda_1 N_{10} e^{-\lambda_1 t}$. Solving this linear differential equation with initial condition $N_2(0) = 0$,we get $N_2(t) = \frac{\lambda_1 N_{10}}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t})$. The activity of $A_2$ is $R_2 = \lambda_2 N_2$. For $R_2$ to be maximum,$\frac{dR_2}{dt} = 0$,which implies $\frac{dN_2}{dt} = 0$. Setting $\lambda_1 N_1 = \lambda_2 N_2$ gives $\lambda_1 N_{10} e^{-\lambda_1 t} = \lambda_2 \left[ \frac{\lambda_1 N_{10}}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t}) \right]$. Simplifying,we get $e^{-\lambda_1 t} = \frac{\lambda_2}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t})$. Rearranging leads to $e^{(\lambda_2 - \lambda_1)t} = \frac{\lambda_2}{\lambda_1}$. Taking the natural logarithm on both sides,$t = \frac{\ln(\lambda_2/\lambda_1)}{\lambda_2 - \lambda_1}$.

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