Then
$\frac{d N_{1}}{d t}=-\lambda_{1} N_{1}(1)$
$\frac{d N_{2}}{d t}=\lambda_{1} N_{1}-\lambda_{2} N_{2}(2)$
From $(1)$
$N_{1}=N_{10} e^{-\lambda_{1} t}$
where $N_{10}$ is the initial number of nuclides $A_{1}$ at time $t=0$
From $(2)$
$\left(\frac{d N_{2}}{d t}+\lambda_{2} N_{2}\right) e^{\lambda_{2} t}=\lambda_{1} N_{10} e^{-\left(\lambda_{1}-\lambda_{2}\right) t}$
$\operatorname{or}\left(N_{2} e^{\lambda_{2} t}\right)=\operatorname{const} \frac{\lambda_{1} N_{10}}{\lambda_{1}-\lambda_{2}} e^{-\left(\lambda_{1}-\lambda_{2}\right) t}$
since $N_{2}=0$ at $t=0$
constant $N_{2}=\frac{\lambda_{1} N_{10}}{\lambda_{1}-\lambda_{2}}$
Thus $=\frac{\lambda_{1} N_{10}}{\lambda_{1}-\lambda_{2}}\left(e^{-\lambda_{2} t-e^{-\lambda_{1} t}}\right)$
(b) The activity of nuclide $A_{2}$ is $\lambda_{2} N_{2} .$ This is maximum when $N_{2}$ is maximum. That happens when
$\frac{d N_{2}}{d t}=0$
This require $\lambda_{2} e^{-\lambda_{2} t_{m}=\lambda_{1} e^{-\lambda_{2} tm}}$
$\operatorname{or} t_{m}=\frac{\ln \left(\lambda_{1} / \lambda_{2}\right)}{\lambda_{1}-\lambda_{2}}$
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