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(C) The decay processes are $A \xrightarrow{\lambda} B$ and $B \xrightarrow{\lambda} C$.The rate of change of the number of $B$ atoms is given by:$\fr

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(C) The decay processes are $A \xrightarrow{\lambda} B$ and $B \xrightarrow{\lambda} C$.The rate of change of the number of $B$ atoms is given by:$\frac{dN_{B}}{dt} = \lambda N_{A} - \lambda N_{B}$Since $N_{A}(t) = N_{A}(0) e^{-\lambda t}$ and $N_{A}(0) = 2 N_{B}(0)$, we have $N_{A}(t) = 2 N_{B}(0) e^{-\lambda t}$.Substituting this into the rate equation:$\frac{dN_{B}}{dt} = \lambda (2 N_{B}(0) e^{-\lambda t}) - \lambda N_{B}$$\frac{dN_{B}}{dt} + \lambda N_{B} = 2 \lambda N_{B}(0) e^{-\lambda t}$This is a linear differential equation. Multiplying by the integrating factor $e^{\lambda t}$:$e^{\lambda t} \frac{dN_{B}}{dt} + \lambda N_{B} e^{\lambda t} = 2 \lambda N_{B}(0)$$\frac{d}{dt} (N_{B} e^{\lambda t}) = 2 \lambda N_{B}(0)$Integrating both sides with respect to $t$:$N_{B} e^{\lambda t} = 2 \lambda N_{B}(0) t + C$At $t=0$, $N_{B} = N_{B}(0)$, so $C = N_{B}(0)$.$N_{B} e^{\lambda t} = N_{B}(0) (1 + 2 \lambda t)$$N_{B}(t) = N_{B}(0) (1 + 2 \lambda t) e^{-\lambda t}$Therefore, $\frac{N_{B}(t)}{N_{B}(0)} = (1 + 2 \lambda t) e^{-\lambda t}$.To find the maximum, set $\frac{d}{dt} (\frac{N_{B}(t)}{N_{B}(0)}) = 0$:$2 \lambda e^{-\lambda t} - \lambda (1 + 2 \lambda t) e^{-\lambda t} = 0$$2 - 1 - 2 \lambda t = 0 \implies 1 = 2 \lambda t \implies t = \frac{1}{2 \lambda}$.At $t = \frac{1}{2 \lambda}$, the value is $(1 + 2 \lambda (\frac{1}{2 \lambda})) e^{-\lambda (\frac{1}{2 \lambda})} = 2 e^{-0.5} \approx 1.21$.This matches the behavior shown in Figure $C$.

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