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(A) The rate of change of the number of nuclei $N$ is given by the differential equation: $\frac{dN}{dt} = \alpha - \lambda N$.At the maximum number o

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$\frac{\alpha}{\lambda}$

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(A) The rate of change of the number of nuclei $N$ is given by the differential equation: $\frac{dN}{dt} = \alpha - \lambda N$.At the maximum number of nuclei,the rate of change is zero,i.e.,$\frac{dN}{dt} = 0$.Setting the equation to zero: $0 = \alpha - \lambda N$.Solving for $N$,we get $N_{max} = \frac{\alpha}{\lambda}$.Note that if the initial number of nuclei $N_0$ is already greater than $\frac{\alpha}{\lambda}$,the number of nuclei will decrease until it reaches the equilibrium value $\frac{\alpha}{\lambda}$. If $N_0 < \frac{\alpha}{\lambda}$,it will increase until it reaches $\frac{\alpha}{\lambda}$.

$A$ radioactive nucleus is being produced at a constant rate $\alpha$ per second. Its decay constant is $\lambda$. If $N_0$ is the number of nuclei at time $t = 0$,then the maximum number of nuclei possible is:

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