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(C) The rate of change of the number of nuclei $N$ is given by the production rate minus the decay rate: $\frac{dN}{dt} = n - \lambda N$.Rearranging t

Vedclass · Accepted Answer

$N = \frac{n}{\lambda} + \left( N_0 - \frac{n}{\lambda} \right) e^{-\lambda t}$

Vedclass · Answer

(C) The rate of change of the number of nuclei $N$ is given by the production rate minus the decay rate: $\frac{dN}{dt} = n - \lambda N$.Rearranging the terms,we get $\frac{dN}{n - \lambda N} = dt$.Integrating both sides with limits from $t = 0$ $(N = N_0)$ to $t = t$ $(N = N)$:$\int_{N_0}^{N} \frac{dN}{n - \lambda N} = \int_{0}^{t} dt$.Using the substitution method,$-\frac{1}{\lambda} [\ln(n - \lambda N)]_{N_0}^{N} = t$.$-\frac{1}{\lambda} \ln\left( \frac{n - \lambda N}{n - \lambda N_0} \right) = t$.$\ln\left( \frac{n - \lambda N}{n - \lambda N_0} \right) = -\lambda t$.$\frac{n - \lambda N}{n - \lambda N_0} = e^{-\lambda t}$.$n - \lambda N = (n - \lambda N_0) e^{-\lambda t}$.$\lambda N = n - (n - \lambda N_0) e^{-\lambda t}$.$N = \frac{n}{\lambda} - \left( \frac{n}{\lambda} - N_0 \right) e^{-\lambda t} = \frac{n}{\lambda} + \left( N_0 - \frac{n}{\lambda} \right) e^{-\lambda t}$.

$A$ radioactive nuclide is produced at a constant rate of $n$ per second (e.g.,by bombarding a target with neutrons). If the number of nuclei at $t = 0$ is $N_0$,the number of nuclei $N$ at time $t$ is given by (where $\lambda$ is the decay constant):

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