At time t=0,a container has N_{0} radioactive | vedclass

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(C) The rate of change of the number of radioactive atoms $N$ is given by the difference between the rate of addition and the rate of decay:$\frac{dN}

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$\frac{c}{\lambda}(1 - \exp(-\lambda T)) + N_0 \exp(-\lambda T)$

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(C) The rate of change of the number of radioactive atoms $N$ is given by the difference between the rate of addition and the rate of decay:$\frac{dN}{dt} = c - \lambda N$Rearranging the terms for integration:$\frac{dN}{c - \lambda N} = dt$Integrating both sides with the initial condition $N = N_0$ at $t = 0$ and $N = N$ at $t = T$:$\int_{N_0}^{N} \frac{dN}{c - \lambda N} = \int_{0}^{T} dt$Let $u = c - \lambda N$,then $du = -\lambda dN$,or $dN = -\frac{du}{\lambda}$:$-\frac{1}{\lambda} [\ln(c - \lambda N)]_{N_0}^{N} = T$$\ln\left(\frac{c - \lambda N}{c - \lambda N_0}\right) = -\lambda T$Taking the exponential of both sides:$\frac{c - \lambda N}{c - \lambda N_0} = e^{-\lambda T}$$c - \lambda N = (c - \lambda N_0)e^{-\lambda T}$$\lambda N = c - (c - \lambda N_0)e^{-\lambda T}$$N = \frac{c}{\lambda}(1 - e^{-\lambda T}) + N_0 e^{-\lambda T}$

At time $t=0$,a container has $N_{0}$ radioactive atoms with a decay constant $\lambda$. In addition,$c$ number of atoms of the same type are being added to the container per unit time. How many atoms of this type are there at $t=T$?

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