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At time $\mathrm{t}=0$, a container has $\mathrm{N}_{0}$ radioactive atoms with a decay constant $\lambda$. In addition, $\mathrm{c}$ numbers of atoms of the same type are being added to the container per unit time. How many atoms of this type are there at $\mathrm{t}=\mathrm{T} ?$
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Verified Answer
The correct answer is:
$\frac{\mathrm{c}}{\lambda}\{1-\exp (-\lambda \mathrm{T})\}+\mathrm{N}_{0} \exp (-\lambda \mathrm{T})$
$\quad \mathrm{N}_{0}$-initial nucleon
at $\mathrm{t}=0, \mathrm{~N}_{0}$
Addition is at a constant rate
$$
\begin{array}{l}
(\lambda \mathrm{N}-\mathrm{C})=-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}} \\
{\int_{0}^{\mathrm{k}} \mathrm{dt}=\int_{\mathrm{N}_{0}}^{\mathrm{N}} \frac{\mathrm{d} \mathrm{N}}{\lambda \mathrm{N}-\mathrm{C}}}
\end{array}
$$
Integrating we get
$$
\begin{array}{l}
\mathrm{N}=\frac{\mathrm{C}}{\lambda}+\frac{\mathrm{e}^{-\lambda t}}{\lambda}\left(\lambda \mathrm{N}_{0}-\mathrm{C}\right) \\
\therefore \mathrm{N}=\frac{\mathrm{C}}{\lambda}\left(1-\mathrm{e}^{-\lambda t}\right)+\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}
\end{array}
$$
at $\mathrm{t}=0, \mathrm{~N}_{0}$
Addition is at a constant rate
$$
\begin{array}{l}
(\lambda \mathrm{N}-\mathrm{C})=-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}} \\
{\int_{0}^{\mathrm{k}} \mathrm{dt}=\int_{\mathrm{N}_{0}}^{\mathrm{N}} \frac{\mathrm{d} \mathrm{N}}{\lambda \mathrm{N}-\mathrm{C}}}
\end{array}
$$
Integrating we get
$$
\begin{array}{l}
\mathrm{N}=\frac{\mathrm{C}}{\lambda}+\frac{\mathrm{e}^{-\lambda t}}{\lambda}\left(\lambda \mathrm{N}_{0}-\mathrm{C}\right) \\
\therefore \mathrm{N}=\frac{\mathrm{C}}{\lambda}\left(1-\mathrm{e}^{-\lambda t}\right)+\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}
\end{array}
$$
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