A radioactive nuclide is produced at the cons | vedclass

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Question

$\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{n}-\lambda \mathrm{N}$

$\mathrm{d} \mathrm{N}=(\mathrm{n}-\lambda \mathrm{N}) \mathrm{dt}$

$\

Vedclass · Accepted Answer

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

Vedclass · Answer

$\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{n}-\lambda \mathrm{N}$

$\mathrm{d} \mathrm{N}=(\mathrm{n}-\lambda \mathrm{N}) \mathrm{dt}$

$\int_{N_{0}}^{N} \frac{d N}{n-\lambda N}=\int_{0}^{t} d t \Rightarrow-\frac{1}{\lambda} \int_{N_{0}}^{N} \frac{-\lambda d N}{n-\lambda N}=t$

$\Rightarrow-\frac{1}{\lambda}\left[\log _{\mathrm{e}}(\mathrm{n}-\lambda \mathrm{N})\right]_{\mathrm{N}_{0}}^{\mathrm{N}}=\mathrm{t}$

$\Rightarrow-\frac{1}{\lambda}\left[\log _{e}\left(\frac{\mathrm{n}-\lambda \mathrm{N}}{\mathrm{n}-\lambda \mathrm{N}_{0}}\right)\right]=\mathrm{t}$

$\Rightarrow \lambda \mathrm{t}=\left[\log _{\mathrm{e}}\left(\frac{\mathrm{n}-\lambda \mathrm{N}_{0}}{\mathrm{n}-\lambda \mathrm{N}}\right)\right]$

$e^{\lambda t}=\frac{n-\lambda N_{0}}{n-\lambda N}$

$\mathrm{n}-\lambda \mathrm{N}=\left(\mathrm{n}-\lambda \mathrm{N}_{0}\right) \mathrm{e}^{-\lambda \mathrm{t}}$

$\frac{\mathrm{n}}{\lambda}-\left(\frac{\mathrm{n}}{\lambda}-\mathrm{N}_{0}\right) \mathrm{e}^{-\lambda \mathrm{t}}=\mathrm{N}$

A radioactive nuclide is produced at the constant rate of $n$ per second (say, by bombarding a target with neutrons). The expected number $N$ of nuclei in existence $t\, seconds$ after the number is $N_0$ is given by Where $\lambda $ is the decay constant of the sample

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