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Nuceli of radioactive element \mathrm{A} are being produced at a constant rate \mathrm{\alpha} . The element has a decay constant \mathrm{\lambda}. At time \mathrm{t=0}, there are No nuclei of the element. The number \mathrm{N} of nuclei of \mathrm{A} at time \mathrm{t -}

Option: 1

\mathrm{\frac{\alpha}{\lambda}\left(1+e^{-\lambda t}\right)+N_0 e^{-\lambda t}}


Option: 2

\mathrm{\frac{\alpha}{\lambda}\left(1-e^{-\lambda t}\right)+N_0 e^{-\lambda t}}


Option: 3

\mathrm{\frac{\alpha}{\lambda}\left(1-e^{-\lambda t}\right)+\frac{N_0}{2} e^{-\lambda t}}


Option: 4

None 


Answers (1)

best_answer

The rate of formation of nuclei is \mathrm{=\alpha}
The rate of decay of nuclei \mathrm{=\lambda N}
\mathrm{\begin{aligned} & \frac{d N}{d t}=(\alpha-\lambda N) \\ & \frac{d N}{(\alpha-\lambda N)}=d t \end{aligned}}
\mathrm{\int_{N_0}^N \frac{d N}{(\alpha-\lambda N)}=\int_0^t d t}
\mathrm{\left|\frac{\ln (\alpha-\lambda N)}{-\lambda}\right|_{N_0}^N=t}
\mathrm{\ln (\alpha-\lambda N)-\ln \left(\alpha-\lambda N_0\right)=-\lambda t}
\mathrm{\begin{aligned} & \ln \left(\frac{\alpha-\lambda N}{\alpha-\partial N_0}\right)=-\partial t \\ & \frac{\alpha-\partial N}{\alpha-\lambda N_0}=e^{-\lambda t} \end{aligned}}
\mathrm{N=\frac{\alpha}{\lambda}\left(1-e^{-\lambda t}\right)+N_0 e^{-\lambda t}}

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SANGALDEEP SINGH

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