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  • Nuccli of radioactive element A are being produced at a constant rate . The element ha

Nuccli of radioactive element A are being produced at a constant rate \mathrm{\alpha }. The element has a decay constant \mathrm{\lambda }. At time t=0, there are \mathrm{N_0 } nuclei of the element. The number N of nuclei of A at time t -

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

None


Answers (1)

best_answer

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

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