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  • A radioactive substance is being produced at a constant rate of 200 nuclei/s. The decay constant of the substance is <img alt="1 \mathrm{~s}^{-1}" src="/latex-image/?1%20%5Cmathrm%7B%7Es%7D%5E%7B-1

A radioactive substance is being produced at a constant rate of 200 nuclei/s. The decay constant of the substance is 1 \mathrm{~s}^{-1}. After what time the number of radioactive nuclei will become 100 . Initially there are no nuclei present?

Option: 1

1 \mathrm{~s}


Option: 2

\frac{1}{\ln (2)} \mathrm{s}


Option: 3

\ln (2) \mathrm{s}


Option: 4

2 \mathrm{~s}


Answers (1)

best_answer

Let N be the number of nuclei at any time t. Then

\frac{\mathrm{dN}}{\mathrm{dt}}=200-\lambda \mathrm{N}

\therefore \int_0^{\mathrm{N}} \frac{\mathrm{dN}}{200-\lambda \mathrm{N}}=\int_0^{\mathrm{t}} \mathrm{dt}

\text { or } \mathrm{N}=\frac{200}{\lambda}\left(1-\mathrm{e}^{-\lambda \mathrm{t}}\right)

\text { Given that } \mathrm{N}=100 \text { and } \lambda=1 \mathrm{~s}^{-1}

\therefore 100=200\left(1-\mathrm{e}^{-t}\right)

\text { or } \mathrm{e}^{-\mathrm{t}}=\left(\frac{1}{2}\right)

\therefore \mathrm{t}=\ln (2) \mathrm{sec} \text {. }

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Nehul

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