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  • A radioactive substance with decay constant of  is being produced at a constant rate of 50 nuclei per

A radioactive substance with decay constant of 0.5 \mathrm{~s}^{-1} is being produced at a constant rate of 50 nuclei per second. If there are no nuclei present initially, the time (in second) after which 25 nuclei will be present is:

Option: 1

1


Option: 2

2 \ln \left(\frac{4}{3}\right)


Option: 3

\ln 2


Option: 4

\ln \left(\frac{4}{3}\right)


Answers (1)

best_answer

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

\frac{\mathrm{dN}}{\mathrm{dt}}=50-\lambda \mathrm{N} \text { or } \frac{\mathrm{dN}}{50-\lambda \mathrm{N}}=\mathrm{dt}

Integrate both sides, we get

\int_0^{\mathrm{N}} \frac{\mathrm{dN}}{50-\lambda \mathrm{N}}=\int_0^{\mathrm{t}} \mathrm{dt} \Rightarrow-\frac{1}{\lambda}[\ln (50-\lambda \mathrm{N})]_0^{\mathrm{N}}=\mathrm{t}

\ln \left(\frac{50-\lambda N}{50}\right)=-\lambda t

\frac{50-\lambda \mathrm{N}}{50}=\mathrm{e}^{-\lambda \mathrm{t}} ; 1-\frac{\lambda \mathrm{N}}{50}=\mathrm{e}^{-\lambda \mathrm{t}}

\mathrm{N}=\frac{50}{\lambda}\left(1-\mathrm{e}^{-\lambda \mathrm{t}}\right)

\text { As } \mathrm{N}=25 \text { and } \lambda=0.5 \mathrm{~s}^{-1}

25=\frac{50}{0.5}\left(1-\mathrm{e}^{-0.5 \mathrm{t}}\right), \frac{1}{4}=1-\mathrm{e}^{-0.5 \mathrm{t}}

\mathrm{e}^{-0.5 \mathrm{t}}=\frac{3}{4} \therefore \quad \mathrm{t}=2 \ln \left(\frac{4}{3}\right)

Posted by

jitender.kumar

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