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JEE Main 201411 AprilOnline
Exam
JEE Main
Year
2014
Date
11 April
Mode
Online
Subject
Physics
Chapter
Nuclear Physics

JEE Main 2014 · 11 April · Online · Physics · Nuclear Physics

A radioactive nuclei with decay constant 0.5 / s is being produced...

A radioactive nuclei with decay constant 0.5/s0.5 / \mathrm{s} is being produced at a constant rate of 100 nuclei/s. If at t=0t=0 there were no nuclei, the time when there are 50 nuclei is:

Options

  1. A1 s1 \mathrm{~s}
  2. B2ln⁡(43)s2 \ln \left(\frac{4}{3}\right) \mathrm{s}
  3. Cln⁡2 s\ln 2 \mathrm{~s}
  4. Dln⁡(43)s\ln \left(\frac{4}{3}\right) \mathrm{s}

Correct answer

B. 2ln⁡(43)s2 \ln \left(\frac{4}{3}\right) \mathrm{s}

Step-by-step solution

Let NN be the number of nuclei at any time tt then, dNdt=100−λN or  ∫0NdN(100−λN)=∫0tdt−1λ[log⁡(100−λN)]0N=tlog⁡(100−λN)−log⁡100=−λtlog⁡100−λN100=−λt100−λN100=e−λt1−λN100=e−λt\begin{aligned} \frac{d N}{d t}=100-\lambda N \\ \text{\text{ or } } \int_0^N \frac{d N}{(100-\lambda N)}=\int_0^t d t \\ &-\frac{1}{\lambda}[\log (100-\lambda N)]_0^N=t \\ &\log (100-\lambda N)-\log 100=-\lambda t \\ &\log \frac{100-\lambda N}{100}=-\lambda t \\ \frac{100-\lambda N}{100}= e^{-\lambda t} \\ &1-\frac{\lambda N}{100}= e^{-\lambda t} \end{aligned}N=100λ(1−e−λt) As, N=50 and  λ=0.5/sec⁡∴50=1000.5(1−e−0.5t)\begin{aligned}&N=\frac{100}{\lambda}\left(1-e^{-\lambda} t\right) \\ \text{ As, } N=50 \text{\text{ and } } \lambda=0.5 / \mathrm\sec\\ &\therefore 50=\frac{100}{0.5}\left(1-e^{-0.5 t}\right) \end{aligned}Solving we get, t=2 \ln \left(\frac{4}{3}\right) \mathrm\sec
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