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JEE Main 2014 (11 Apr, Shift 1) · Law of Radioactive Decay

Law of Radioactive Decay question from JEE Main 2014 (11 Apr, Shift 1)

Q96JEE Main 2014 · 11 APR · SHIFT 1Difficult

Law of Radioactive Decay

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

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Answer: (b)

  1. dNdt=R−λN⇒N=Rλ(1−e−λt)\dfrac{dN}{dt} = R - \lambda N \Rightarrow N = \dfrac{R}{\lambda}(1 - e^{-\lambda t})
  2. Rλ=1000.5=200\dfrac{R}{\lambda} = \dfrac{100}{0.5} = 200, so 50=200(1−e−0.5t)⇒e−0.5t=3450 = 200(1 - e^{-0.5t}) \Rightarrow e^{-0.5t} = \dfrac{3}{4}
  3. t=2ln⁡(43)t = 2\ln\left(\dfrac{4}{3}\right) s
  4. So the answer is (b) 2ln⁡(43)2\ln\left(\dfrac{4}{3}\right) s.
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