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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Nuclear Physics

JEE Main 2025 · 23 January · Morning Shift · Physics · Nuclear Physics

A radioactive nucleus n_2 has 3 times the decay constant as...

A radioactive nucleus n2n_2 has 3 times the decay constant as compared to the decay constant of another radioactive nucleus n1n_1. If initial number of both nuclei are the same, what is the ratio of number of nuclei of n2n_2 to the number of nuclei of n1n_1, after one half-life of n1n_1 ?

Options

  1. A1/81 / 8
  2. B88
  3. C44
  4. D1/41 / 4

Correct answer

D. 1/41 / 4

Step-by-step solution

N2=N0e−3λtN1=N0e−λtN2N1=e−2λtthalf lifeofN tt=ln⁡2λn1N02=N0e−λt\begin{aligned} & \mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-3 \lambda t} \\ & \mathrm{ N}_1=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\ & \frac{\mathrm{N}_2}{\mathrm{ N}_1}=\mathrm{e}^{-2 \lambda \mathrm{t}} \\ & \mathrm{t}_{\text{half lifeofN }} \mathrm{t} \\ & \mathrm{t}=\frac{\ln 2}{\lambda}{ }_{\mathrm{n}} 1 \\ & \frac{\mathrm{ N}_0}{2}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}\end{aligned}λt=ln⁡2t=ln⁡2λ=e−2λln⁡2λN2N1=14\begin{aligned} & \lambda \mathrm{t}=\ln 2 \\ & \mathrm{t}=\frac{\ln 2}{\lambda} \\ & =\mathrm{e}^{-2 \lambda \frac{\ln 2}{\lambda}} \\ & \frac{\mathrm{ N}_2}{\mathrm{ N}_1}=\frac{1}{4}\end{aligned}$
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