Q144 / 200 · NEET 2022 (Phase 2) · Physics

At any instant, two elements $X_1$ and $X_2$ have same number of radioactive atoms. If the decay constant of $X_1$ and $X_2$ are $10 \lambda$ and $\lambda$ respectively. Then the time when the ratio of their atoms becomes $\frac{1}{e}$ respectively will be:

  • A
    $\frac{1}{11 \lambda}$
  • Correct answer
    B
    $\frac{1}{9 \lambda}$
  • C
    $\frac{1}{6 \lambda}$
  • D
    $\frac{1}{5 \lambda}$

Solution

By radioactivity law, we have, $\mathrm{N}=\mathrm{N}_o e^{-\lambda t}$ (where, the symbols have their usual meanings) Hence, for two elements, $\frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{N}_o e^{-10 \lambda t}}{\mathrm{~N}_o e^{-\lambda t}}=\frac{1}{e}$ Hence, $e^{-9 \lambda t}=e$ $t=\frac{1}{9 \lambda}$

Hence, the correct answer is option: 'B'.

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