Q126 / 150 · MHT CET 2022 (05 Aug Shift 2) · Physics

For a substance, fraction of its initial quantify $\left(\mathrm{N}_0\right)$ which will disintegrate in its average life time is about $(\mathrm{e}=2.71)$

  • A
    $\left(\frac{1}{3}\right) \mathrm{N}_0$
  • B
    $\left(\frac{1}{2}\right) \mathrm{N}_0$
  • Correct answer
    C
    $\left(\frac{2}{3}\right) \mathrm{N}_0$
  • D
    $(0.9) \mathrm{N}_0$

Solution

For a radioactive substance, $\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda t}$ Given the mean life, $\mathrm{t}=\frac{1}{\lambda}$ $\therefore \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}=\frac{\mathrm{N}_0}{\mathrm{e}}=\frac{\mathrm{N}_0}{2.71}=0.37 \mathrm{~N}_0$ $0.37 \mathrm{~N}_0$ is the amount of radioactive substance left after time $\tau$. $\therefore$ The amount disintegrated in time $\tau$ $=\mathrm{N}_0-0.37 \mathrm{~N}_0=0.63 \mathrm{~N}_0=\frac{2}{3} \mathrm{~N}_0$

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

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