MCQ
For a radioactive substance, fraction of its initial quantity $\left(N_0\right)$ which will disintegrate in its average life time is about $(e=2.71)$
  • A
    $\left(\frac{1}{3}\right) N_0$
  • $\left(\frac{2}{3}\right) N_0$
  • C
    $(0.9) N_0$
  • D
    $\left(\frac{1}{2}\right) N_0$

Answer

Correct option: B.
$\left(\frac{2}{3}\right) N_0$
(b) : Let the amount of radioactive substance initially be ' $N_0$ ' and amount remaining be ' $N$.
As $N=N_0 e^{-\lambda t}$
$t=\tau=\frac{1}{\lambda} \Rightarrow N=N_0 e^{-\frac{\lambda}{\lambda}}=\frac{N_0}{e}$
Fraction of remaining substance, $\frac{N}{N_0}=\frac{1}{e}$
Fraction of disintegrated substance $=1-\frac{N}{N_0}$
$
=1-\frac{1}{e}=1-\frac{1}{2.71}=\frac{1.71}{2.71}=\frac{2}{3}
$

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