In a radioactive material, fraction of active material remaining after time

Question:

In a radioactive material, fraction of active material remaining after time $t$ is $9 / 16$. The fraction that was remaining after $\mathrm{t} / 2$ is :

  1. $\frac{3}{4}$

  2. $\frac{7}{8}$

  3. $\frac{4}{5}$

  4. $\frac{3}{5}$


Correct Option: 1

Solution:

First order decay

$\mathrm{N}(\mathrm{t})=\mathrm{N}_{0} \mathrm{e}^{-\mathrm{i} \mathrm{t}}$

Given $\mathrm{N}(\mathrm{t}) / \mathrm{N}_{0}=9 / 16=\mathrm{e}^{-\lambda \mathrm{t}}$

Now, $\mathrm{N}(\mathrm{t} / 2)=\mathrm{N}_{0} \mathrm{e}^{-2 \mathrm{t} / 2}$

$\frac{\mathrm{N}(\mathrm{t} / 2)}{\mathrm{N}_{0}}=\sqrt{\mathrm{e}^{-\lambda . t}}=\sqrt{9 / 16}$

$\mathrm{N}(\mathrm{t} / 2)=3 / 4 \mathrm{~N}_{0}$

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