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 :
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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