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The half-lives of two radioactive nuclides $A$ and $B$ are 1 and $2 \mathrm{~min}$ respectively. Equal weights of $A$ and $B$ are taken separately and allowed to disintegrate for $4 \mathrm{~min}$. What will be the ratio of weights of $A$ and $B$ disintegrated?
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The correct answer is:
5:4
For, $A t_{1 / 2}=1 \mathrm{~min}$
$\therefore$ part of $A$ remained after 4 minutes
$\text { (i.e., } 4 \text { half- life) will be }=\frac{1}{16}$
$\therefore$ part of $A$ disintegrated in the same period $=\frac{15}{16}$
For $B, t_{1 / 2}=2 \mathrm{~min}$
$\therefore$ Part of $B$ remained after 4 minutes (i.e., 2 half life)
$=\frac{1}{4}$
$\therefore$ part of $B$ disintegrated in the same time $=\frac{3}{4}$
Hence, ratio of disintegrated weights of $A$ and
$B=\frac{15}{16}: \frac{3}{4}=15: 12=5: 4$
$\therefore$ part of $A$ remained after 4 minutes
$\text { (i.e., } 4 \text { half- life) will be }=\frac{1}{16}$
$\therefore$ part of $A$ disintegrated in the same period $=\frac{15}{16}$
For $B, t_{1 / 2}=2 \mathrm{~min}$
$\therefore$ Part of $B$ remained after 4 minutes (i.e., 2 half life)
$=\frac{1}{4}$
$\therefore$ part of $B$ disintegrated in the same time $=\frac{3}{4}$
Hence, ratio of disintegrated weights of $A$ and
$B=\frac{15}{16}: \frac{3}{4}=15: 12=5: 4$
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