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Samples of two radioactive nuclides, X and Y, each have equal activity $A_{o}$ at time t = 0. X has a half life of 24 years and Y has a half life of 16 years. The samples are mixed together. What will be the ratio of their activity in the mixture at t = 48 years?

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$2$

$\frac{3}{8}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

In 48 hours no of half life time of X = 48/24=2

$So activity of X = \frac{A_{0}}{2^{2}} = \frac{A_{0}}{4} similarly no of half life time for Y \frac{48}{16} = 3 Activiity becomes \frac{A_{0}}{8} So there activity ratio \frac{A_{0}}{4} / \frac{A_{0}}{8} = 2$

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