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Q.

If T is the half-life of a radioactive material, then the fraction that would remain after a time $\frac{T}{2}$ is $\frac{1}{\sqrt{x}}$ . The value of $x$ is

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Detailed Solution

$N = \frac{N_{0}}{2^{n}} = \frac{N_{0}}{2^{1 / 2}} = \frac{N_{0}}{\sqrt{2}} = 0 .707 N_{0}$

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