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The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:
Detailed Solution
$\frac{A}{A_0} = \left(\frac{1}{2}\right)^{t/T_{1/2}} = \left(\frac{1}{2}\right)^{150/100} = \left(\frac{1}{2}\right)^{3/2}$
$= \frac{1}{2\sqrt{2}}$