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Q.

The half life of a radioactive nuclide is 20 hrs. The fraction of the original activity that will remain after 40 hrs is

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a

$\frac{1}{6}$

b

$\frac{1}{8}$

c

$\frac{1}{2}$

d

$\frac{1}{4}$

answer is C.

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Detailed Solution

fraction $= \frac{R}{R_{0}} = {(\frac{1}{2})}^{\frac{t}{t_{h}}} = {(\frac{1}{2})}^{\frac{40}{20}} = \frac{1}{4}$

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