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NEET PHYSICSMedium

The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be

A

12\frac{1}{2}

B

122\frac{1}{2\sqrt{2}}

C

23\frac{2}{3}

D

232\frac{2}{3\sqrt{2}}

Step-by-Step Solution

The activity AA is given by A=A0(12)t/T1/2A = A_0 (\frac{1}{2})^{t/T_{1/2}}. Given t=150t = 150 hours and T1/2=100T_{1/2} = 100 hours, AA0=(12)150/100=(12)3/2=122\frac{A}{A_0} = (\frac{1}{2})^{150/100} = (\frac{1}{2})^{3/2} = \frac{1}{2\sqrt{2}}.

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