Question: 27. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be :
(1) \frac{1}{2}
(2) \frac{1}{2\sqrt{2}}
(3) \frac{2}{3}
(4) \frac{2}{3\sqrt{2}}
Answer: Option (2)
Explanation:
The activity of a radioactive nuclide is directly proportional to the number of undecayed nuclei present.
The relation between remaining activity and time is given by
\frac{A}{A_0}=\left(\frac{1}{2}\right)^{t/T_{1/2}}.
Here, the half-life is T_{1/2}=100\,\text{hours}
and the given time is t=150\,\text{hours}.
Substituting these values,
\frac{A}{A_0}=\left(\frac{1}{2}\right)^{150/100}.
This gives
\frac{A}{A_0}=\left(\frac{1}{2}\right)^{3/2}.
Writing \left(\frac{1}{2}\right)^{3/2}=\frac{1}{2\sqrt{2}},
the fraction of original activity remaining after 150 hours is \frac{1}{2\sqrt{2}}.
Therefore, the correct answer is Option (2).