\( \displaystyle\frac{1}{16} \)th of the initial amount of a certain radio-active isotope remains undecayed after two hours. The half life of the isotope would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 30 minutes
In each half-life the amount of a radioactive substance halves, so after n half-lives the fraction left is \( \left(\frac{1}{2}\right)^{n} \).
\( \displaystyle \left(\frac{1}{2}\right)^{n} = \frac{1}{16} = \left(\frac{1}{2}\right)^{4} \Rightarrow n = 4 \)
Four half-lives take two hours, that is 120 minutes.
\( \displaystyle T_{1/2} = \frac{120\,\mathrm{min}}{4} = 30\,\mathrm{min} \)