A radioactive substance decays to \( \displaystyle\frac{1}{32}^{\text{th}} \) of its initial activity in 30 days. Its half life period will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6 days
The activity of a radioactive sample halves in every half-life \( T_{1/2} \), so after a time t it is \( A = A_0\left(\frac{1}{2}\right)^{t/T_{1/2}} \).
\( \displaystyle \frac{A}{A_0} = \frac{1}{32} = \left(\frac{1}{2}\right)^{5} \)
The activity has halved 5 times, so
\( \displaystyle \frac{t}{T_{1/2}} = 5 \Rightarrow T_{1/2} = \frac{30\,\mathrm{days}}{5} = 6\,\mathrm{days} \)