A radioactive isotope has half-life of ‘K’ years. How long will it take, so that activity reduces to 6.25% of its original value?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4 K
The activity of a radioactive sample falls to half in every half-life, so after n half-lives \( A = A_0\left(\frac{1}{2}\right)^{n} \).
\( \displaystyle \frac{A}{A_0} = \frac{6.25}{100} = \frac{1}{16} = \left(\frac{1}{2}\right)^{4} \Rightarrow n = 4 \)
\( t = n\,T_{1/2} = 4K \)
Shortcut: Halve step by step: 100% → 50% → 25% → 12.5% → 6.25%. That is four halvings, so 4K.