A radioactive isotope has a half life of \( T \) years. How long (in years) will it take the activity to reduce to 3.125% of its original value?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5T \)
The half-life is the time in which the activity of a radioactive sample falls to half its value. After \( n \) half-lives the activity is
\( \displaystyle A = A_0 \left(\frac{1}{2}\right)^{n} \)
The activity has to fall to 3.125% of \( A_0 \):
\( \displaystyle \frac{A}{A_0} = \frac{3.125}{100} = \frac{1}{32} = \left(\frac{1}{2}\right)^{5} \)
So \( n = 5 \) half-lives, and the time taken is
\( \displaystyle t = nT = 5T \)