A radioactive sample decays 7/8 times its initial quantity in 21 minutes. The half life of the sample will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 7 min
By the radioactive decay law, the amount left halves in every half-life. After \( n \) half-lives
\( \displaystyle N = N_0\left(\frac{1}{2}\right)^{n} \)
The sample has lost \( 7/8 \) of its initial quantity, so the fraction left is
\( \displaystyle \frac{N}{N_0} = 1 - \frac{7}{8} = \frac{1}{8} = \left(\frac{1}{2}\right)^{3} \)
So \( n = 3 \) half-lives have passed in 21 min.
\( \displaystyle T_{1/2} = \frac{t}{n} = \frac{21\,\mathrm{min}}{3} = 7\,\mathrm{min} \)