A piece of ancient wood shows an activity of 3.9 disintegrations per second per gram of \( {}^{14}\mathrm{C} \). The age of wood, if half life of \( {}^{14}\mathrm{C} \) is 5568 years, and activity of fresh \( {}^{14}\mathrm{C} \) is 15.6 disintegrations per second per gram, is (take \( \log_e 2 \) = 0.693).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 11,136 years
The activity of a radioactive sample falls exponentially, \( A = A_0e^{-\lambda t} \), with the decay constant \( \lambda = \frac{\ln 2}{T_{1/2}} \). Taking the ratio of the fresh and the present activities gives the age.
\( \displaystyle \lambda = \frac{0.693}{5568\,\mathrm{yr}} \approx 1.245\times10^{-4}\,\mathrm{yr^{-1}} \)
\( \displaystyle \frac{A_0}{A} = e^{\lambda t} = \frac{15.6}{3.9} = 4 \)
\( \displaystyle t = \frac{1}{\lambda}\ln 4 = \frac{2\ln 2}{\lambda} = 2\times5568 = 11136\,\mathrm{yr} \)
Shortcut: The ratio \( A_0/A = 4 = 2^{2} \), so the activity has halved twice, and the age is two half-lives, \( 2\times5568 = 11136 \) years. No logarithm is needed.