The half life for radioactive decay of C-14 is 5730 years. A sample of tree had only 80% of C-14 left. Estimate the age of tree. \((\log 5=0.698) (\log 4=0.602)\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Approx. 1800 years
Radioactive decay follows first order kinetics:
\( \displaystyle k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730\ \text{years}} = 1.209 \times 10^{-4}\ \text{year}^{-1} \)
\( \displaystyle \log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} = \log\frac{100}{80} = \log 5 - \log 4 = 0.698 - 0.602 = 0.096 \)
\( \displaystyle t = \frac{2.303}{k}\log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} = \frac{2.303}{1.209 \times 10^{-4}} \times 0.096 \approx 1828\ \text{years} \)
The tree is about 1800 years old.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Exercises – “… The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.”