The half-life for the radioactive decay of carbon-14 is 5730 years and decay is first order. An artifact contains 25% of the C-14 of a living tree. What is its approximate age?

1. Radioactive decay follows first order kinetics with constant half-life. 2. 25% remaining means $\left(\tfrac{1}{2}\right)^n = 0.25$, so $n = 2$ half-lives. 3. Age $= 2\times5730 = 11460\ \text{years}$. 4. Distractor A is one half-life (50% remaining); distractor B is three half-lives (12.5%). _Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.25_