The half-life for radioactive decay of $^{14}C$ is $5730$ years. An archaeological artifact containing wood had only $80 \%$ of the $^{14}C$ found in a living tree. Estimate the age of the sample.

(N/A) The radioactive decay follows first-order kinetics.
Given: $t_{1/2} = 5730 \text{ years}$,$[R]_0 = 100$,$[R] = 80$.
First,calculate the decay constant $k$:
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \text{ years}^{-1}$.
Using the first-order integrated rate equation:
$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$
$t = \frac{2.303}{0.693 / 5730} \times \log \left( \frac{100}{80} \right)$
$t = \frac{2.303 \times 5730}{0.693} \times \log(1.25)$
$t \approx 19039.5 \times 0.0969 \approx 1845 \text{ years}$.
Thus,the age of the sample is approximately $1845 \text{ years}$.