The ratio of $$\frac{^{14}C}{^{12}C}$$ in a piece of wood is $$\frac{1}{8}$$ part that of atmosphere. If half life of $$^{14}C$$ is $$5730$$ years, the age of wood sample is _____ years.

The amount of a radioactive isotope remaining after time $$t$$ follows the decay formula $$N = N_0\left(\frac{1}{2}\right)^{t/T_{1/2}},$$ where $$N_0$$ is the initial amount, $$N$$ is the current amount, and $$T_{1/2}$$ is the half-life.

In this wood sample the ratio of $$\,^{14}C/\,^{12}C$$ has decreased to $$\tfrac{1}{8}$$ of the atmospheric value, so we set $$\frac{N}{N_0} = \frac{1}{8} = \left(\frac{1}{2}\right)^{t/T_{1/2}}.$$

Noting that $$\frac{1}{8} = \left(\frac{1}{2}\right)^3$$ shows that $$\tfrac{t}{T_{1/2}} = 3,$$ which means three half-lives have elapsed.

Since the half-life $$T_{1/2}$$ is 5730 years, the age of the sample is $$t = 3 \times 5730 = 17190 \text{ years}.$$

The answer is 17190 years.