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

<p>The given problem involves calculating the age of a wood sample using the carbon-14 dating method. Carbon-14 ($^{14}C$) is a radioactive isotope of carbon that decays over time, and its ratio compared to carbon-12 ($^{12}C$) can be used to date organic materials. The ratio of $\frac{^{14}C}{^{12}C}$ in the wood is $\frac{1}{8}$th of that in the atmosphere, suggesting the $^{14}C$ has decayed. The half-life of $^{14}C$ is given as 5730 years, which is the time for half the $^{14}C$ to decay.</p><p>To solve for the age of the wood, we use the formula for radioactive decay:</p>$N = N_0 e^{-\lambda t},$<p>where $N$ is the remaining amount of $^{14}C$, $N_0$ is the original amount of $^{14}C$, $\lambda$ is the decay constant, and $t$ is the time (age of the wood).</p><p>Since the ratio $\frac{N}{N_0} = \frac{1}{8}$, we infer that $N = \frac{N_0}{8}$. Substituting into the decay equation and solving for $t$, first we find the decay constant $\lambda$:</p>$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730}.$<p>Substitute $\lambda$ and rearrange to solve for $t$:</p>$$t = \frac{\ln(8)}{\lambda} = \frac{\ln(8)}{0.693} \times 5730 = 3 \times 5730 = 17190 \text{ years}.$$</p><p>This calculation shows that the wood sample is 17190 years old.</p>