A burnt plece of wood found at an archaeological site contained $0.0378 \mathrm{~g}$ of ${ }^{14} \mathrm{C}\left(\mathrm{C}_{\frac{1}{2}}=5.70 \times 10^{3}\right.$ years $)$. If it was determined that the wood was about 17.100 vears old, how much ${ }^{14} \mathrm{C}$ was present before it was burned? Be sure your answer has the correct number of significant figures.

Step 1 ：Given the half-life of 14C is \(5.70 \times 10^3\) years, we can calculate the decay constant \(\lambda\) using the formula \(\lambda = \frac{\ln(2)}{T_{1/2}}\).

Step 2 ：Substituting the given half-life into the formula, we get \(\lambda = \frac{\ln(2)}{5.70 \times 10^3 \text{ years}} = 1.22 \times 10^{-4} \text{ years}^{-1}\).

Step 3 ：The amount of a radioactive isotope remaining after a certain time can be calculated using the formula \(N = N_0 \times e^{-\lambda t}\), where \(N\) is the final amount of the isotope, \(N_0\) is the initial amount of the isotope, \(\lambda\) is the decay constant, and \(t\) is the time elapsed.

Step 4 ：Substituting the given time (17,100 years), the final amount of 14C (0.0378 g), and the calculated decay constant into the formula, we get \(0.0378 \text{ g} = N_0 \times e^{-1.22 \times 10^{-4} \text{ years}^{-1} \times 17,100 \text{ years}}\).

Step 5 ：Solving for \(N_0\), we get \(N_0 = \frac{0.0378 \text{ g}}{e^{-1.22 \times 10^{-4} \text{ years}^{-1} \times 17,100 \text{ years}}} = \frac{0.0378 \text{ g}}{0.135} = 0.280 \text{ g}\).

Step 6 ：\(\boxed{N_0 = 0.280 \text{ g}}\) is the initial amount of 14C.