Question

Planks from an old boat were found in a cave near the shore. Analysis of the boat's wood gave 13.6 disintegrations/ \(\min / \mathrm{g}\) of carbon. The half-life of \(\mathrm{C}-14\) is 5730 years. Analysis of a tree cut down when the cave was found showed 15.3 disintegrations/min/g of C-14. In what century was the boat built?

Short answer

Answer: The boat was built around the 11th century.

Step-by-step solution

Determine the decay constant \(\lambda\)

The half-life of carbon-14 is given as 5730 years. We can use the decay constant formula: \(\lambda = \frac{ln(2)}{t_{1/2}}\) Where \(t_{1/2}\) is the half-life. Plugging in the value for carbon-14: \(\lambda = \frac{ln(2)}{5730} \approx 1.209 \times 10^{-4}\ \mathrm{year^{-1}}\)

Calculate the initial and remaining ratios

We are given the disintegrations per minute per gram of carbon: - For the wooden planks: \(13.6\ (\mathrm{dpm/g})\) - For the tree (when the cave was found): \(15.3\ (\mathrm{dpm/g})\) Let's define the ratio between remaining and initial disintegrations as: \(R = \frac{N_t}{N_0} = \frac{13.6}{15.3} = 0.8889\)

Find the elapsed time

We have the decay formula, \(N_t = N_0 e^{-\lambda t}\), and we know the ratio \(R\). We can rewrite the decay formula in terms of elapsed time and the ratio \(R\): \(\frac{N_t}{N_0} = e^{-\lambda t}\) Substitute \(R\) into the equation: \(0.8889 = e^{(-1.209 \times 10^{-4})t}\) Now, we'll take the natural logarithm of both sides to isolate \(t\): \(ln(0.8889) = -1.209 \times 10^{-4}t\) And then, solve for \(t\): \(t = \frac{ln(0.8889)}{-1.209 \times 10^{-4}} \approx 904.962\ \mathrm{years}\)

Determine the century in which the boat was built

Since the age of the wooden planks is approximately 905 years, we will calculate the century in which the boat was built. Let's assume the cave was found in the year 2000. Century on which the boat was built: \((2000-905) = 1095\ \mathrm{A.D.}\) So, the boat was built around the 11th century.