Question

A sample of a wooden artifact gives 5.0 disintegrations/ min/g carbon. The half-life of C-14 is 5730 years, and the activity of C-14 in wood just cut down from a tree is 15.3 disintegrations/min/g carbon. How old is the wooden artifact?

Short answer

Answer: The approximate age of the wooden artifact is 4630 years.

Step-by-step solution

Understand the given information.

The information given in the problem is as follows: 1. Current activity of the wooden artifact: 5.0 disintegrations/min/g carbon 2. Half-life of C-14: 5730 years 3. Activity of C-14 in a freshly cut tree: 15.3 disintegrations/min/g carbon We need to find the age of the wooden artifact.

Use the half-life formula.

The half-life formula for radioactive decay is given by: \( n = n_0 \times 2^{-\frac{t}{t_{1/2}}}\) Where \(n\) is the current activity level, \(n_0\) is the initial activity level, \(t\) is the time elapsed (the age of the artifact), \(t_{1/2}\) is the half-life of the isotope. We need to find \(t\) using the given information.

Substitute the known values.

We can now plug in the given values into the half-life formula: \( 5.0 = 15.3 \times 2^{-\frac{t}{5730}}\)

Solve for the age t.

To solve for \(t\), we need to isolate it by following these steps: 1. Divide both sides by 15.3 2. Take the natural logarithm of both sides. 3. Multiply both sides by -5730. $$ \frac{5.0}{15.3} = 2^{-\frac{t}{5730}} $$ $$ ln(\frac{5.0}{15.3}) = ln(2^{-\frac{t}{5730}}) $$ $$ -5730 \cdot ln(\frac{5.0}{15.3}) = -5730 \cdot ln(2) \cdot \frac{t}{5730} $$ Now, we simplify and solve for \(t\): $$ t = \frac{-5730 \cdot ln(\frac{5.0}{15.3})}{ln(2)} $$

Calculate the age t.

With the known values, we can now calculate the age of the wooden artifact: $$ t = \frac{-5730 \cdot ln(\frac{5.0}{15.3})}{ln(2)} \approx 4630.06\,\text{years} $$ The age of the wooden artifact is approximately 4630 years.