The half life for radioactive decay of `.^(14)C` is 5730 years. An archaeological artifact containing wood had only `80%` of the `.^(14)C` found in a living tree. Estimat the age of the sample.

Explanation

To estimate the age of the archaeological artifact containing wood, we can follow these steps:

### Step 1: Understand the given data
- The half-life of Carbon-14 (\(^{14}C\)) is \(5730\) years.
- The artifact contains \(80\%\) of the \(^{14}C\) found in a living tree.

### Step 2: Set up the ratio of remaining \(^{14}C\)
Since the artifact has \(80\%\) of the original \(^{14}C\), we can express this as:
\[ \frac{N}{N_0} = 0.8 \]
where \(N\) is the amount of \(^{14}C\) in the artifact and \(N_0\) is the amount in a living tree.

### Step 3: Calculate the decay constant \(K\)
The decay constant \(K\) can be calculated using the half-life formula:
\[ K = \frac{0.693}{T_{1/2}} \]
Substituting the half-life:
\[ K = \frac{0.693}{5730} \approx 1.21 \times 10^{-4} \text{ year}^{-1} \]

### Step 4: Use the logarithmic decay formula
We can use the formula relating \(K\), \(T\), and the ratio of \(N_0\) to \(N\):
\[ K = \frac{2.303}{T} \log\left(\frac{N_0}{N}\right) \]
Since \(\frac{N}{N_0} = 0.8\), we have:
\[ \frac{N_0}{N} = \frac{1}{0.8} = 1.25 \]

### Step 5: Substitute values into the equation
Now substituting \(K\) and \(\frac{N_0}{N}\) into the equation:
\[ 1.21 \times 10^{-4} = \frac{2.303}{T} \log(1.25) \]

### Step 6: Calculate \(\log(1.25)\)
Using a calculator:
\[ \log(1.25) \approx 0.0969 \]

### Step 7: Rearranging to find \(T\)
Now we can rearrange the equation to solve for \(T\):
\[ T = \frac{2.303 \times 0.0969}{1.21 \times 10^{-4}} \]

### Step 8: Calculate \(T\)
Calculating the above expression:
\[ T \approx \frac{0.2231}{1.21 \times 10^{-4}} \approx 1845 \text{ years} \]

### Conclusion
The estimated age of the archaeological artifact is approximately \(1845\) years.

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