A piece of wood from an archaeological source shows a `.^(14)C` activity which is `60%` of the activity found in fresh wood today. Calculate the age of the archaeological sample. (`t_(1//2)` for `.^(14)C = 5570` year)

Explanation

To calculate the age of the archaeological sample, we can follow these steps:

### Step 1: Determine the decay constant (λ)
The decay constant (λ) can be calculated using the half-life (t₁/₂) of Carbon-14, which is given as 5570 years. The formula for the decay constant is:

\[ \lambda = \frac{0.693}{t_{1/2}} \]

Substituting the value of the half-life:

\[ \lambda = \frac{0.693}{5570} \approx 1.243 \times 10^{-4} \text{ year}^{-1} \]

### Step 2: Set up the relationship between initial and remaining activity
We know that the activity of the archaeological sample (Nₜ) is 60% of the activity of fresh wood today (N₀):

\[ N_t = 0.60 N_0 \]

### Step 3: Use the radioactive decay formula
The formula relating the initial and remaining quantities over time is:

\[ N_t = N_0 e^{-\lambda t} \]

We can rearrange this to find the time (t):

\[ \frac{N_t}{N_0} = e^{-\lambda t} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{N_t}{N_0}\right) = -\lambda t \]

### Step 4: Substitute the values
Substituting \(N_t = 0.60 N_0\):

\[ \ln(0.60) = -\lambda t \]

Now substituting for λ:

\[ \ln(0.60) = -\left(1.243 \times 10^{-4}\right) t \]

### Step 5: Solve for time (t)
Now we can solve for t:

\[ t = -\frac{\ln(0.60)}{1.243 \times 10^{-4}} \]

Calculating \(\ln(0.60)\):

\[ \ln(0.60) \approx -0.5108 \]

Substituting this value:

\[ t = -\frac{-0.5108}{1.243 \times 10^{-4}} \approx 4115.5 \text{ years} \]

### Step 6: Final calculation and rounding
Rounding to the nearest year, we find:

\[ t \approx 4116 \text{ years} \]

### Conclusion
The age of the archaeological sample is approximately **4116 years**.

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