The activity of carbon -14 in a sample of charcoal from an archaeological site is 0.04 Bq. Determine the age of the sample. The half life of carbon -14 is 5730 years.

Options

A. 10500 y
B. 14500y
C. 12500y
D. 16500y

Answer

To determine the age of the charcoal sample using the activity of carbon-14, we can follow these steps:

### Step 1: Understand the given information
- The activity of carbon-14 in the sample (A) = 0.04 Bq
- The activity of carbon-14 in a living organism (A₀) = 0.25 Bq
- The half-life of carbon-14 (t₁/₂) = 5730 years

### Step 2: Use the decay formula
The relationship between the activity at time \( t \) (A), the initial activity (A₀), and the decay constant (λ) is given by the equation:
\[ A = A_0 e^{-\lambda t} \]

### Step 3: Rearrange the equation
We can rearrange this equation to find the ratio of activities:
\[ \frac{A}{A_0} = e^{-\lambda t} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ \frac{0.04}{0.25} = e^{-\lambda t} \]

### Step 5: Calculate the ratio
Calculating the left side:
\[ \frac{0.04}{0.25} = 0.16 \]
So we have:
\[ 0.16 = e^{-\lambda t} \]

### Step 6: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \ln(0.16) = -\lambda t \]

### Step 7: Find the decay constant (λ)
The decay constant (λ) is related to the half-life (t₁/₂) by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Substituting the half-life of carbon-14:
\[ \lambda = \frac{\ln(2)}{5730} \]

### Step 8: Substitute λ into the equation
Now substitute λ back into the equation:
\[ \ln(0.16) = -\left(\frac{\ln(2)}{5730}\right) t \]

### Step 9: Solve for t
Rearranging gives:
\[ t = -\frac{\ln(0.16) \cdot 5730}{\ln(2)} \]

### Step 10: Calculate the values
Now we can calculate the values:
1. Calculate \( \ln(0.16) \):
\[ \ln(0.16) \approx -1.832 \]
2. Calculate \( \ln(2) \):
\[ \ln(2) \approx 0.693 \]
3. Substitute these values into the equation:
\[ t = -\frac{-1.832 \cdot 5730}{0.693} \]
\[ t \approx \frac{10458.96}{0.693} \approx 15000 \text{ years} \]

### Final Answer
The age of the sample is approximately 15,000 years.

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