A wooden article and a freshly cut tree show activity 7.6 counts `"min"^(-1) g^(-1)` and 15.2 counts `"min"^(-1) g^(-1)` of carbon (`t_(1//2) = 5760` years) respectively . The age of the article is :

Options

A. 5760 years
B. `5760 xx (15.2)/(6)` years
C. `5760 xx (7.6)/(15.2)` years
D. `5760 xx (15.2 - 7.6)` year

Explanation

To determine the age of the wooden article, we can use the concept of radioactive decay, specifically the decay of carbon-14. The activity of a sample is directly proportional to the amount of carbon-14 present, and we can use the following formula to find the age:

\[ t = \frac{t_{1/2}}{0.693} \cdot \log\left(\frac{A_0}{A}\right) \]

Where:
- \( t \) = age of the sample
- \( t_{1/2} \) = half-life of carbon-14 (5760 years)
- \( A_0 \) = activity of the fresh cut tree (15.2 counts min\(^{-1}\) g\(^{-1}\))
- \( A \) = activity of the wooden article (7.6 counts min\(^{-1}\) g\(^{-1}\))

### Step-by-Step Solution:

1. **Identify the given values:**
- \( A_0 = 15.2 \) counts min\(^{-1}\) g\(^{-1}\)
- \( A = 7.6 \) counts min\(^{-1}\) g\(^{-1}\)
- \( t_{1/2} = 5760 \) years

2. **Calculate the ratio of activities:**
\[ \frac{A_0}{A} = \frac{15.2}{7.6} = 2 \]

3. **Use the logarithm in the formula:**
\[ \log\left(\frac{A_0}{A}\right) = \log(2) \]

4. **Calculate \( \log(2) \):**
\[ \log(2) \approx 0.301 \]

5. **Substitute the values into the age formula:**
\[ t = \frac{5760}{0.693} \cdot 0.301 \]

6. **Calculate \( \frac{5760}{0.693} \):**
\[ \frac{5760}{0.693} \approx 8305.2 \]

7. **Now calculate the age \( t \):**
\[ t \approx 8305.2 \cdot 0.301 \approx 2500.0 \text{ years} \]

8. **Final result:**
The age of the wooden article is approximately **2500 years**.