The fossil bone has a `.^(14)C` : `.^(12)C` ratio, which is `[(1)/(16)]` of that in a living animal bone. If the half -life of `.^(14)C` is `5730` years, then the age of the fossil bone is :

Options

A. 11460 years
B. 17190 years
C. 22920 years
D. 45840 years

Explanation

To find the age of the fossil bone based on the given ratio of carbon isotopes and the half-life of carbon-14, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Ratio**:
- The ratio of \( ^{14}C \) to \( ^{12}C \) in the fossil bone is given as \( \frac{1}{16} \) of that in a living animal bone.
- Let’s denote the initial amount of \( ^{14}C \) in a living animal bone as \( N_0 \). Therefore, in the fossil bone, the amount of \( ^{14}C \) is \( \frac{N_0}{16} \).

2. **Using the Half-life Formula**:
- The relationship between the remaining quantity of a radioactive substance and its half-life can be expressed as:
\[ N = N_0 \left( \frac{1}{2} \right)^n \]
- Here, \( N \) is the remaining quantity, \( N_0 \) is the initial quantity, and \( n \) is the number of half-lives that have passed.

3. **Setting Up the Equation**:
- From the information given, we can set up the equation:
\[ \frac{N_0}{16} = N_0 \left( \frac{1}{2} \right)^n \]
- Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{16} = \left( \frac{1}{2} \right)^n \]

4. **Expressing \( \frac{1}{16} \) as a Power of 2**:
- We know that \( \frac{1}{16} = \frac{1}{2^4} \). Thus, we can rewrite the equation as:
\[ \frac{1}{2^4} = \left( \frac{1}{2} \right)^n \]
- This implies that:
\[ n = 4 \]

5. **Calculating the Age of the Fossil**:
- The age of the fossil can be calculated using the formula:
\[ \text{Age} = n \times \text{half-life} \]
- Given that the half-life of \( ^{14}C \) is \( 5730 \) years, we can substitute the values:
\[ \text{Age} = 4 \times 5730 = 22920 \text{ years} \]

### Final Answer:
The age of the fossil bone is **22920 years**.