To determine the age of the bone, we will use the concept of half-life, which is the time required for half of the radioactive substance to decay. In the case of Carbon-14 (C-14), its half-life is approximately 5730 years.
The steps to solve the problem are as follows:
1. **Determine the initial quantity of C-14**: According to the information provided, the bone should have contained about 100 grams of C-14 when the animal was alive.
2. **Find the current quantity of C-14**: The bone now contains 12.5 grams of C-14.
3. **Calculate the number of half-lives that have elapsed**: To find out how many half-lives have passed since the bone had 100 grams of C-14, we use the relationship that after one half-life, the substance will have half its initial quantity, after two half-lives, it will have a quarter of its quantity, and so on. Mathematically, this relationship can be represented by the exponential decay formula:
\[ N = N_0 \cdot \left(\frac{1}{2}\right)^{n} \]
where \( N \) is the final amount of the substance, \( N_0 \) is the initial amount of the substance, and \( n \) is the number of half-lives. We can rearrange this to solve for \( n \):
\[ n = \frac{\log(N / N_0)}{\log(1/2)} \]
In our case, \( N_0 = 100 \text{ grams} \) and \( N = 12.5 \text{ grams} \). Substituting these into the equation:
\[ n = \frac{\log(12.5 / 100)}{\log(1/2)} \]
\[ n = \frac{\log(0.125)}{\log(0.5)} \]
\[ n = \frac{-3}{-1} \]
\[ n = 3 \]
So there have been 3 half-lives since the bone was from a living being.
4. **Calculate the age of the bone**: To find the age of the bone, you simply multiply the number of half-lives by the half-life duration of C-14:
\[ \text{Age of the bone} = \text{number of half-lives} \times \text{half-life of C-14} \]
\[ \text{Age of the bone} = 3 \times 5730 \text{ years} \]
\[ \text{Age of the bone} = 17190 \text{ years} \]
Based on these calculations, the age of the bone is 17190 years old.
