Radioactive carbon dating can determine how long ago an organism lived by measuring how much of thte `""^(14) C` in the sample has decayed. `""^(14)C` is an isotope of carbon that has a half-life of 5,600 years.Half-life is the amount of time it takes for half of the original amount to decay. If a sample of a petrified tree contains 6.25 percent of its original `""^(14)C,` how long ago did the tree die ?

Options

A. 22,400 years
B. 28,000 years
C. 35,000 years
D. 89,600 years

Answer

To determine how long ago the petrified tree died based on the amount of radioactive carbon-14 remaining, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Life**: The half-life of carbon-14 is given as 5,600 years. This means that every 5,600 years, half of the carbon-14 in a sample decays.

2. **Determine Remaining Carbon-14**: The problem states that the sample contains 6.25% of its original carbon-14. This can be expressed as a fraction:
\[ \text{Remaining amount} = \frac{6.25}{100} = \frac{1}{16} \]

3. **Relate Remaining Amount to Half-Lives**: We know that each half-life reduces the amount of carbon-14 by half. We can express the remaining amount in terms of half-lives:
\[ \frac{1}{16} = \left(\frac{1}{2}\right)^n \]
where \( n \) is the number of half-lives that have passed.

4. **Find the Number of Half-Lives**: To find \( n \), we can equate:
\[ \frac{1}{16} = \left(\frac{1}{2}\right)^4 \]
This tells us that 4 half-lives have passed.

5. **Calculate Total Time**: Since each half-life is 5,600 years, we can calculate the total time \( t \) as follows:
\[ t = n \times \text{half-life} = 4 \times 5600 \]
\[ t = 22400 \text{ years} \]

6. **Conclusion**: Therefore, the petrified tree died approximately 22,400 years ago.

### Final Answer:
The tree died approximately **22,400 years ago**.

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