A wood piece is 11460 years old. What is the fraction of `.^(14)C` activity left in the piece (Half-life period of `.^(14)C` is 5730 years)

Options

A. 0.12
B. 0.25
C. 0.5
D. 0.75

Explanation

To find the fraction of Carbon-14 activity left in a wood piece that is 11,460 years old, we can follow these steps:

### Step 1: Understand the Half-Life Concept
The half-life of Carbon-14 (C-14) is the time it takes for half of the radioactive substance to decay. For C-14, this is given as 5730 years.

### Step 2: Determine the Number of Half-Lives
To find out how many half-lives have passed in 11,460 years, we can divide the total time by the half-life period:
\[ \text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} = \frac{11460 \text{ years}}{5730 \text{ years}} = 2 \]

### Step 3: Calculate the Remaining Activity
After each half-life, the remaining activity is halved. Therefore, after:
- 1 half-life (5730 years), the remaining activity is \( \frac{1}{2} \) of the original activity.
- 2 half-lives (11,460 years), the remaining activity is \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \) of the original activity.

### Step 4: Express the Fraction
Thus, the fraction of C-14 activity left in the wood piece after 11,460 years is:
\[ \text{Fraction of activity left} = \frac{1}{4} = 0.25 \]

### Conclusion
The fraction of Carbon-14 activity left in the wood piece is 0.25, or 25%.

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