Half - life for radioactive `.^(14)C` is 5760 years. In how many years 200 mg of `.^(14)C` will be reduced to 25 mg ?

Options

A. 17280 years
B. 23040 years
C. 5760 years
D. 11524 years

Answer

To solve the problem of how long it will take for 200 mg of Carbon-14 to reduce to 25 mg, we can use the concept of half-lives in radioactive decay.

### Step-by-Step Solution:

1. **Understand the Half-Life Concept**:
The half-life of a radioactive substance is the time required for half of the substance to decay. For Carbon-14, the half-life is given as 5760 years.

2. **Determine the Initial and Final Amounts**:
- Initial amount (N₀) = 200 mg
- Final amount (N) = 25 mg

3. **Calculate the Number of Half-Lives**:
We need to find how many times the initial amount (200 mg) needs to be halved to reach the final amount (25 mg).
- First half-life: 200 mg → 100 mg
- Second half-life: 100 mg → 50 mg
- Third half-life: 50 mg → 25 mg

Therefore, it takes 3 half-lives to reduce from 200 mg to 25 mg.

4. **Calculate the Total Time**:
Since each half-life is 5760 years, we can calculate the total time taken for 3 half-lives:
\[ \text{Total Time} = \text{Number of Half-Lives} \times \text{Half-Life Duration} \]
\[ \text{Total Time} = 3 \times 5760 \text{ years} = 17280 \text{ years} \]

5. **Final Answer**:
It will take 17,280 years for 200 mg of Carbon-14 to reduce to 25 mg.

### Summary:
The time required for 200 mg of Carbon-14 to decay to 25 mg is **17,280 years**.

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