20 mg of C-14 has half-life of 5760 yr. 100 mg of sample containing C-14 is reduced to 25 mg in

Options

A. 5760 yr
B. 11520 yr
C. 17280 yr
D. 23040 yr

Explanation

To solve the problem of how long it takes for a sample containing C-14 to reduce from 100 mg to 25 mg, we can follow these steps:

### Step-by-Step Solution:

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

2. **Determine the Initial and Final Amounts**:
- Initial amount (N0) = 100 mg
- Final amount (N) = 25 mg

3. **Calculate the Number of Half-Lives**:
To find out how many half-lives it takes to go from 100 mg to 25 mg, we can use the formula:
\[ N = N_0 \times \left(\frac{1}{2}\right)^n \]
where \(n\) is the number of half-lives.
Rearranging the formula gives:
\[ \left(\frac{1}{2}\right)^n = \frac{N}{N_0} \]
Substituting the values:
\[ \left(\frac{1}{2}\right)^n = \frac{25 \text{ mg}}{100 \text{ mg}} = \frac{1}{4} \]

4. **Solve for n**:
Since \(\frac{1}{4} = \left(\frac{1}{2}\right)^2\), we can conclude that:
\[ n = 2 \]
This means it takes 2 half-lives to reduce the amount from 100 mg to 25 mg.

5. **Calculate the Total Time**:
Now, we can calculate the total time taken using the number of half-lives:
\[ \text{Total time} = n \times t_{1/2} = 2 \times 5760 \text{ years} = 11520 \text{ years} \]

### Final Answer:
The time taken for 100 mg of C-14 to reduce to 25 mg is **11520 years**.

---