10 g of a radioactive isotope is reduced to 1.25 g in 12 years. Therefore half-life period of the isotope is

Options

A. 24 years
B. 4 years
C. 3 years
D. 8 years

Explanation

To determine the half-life period of a radioactive isotope that is reduced from 10 g to 1.25 g in 12 years, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Amounts**:
- Initial amount (A₀) = 10 g
- Final amount (A) = 1.25 g

2. **Determine the Total Time**:
- Total time (t) = 12 years

3. **Use the Formula for Radioactive Decay**:
The amount of a radioactive substance remaining after a certain time can be expressed as:
\[ A = A₀ \left(\frac{1}{2}\right)^n \]
where \( n \) is the number of half-lives that have passed.

4. **Rearranging the Formula**:
To find \( n \), we can rearrange the formula:
\[ \frac{A}{A₀} = \left(\frac{1}{2}\right)^n \]
Substituting the values we have:
\[ \frac{1.25}{10} = \left(\frac{1}{2}\right)^n \]

5. **Calculate the Ratio**:
\[ \frac{1.25}{10} = 0.125 \]

6. **Express 0.125 as a Power of 2**:
We can express 0.125 as:
\[ 0.125 = \frac{1}{8} = \left(\frac{1}{2}\right)^3 \]
This means:
\[ \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 \]
Therefore, \( n = 3 \).

7. **Calculate the Half-Life**:
Since \( n \) represents the number of half-lives, and we know the total time is 12 years, we can find the half-life (T₁/₂):
\[ n = \frac{t}{T_{1/2}} \implies T_{1/2} = \frac{t}{n} \]
Substituting the values:
\[ T_{1/2} = \frac{12 \text{ years}}{3} = 4 \text{ years} \]

### Conclusion:
The half-life period of the radioactive isotope is **4 years**.