A radioactive isotope has a half-life of T years. The time required for its activity reduced to 6.25% of its original activity

Options

A. 4T years
B. T years
C. 3T years
D. 2T years

Answer

To solve the problem of determining the time required for the activity of a radioactive isotope to reduce to 6.25% of its original activity, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Half-Life Concept**:
The half-life (T) of a radioactive isotope is the time required for half of the radioactive atoms in a sample to decay.

2. **Determine the Final Activity**:
We know that the final activity (N) is 6.25% of the original activity (N0). This can be expressed mathematically as:
\[ N = N_0 \times \frac{6.25}{100} = \frac{N_0}{16} \]

3. **Use the Decay Formula**:
The formula for radioactive decay is:
\[ N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \]
where \( t \) is the time elapsed and \( T \) is the half-life.

4. **Set Up the Equation**:
Substitute \( N \) from step 2 into the decay formula:
\[ \frac{N_0}{16} = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \]

5. **Cancel \( N_0 \)**:
Since \( N_0 \) is common on both sides, we can cancel it out (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{16} = \left(\frac{1}{2}\right)^{\frac{t}{T}} \]

6. **Express \( \frac{1}{16} \) as a Power of \( \frac{1}{2} \)**:
We know that:
\[ \frac{1}{16} = \left(\frac{1}{2}\right)^4 \]

7. **Set the Exponents Equal**:
Since the bases are the same, we can set the exponents equal to each other:
\[ \frac{t}{T} = 4 \]

8. **Solve for \( t \)**:
Multiply both sides by \( T \) to find \( t \):
\[ t = 4T \]

### Conclusion:
The time required for the activity to reduce to 6.25% of its original value is \( 4T \).

### Final Answer:
The correct option is \( 4T \).