A radioactive isotope has a half-life of T years. After how much time is its activity reduced to `6.25%` of its original activity ? Given `T _(1//2)=T`

Options

A. 6T
B. 8T
C. 4T
D. 5 T

Answer

To solve the problem of determining the time after which the activity of a radioactive isotope is reduced to 6.25% of its original activity, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We are given that the half-life of the radioactive isotope is \( T \) years.
- We need to find the time \( t \) when the activity \( A(t) \) is reduced to 6.25% of its original activity \( A_0 \).

2. **Expressing the Activity Reduction:**
- The activity at time \( t \) can be expressed as:
\[ A(t) = A_0 \times \left( \frac{1}{2} \right)^{n} \]
- Here, \( n \) is the number of half-lives that have passed.

3. **Setting Up the Equation:**
- We know that 6.25% can be expressed as:
\[ 6.25\% = \frac{6.25}{100} = 0.0625 \]
- Therefore, we set up the equation:
\[ A(t) = 0.0625 A_0 \]
- Substituting into the activity equation:
\[ 0.0625 A_0 = A_0 \times \left( \frac{1}{2} \right)^{n} \]

4. **Cancelling \( A_0 \):**
- Since \( A_0 \) is common on both sides, we can cancel it out:
\[ 0.0625 = \left( \frac{1}{2} \right)^{n} \]

5. **Expressing 0.0625 as a Power of 1/2:**
- We can rewrite 0.0625 as:
\[ 0.0625 = \left( \frac{1}{2} \right)^{4} \]
- Thus, we have:
\[ \left( \frac{1}{2} \right)^{n} = \left( \frac{1}{2} \right)^{4} \]
- This implies:
\[ n = 4 \]

6. **Calculating the Time:**
- Since \( n \) represents the number of half-lives, the total time \( t \) is given by:
\[ t = n \times T = 4 \times T \]

### Final Answer:
The time after which the activity is reduced to 6.25% of its original activity is:
\[ t = 4T \]