Activity of a radioactive substance decreases from 8000 Bq to 1000 Bq in 9 days. What is the half-life and average life of the radioactive substance?

Explanation

To solve the problem, we need to find both the half-life and the average life of a radioactive substance given that its activity decreases from 8000 Bq to 1000 Bq in 9 days.

### Step-by-Step Solution:

1. **Understanding the decay of activity**:
The activity \( A \) of a radioactive substance can be expressed as:
\[ A = A_0 \left( \frac{1}{2} \right)^n \]
where \( A_0 \) is the initial activity, \( A \) is the final activity, and \( n \) is the number of half-lives.

2. **Setting up the equation**:
Given:
- Initial activity \( A_0 = 8000 \, \text{Bq} \)
- Final activity \( A = 1000 \, \text{Bq} \)
- Time period \( t = 9 \, \text{days} \)

We can set up the equation:
\[ 1000 = 8000 \left( \frac{1}{2} \right)^n \]

3. **Solving for \( n \)**:
Dividing both sides by 8000:
\[ \frac{1000}{8000} = \left( \frac{1}{2} \right)^n \]
Simplifying the left side:
\[ \frac{1}{8} = \left( \frac{1}{2} \right)^n \]
Recognizing that \( \frac{1}{8} = \left( \frac{1}{2} \right)^3 \), we can equate the exponents:
\[ n = 3 \]

4. **Finding the half-life**:
The total time for \( n \) half-lives is given as:
\[ n \cdot t_{1/2} = 9 \, \text{days} \]
Substituting \( n = 3 \):
\[ 3 \cdot t_{1/2} = 9 \]
Solving for \( t_{1/2} \):
\[ t_{1/2} = \frac{9}{3} = 3 \, \text{days} \]

5. **Calculating the average life**:
The average life \( T_{avg} \) is related to the half-life by the formula:
\[ T_{avg} = \frac{t_{1/2}}{\ln 2} \]
Using \( \ln 2 \approx 0.693 \):
\[ T_{avg} = \frac{3}{0.693} \approx 4.32 \, \text{days} \]

### Final Answers:
- Half-life \( t_{1/2} = 3 \, \text{days} \)
- Average life \( T_{avg} \approx 4.32 \, \text{days} \)