The activity of a radioactive substance becomes from 8000 Bq to 1000 Bq in 12 days . What is the half-life of the radioactive substance ?

Options

A. 4 days
B. 6 days
C. 2 days
D. 3 days

Answer

To find the half-life of a radioactive substance given that its activity decreases from 8000 Bq to 1000 Bq in 12 days, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the initial and final activities**:
- Initial activity, \( A_0 = 8000 \, \text{Bq} \)
- Final activity, \( A_t = 1000 \, \text{Bq} \)
- Time interval, \( t = 12 \, \text{days} \)

2. **Use the radioactive decay formula**:
The activity of a radioactive substance at any time \( t \) can be expressed as:
\[ A_t = A_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

3. **Substitute the known values into the formula**:
\[ 1000 = 8000 e^{-\lambda \cdot 12} \]

4. **Rearrange the equation**:
Divide both sides by 8000:
\[ \frac{1000}{8000} = e^{-\lambda \cdot 12} \]
Simplifying the left side gives:
\[ \frac{1}{8} = e^{-\lambda \cdot 12} \]

5. **Take the natural logarithm of both sides**:
\[ \ln\left(\frac{1}{8}\right) = -\lambda \cdot 12 \]

6. **Use properties of logarithms**:
\[ \ln\left(\frac{1}{8}\right) = -\ln(8) = -\ln(2^3) = -3\ln(2) \]
Therefore, we have:
\[ -3\ln(2) = -\lambda \cdot 12 \]

7. **Solve for the decay constant \( \lambda \)**:
\[ 3\ln(2) = \lambda \cdot 12 \]
\[ \lambda = \frac{3\ln(2)}{12} = \frac{\ln(2)}{4} \]

8. **Find the half-life \( T_{1/2} \)**:
The half-life is related to the decay constant by the formula:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting the value of \( \lambda \):
\[ T_{1/2} = \frac{\ln(2)}{\frac{\ln(2)}{4}} = 4 \, \text{days} \]

### Final Answer:
The half-life of the radioactive substance is **4 days**.