A sample of radioactive substance loses half of its activity in 4 days. The time in which its activity is reduced to 5% is

Options

A. (a) 12 days
B. (b) 8.3 days
C. (c) 17.3 days
D. (d) None of these

Explanation

To solve the problem of how long it takes for the activity of a radioactive substance to reduce to 5% of its original value, given that it loses half of its activity in 4 days, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Half-Life**:
The half-life (T_half) of the substance is given as 4 days. This means that in 4 days, the activity of the substance will reduce to half of its initial value.

2. **Calculate the Decay Constant (λ)**:
The decay constant (λ) can be calculated using the formula:
\[ λ = \frac{\ln(2)}{T_{half}} \]
Substituting the value of T_half:
\[ λ = \frac{\ln(2)}{4 \text{ days}} \]

3. **Set Up the Activity Equation**:
The activity A at time t can be expressed as:
\[ A = A_0 e^{-λt} \]
where \( A_0 \) is the initial activity.

4. **Determine the Final Activity**:
We want to find the time when the activity reduces to 5% of its initial value:
\[ A = 0.05 A_0 \]
Substituting this into the activity equation gives:
\[ 0.05 A_0 = A_0 e^{-λt} \]

5. **Cancel \( A_0 \)**:
Since \( A_0 \) is common on both sides, we can cancel it out:
\[ 0.05 = e^{-λt} \]

6. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln(0.05) = -λt \]

7. **Substitute for λ**:
Substitute \( λ = \frac{\ln(2)}{4} \):
\[ \ln(0.05) = -\frac{\ln(2)}{4} t \]

8. **Solve for t**:
Rearranging gives:
\[ t = -\frac{4 \ln(0.05)}{\ln(2)} \]

9. **Calculate \( \ln(0.05) \) and \( \ln(2) \)**:
Using approximate values:
\[ \ln(0.05) \approx -2.9957, \quad \ln(2) \approx 0.693 \]
Substituting these values:
\[ t = -\frac{4 \times (-2.9957)}{0.693} \approx \frac{11.9828}{0.693} \approx 17.3 \text{ days} \]

### Final Answer:
The time in which the activity is reduced to 5% is approximately **17.3 days**.

---