The mean life time of a radionuclide, if the activity decrease by `4%` for every `1 h`, would b e(product is non-radioactive, i.e., stable)

Options

A. `25h`
B. `1.042 h`
C. `2h`
D. `30h`

Explanation

To solve the problem, we need to determine the mean lifetime of a radionuclide given that its activity decreases by 4% every hour.

### Step-by-Step Solution:

1. **Understanding Activity Decrease**:
The activity of a radionuclide decreases by 4% every hour. This means that after 1 hour, the remaining activity is 96% of the original activity.

2. **Expressing Activity in Terms of Decay Constant**:
The activity \( A \) of a radionuclide is related to the number of radioactive nuclei \( N \) and the decay constant \( \lambda \) by the formula:
\[ A = \lambda N \]
Since the activity decreases by 4%, we can express this mathematically as:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( A_0 \) is the initial activity.

3. **Setting Up the Equation**:
After 1 hour (which is 3600 seconds), the activity is:
\[ A(3600) = 0.96 A_0 \]
Substituting into the decay equation gives:
\[ 0.96 A_0 = A_0 e^{-\lambda \cdot 3600} \]
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ 0.96 = e^{-\lambda \cdot 3600} \]

4. **Taking the Natural Logarithm**:
To solve for \( \lambda \), take the natural logarithm of both sides:
\[ \ln(0.96) = -\lambda \cdot 3600 \]
Rearranging gives:
\[ \lambda = -\frac{\ln(0.96)}{3600} \]

5. **Calculating the Decay Constant**:
Now, calculate \( \ln(0.96) \):
\[ \ln(0.96) \approx -0.04082 \]
Thus,
\[ \lambda \approx -\frac{-0.04082}{3600} \approx 1.134 \times 10^{-5} \text{ s}^{-1} \]

6. **Finding the Mean Lifetime**:
The mean lifetime \( \tau \) is given by:
\[ \tau = \frac{1}{\lambda} \]
Substituting the value of \( \lambda \):
\[ \tau \approx \frac{1}{1.134 \times 10^{-5}} \approx 88200 \text{ seconds} \]

7. **Converting to Hours**:
To convert seconds to hours:
\[ \tau \approx \frac{88200}{3600} \approx 24.5 \text{ hours} \]

### Final Answer:
The mean lifetime of the radionuclide is approximately **24.5 hours**.