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 find the mean lifetime of a radionuclide given that its activity decreases by 4% every hour, we can follow these steps:

### Step 1: Understand the decay rate
The activity of a radionuclide decreases by 4% per hour. This means that after one hour, the remaining activity is 96% of the original activity.

### Step 2: Express the decay mathematically
If \( N_0 \) is the initial activity, after one hour, the activity \( N \) can be expressed as:
\[ N = N_0 \times (1 - 0.04) = N_0 \times 0.96 \]

### Step 3: Relate activity to decay constant
The relationship between the decay constant \( \lambda \) and the activity is given by:
\[ \frac{dN}{dt} = -\lambda N \]
Given that the activity decreases by 4% in one hour (3600 seconds), we can express this as:
\[ \frac{dN}{dt} = -0.04 N_0 / 3600 \]

### Step 4: Set up the equation
From the decay equation, we can equate the two expressions for \( \frac{dN}{dt} \):
\[ -\lambda N = -\frac{0.04 N_0}{3600} \]
This simplifies to:
\[ \lambda = \frac{0.04}{3600} \]

### Step 5: Calculate the decay constant
Calculating \( \lambda \):
\[ \lambda = \frac{0.04}{3600} = 1.11 \times 10^{-5} \, \text{s}^{-1} \]

### Step 6: Find the mean lifetime
The mean lifetime \( \tau \) of a radionuclide is given by:
\[ \tau = \frac{1}{\lambda} \]
Substituting the value of \( \lambda \):
\[ \tau = \frac{1}{1.11 \times 10^{-5}} \approx 90000 \, \text{s} \]

### Step 7: Convert seconds to hours
To convert seconds into hours, we divide by the number of seconds in an hour (3600 seconds):
\[ \tau \text{ (in hours)} = \frac{90000}{3600} \approx 25 \, \text{hours} \]

### Final Answer
The mean lifetime of the radionuclide is approximately **25 hours**.

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