In one average life (assuming the decay to proceed according to radioactive law)

Options

A. Half the active nuclei decay
B. Less than half the active nuclei decay
C. More than half the active nuclei decay
D. All the nuclei decay

Answer

To solve the question regarding radioactive decay and average life, we will follow these steps:

### Step 1: Understand the concept of average life
The average life (or mean life) of a radioactive substance is defined as the average time that a nucleus of the substance will exist before it decays. It is denoted by \( t_{avg} \) and is related to the decay constant \( \lambda \) by the formula:
\[ t_{avg} = \frac{1}{\lambda} \]

### Step 2: Use the radioactive decay law
The number of undecayed nuclei \( N \) at time \( t \) is given by the equation:
\[ N = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei and \( e \) is the base of the natural logarithm (approximately 2.718).

### Step 3: Substitute \( t \) with \( t_{avg} \)
Now, we will substitute \( t \) with \( t_{avg} \):
\[ N = N_0 e^{-\lambda \cdot t_{avg}} = N_0 e^{-\lambda \cdot \frac{1}{\lambda}} = N_0 e^{-1} \]

### Step 4: Calculate \( N \)
Since \( e^{-1} \) is approximately \( \frac{1}{e} \) (or about 0.3679), we can express the remaining number of nuclei as:
\[ N \approx \frac{N_0}{e} \approx \frac{N_0}{2.718} \]

### Step 5: Compare \( N \) with \( N_0 \)
Now, we can compare \( N \) with \( N_0 \):
\[ N_0 > N \quad \text{(since } \frac{N_0}{e} < \frac{N_0}{2} \text{)} \]
This implies that the number of undecayed nuclei after one average life is less than the initial number of nuclei.

### Step 6: Conclusion
Since more than half of the active nuclei have decayed after one average life, the correct answer to the question is:
**More than half of the active nuclei decay.**