`._(92)U^(235)` is `alpha` (alpha) active. Then in a large quantity of the element:

Options

A. The probability of a nucleus disintegrating during one second is lower in the first half life and greater in the fifth life
B. The probability of a nuelcues disintegrating during one second remains constant for all time.
C. Quite an appreciable quantity of `U^(235)` will remain even after the average life
D. The energy of the emitted `'alpha'` paticle is less than the disintegration energy of the `U_(235)` nucleus.

Explanation

To solve the question regarding the alpha activity of Uranium-235 (U-235), we will analyze the statements provided and determine their correctness step by step.

### Step-by-Step Solution:

1. **Understanding Alpha Activity**:
U-235 is known to be alpha active, meaning it undergoes alpha decay, emitting alpha particles. This decay process is characterized by a specific half-life, which is the time required for half of the radioactive nuclei in a sample to disintegrate.

2. **Analyzing Statement 1**:
The statement claims that "the probability of a nucleus disintegrating during 1 second is lower in the first half-life and greater in the fifth half-life."
- The disintegration rate is given by the equation:
\[ \frac{dn}{dt} = -n_0 \lambda e^{-\lambda t} \]
where \( n_0 \) is the initial number of nuclei, \( \lambda \) is the decay constant, and \( t \) is time.
- In the early stages (first half-life), fewer nuclei have decayed, so the probability of disintegration is lower. As time progresses (fifth half-life), more nuclei have decayed, leading to a higher probability of disintegration.
- **Conclusion**: Statement 1 is correct.

3. **Analyzing Statement 2**:
The statement claims that "the probability of a nucleus disintegrating during one second remains constant for all time."
- This is incorrect because the decay process is exponential. The probability of disintegration does not remain constant; it decreases over time as fewer nuclei are available to decay.
- **Conclusion**: Statement 2 is incorrect.

4. **Analyzing Statement 3**:
The statement discusses the amount of U-235 remaining after the average life.
- The average life \( T \) is given by \( T = 0.693 \times t_{1/2} \). After one average life, approximately 63.2% of the original nuclei remain (since 36.8% would have decayed).
- Thus, a significant amount of U-235 will still be present after the average life.
- **Conclusion**: This statement is incorrect.

5. **Analyzing Statement 4**:
The statement claims that "the energy of the emitted alpha particle is less than the disintegration energy of the U-235 nucleus."
- If the emitted alpha particle's energy were less than the disintegration energy, energy would need to be supplied for the decay to occur, which contradicts the nature of radioactive decay (which releases energy).
- Therefore, the emitted alpha particle's energy must be greater than the disintegration energy.
- **Conclusion**: This statement is incorrect.

### Final Conclusion:
- The only correct statement is Statement 1.