Let `barv,v_(rms) and v_p` respectively denote the mean speed. Root mean square speed, and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. Then

Options

A. no molecule can have a speed greater than `sqrt2v_(rms)`
B. no molecule can have speed less than `v_p//sqrt2`
C. `v_pltbarvltv_("rms")`
D. the average kinetic energy of a molecule is `3//4mv_p^2`

Explanation

To solve the question regarding the mean speed (v̄), root mean square speed (v_rms), and most probable speed (v_p) of molecules in an ideal monoatomic gas at absolute temperature T, we will derive the relationships and analyze the options step by step.

### Step 1: Define the speeds
1. **Mean Speed (v̄)**: The mean speed of molecules in an ideal monoatomic gas is given by:
\[ v̄ = \sqrt{\frac{8RT}{\pi m}} \]
where R is the universal gas constant, T is the absolute temperature, and m is the mass of a molecule.

2. **Root Mean Square Speed (v_rms)**: The root mean square speed is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{m}} \]

3. **Most Probable Speed (v_p)**: The most probable speed is given by:
\[ v_p = \sqrt{\frac{2RT}{m}} \]

### Step 2: Analyze the relationships between the speeds
From the equations derived:
- It is known that:
\[ v_p < v̄ < v_{rms} \]
This means that the most probable speed is the lowest, followed by the mean speed, and the root mean square speed is the highest.

### Step 3: Analyze the options
1. **Option 1**: "No molecules can have the speed greater than \( \sqrt{2} v_{rms} \)".
- This is correct because the distribution of speeds in an ideal gas does not allow for speeds greater than this threshold.

2. **Option 2**: "No molecules can have a speed less than \( \frac{v_p}{\sqrt{2}} \)".
- This is also correct since the most probable speed represents a peak in the distribution, and speeds below this value are less likely.

3. **Option 3**: "The average kinetic energy of a molecule is \( \frac{3}{4} m v_p^2 \)".
- To analyze this, we can relate the average kinetic energy (E) to the speeds:
\[ E = \frac{1}{2} m v_{rms}^2 = \frac{3}{2} \cdot \frac{1}{2} m v_p^2 \]
- This implies that the average kinetic energy can also be expressed in terms of \( v_p \), confirming that this statement is correct.

### Conclusion
All the options provided in the question are correct based on the relationships derived from the speeds of the molecules in an ideal monoatomic gas.

### Final Answer
All options are correct.