The ratio of r.m.s speed to the r.m.s angular speed of a diatomic gas at certain temperature is (assume m =mass of one molecule, M=molecular mass, I=moment of inertia of the molecules )

Options

A. `sqrt(3/2)`
B. `sqrt((3I)/(2M))`
C. `sqrt((3I)/(2m))`
D. 1

Explanation

To find the ratio of the root mean square (r.m.s) speed to the r.m.s angular speed of a diatomic gas at a certain temperature, we can follow these steps:

### Step 1: Understand the Degrees of Freedom
A diatomic gas has:
- 3 translational degrees of freedom (movement in x, y, and z directions).
- 2 rotational degrees of freedom (rotation about two axes perpendicular to the line connecting the two atoms).

### Step 2: Write the Kinetic Energy for Translational Motion
The kinetic energy associated with translational motion is given by:
\[ KE_{trans} = \frac{1}{2} mv^2 \]
where \( m \) is the mass of one molecule and \( v \) is the r.m.s speed.

### Step 3: Relate Kinetic Energy to Temperature
For a gas with \( f \) degrees of freedom, the average kinetic energy per degree of freedom is given by:
\[ \frac{1}{2} kT \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature. For 3 translational degrees of freedom:
\[ KE_{trans} = \frac{3}{2} kT \]

### Step 4: Set the Two Expressions for Kinetic Energy Equal
Setting the translational kinetic energy equal to the expression in terms of temperature:
\[ \frac{1}{2} mv^2 = \frac{3}{2} kT \]

### Step 5: Write the Kinetic Energy for Rotational Motion
The kinetic energy associated with rotational motion is given by:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the r.m.s angular speed.

### Step 6: Relate Rotational Kinetic Energy to Temperature
For the 2 rotational degrees of freedom:
\[ KE_{rot} = kT \]

### Step 7: Set the Two Expressions for Rotational Kinetic Energy Equal
Setting the rotational kinetic energy equal to the expression in terms of temperature:
\[ \frac{1}{2} I \omega^2 = kT \]

### Step 8: Divide the Two Kinetic Energy Equations
Now, we will divide the translational kinetic energy equation by the rotational kinetic energy equation:
\[ \frac{\frac{1}{2} mv^2}{\frac{1}{2} I \omega^2} = \frac{\frac{3}{2} kT}{kT} \]
This simplifies to:
\[ \frac{mv^2}{I \omega^2} = \frac{3}{2} \]

### Step 9: Rearrange to Find the Ratio
Rearranging gives us:
\[ \frac{v^2}{\omega^2} = \frac{3I}{2m} \]
Taking the square root of both sides:
\[ \frac{v}{\omega} = \sqrt{\frac{3I}{2m}} \]

### Final Result
Thus, the ratio of the r.m.s speed to the r.m.s angular speed of a diatomic gas is:
\[ \frac{v}{\omega} = \sqrt{\frac{3I}{2m}} \]