The ratio of average translational K.E. and average rotational K.E. of a linear polyatomic molecule at temperature T is:

Options

A. `3`
B. `5`
C. `3//2`
D. `7//5`

Explanation

To find the ratio of average translational kinetic energy (K.E.) to average rotational kinetic energy of a linear polyatomic molecule at temperature T, we can follow these steps:

### Step 1: Understand Degrees of Freedom
A linear polyatomic molecule has degrees of freedom associated with both translational and rotational motion.

- **Translational Degrees of Freedom (F_trans)**: For a linear molecule, there are 3 translational degrees of freedom (movement in x, y, and z directions).
- **Rotational Degrees of Freedom (F_rot)**: For a linear molecule, there are 2 rotational degrees of freedom (rotation about two axes perpendicular to the molecular axis).

### Step 2: Write the Formula for Average K.E.
The average kinetic energy associated with translational and rotational motion can be expressed using the formula:

\[ \text{Average K.E.} = \frac{F}{2} k T \]

where \( F \) is the degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the temperature.

### Step 3: Calculate Average Translational K.E.
Using the degrees of freedom for translational motion:

\[ K.E._{trans} = \frac{3}{2} k T \]

### Step 4: Calculate Average Rotational K.E.
Using the degrees of freedom for rotational motion:

\[ K.E._{rot} = \frac{2}{2} k T = k T \]

### Step 5: Find the Ratio of K.E.s
Now, we need to find the ratio of average translational K.E. to average rotational K.E.:

\[ \text{Ratio} = \frac{K.E._{trans}}{K.E._{rot}} = \frac{\frac{3}{2} k T}{k T} \]

### Step 6: Simplify the Ratio
The \( k T \) terms cancel out:

\[ \text{Ratio} = \frac{3/2}{1} = \frac{3}{2} \]

### Final Answer
Thus, the ratio of average translational K.E. to average rotational K.E. for a linear polyatomic molecule at temperature T is:

\[ \text{Ratio} = \frac{3}{2} \quad \text{or} \quad 3:2 \]