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 use the principles of the equipartition of energy.

### Step-by-Step Solution:

1. **Understanding Degrees of Freedom**:
- A linear polyatomic molecule has 3 translational degrees of freedom (movement along the x, y, and z axes).
- For rotational motion, a linear molecule has 2 rotational degrees of freedom (it can rotate about two perpendicular axes that are perpendicular to the molecular axis).

2. **Applying the Equipartition Theorem**:
- According to the equipartition theorem, each degree of freedom contributes an average energy of \(\frac{1}{2} k_B T\) to the total energy, where \(k_B\) is the Boltzmann constant and \(T\) is the temperature in Kelvin.

3. **Calculating Average Translational K.E.**:
- For translational motion, the total average translational kinetic energy \(E_t\) is given by:
\[ E_t = \text{(number of translational degrees of freedom)} \times \frac{1}{2} k_B T = 3 \times \frac{1}{2} k_B T = \frac{3}{2} k_B T \]

4. **Calculating Average Rotational K.E.**:
- For rotational motion, the total average rotational kinetic energy \(E_r\) is given by:
\[ E_r = \text{(number of rotational degrees of freedom)} \times \frac{1}{2} k_B T = 2 \times \frac{1}{2} k_B T = k_B T \]

5. **Finding the Ratio of K.E.**:
- Now, we can find the ratio of average translational kinetic energy to average rotational kinetic energy:
\[ \text{Ratio} = \frac{E_t}{E_r} = \frac{\frac{3}{2} k_B T}{k_B T} = \frac{3}{2} \]

6. **Conclusion**:
- Therefore, the ratio of average translational K.E. to average rotational K.E. of a linear polyatomic molecule at temperature T is \( \frac{3}{2} \).

### Final Answer:
The ratio of average translational K.E. to average rotational K.E. of a linear polyatomic molecule at temperature T is \( \frac{3}{2} \).