Question

The translational degrees of freedom (\(f_t\)) and rotational degrees of freedom (\(f_r\)) of \( \text{CH}_4 \) molecule are:

• A. \( f_t = 2 \, \text{and} \, f_r = 2 \)
• B. \( f_t = 3 \, \text{and} \, f_r = 3 \)
• C. \( f_t = 3 \, \text{and} \, f_r = 2 \)
• D. \( f_t = 2 \, \text{and} \, f_r = 3 \)

Answer 1

The Correct Option is B

Since CH₄ is polyatomic Non-Linear D.O.F of CH₄:
T. DOF = 3 R DOF = 3

The molecule CH₄ (methane) is a polyatomic molecule with a non-linear structure.

For non-linear polyatomic molecules:
The translational degrees of freedom (f_t) are 3, corresponding to motion along the x, y, and z axes.

The rotational degrees of freedom (f_r) are also 3, as the molecule can rotate about three mutually perpendicular axes.

Thus, for CH₄, we have:
\[ f_{t} = 3 \, \text{and} \, f_{r} = 3. \]

Answer 2

To determine the translational (\(f_t\)) and rotational (\(f_r\)) degrees of freedom for the \( \text{CH}_4 \) molecule, we can use some fundamental principles of molecular motion in physics.

1. Molecules can move through space, which accounts for the translational degrees of freedom. Each molecule can move in three dimensions: x, y, and z axes. Hence, all molecules have 3 translational degrees of freedom, regardless of their structure. Therefore, for \( \text{CH}_4 \), \( f_t = 3 \).
2. Next, we consider the rotational degrees of freedom. This corresponds to the molecule's ability to rotate around its center of mass. For linear molecules, there are 2 rotational degrees of freedom, but for non-linear molecules, such as \( \text{CH}_4 \), there are 3 rotational degrees of freedom. \( \text{CH}_4 \) being a non-linear tetrahedral molecule will exhibit rotation about all three spatial axes.

Thus, for a non-linear molecule like \( \text{CH}_4 \):

• Translational degrees of freedom, \( f_t = 3 \).
• Rotational degrees of freedom, \( f_r = 3 \).

This means the correct option is \( f_t = 3 \, \text{and} \, f_r = 3 \).

Therefore, the answer is:

\( f_t = 3 \, \text{and} \, f_r = 3 \)