The translational degrees of freedom (\(f_t\)) and rotational degrees of freedom (\(f_r\)) of \( \text{CH}_4 \) molecule are:
- \( f_t = 2 \, \text{and} \, f_r = 2 \)
- \( f_t = 3 \, \text{and} \, f_r = 3 \)
- \( f_t = 3 \, \text{and} \, f_r = 2 \)
- \( f_t = 2 \, \text{and} \, f_r = 3 \)
The Correct Option is B
Approach Solution - 1
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 (ft) are 3, corresponding to motion along the x, y, and z axes.
The rotational degrees of freedom (fr) are also 3, as the molecule can rotate about three mutually perpendicular axes.
Thus, for CH4, we have:
\[ f_{t} = 3 \, \text{and} \, f_{r} = 3. \]
Approach Solution -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.
- 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 \( \text{CH}_4 \), \( f_t = 3 \).
- 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:
\( f_t = 3 \, \text{and} \, f_r = 3 \)
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