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A diatomic molecule can be thought of as a dumbbell: two masses joined together by an ideal spring. The system can oscillate with a frequency w, but it can also rotate about the center of mass. If the molecule rotates what happens to (a) x eq the equilibrium separation? (A) x eq decreases. (B) x eq remains the same. (C) x eq increases. (b) ω, the vibrational frequency? (A) ω decreases. (B) ω remains the same. (C) ω increases.

A diatomic molecule can be thought of as a dumbbell: two masses joined together by an ideal spring. The system can oscillate with a frequency w, but it can also rotate about the center of mass. If the molecule rotates what happens to
(a) xeq the equilibrium separation?
(A) xeq decreases.
(B) xeq remains the same.
(C) xeq increases.
(b) ω, the vibrational frequency?
(A) ω decreases.
(B) ω remains the same.
(C) ω increases.

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
7 years ago
(a) The correct option is (B).
In case of diatomic molecule, if it rotates, the centrifugal force will act on both the molecule equally. So the position of the center of mass will remain unchanged. Thus if the molecule rotates the equilibrium separation xeq will remain the same. Therefore option (B) is correct.
(b) The correct option is (B).
The value center of mass of the system will not change even the molecule rotates about the center of mass. Since the vibrational frequency w depends upon the center of mass, but the value of center mass remains constant, thus the vibrational frequency w remains the same. Therefore option (B) is correct.

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