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**Classical Model of Rigid Rotor**

A particle rotating around a fixed point, as shown below, has angular momentum and rotational kinetic energy (“rigid rotor”) The classical kinetic energy is given by: If the particle is rotating about a fixed point at radius r with a frequency ʋ (s−1 or Hz), the velocity of the particle is given by: where ω is the angular frequency (rad s−1 or rad Hz). The rotational kinetic energy can be now expressed as: Also where

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**Note that there is no potential energy involved in free rotation.**

Consider a classical rigid rotor corresponding to a diatomic molecule. Here we consider only rotation restricted to a 2-D plane where the two masses (i.e., the nuclei) rotate about their center of mass. The rotational kinetic energy for diatomic molecule in terms of angular momentum Note that there is no potential energy involved in free rotation.

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**Momentum Summary Classical QM Linear Momentum Energy Rotational**

(Angular) Momentum Energy

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Angular Momentum

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Angular Momentum

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Angular Momentum

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Angular Momentum

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**Two-Dimensional Rotational Motion**

Polar Coordinates y r f x

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**Two-Dimensional Rotational Motion**

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**Two-Dimensional Rigid Rotor**

Assume r is rigid, ie. it is constant

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**Two-Dimensional Rigid Rotor**

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Solution of equation

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Energy and Momentum As the system is rotating about the z-axis

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**Two-Dimensional Rigid Rotor**

18.0 12.5 E 8.0 4.5 2.0 0.5 Only 1 quantum number is require to determine the state of the system.

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**Spherical coordinates**

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**Spherical polar coordinate**

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**Hamiltonian in spherical polar coordinate**

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**Rigid Rotor in Quantum Mechanics**

Transition from the above classical expression to quantum mechanics can be carried out by replacing the total angular momentum by the corresponding operator: Wave functions must contain both θ and Φ dependence: are called spherical harmonics

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**Schrondinger equation**

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Two equations

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**Solution of second equation**

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**Solution of First equation**

Associated Legendre Polynomial

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**Associated Legendre Polynomial**

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For l=0, m=0

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**First spherical harmonics**

Spherical Harmonic, Y0,0

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l= 1, m=0

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l= 1, m=0 θ cos2θ 1 30 3/4 45 1/2 60 1/4 90

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**l=2, m=0 θ cos2θ 3cos2θ-1 1 2 30 3/4 (9/4-1)=5/4 45 1/2 (3/2-1)=1/2 60**

1 2 30 3/4 (9/4-1)=5/4 45 1/2 (3/2-1)=1/2 60 1/4 (3/4-1)=-1/4 90 -1

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l = 1, m=±1 Complex Value?? If Ф1 and Ф2 are degenerateeigenfunctions, their linear combinations are also an eigenfunction with the same eigenvalue.

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l=1, m=±1 Along x-axis

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**Three-Dimensional Rigid Rotor States**

3 2 1 6.0 -1 -2 -3 E 2 1 3.0 -1 -2 1 1.0 -1 0.5 Only 2 quantum numbers are required to determine the state of the system.

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**Rotational Spectroscopy**

J : Rotational quantum number Rotational Constant

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**Rotational Spectroscopy**

Wavenumber (cm-1) Rotational Constant Line spacing v Dv Frequency (v)

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Bond length To a good approximation, the microwave spectrum of H35Cl consists of a series of equally spaced lines, separated by 6.26*1011 Hz. Calculate the bond length of H35Cl.

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