Question

For a particular diatomic molecule, the vibrational frequency is \omega = 1,821.4 cm–1, and its rotational constant is = 5.47 cm–1. If the molecule can be treated as a harmonic oscillator and rigid rotor, then E(n, j) = \omega (n + 1/2) + B j(j + 1). Use this information to calculate the energy of the state E(0,26) in units of cm–1.

Name: for a particular diatomic molecule the vibrational frequency is omega 18214 cm1 and its rotational constant is 547 cm1 if the molecule can be treated as a harmonic oscillator and rigid rotor 77514
Uploaded: 2023-11-23T10:11:46-08:00
Duration: 3 min 22 s
Channel: Adi S
Description: for a particular diatomic molecule the vibrational frequency is omega 18214 cm1 and its rotational constant is 547 cm1 if the molecule can be treated as a harmonic oscillator and rigid rotor 77514

          For a particular diatomic molecule, the vibrational frequency is \omega  = 1,821.4 cm–1, and its rotational constant is  = 5.47 cm–1. If the molecule can be treated as a harmonic oscillator and rigid rotor, then E(n, j) =  \omega  (n + 1/2) + B j(j + 1). Use this information to calculate the energy of the state E(0,26) in units of cm–1.

Added by Joshua D.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Adi S

11/23/2023

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For a particular diatomic molecule, the vibrational frequency is \omega = 1,821.4 cm–1, and its rotational constant is = 5.47 cm–1. If the molecule can be treated as a harmonic oscillator and rigid rotor, then E(n, j) = \omega (n + 1/2) + B j(j + 1). Use this information to calculate the energy of the state E(0,26) in units of cm–1.