Rotational and Vibrational Spectra Spectroscopy 1: Rotational and Vibrational Spectra CHAPTER 13

Vibrations of Diatomic Molecules

Fig 9.21 Energy levels of a harmonic oscillator Solving Gives: where: Zero point energy

Fig 13.26 Molecular potential energy curve: approximates parabola only near bottom of well k ≡ force constant

From the Schrodinger equation: Fig 13.27 The force constant is a measure of the curvature of the potential energy From the Schrodinger equation: stiff bond where: v = 0, 1, 2,... Vibrational terms: flexible bond

Fig 13.28 The oscillation of a molecule may result in an oscillating dipole Gross selection rule: The electric dipole of molecule must change when atoms are displaced Such vibrations are said to be: infrared active Specific selection rule: Δv = ±1 If the dipole does not change: infrared inactive

Fig 13.29 Difference between well-depth De and dissociation energy D0 Zero-point energy

(accounts for anharmonicity) Fig 13.30 Morse potential energy curve (accounts for anharmonicity) Morse potential energy: where: Permitted energy levels: G(v) = (v+½) ṽ - (v+½)2χe ṽ where: anharmonicity constant

Fig 13. 31 Dissociation energy is the sum of energy level Fig 13.31 Dissociation energy is the sum of energy level separations up to its dissociation limit

Vibration-Rotation Spectra Each line of a high-resolution vibrational spectrum consists of many closely-space lines Molecular spectra often called band spectra Structure is due to rotational transitions accompanying a specific vibrational transition A molecular rotation is enhanced or retarded by a vibrational transition When vibrational transition occurs, ΔJ changes by ±1 and sometimes 0 when ΔJ = 0 is allowed

Fig 13.34 High resolution vibration-rotation spectrum of HCl for a v + 1 ← v transition Combined vib-rot terms, S: S(v, J) = G(v) + F(J) = (v+½) ṽ + BJ(J+1)

S(v+1, J+1) – S(v,J) = ṽ + 2B(J+1) Fig 13.35 Formation of P, Q, and R branches in vib-rot spectra When v+1 ← v occurs, ΔJ = ±1 (and sometimes 0) S(v+1, J-1) – S(v,J) = ṽ - 2BJ S(v+1, J) – S(v,J) = ṽ S(v+1, J+1) – S(v,J) = ṽ + 2B(J+1)