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

2. Vibrations of Diatomic Molecules

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

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

5. From the Schrodinger equation:
Fig 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

6. 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

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

8. (accounts for anharmonicity)
Fig 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

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

10. 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

11. 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)

12. S(v+1, J+1) – S(v,J) = ṽ + 2B(J+1)
Fig 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)