Post on 18-Dec-2015
Light behaves like an oscillating electromagnetic field. The electric field interacts with charges. Two charges separated in space represent a dipole moment which can interact with an electric field. Energy can only be taken or added to the electric field in units of hn (photons).
Assumed knowledge
Learning outcomes from lecture 5• Be able to draw the wavefunctions for the first few solutions to the Schrödinger
equation for the harmonic oscillator• Be able to calculate the energy separation between the vibrational levels for the
harmonic oscillator• Be able to explain the dependence of this separation on the masses of the
atoms and the strength of the bond• Be aware of and be able to calculate the zero point energy of a vibrating
molecule• Be able to draw the potential energy curve for a real molecule and explain why
it is different to the harmonic potential
Revision: The Electromagnetic Spectrum
Spectral range
l (nm) n (Hz) (cm-1) Energy (kJ/mol)
Spectroscopy
Radio ~1 x 109 ~108 ~0.03 ~10-8 NMR/ESR
Microwave ~100,000 ~1012 ~30 ~10-2 Rotational
Infrared ~1000 ~1014 ~3,000 ~103 Vibrational
Visible 400-750 4-6 x 1014 14,000-25,000
1 – 3x105 Electronic
Ultraviolet 100-400 ~1015 ~40,000 ~5x105 Electronic
X ray <100 >1016 >100,000 >106 Core electronic
~
Quantity Symbol SI Unit Common Unit
Energy E J kJ/mol
Frequency n s-1 or Hz
s-1 or Hz
Wavelength l m nm or mm
Wavenumber m-1 cm-1
Revision: Fundamental equations
hE
c
c
1~ ~
Constant Symbol Value
Speed of light c 3.00 x 108 m/s
Planck constant h 6.626 x 10-34 Js
Classical absorption of light
Because light is an oscillating electromagnetic field, it can cause charges to oscillate. If the charge can oscillate in resonance with the field then energy can be absorbed.
Alternatively, an oscillating charge can emit radiation with frequency in resonance with the original oscillation.
For example, in a TV antenna, the oscillating EM field broadcast by the transmitter causes the electrons in your antenna to oscillate at the same frequency.
Classical absorption of lightWhat about a molecule as an antenna?
How can we get oscillating charges in a molecule?
+ -+-
The Quantum Harmonic OscillatorSolution of Schrödinger equation:
hnn )( 21 S.I. units
)v()v( 21G G(v) & w in cm-1
G(v) is the “energy” from the bottom of the well and w is the “harmonic frequency”
2
1
2n
khn
Oscillation (vibrational) frequencyn = 0,1,2,…
k
2
1But:
Need to know!
( ) ( )e
ee D
G4
+v+v=)v(2
22
12
1
( ) ( ) eee xG 22
12
1 +v+v=)v(
This one you DO have to know how to use
all in wavenumber (cm-1) units:
In spectroscopy, we tend to use the letter “vee” to indicate the quantum number for vibration. The vibration frequency is indicated by we in cm-1. When solving the general quantum mechanical problem we used the letter n, to minimize confusion with “nu”, the vibrational frequency in s-1.
The Morse Anharmonic Oscillator
The Morse energy levels
eee xG 22
12
1 vv)v(
r
V (r
)
Harmonic term
Anharmonic term
Harmonic frequency
Anharmonicity constant
wexe is usually positive, so vibrational levels get closer together
k
2
1
k
c2
1=
The force constant for the AHO is the same as for the harmonic case:
or
Different levels of approximation…• General AHO:
...vvv)v(3
212
21
21 eeeee yxG
• Morse oscillator:
eee xG 22
12
1 vv)v(
• Harmonic oscillator:
21v)v( G
The level of approximation to use depends on:i) the information you haveii) the information you need
The important equations!
( ) ( )e
ee D
G4
+v+v=)v(2
22
12
1
( ) ( ) eee xG 22
12
1 +v+v=)v(
The ones you DO have to know how to use
• In wavenumber (cm-1) units:
ee
ee xD
4
2
ee
ee xD
4
2
This is the energy from the bottom of the well to the dissociation limit.
r
V (
r)
De
D0
G(0)
(Zero-point energy)
D0 = De – G(0)
But we know about zero-point energy, therefore slightly less energy is required to break the bond.
We can estimate the bond dissociation energy from spectroscopic measurements!
Dissociation energy
Selection rules• All forms of spectroscopy have a set of selection rules
that limit the number of allowed transitions.• Selection rules arise from the resonance condition, which
may be expressed as a transition dipole moment:
0)( )()( 1*221 drrrr μμ
Upper state
lower state
molecular dipole
vibrational coordinate
• Selection rules tell us when this integral is zero
selection rule:Dv=±1
Dv=+1: absorptionDv=-1: emission
Selection rules limit the number of allowed transitions
Harmonic Oscillator Selection Rules
If an oscillator has only one frequency associated with it, then it can only interact with radiation of that frequency.
Transitions arising from v0 are called “hot bands”(Their intensity is strongly
temp. dependent)
Much of IR spectroscopy can understood from just these two results of the quantum harmonic oscillator:
E = (v+½)hn and Dv = ±1
Thermal populationAt normal temperatures, only the lowest vibrational state(v =0 ) is usually populated, therefore, only the first transition is typically seen.
2000 4000 6000 8000
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
63
52 c
m-1
(seco
nd o
vert
one)
42
60 c
m-1
(first overt
one)
21
43 c
m-1, (
fundam
enta
l) CO
x 100x 10
Ab
sorb
an
ce
Wavenumber (cm-1)
More problems with harmonic model….
COWhat are these?
What are the new peaks?• Three peaks…
i. 2143 cm-1
ii. 4260 cm-1
iii. 6352 cm-1
v=0
v=1
v=2
v=3
v=4
These are almost 1 : 2 : 3
which suggests transitions might be
0 10 20 3
Anharmonic oscillator (A.H.O.) selection rules:
H.O.
A.H.O.
There are none!
But!...Harmonic and anharmonic models are very similar at low energy, so selection rules of AHO converge on HO as the anharmonicity becomes less:
A.H.O. selection rule:Dv=±1,±2, ±3
Intensity gets weaker and weaker (typically 10× weaker for each)
Anharmonic oscillator (A.H.O.)
A.H.O. selection rule:Dv= ±1,±2, ±3
Dv = 1 : fundamentalDv = 2 : first overtoneDv = 3 : second overtone, etc
• Consider the infrared absorption spectrum of CO below.a) From the wavenumber measurements on the spectrum, assign the
spectrum, hence determine the harmonic frequency, we (in cm-1) and the anharmonicity constant wexe (in cm-1).
b) Estimate the bond dissociation, D0, for this molecule. There is no absorption below 2000 cm-1.
Fundamental:2143 = G(1) – G(0),
Overtone:4260 = G(2) – G(0)
Typical Exam Question
Using the spectra to get information…
G(1)-G(0) = [(1.5)we – (1.5)2 wexe] - [(0.5)we – (0.5)2 wexe]
2143 = we – 2wexe …(1)
G(2)-G(0) = [(2.5)we – (2.5)2 wexe] - [(0.5)we – (0.5)2 wexe]
4260 = 2we – 6wexe …(2)
Two simultaneous equations (simple to solve)
→ we = 2169 cm-1, and wexe = 13 cm-1
eee xG 22
12
1 vv)v(
Using the spectra to get information…
1-2
cm 500,90134
)2169(
eD
1-
0
cm 400,891080500,90
)0(
GDD e
89,400 cm-1 = 1069 kJ/molc.f. exp. value: 1080 kJ/mol
r
V (
r)
De
D0
G(0)
(Zero-point energy)
Why the difference?Remember Morse is still an approx. to the true intermolecular potential. Still 2% error is pretty good for just 2 measurements!
ee
ee xD
4
2
Equations to know how to use…
eee xG 22
12
1 vv)v(
)v()v( 21G
k
2
1m
m1 × m2
m1 + m2
ee
ee xD
4
2)0(0 GDD e
Learning outcomes• Be able to manipulate and use the key equations given in the
green box at the end of the lecture.• Utilize the harmonic oscillator and anharmonic oscillator as a
model for the energy level structure of a vibrating diatomic molecule.
Next lecture
• The vibrational spectroscopy of polyatomic molecules.
Week 11 homework• Work through the tutorial worksheet and the practice problems at
the end of the lectures and check your answers with those available online
• Play with the “IR Tutor” in the 3rd floor computer lab and with the online simulations:
http://assign3.chem.usyd.edu.au/spectroscopy/index.php
Practice Questions1. Which of the following diatomic molecules will exhibit an infrared
spectrum?a) HBr b) H2 c) CO d) I2
2. An unknown diatomic oxide has a harmonic vibrational frequency of ω = 1904 cm−1 and a force constant of 1607 N m −1. Identify the molecule.b) CO b) BrO c) NO d) 13CO
3. As the energy increases, the vibrational level spacing for a harmonic oscillator is:c) increases b) decreases c) stays constant
4. As the energy increases, the vibrational level spacings for a Morse oscillator usually:d) increase b) decrease c) stay constant
5. As the energy increases, the vibrational level spacings for an anharmonic oscillator usually:e) increase b) decrease c) stay constant
Practice Questions6. For a Morse oscillator the observed dissociation energy, D0, is related to
the equilibrium vibrational frequency and the anharmonicity by the following expression:a) ωe
2/4ωexe b) [ωe2/4ωexe]-G(0) c) (v+½)ωe+(v+½)2ωexe d) (v+½)ωe
7. Which of the following statements about the classical and quantum harmonic oscillator (HO) are true (more than one possible answer here)?a) The classical HO frequency is continuous, whereas the quantum frequency is discrete.b) The classical HO has continuous energy levels, whereas the quantum HO levels are discrete.c) The classical HO depends on the force constant, but the quantum HO does not.d) The classical HO may have zero energy, but the quantum HO may not.e) The classical HO does allow the bond to break, whereas the quantum HO does.
Practice Questions8. Which of the following statements correctly describe features of the
quantum Morse oscillator and energy levels associated with it?a) The vibrational energy levels get more closely spaced with increasing v.b) The vibrational energy levels approach a continuum as the dissociation energy is approached.c) The Morse oscillator and HO are nearly the same at very low vd) The Morse potential is steeper than the HO for r < re
e) The Morse potential exactly describes the interatomic potential.