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Electromagnetic Spectrum
Near Infrared
Thermal Infrared
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Solution of Schrӧdinger Equation for Quantum Harmonic Oscillator
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Harmonic Oscillator
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Hermite polynomial
• Recurrence
Relation: A Hermite
Polynomial at one
point can be
expressed by
neighboring Hermite
Polynomials at the
same point.
xnHxxHxH
xdx
dexH
nnn
n
nxn
n
11
22
22
exp1
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Quantum Mechanical Linear Harmonic Oscillator
It is interesting to calculate probabilities Pn(x) for finding a
harmonically oscillating particle with energy En at x; it is easier
to work with the coordinate q; for n=0 we have:
2
1/ 2
/ 2
2 !
x
n nnx e H x
n
2 2
2 2
2 2
2 2
1/ 22/ 2
0 0 0 0
1/ 2 22/ 2
1 1 1 1
221/ 2
22 / 2
2 2 2 2
231/ 2
23 / 2
3 3 3 3
1 1
2 2
2 112 1
2 2
2 312 3
3 3
q q
q q
q q
q q
q A e P q q e
qq A qe P q q e
qq A q e P q q e
q qq A q q e P q q e
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IR Stretching Frequencies of two bonded atoms:
= frequency
k = spring strength (bond stiffness)
= reduced mass (~ mass of largest atom)
What Does the Frequency, , Depend On?
kh
hE clas2
is directly proportional to the strength of the bonding between
the two atoms ( k)
is inversely proportional to the reduced mass of the two atoms (v 1/)
51
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Stretching Frequencies
• Frequency decreases with increasing atomic weight.
• Frequency increases with increasing bond energy.
52
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IR spectroscopy is an important tool in structural determination of
unknown compound
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IR Spectra: Functional Grps
12
Alkane
Alkene
Alkyne
-C-H C-C
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13
IR: Aromatic Compounds
(Subsituted benzene “teeth”)
C≡C
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14
IR: Alcohols and Amines
CH3CH2OH
Amines similar to OH
O-H broadens with Hydrogen bonding
N-H broadens with Hydrogen bonding
C-O
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Question: A strong absorption band of infrared radiation is observed for 1H35Cl at 2991 cm-1. (a) Calculate the force constant, k, for this molecule. (b) By what factor do you expect the frequency to shift if H is replaced by D? Assume the force constant to be unaffected by this
substitution. [516.3 Nm-1; 0.717]
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CO2, A greenhouse gas ?
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Electromagnetic Spectrum
• Over 99% of solar radiation is in the UV, visible, and near infrared bands
• Over 99% of radiation emitted by Earth and the atmosphere is in the thermal IR band (4 -50 µm)
Near Infrared
Thermal Infrared
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What are the Major Greenhouse Gases?
N2 = 78.1% O2 = 20.9%
H20 = 0-2%
Ar + other inert gases = 0.936%
CO2 = 370ppm
CH4 = 1.7 ppm
N20 = 0.35 ppm
O3 = 10^-8
+ other trace gases
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PY3P05
Molecular vibrations • The lowest vibrational transitions
of diatomic molecules approximate the quantum harmonic oscillator and can be used to imply the bond force constants for small oscillations.
• Transition occur for v = ±1
• This potential does not apply to energies close to dissociation energy.
• In fact, parabolic potential does not allow molecular dissociation.
• Therefore more consider anharmonic oscillator.
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Intensity of spectral lines
• The transition probability between the two states (selection rules)
ififfi d ˆˆ
Only if this integral is non-zero, the transition is allowed
Transition dipole moment
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Selections rules
Electric dipole moment operator
The probability for a vibrational transition to occur, i.e. the intensity of
the different lines in the IR spectrum, is given by the transition dipole
moment fi between an initial vibrational state i and a vibrational final
state f :
ififfi d ˆˆ ...
2
1)( 2
0
2
2
0
0
xx
xx
x
The electric dipole moment operator depends on the location of all electrons and nuclei, so its varies with the modification in the intermolecular distance “x”. 0 is the permanent dipole moment for the molecule in the equilibrium position Re
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The two states i and f are
orthogonal.
Because they are solutions of the
operator H which is Hermitian
0
The higher terms can
be neglected for small
displacements of the
nuclei
...2
1 2
0
2
2
0
0
dxx
dxx
d ifififfi
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dxx
iffi
0
In order to have a vibrational
transition visible in IR
spectroscopy: the electric dipole
moment of the molecule must
change when the atoms are
displaced relative to one another.
Such vibrations are “ infrared
active”. It is valid for polyatomic
molecules.
First condition: fi= 0, if ∂/ ∂x = 0 Second condition: 0 dx if
By introducing the
wavefunctions of the
initial state i and final
state f , which are the
solutions of the SE for an
harmonic oscillator, the
following selection rules is
obtained: = ±1
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Note 1: Vibrations in homonuclear diatomic molecules do not
create a variation of not possible to study them with IR
spectroscopy.
Note 2: A molecule without a permanent dipole moment can be
studied, because what is required is a variation of with the
displacement. This variation can start from 0.
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Spectroscopic selection rule tell us nothing about the
intensities.
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Vibrational modes of CO2
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PY3P05
Anharmonic oscillator • A molecular potential energy curve
can be approximated by a parabola near the bottom of the well. The parabolic potential leads to harmonic oscillations.
• At high excitation energies the parabolic approximation is poor (the true potential is less confining), and does not apply near the dissociation limit.
• Must therefore use a asymmetric potential. E.g., The Morse potential:
where De is the depth of the potential minimum and
V hcDe
1ea(RRe ) 2
a 2
2hcDe
1/ 2
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PY3P05
Anharmonic oscillator
• The Schrödinger equation can be solved for the Morse potential, giving permitted energy levels:
where xe is the anharmonicity constant:
• The second term in the expression for E increases
• with v => levels converge at high quantum numbers.
• The number of vibrational levels for a Morse
oscillator is finite:
v = 0, 1, 2, …, vmax
eeffe
e
Dm
ax
hcxhcE
4
~
2
,...2,1,0;~
2
1~
2
1
2
max
2
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Energy Levels: Basic Ideas
About 15 micron radiation Basic Global Warming: The C02 dance …
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Raman Spectra
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Selection rule for Raman
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Intensity of Raman lines
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Vibrational-Rotational Spectra
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Raman Spectra
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Sources of light
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Absorption Experiment
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Dispersing Element
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Resolving Power
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Resolution
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Diffraction grating
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Blazed Grating
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Calculation
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FT
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Interferometer
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Emission
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Absorption at single wavelength
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http://www.colorado.edu/chemistry/volkamer/teaching/lectures/ Lecture%209%20-%20Light%20sources.pdf
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Questions
• Q4. (i) To a crude first approximation, a electron in linear polyene may be considered to be a particle in a one-dimensional box. The polyene in β- carotene contains 22 conjugated C atoms and the average internuclear distance is 140 pm. Each state upto n = 11 is occupied by two electrons. Calculate (a) the separation energy between the ground state and the first excited state in which one electron occupies the state with n = 12 and (b) The frequency of the radiation required to produce a transition between these two states. (8+2)
• (ii) When β- carotene is oxidized, it breaks into half and forms two molecules of retinal (vitamin A) which is a precursor to the pigment in the retina responsible for vision. The conjugated system for retinal consists of 11 C atoms and one O atom. In the ground state of retinal, each level upto n = 6 is occupied by 2 electrons. Treating everything else to be similar repeat calculations for parts (a) and (b) of the previous problem keeping in mind that in this case the first excited state has one electron in the n = 7 state.
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• Qa. What is the value of n of a particle in a one-dimensional box such that separation between neighbouring levels is equal to ½ kT.
• Qb.