Electromagnetic Spectrum
Near Infrared
Thermal Infrared
Solution of Schrӧdinger Equation for Quantum Harmonic Oscillator
Harmonic Oscillator
Hermite polynomial
• Recurrence
Relation: A Hermite
Polynomial at one
point can be
expressed by
neighboring Hermite
Polynomials at the
same point.
H n x    1 e
n
x2

dn
2
exp

x
dx n

H n 1 x   2 xH n x   2nH n 1 x 
Quantum Mechanical Linear Harmonic Oscillator
1/ 2

 
 n  x    n

 2 n!  
e x
2
/2


Hn x 
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:
1/ 2
 1 
 0  q   A0 

  
e q
2
/2
1/ 2
 2 
 1  q   A1 

  
qe
1/ 2
 1 
 2  q   A2 

2  
1/ 2
 1 
 3  q   A3 

3  
 q2 / 2
 2q
 2q
2
3
 P0  q    0  q  
1
 P1  q    1  q  
2q 2
2
2
 1 e  q
2
/2
 3q  e  q
2


eq
eq
 P2  q    2  q 
/2
2
2
 P3  q    3  q 
2
 2q

 1
2
2 
2
 2q

3
2
eq
 3q 
3 
2
2
eq
2
IR Stretching Frequencies of two bonded atoms:
What Does the Frequency, , Depend On?
E  h clas
h

2
k

 = frequency
k = spring strength (bond stiffness)
 = reduced mass (~ mass of largest atom)
 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
Stretching Frequencies
• Frequency decreases with increasing atomic weight.
• Frequency increases with increasing bond energy.
52
IR spectroscopy is an important
tool in structural determination of
unknown compound
IR Spectra: Functional Grps
Alkane
-C-H
C-C
Alkene
Alkyne
12
IR: Aromatic Compounds
(Subsituted benzene “teeth”)
C≡C
13
IR: Alcohols and Amines
O-H broadens with Hydrogen bonding
CH3CH2OH
C-O
Amines similar to OH
N-H broadens with Hydrogen bonding
14
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]
CO2, A greenhouse gas ?
Electromagnetic Spectrum
Near Infrared
Thermal Infrared
•
•
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)
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
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
anharmonic oscillator.
consider
PY3P05
Intensity of spectral lines
• The transition probability between the two
states (selection rules)
Transition dipole moment
 fi    f ˆ i d   f ˆ i
Only if this integral is non-zero, the transition
is allowed
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 :
 fi    f ˆ i d   f ˆ i
1  2  2
  
 ( x)   0    x   2  x  ...
2  x 0
 x 0
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
1  2 
  
 fi   0   f i d      f xi d   2    f x 2i d  ...
2  x 0
 x 0
0
The two states i and f are
orthogonal.
Because they are solutions of the
operator H which is Hermitian
The higher terms can
be neglected for small
displacements of the
nuclei
  
 fi      f xi d
 x 0
First condition: fi= 0, if ∂/ ∂x = 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.
Second condition:

f
x i d  0
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
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.
Spectroscopic selection rule tell us nothing about the
intensities.
Vibrational modes of CO2
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:
V  hcDe 1 ea(RRe ) 
2
where De is the depth of the potential
minimum and
฀
  2 1/ 2
a  

2hcDe 
PY3P05
Anharmonic oscillator
•
The Schrödinger equation can be solved for the Morse potential, giving permitted
energy levels:
2
1
1


E     hc~     hcxe~ ;   0,1,2,... max
2
2


~
a 2

xe 
2meff 4 De
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
PY3P05
Energy Levels: Basic Ideas
Basic Global Warming: The C02 dance …
About 15 micron radiation
Raman Spectra
Selection rule for Raman
Intensity of Raman lines
Vibrational-Rotational Spectra
Raman Spectra
Sources of light
Absorption Experiment
Dispersing Element
Resolving Power
Resolution
Diffraction grating
Blazed Grating
Calculation
FT
Interferometer
Emission
Absorption at single wavelength
http://www.colorado.edu/chemistry/volkamer/teaching/lectures/
Lecture%209%20-%20Light%20sources.pdf
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.
• 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.