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.

xnHxxHxH

xdx

dexH

nnn

n

nxn

n

11

22

22

exp1

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

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

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

12

Alkane

Alkene

Alkyne

-C-H C-C

13

IR: Aromatic Compounds

(Subsituted benzene “teeth”)

C≡C

14

IR: Alcohols and Amines

CH3CH2OH

Amines similar to OH

O-H broadens with Hydrogen bonding

N-H broadens with Hydrogen bonding

C-O

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

• 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

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

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.

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

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

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

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

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

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

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

Energy Levels: Basic Ideas

About 15 micron radiation Basic Global Warming: The C02 dance …

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.