Raman spectroscopy for the college students

VinothPanneer 12 views 81 slides Feb 26, 2025
Slide 1
Slide 1 of 81
Slide 1
1
Slide 2
2
Slide 3
3
Slide 4
4
Slide 5
5
Slide 6
6
Slide 7
7
Slide 8
8
Slide 9
9
Slide 10
10
Slide 11
11
Slide 12
12
Slide 13
13
Slide 14
14
Slide 15
15
Slide 16
16
Slide 17
17
Slide 18
18
Slide 19
19
Slide 20
20
Slide 21
21
Slide 22
22
Slide 23
23
Slide 24
24
Slide 25
25
Slide 26
26
Slide 27
27
Slide 28
28
Slide 29
29
Slide 30
30
Slide 31
31
Slide 32
32
Slide 33
33
Slide 34
34
Slide 35
35
Slide 36
36
Slide 37
37
Slide 38
38
Slide 39
39
Slide 40
40
Slide 41
41
Slide 42
42
Slide 43
43
Slide 44
44
Slide 45
45
Slide 46
46
Slide 47
47
Slide 48
48
Slide 49
49
Slide 50
50
Slide 51
51
Slide 52
52
Slide 53
53
Slide 54
54
Slide 55
55
Slide 56
56
Slide 57
57
Slide 58
58
Slide 59
59
Slide 60
60
Slide 61
61
Slide 62
62
Slide 63
63
Slide 64
64
Slide 65
65
Slide 66
66
Slide 67
67
Slide 68
68
Slide 69
69
Slide 70
70
Slide 71
71
Slide 72
72
Slide 73
73
Slide 74
74
Slide 75
75
Slide 76
76
Slide 77
77
Slide 78
78
Slide 79
79
Slide 80
80
Slide 81
81

About This Presentation

To know about the raman spectroscopy


Slide Content

Introduction to Basics of Introduction to Basics of
Raman SpectroscopyRaman Spectroscopy
Chandrabhas NarayanaChandrabhas Narayana
Chemistry and Physics of Materials
Jawaharlal Nehru Centre for Advanced Scientific
Research, Jakkur P.O., Bangalore 560064, India
[email protected]
http://www.jncasr.ac.in/cbhas
Lecture at Lecture at MASTANI Summer School, IISER, Pune, June 30, 2014 to July 12, 2014MASTANI Summer School, IISER, Pune, June 30, 2014 to July 12, 2014
July 11, 2014July 11, 2014

What happens when light falls
on a material?
Transmission
Reflection
Absorption
Luminescence
Elastic Scattering
Inelastic Scattering

Raman SpectroscopyRaman Spectroscopy
1 in 101 in 10
77
photons is scattered inelastically photons is scattered inelastically
Infrared
(absorption)
Raman
(scattering)
v” = 0
v” = 1
virtual
state
E
x
c
i
t
a
t
i
o
n
S
c
a
t
t
e
r
e
d
Rotational RamanRotational Raman
Vibrational RamanVibrational Raman
Electronic RamanElectronic Raman

Raman visible through unaided eye

Raman, Fluorescence and IRRaman, Fluorescence and IR
Scattering
Absorption
and emission
Absorption

Concept of normal modes in a
molecule
•There are 3N possible movements in a molecule made of N
atoms, each of which moving in one of three directions, x, y and
z.
–There are three transitional movements: all atoms in the
molecule moving in x, y or z direction at the same time.
–There are three rotational movements around x, y or z-axis
•Linear molecules are exceptions because two axes that
are perpendicular to the molecular axis are identical.
–The rest of movements are vibrational movements
•For linear molecules, 3N – 5 movements
•For non-linear molecules, 3N – 6 movements
–All vibrational movements of the sample can be described as
linear combinations of vibrational normal modes.

Vibrations in MoleculesVibrations in Molecules
HClHCl HFHF
HH
22OO
NHNH
33
SFSF
66
Sym. StretchingSym. Stretching
Asym. StretchingAsym. StretchingSym. BendingSym. Bending
Asym. BendingAsym. Bending

1
= 3835 cm
-1

2
= 1648 cm
-1

3
= 3939 cm
-1
 = 2991 cm
-1
 = 4139 cm
-1

1 = 3505.7 cm
-1

2
= 1022 cm
-1

3
= 3573.1 cm
-1

1
= 774.55 cm
-1

4
= 523.56 cm
-1

3
= 947.98 cm
-1

5
= 643.35 cm
-1

6
= 348.08 cm
-1

2
= 615.02 cm
-1

4 = 1689.7 cm
-1
8086 cm
-1
= 1 eV

Vibrational SpectroscopyVibrational Spectroscopy
r
e
= equilibrium distance between A and B
r
e
For a diatomic molecule (A-B), the bond between the two
atoms can be approximated by a spring that restores the
distance between A and B to its equilibrium value. The
bond can be assigned a force constant, k (in Nm
-1
; the
stronger the bond, the larger k) and the relationship
between the frequency of the vibration, , is given by the
relationship:
D
AB
r
AB
0
D
AB = energy required to dissociate into A and B atoms



k
2

c
k

or, more typically
where , c is the speed of light,  is the frequency in “wave
numbers” (cm
-1
) and  is the reduced mass (in amu) of A
and B given by the equation:



m m
m m
A B
A B

Vibrational Spectroscopy
Molecule  (cm
-1
) k (N/m)  (amu)
HF 3962 878 19/20
HCl 2886 477 35/36 or 37/38
HBr 2558 390 79/80 or 81/82
HI 2230 290 127/128
Cl
2
557 320 17.5
Br
2
321 246 39.5
CO 2143 1855 6.9
NO 1876 1548 7.5
N
2
2331 2240 7
2

c
k

can be rearranged to solve for k (in N/m):k 

58910
52
. 
For a vibration to be active (observable) in an infrared (IR) spectrum, the
vibration must change the dipole moment of the molecule. (the vibrations for
Cl
2, Br
2, and N
2 will not be observed in an IR experiment)
For a vibration to be active in a Raman spectrum, the vibration must change
the polarizability of the molecule.

Classical Picture of RamanClassical Picture of Raman
Stokes Raman
Anti-Stokes Raman
Induced Polarization Polarizability

Mutually exclusive principleMutually exclusive principle

max 0
max max 0
max max 0
( ) cos2
1
cos2 ( )
2
1
cos2 ( )
2
equil
z zz
zz
vib
zz
vib
t E t
d
r E t
dr
d
r E t
dr
  

  

  
 
  
 



www.andor.comwww.andor.com
Selection rule: Selection rule: v = ±1v = ±1
Overtones: Overtones: v = ±2, ±3, …v = ±2, ±3, …
Raman ScatteringRaman Scattering
Must also have a change in polarizabilityMust also have a change in polarizability
Classical Description does not suggest any difference Classical Description does not suggest any difference
between Stokes and Anti-Stokes intensitiesbetween Stokes and Anti-Stokes intensities
1
0
vib
h
kT
N
e
N


Calculate the ratio of Anti-Stokes to Stokes scattering Calculate the ratio of Anti-Stokes to Stokes scattering
intensity when T = 300 K and the vibrational frequency is intensity when T = 300 K and the vibrational frequency is
1440 cm1440 cm
-1-1
..
Are you getting the concept?Are you getting the concept?
h = 6.63 x 10h = 6.63 x 10
-34-34
Js Js
k = 1.38 x 10k = 1.38 x 10
-23-23
J/K J/K
1
0
vib
h
kT
N
e
N


 ~ 0.5

Energy diagram and Energy diagram and
Quantum pictureQuantum picture
Vibrational statesElectronic states
Virtual states
g
ex
photon
<e
g
,p
2
|H
er
|p
2
,e
b
> <e
b
,p
2
|H
ep
|p
1
,e
a
> <e
a
,p
1
|H
er
|p
1
,e
g
>
|E
s-E
b|x|E
i-E
a|
a,b
Raman cross section
If E
i = E
a or E
s =
E
b
We have Resonance Raman effect

Intensity of Normal Raman Peaks
The intensity or power of a normal Raman peak
depends in a complex way upon
•the polarizability of the molecule,
•the intensity of the source, and
•the concentration of the active group.
•The power of Raman emission increases with
the fourth power of the frequency of the
source; - photodecomposition is a problem.
• Raman intensities are usually directly
proportional to the concentration of the active
species.

Raman Depolarization Ratios
Polarization is a property of a beam of
radiation and describes the plane in
which the radiation vibrates. Raman
spectra are excited by plane-polarized
radiation. The scattered radiation is
found to be polarized to various degrees
depending upon the type of vibration
responsible for the scattering.

Raman Depolarization Ratios
The depolarization ratio p is defined as
Experimentally, the depolarization ratio may be
obtained by inserting a polarizer between the
sample and the monochromator. The
depolarization ratio is dependent upon the
symmetry of the vibrations responsible for
scattering.
p
I
I


Raman Depolarization Ratios
Polarized band: p = < 0.76 for totally
symmetric modes (A
1g
)
Depolarized band: p = 0.76 for B
1g
and B
2g

nonsymmetrical vibrational modes
Anomalously polarized band: p = > 0.76
for A
2g
vibrational modes

Raman spectra of CClRaman spectra of CCl
44
Isotope effect
Cl has two isotopes
35
Cl and
37
Cl
Relative abundance is 3:1

CClCCl
44 Spectra Spectra
•461.5 cm
-1
is due to
35
Cl
4C
•458.4 cm
-1
is due to
35
Cl
3
37
ClC
•455.1 cm
-1
is due to
35
Cl
2
37
Cl
2C
•What are the two question
marks?
•Why are these bands weak?

Raman Spectra of Methanol and EthanolRaman Spectra of Methanol and Ethanol
20000
15000
10000
5000
0
500 1000 1500 2000 2500 3000 3500
OH
stretching
CH
stretching
CO
stretching
CH
3
deformation
Raman Shift (cm
-1
)
R
a
m
a
n

I
n
t
e
n
s
it
y

(
a
r
b
it
r
a
r
y

u
n
it
)
CCO
stretching
CH
3
and CH
2
deformation
Significant identification of alcohols which differ just in one CH
2
-group
CASR

Peak position – Chemical identity – Peak position – Chemical identity –
Similar StructuresSimilar Structures
3,4-Methylenedioxymethamphetamine (MDMA) Methamphetamine
500 1000 1500 2000 2500 3000 3500
Raman Shift (cm
-1
)

R
a
m
a
n

I
n
t
e
n
s
it
y

(
a
r
b
it
r
a
r
y

u
n
it
)
CASR
ecstasy

1200
1000
800
600
400
200
0
I
n
t
e
n
s
i
t
y
(
c
o
u
n
t
s
/
s
)
Wa venumber (cm-1)
1000
800
600
400
200
I
n
t
e
n
s
i
t
y
(
c
o
u
n
t
s
/
s
)
Wavenumber (cm-1)
Mg - SO
4
Na
2
- SO
4
The Mass Effect on Raman SpectraThe Mass Effect on Raman Spectra
Significant identification of salts (SO
4
2-
) which
differ just in the metal ion employed
CASR

Peak positions – Chemical identity Peak positions – Chemical identity
DiasteromersDiasteromers
PseudoephedrineEphedrine
500 1000 1500 2000 2500 3000 3500
Raman Shift (cm
-1
)
R
a
m
a
n

I
n
t
e
n
s
it
y

(
a
r
b
it
r
a
r
y

u
n
it
)
CASR

Peak Position – Crystal Phases –
Polymorphs
200
400
600
800
1 000
1 200
1 400
1 600
1 800
2 000
2 200
2 400
I
n
t
e
n
s
i
t
y

(
c
n
t
)
200 400 600 800 1 000 1 200 1 400
Raman Shift (cm
-1
)
Both Anatase and Rutile are TiO
2
but of different
polymorphic forms - identical chemical composition,
different crystalline structures.
Rutile
Anatase

Peak Shift – Stress and StrainPeak Shift – Stress and Strain
Nasdala, L., Harris, J.W. & Hofmeister, W. (2007): Micro-spectroscopy of diamond. Asia Oceania
Geosciences Society, 4th Annual Meeting, Bangkok, Thailand, August, 2007.
Nasdala, L., Raman barometry of mineral inclusions in diamond crystals. Mitt. Österr. Miner. Ges. 149
(2004)
CASR
Larnite ( – Ca
2SiO
4) inclusion in Diamond

Bandwidth – Crystallinity – Bandwidth – Crystallinity –
Structural order/disorderStructural order/disorder
Raman spectra of zircon, showing typical amorphous (blue)
and crystalline (red) spectra.
CASR

Intensity – ConcentrationIntensity – Concentration
4-Nitrophenol dissolved in CH
2
Cl
2
0
500
1 000
1 500
2 000
2 500
3 000
3 500
I
n
t
e
n
s
it
y

(
c
n
t
/
s
e
c
)
1 200 1 400 1 600
Raman Shift (cm
-1
)
4-Nitrophenol in CH2Cl2_0,1 M
4-Nitrophenol in CH2Cl2_0,01 M
4-Nitrophenol in CH2Cl2_0,001 M
1
3
4
1
.
0
CASR

Raman technique – what Raman technique – what
requirements are needed? requirements are needed?
Requirements for Raman technique to determine peak position, peak shift,
bandwidth and intensity
- Laser Excitation
- Reduction of stray light
- Collecting Optics
- Spectral resolution and spectral coverage
- Spatial resolution and confocality
- Sensitivity: subject to detector
- Light flux: subject to dispersion
CASR

What do we need to make a What do we need to make a
Raman measurement ?Raman measurement ?
Laser
Sample
Filter
Grating
Detector
•Rejection filter
(A way of removing the
scattered light that is not
shifted( changed in colour).
(A way of focusing the laser onto
the sample and then collecting the
Raman scatter.)
•Sampling optics
•Monochromatic Light source
typically a laser
•Spectrometer and detector
(often a single grating spectrometer
and CCD detector.)
CASR

Demonstration of the very high Demonstration of the very high
spectral resolution obtained in the spectral resolution obtained in the
triple additive modetriple additive mode
8000
6000
4000
2000
I
n
t
e
n
s
i
t
y

(
a
.
u
.
)
1520 1540 1560 1580
Wavenumber (cm-1)
Triple additive mode
Slit widths= 30 m
Rotation-Vibration Spectrum of O
2
Triple subtractive mode . Slit=30 m
CASR


Laser excitation – Laser Selection to Laser excitation – Laser Selection to
avoid fluorescence avoid fluorescence
Laser
wavelength, 
1
Raman shift, 
1
-1
+
Laser wavelength, 
3 Raman shift, 
3
-1
+
Laser wavelength, 
3 Fluorescence
Laser wavelength: 
3
< 
2
< 
1
CASR

Laser excitation – Laser selection to Laser excitation – Laser selection to
avoid fluorescence avoid fluorescence
0
10 000
20 000
30 000
40 000
50 000
60 000
I
n
t
e
n
s
it
y
(
c
n
t
)
600 800 1 000
Wavelength (nm)
Green spectrum: 532 nm laser
Red spectrum: 633 nm laser
Dark red spectrum: 785 nm laser
Fluorescence is wavelength dependent
Ordinary Raman is wavelength independent
0
10 000
20 000
30 000
40 000
50 000
60 000
I
n
t
e
n
s
it
y
(
c
n
t
)
1 000 2 000 3 000
Raman Shift (cm
-1
)
CASR

Commercial Hand Cream
785 nm – 633 nm – 473 nm
x1 0
3
5
10
15
20
25
30
35
40
45
I
n
t
e
n
s
i
t
y

(
c
n
t
)
50 0 1 000 1 500 2 000 2 500 3 000 3 500
Raman Shift (cm
-1
)
Reduction of Fluorescence
CASR
Laser excitation – Laser selection to Laser excitation – Laser selection to
avoid fluorescence avoid fluorescence

Laser excitation – laser radiation Laser excitation – laser radiation
power power
Laser wavelength: 473 nm
Laser power at sample: 25.5 mW
ObjectiveN.A. Laser
spot size
(µm)
Radiation
power
(kW/cm
2
)
100× 0.90 0.64 ~7900
50× 0.75 0.77 ~5400
10× 0.25 2.31 ~600
Laser wavelength: 633 nm
Laser power at sample: 12.6 mW
ObjectiveN.A.Laser
spot
size
(µm)
Radiatio
n power
(kW/cm
2
)
100× 0.90 0.85 ~2200
50× 0.75 1.03 ~1500
10× 0.25 3.09 ~200
CASR

Laser excitation – laser radiation Laser excitation – laser radiation
powerpower
•Keep in mind: the usage of high numerical
objective lenses causes a very small spot
size of the laser which results in a high
power density
•To avoid sample burning radiation power
has to be adapted INDIVIDUALY to the
sample
CASR

Collecting OpticsCollecting Optics
Sampling volume
Small for high N.A. lens
Large for low N.A. lens
Laser spot size
Small for high N.A. lens
Large for low N.A. lens
Collection solid angle
Large for high N.A. lens
Small for low N.A. lens
High N.A. lens
θ
Low N.A. lens
θ
NA = n · sin ()
n: refraction index
: aperture angle
W
o
r
k
i
n
g

d
i
s
t
a
n
c
eW
o
r
k
i
n
g

d
i
s
t
a
n
c
e
CASR

Collecting Optics – Overview on Collecting Optics – Overview on
common objectivescommon objectives
Objective N.A.
Working distance
[mm]
x100 0.90 0.21
x50 0.75 0.38
x10 0.25 10.6
x100 LWD 0.80 3.4
x50 LWD 0.50 10.6
CASR

Collecting Optics – what objective Collecting Optics – what objective
should be used?should be used?
A distinction between opaque and transparent samples has to be made
For opaque samples, high N.A. lens works better because there is almost no
penetration of the laser into the sample. High N.A. lens enables
- High laser power density (mW/m
3
)  increases sensitivity
- Wide collection solid angle  increases sensitivity
0
5 000
10 000
15 000
20 000
25 000
30 000
I
n
t
e
n
s
it
y

(
c
n
t
/
s
e
c
)
460 480 500 520 540 560 580 600 620
Raman Shif t (cm
-1
)
Silicon
x100 – NA = 0.9 – 31.350 C/s
x50 – NA = 0.75 - 21.995 C/s
x10 – NA = 0.25 - 1.462 C/s
100 %
70 %
5 %
-20
0
20
Y

(
µ
m
)
0
X (µm)
10 µm
Example for an opaque sample:
Silicon wafer
CASR

Collecting Optics – what objective Collecting Optics – what objective
should be used?should be used?
A distinction between opaque and transparent samples has to be made
For transparent samples, low N.A. lens works better because of penetration of
the laser into the sample. Low N.A. lens enables
- Large sampling volume  increases sensitivity
x10
3
0
2
4
6
8
10
12
14
I
n
te
n
s
ity
(
c
n
t)
740 760 780 800 820 840 860 880
Raman Shift (cm
-1
)
cyclo_100xLWD
cyclo_macro
Sample: Cyclohexane
Instrument: ARAMIS
Red: x100LWD, 7,000 cts/s
Blue: Macro lens, 14,500 cts/s
100 %
48 %
CASR

Spectral resolution and spectral Spectral resolution and spectral
coveragecoverage
Slit
D
e
t
e
c
t
o
r
Focal Length
Collimating
mirror
Focusing mirror
Grating
Schematic diagram of a Czerny-Turner spectrograph
CASR

Spectral resolution and spectral Spectral resolution and spectral
coveragecoverage
•Spectral resolution is a function of 1. dispersion, 2. widths of entrance slit
and 3. pixel size of the CCD
•Dispersion is the relation between refraction of light according to the
wavelength of light
•Dispersion is a function of the 1. focal length of spectrograph the 2.
groove density of the grating and 3. the excitation wavelength
•In general, long focal length and high groove density grating offer
high spectral resolution.
CASR

Dispersion as a function of the focal Dispersion as a function of the focal
lengthlength
Same grating
Same excitation wavelength
C
C
D

D
e
t
e
c
t
o
r
C
C
D

D
e
t
e
c
t
o
r
Long focal length
Short focal length
CASR

Dispersion as a function of the focal Dispersion as a function of the focal
length vis-a vis wavelengthlength vis-a vis wavelength
Dispersion in cm
-1
/ pixel
1800 gr/mm Grating
LabRAM (F = 300 mm)
LabRAM HR (F = 800 mm)
CASR

20040060080010001200140016001800
244 - 269 nm (25 nm)
325 - 371 nm (46 nm)
457 - 553 nm (96 nm)
488 - 599 nm (111 nm)
514 - 639 nm (125 nm)
532 - 667 nm (135 nm)
633 - 833 nm (200 nm)
785 - 1119 nm (334 nm)
830 - 1210 nm (380 nm)
1064 - 1768 nm (704 nm)
Horizontal lines indicate a relative Raman Shift of 3800 cm
-1


Wavelength [nm]
Dispersion as a function of excitation Dispersion as a function of excitation
wavelengthwavelength
Long wavelengthShort wavelength
Same focal length
Same grating
C
C
D

D
e
t
e
c
t
o
r
C
C
D

D
e
t
e
c
t
o
r
CASR

Spectral coverage - dependence from Spectral coverage - dependence from
excitation wavelengthexcitation wavelength
Length of CCD Chip
x10
3
0
2
4
6
8
10
12
14
16
18
20
22
I
n
t
e
n
s
i
t
y

(
c
n
t
/
s
e
c
)
500 600 700 800 900 1 000
Wavelength (nm)
Relative Raman shift of 3100 cm
-1
corresponds to 81 nm
Relative Raman shift of 3100 cm
-1
corresponds to 154 nm
Relative Raman shift of 3100 cm
-1
corresponds to 252 nm
473 nm – 633 nm – 785 nm
Same focal length
Same grating
Length of CCD Chip
Length of CCD Chip
CASR

Dispersion as a function of groove densityDispersion as a function of groove density
High density groove grating
Low density groove grating
C
C
D

D
e
t
e
c
t
o
r
C
C
D

D
e
t
e
c
t
o
r
Same focal length
Same excitation wavelength
CASR

Spectral resolution as a function slit Spectral resolution as a function slit
widthwidth
Slit Slit Slit
One of parameters that determines the spectral resolution is the entrance
slit width. The narrower the slit, the narrower the FWHM (full width at half
maximum), and higher the spectral resolution.
When recording a line whose natural width is smaller than the
monochromator’s resolution, the measured width will reflect the
spectrograph’s resolution.
CASR

Spectral resolution as a function of pixel Spectral resolution as a function of pixel
sizesize
•Because a CCD detector
is made of very small
pixels, each pixel serves
as an exit slit (pixel size =
exit slit width)
•For the same size CCDs,
the CCD with a larger
number of smaller pixels
produces a larger number
of spectral points closer to
each other increasing the
limiting spectral resolution
and the sampling
frequency
•26 m pixel vs. 52 m
pixel (simulation)
I
n
t
e
n
s
i
t
y
600 650 700
Raman Shift (cm
-1
)
D
e
t
e
c
t
o
r
D
e
t
e
c
t
o
r
CASR

Choice of Laser for RamanChoice of Laser for Raman
The choice of laser will influence
different parameters:
• Signal Intensity: (1/
4
rule applies to
Raman intensity.
• Probing volume: spot size and
material penetration depth.
• Fluorescence: may overflow Raman
signal.
• Enhancement: some bounds only
react strongly at a certain wavelength.
• Coverage range and Resolution:
grating dispersion varies along the
spectral range.
0,001
0,01
0,1
1
10
100
200300400500600700800900
P
e
n
e
t
r
a
t
i
o
n

d
e
p
t
h

(
µ
m
)
Wavelength (nm)
Silicium
CASR

Spatial resolution: penetration depthSpatial resolution: penetration depth
0
500
1000
1500
2000
2500
3000
244 nm325 nm457 nm488 nm514 nm633 nm
Wavelength [nm]
D
e
p
t
h

p
e
n
e
t
r
a
t
i
o
n

[
n
m
]
 General: The larger the excitation wavelength,
the deeper the penetration.
 The exact values depend on material.
Penetration depth in Silicon
0
2
4
6
8
10
12
244 nm 325 nm
CASR

250 300 350 400 450 500 550
I
n
t
e
n
s
it
y

[a
.
u
.
]
Raman shift (cm
-1
)
785 nm
Spatial resolution: penetration depthSpatial resolution: penetration depth
Uniform layer of SiGe
Gradient SiGe layer
Pure Si substrate
Strained silicon layer
250 300 350 400 450 500 550
I
n
t
e
n
s
it
y

[a
.
u
.
]
Raman shift (cm
-1
)
325 nm 488 nm 785 nm
 The higher the excitation wavelength, the deeper the penetration.
488 nm
250 300 350 400 450 500 550
I
n
t
e
n
s
it
y

[a
.
u
.
]
Raman shift (cm
-1
)
325 nm
Strained Si
of top layer
Si of silicon
substrate
Si of SiGe
layer
~nm
~nm
~µm
CASR

Spatial resolution: penetration depthSpatial resolution: penetration depth
EXAMPLE
325nm laser results
The strain is not homogenous.
A characteristic, cross-hatch pattern is observed.
CASR

D = 1.22  / NA
Laser spot size D is defined by the Rayleigh criterion:
excitation wavelength ()
objective numerical aperture (NA)
With NA=n sinα
Spatial resolution and spot sizeSpatial resolution and spot size
Spatial resolution is half of the laser spot diameter
CASR

Sample
Length of
Laser Focus
Nearly closed
confocal apertur
P P '
P '
P P 'P
Image P ' of laser focus P matches
exactly the confocal hole.
Confocality and Spatial ResolutionConfocality and Spatial Resolution
CASR

CASR
Axial resolution of a Confocal Raman Microprobe Axial resolution of a Confocal Raman Microprobe

Confocal z-scan against silicon
with different hole apertures

exc
= 633 nm
Sampling
Volume
Wide Hole
Laser focus
waist diameter
Depth of
laser focus
(d.o.f)
Narrow Hole
Narrow Hole:
Collecting Raman radiation that
originates only from within a
diffraction limited laser focal
volume with a dimension of:
Focus waist diameter ~ 1.22  / NA
Depth of laser focus ~ 4  / (NA)
2
Confocality and spatial Confocality and spatial ResolutionResolution
CASR

Example of application using the confocality Example of application using the confocality
principle : depth profile on a multilayer polymer principle : depth profile on a multilayer polymer
sample sample
5000
4000
3000
2000
1000
0
I
n
t
e
n
s
i
t
y

(
a
.
u
.
)
1000 1200 1400 1600
Wavenumber (cm-1)
3000
2500
2000
1500
1000
500
0
I
n
t
e
n
s
i
t
y

(
a
.
u
.
)
1000 1200 1400 1600
Wavenumber (cm-1)
75 m
Polyethylene
Polyethylene
nylon
z
x
CASR

Thank you for your attention!

Symmetry – Identity (E)Symmetry – Identity (E)
Identity operation (E)
This is a very important symmetry operation which is
where the molecule is rotated by 360º. In otherwords a
full rotation or doing nothing at all.
This appears in all molecules!!!

Symmetry – Rotation (CSymmetry – Rotation (C
nn))
Rotations (C
n)
These are rotations about the axes of symmetry. n
denotes 360º divided by the number for the rotation.
C
2
= 180ºC
3
= 120ºC
4
= 90ºC
5
= 72ºC
6
= 60º

Symmetry – Symmetry – Reflections (Reflections ())
Reflections (
h, 
d and 
v) These are reflections in a
symmetry planes (x, y and z).

h
- Horizontal
Plane
 (y)
perpendicular
to the highest
rotation axis

v
- Vertical
Plane
 (z)
parallel
to the highest
rotation axis

d
- Diagonal
(dihedral) Plane (x)
the Diagonal Plane
that bisects two
axes

Symmetry – Inversion (i)Symmetry – Inversion (i)
Inversion centre (i)
These are where the molecule can be inverted through the
centre of the molecule (atom or space).

Symmetry – Improper Rotation (SSymmetry – Improper Rotation (S
nn))
Improper rotations (S
n)
These are rotations about the axes of symmetry
followed by reflections.

Vibrational SpectroscopyVibrational Spectroscopy
For polyatomic molecules, the situation is more complicated because there are more
possible types of motion. Each set of possible atomic motions is known as a mode.
There are a total of 3N possible motions for a molecule containing N atoms because
each atom can move in one of the three orthogonal directions (i.e. in the x, y, or z
direction).
Translation
al modes
Rotational
modes
A mode in which all the atoms are moving in the same
direction is called a translational mode because it is
equivalent to moving the molecule - there are three
translational modes for any molecule.
A mode in which the atoms move to rotate (change the
orientation) the molecule called a rotational mode - there
are three rotational modes for any non-linear molecule
and only two for linear molecules.
The other 3N-6 modes (or 3N-5 modes for a linear molecule) for a molecule
correspond to vibrations that we might be able to observe experimentally. We must
use symmetry to figure out what how many signals we expect to see and what atomic
motions contribute to the particular vibrational modes.

Vibrational Spectroscopy and SymmetryVibrational Spectroscopy and Symmetry
1.Determine the point group of the molecule.
2.Determine the Reducible Representation, 
tot, for all possible motions of
the atoms in the molecule.
3.Identify the Irreducible Representation that provides the Reducible
Representation.
4.Identify the representations corresponding to translation (3) and rotation
(2 if linear, 3 otherwise) of the molecule. Those that are left correspond
to the vibrational modes of the molecule.
5.Determine which of the vibrational modes will be visible in an IR or
Raman experiment.
We must use character tables to determine how many signals we will see
in a vibrational spectrum (IR or Raman) of a molecule. This process is
done a few easy steps that are similar to those used to determine the
bonding in molecules.

Finding the Point GroupFinding the Point Group

Example, the vibrational modes in water.
The point group is C
2v
so we must use the appropriate character table
for the reducible representation of all possible atomic motions, 
tot
. To
determine 
tot
we have to determine how each symmetry operation
affects the displacement of each atom the molecule – this is done by
placing vectors parallel to the x, y and z axes on each atom and
applying the symmetry operations. As with the bonds in the previous
examples, if an atom changes position, each of its vectors is given a
value of 0; if an atom stays in the same place, we have to determine
the effect of the symmetry operation of the signs of all three vectors.
The sum for the vectors on all atoms is placed into the reducible
representation.
Vibrational Spectroscopy and SymmetryVibrational Spectroscopy and Symmetry
Make a drawing of the molecule and add in vectors
on each of the atoms. Make the vectors point in
the same direction as the x (shown in blue), the y
(shown in black) and the z (shown in red) axes.
We will treat all vectors at the same time when we
are analyzing for molecular motions.
H
O
H
HOH
top view

The E operation leaves everything where it is so all nine
vectors stay in the same place and the character is 9.
The C
2
operation moves both H atoms so we can ignore the
vectors on those atoms, but we have to look at the vectors on
the oxygen atom, because it is still in the same place. The
vector in the z direction does not change (+1) but the vectors
in the x, and y directions are reversed (-1 and -1) so the
character for C
2
is -1.
The 
v
(xz) operation leaves each atom where it was so we
have to look at the vectors on each atom. The vectors in the
z and x directions do not move (+3 and +3) but the vectors in
the y direction are reversed (-3) so the character is 3.
The ’
v
(yz) operation moves both H atoms so we can ignore
the vectors on those atoms, but we have to look at the vectors
on the oxygen atom, because it is still in the same place. The
vectors in the z and y directions do not move (+1 and +1) but
the vectors in the x direction is reversed (-1) so the character
is 1.
C
2V
E C
2

v
(xz) ’
v
(yz)

tot
9 -1 3 1
Example, the vibrational modes in water.
Vibrational Spectroscopy and SymmetryVibrational Spectroscopy and Symmetry
HOH
C
2
H
O
H
HOH
HOH
HOH

v
(xz)
HOH
HOH
’
v (yz)
z
y
x

C
2V
E C
2

v
(xz)’
v
(yz)
A
1 1 1 1 1 z x
2
,y
2
,z
2
A
2
1 1 -1 -1 R
z
xy
B
1
1 -1 1 -1 x, R
y
xz
B
2
1 -1 -1 1 y, R
x
yz
From the 
tot and the character table, we can figure out the number and types of modes
using the same equation that we used for bonding:
    n
1
order
# of operations in class(character of RR)character of X
X  
 n
1
4
A
1
    191111131111
This gives:
 n
1
4
B
1
     191111131111
 n
1
4
B
2
    191111131111 n
1
4
A
2
     191111131111
Which gives: 3 A
1
’s, 1 A
2
, 3 B
1
’s and 2 B
2
’s or a total of 9 modes, which is what we
needed to find because water has three atoms so 3N = 3(3) =9.
Vibrational Spectroscopy and SymmetryVibrational Spectroscopy and Symmetry
C
2V
E C
2

v
(xz) ’
v
(yz)

tot
9 -1 3 1

Vibrational Spectroscopy and SymmetryVibrational Spectroscopy and Symmetry
Now that we have found that the irreducible representation for 
tot
is (3A
1
+ A
2
+ 3B
1
+
2B
2
), the next step is to identify the translational and rotational modes - this can be
done by reading them off the character table! The three translational modes have
the symmetry of the functions x, y, and z (B
1
, B
2
, A
1
) and the three rotational modes
have the symmetry of the functions R
x, R
y and R
z (B
2, B
1, A
2).
Translation
al modes
Rotational
modes
The other three modes (3(3)-6 = 3) that are left over for
water (2A
1
+ B
1
) are the vibrational modes that we might be
able to observe experimentally. Next we have to figure out if
we should expect to see these modes in an IR or Raman
vibrational spectrum.
C
2V
E C
2

v
(xz)’
v
(yz)
A
1
1 1 1 1 z x
2
,y
2
,z
2
A
2
1 1 -1 -1 R
z
xy
B
1 1 -1 1 -1 x, R
y xz
B
2
1 -1 -1 1 y, R
x
yz

Remember that for a vibration to be observable in an IR
spectrum, the vibration must change the dipole moment of the
molecule. In the character table, representations that change the
dipole of the molecule are those that have the same symmetry
as translations. Since the irreducible representation of the
vibrational modes is (2A
1
+ B
1
) all three vibrations for water will
be IR active (in red) and we expect to see three signals in the
spectrum.
For a vibration to be active in a Raman spectrum, the vibration
must change the polarizability of the molecule. In the character
table, representations that change the polarizability of the
molecule are those that have the same symmetry as the second
order functions (with squared and multiplied variables). Thus all
three modes will also be Raman active (in blue) and we will see
three signals in the Raman spectrum.
Vibrational Spectroscopy and SymmetryVibrational Spectroscopy and Symmetry
C
2V
E C
2

v
(xz)’
v
(yz)
A
1 1 1 1 1 z x
2
,y
2
,z
2
A
2
1 1 -1 -1 R
z
xy
B
1
1 -1 1 -1 x, R
y
xz
B
2
1 -1 -1 1 y, R
x
yz
The three vibrational
modes for water. Each
mode is listed with a 
(Greek letter ‘nu’) and a
subscript and the
energy of the vibration is
given in parentheses. 
1

is called the “symmetric
stretch”, 
3 is called the
“anti-symmetric stretch”
and 
2
is called the
“symmetric bend”.

The Geometry of the Sulfur Dioxide The Geometry of the Sulfur Dioxide
Molecule Molecule

C
s
structure: 3 normal modes, all
having A' symmetry
The C
s
structure should have 3 IR active fundamental
transitions. These three fundamental transitions also should
be Raman active. We would expect to observe three strong
peaks in the IR and three strong peaks in the Raman at the
same frequency as in the IR. All of the Raman lines would be
polarized because they are totally symmetric (A' symmetry).

C
2v
structure: 3 normal modes, two
with A
1 symmetry, one with B
2
 
The C
2v structure should have 3 IR active fundamental
transitions. These three fundamental transitions also should
be Raman active.We would expect to observe three strong
peaks in the IR and three strong peaks in the Raman at the
same frequency as in the IR.
Two of the Raman lines are totally symmetric (A
1
symmetry)
and would be polarized. One Raman line would be
depolarized.

The D
ooh structure should have two IR active
fundamental transitions. It will have one Raman
active fundamental transition at a different
frequency than either of the IR peaks.. The
Raman line will be polarized.

Experimental ObservationExperimental Observation
Fundamental2 1 3
IR (cm
-1
) 51911511336
Raman (cm
-1
)52411511336
The experimental infrared and Raman bands of liquid
and gaseous sulfur dioxide have been reported in a
book by Herzberg
7
. Only the strong bands
corresponding to fundamental transitions are shown
below. The polarized Raman bands are in red.

ConclusionConclusion
The existence of three experimental bands in the IR
and Raman corresponding to fundamental transitions
weighs strongly against the symmetrical linear (D
ooh)
structure. We usually do not expect more strong bands
to exist than are predicted by symmetry.
Group theory predicts that both bent structures would
have three fundamental transitions that are active in
both the IR and Raman. However all three of the
Raman lines would be polarized if the structure were
unsymmetrical (C
s
symmetry). The fact that one
Raman line is depolarized indicates that the structure
must be bent and symmetrical (C
2v
symmetry).
The sulfur dioxide molecule has C
2v
symmetry.

Problems with Raman:
a)Very Weak – for every 10
6
photons only 1
photon Raman
a)Resonant Raman not feasible with every sample.
b)Absorption a better process than scattering

Raman SpectrometersRaman SpectrometersMicro–Raman setupMicro–Raman setup
International and National Patent (2007), G.V. Pavan Kumar et al Current Science (2007) 93, 778.
Stage
Objective lens
Dichroic
Mirror
Camera
Edge filter
Focusing lens
Computer
Mono-
chromator
CCD
Optical fiber
LASER
Tags