This document provides an overview of Raman spectroscopy. It discusses the discovery of Raman spectroscopy and how it is used to observe vibration, rotational, and other low-frequency modes in a system. It also describes key aspects of Raman spectroscopy including the instrumentation, principle, types of molecules that show Raman spectra, quantum and classical theories, and applications to analyze rotational, vibrational, and pure rotational Raman spectra of molecules. In summary, the document serves as an introduction to Raman spectroscopy and its use in chemistry to identify molecules based on their unique Raman fingerprint.

1. RAMAN
SPECTROSCOPY
Dr. Mahesh Kumar (Assistant Professor)
Mukand Lal National College, Yamuna Nagar, Haryana

2. TOPICS:
 Introduction
(Stokes, Antistokes & Rayleigh Line)
 Instrumentation
 Principle
 Instrumentation
 Type of molecule that show Raman Spectra (Raman Active)
 Mutual exclusion principle
 Quantum theories
 Classical theories
 Pure rotational Raman spectra of linear molecules
 vibrational Raman spectra
 Polarization and depolarization of Raman lines

3. INTRODUCTION
 Raman spectroscopy was discovered by C. V. Raman in
1928.
 It is a spectroscopic technique used to observe vibration ,
rotational, and other low-frequency modes in a system.
 Raman spectroscopy is commonly used in chemistry to
provide a fingerprint by which molecules can be
identified.
 When the radiation pass through the transparent medium,
the species present scatter a fraction of the beam in all
direction.
 The scattered radiations are observed at right angle to the
incident.

4.  The scattered radiations are found to have different frequencies as
compared to incident freq.
Let νi = freq. of incident rad.
νs = freq. of incident rad.
 If νs > νi then lines obtained are called “Anti-Stokes Lines”
 If νs < νi then lines obtained are called “Stokes Lines”
 If νs= νi then lines obtained are called “Rayleigh Lines”
Raman Spectra:
INTRODUCTION CONTINUED..
Stokes Lines
νs < νi
Anti-Stokes Lines
νs > νi
Rayleigh Lines
νs = νi
Raman Shift: The modulus of difference νs – νi is called Raman Shift

6. 1) Source:
 The sources used in modern Raman spectrometry are nearly always lasers
because their high intensity is necessary to produce Raman scattering of
sufficient intensity to be measured with a reasonable signal-to-noise ratio
e.g. laser source are used
2. Sample
Solid, liquid or gas may be used as scattered can be obtained for any state of the substance.
This is major advantage of Raman Spectra over Rot. & Vibrational Spectra.

7. TYPE OF MOLECULE THAT SHOW RAMAN
SPECTRA (RAMAN ACTIVE MOLECULES)
 Raman spectroscopy occurs as a result of a molecular
vibration causing a "Change in Polarizability" of the molecule
and molecule must be “Anisotropically Polarizable”.
 Polarizability depend upon direction of applied electric field
e.g. in case of H2 , the distortion produced is more when
electric field is applied parallel to bond axis than when it is
applied perpendicular it and we write α║>α┴ Such molecule
are said to be “Anisotropically Polarizable”.
 So homonuclear diatomic molecules like H2, N2, O2 etc. will be
Raman Active Molecules.
 In case molecules spherically symmetric molecules like SF6,
CCl4 same polarizability is produced whatever be the direction
of applied electric field. Such molecule are called “Isotropically
Polarizable” & such molecule will be “Raman Inactive”.

8. RULE OF MUTUAL EXCLUSION
 According to this Rule:
 If a molecule has centre of symmetry then IR active vib.
will be Raman Inactive and vice versa. eg: CO2

10. QUANTUM THEORY OF RAMAN SCATTERING
v = 2
v = 1
v = 0
Any virtual excited state
Vibrational Energy Levels
hνi
(i) Rayleigh Line
There is no net absorption or
emission of light the
Output = Input
hνs = hνi
νs = νi
(Rayleigh Line)
hνs

11. QUANTUM THEORY OF RAMAN SCATTERING
v = 2
v = 1
v = 0
Any virtual excited state
Vibrational Energy Levels
hνi
(ii) Anti-Stokes Line
There is net emission of rad. So
Scattered energy = Input + Emitted Energy
hνs = hνi + hνo
νs > νi
(Anti-Stokes Line)
Net emission hνo

12. QUANTUM THEORY OF RAMAN SCATTERING
v = 2
v = 1
v = 0
Any virtual excited state
Vibrational Energy Levels
hνi
(iii) Stokes Line
There is net absorption of rad. So
Scattered energy = Input - Absorbed Energy
hνs = hνi - hνo
νs < νi
(Stokes Line)
Net Absorbed
enerygy i.e. hνo

13. POLARIZABILITY OR CLASSICAL THEORY OF
RAMAN SPECTRA
 It is based on the polarizability of molecule in presence of
electric field component of incident radiations.
 Let a molecule is placed in a electric field then induced dipole
moment (µ) or polarizability produced is given by
 Where E = electric field acting on molecule
α = Polarizability constant
o The electric field of incident rad. is a sine wave function as
Where: Eo = Strength of applied electric field
vi = Frequency of incident radiation at time ‘t’
)........(iE 
).......(20 iitSinEE i

14.  So putting value of E in eq (i), we get
 Due to incident rad, there will be vibrations in a
molecule then
 Where αo = Eqm. polarizability constant
β= Degree of polarizability
vvib = vibrational frequency at time ‘t’
).......(20 iiitSinE i 
)(..........20 ivtSin vib 
Now eq (iii) becomes
  tSinEtSin ivib  22 00 

15. )2)(2(2 000 tSintSinEtSinE vibii  
 tCostCos
E
tSinE vibivibii )(2)(2
2
2 0
00 

 
Case (i): if β = 0 then
tSinE i 200
So µ is dependent on frequency of incident radiations only and So the
scattered rad. will have same frequency as that of incident i.e. the line
obtained will be corresponding to the Rayleigh line.
Case (ii): if β ≠ 0 then µ is dependent on vi and vvib
tSinE i 200
So the scattered frequency will be either vi+vvib or vi ― vvib
If vs= vi + vvib then, vs> vi Anti-Stokes Line
If vs= vi ― vvib then, vs< vi Stokes Line

16. QUANTUM THEORY OF PURE ROTATIONAL
RAMAN SPECTRA
Selection Rule: J = ±2
 If J = +2 (Stokes Line)
 If J = -2 (Anti-Stokes Line)
As ῡ = BJ(J+1) cm-1
The energy absorbed during the transition from JJ’ i.e.
ῡ = B(J’)(J’+1) – B(J)(J+1) (i)
 Case (i): If J = +2 then from eq (i), the energy
absorbed from transition JJ’ (J+2) will be
ῡ = B(J+2)(J+3) – B(J)(J+1)
ῡ = B(4J+6) cm-1

17.  Case (ii): If J = -2 then from eq (i), the energy absorbed from
transition J’(J+2) J’ will be
ῡ = B(J+2)(J+3) – B(J)(J+1)
ῡ = B(4J+6) cm-1
 In Raman spectra, the frequency of scattered radiation is
observed. Now, the scattered frequency will be
ῡs = ῡi ± ṽ
Case (a): If J = +2 then ῡs = ῡi -  ῡ (Stokes Line)
ῡs = ῡi – B(4J+6) cm-1
ῡs for J=0 to J’=2; ῡs = ῡi – 6B cm-1
ῡs for J=1 to J’=3; ῡs = ῡi – 10B cm-1
ῡs for J=3 to J’=5; ῡs = ῡi – 14B cm-1 & so on
Case (b): If J = -2 then ῡs = ῡi +  ῡ (Anti-Stokes Line)
ῡs = ῡi + B(4J+6) cm-1
ῡs for J’=2 to J=0; ῡs = ῡi + 6B cm-1
ῡs for J’=3 to J=1; ῡs = ῡi + 10B cm-1
ῡs for J’=4 to J=2; ῡs = ῡi + 14B cm-1 & so on

18. J=0
J=1
J=2
J=4
J=3
20
31
42
13
02
24
Fig:1 Rotational Transitions in Raman Spectra
J = +2 J = -2
20
31
4202
13
24
J = -2 (Anti-Stokes)J = +2 (Stokes)
ῡs
Fig: 2 Pure Rotational Raman Spectra
6B6B 4B 4B4B 4B
J = 0

19. CONCLUSIONS
 Stokes line appear at lower energy level i.e. 6B cm-
1 as that required for Rayleigh Line.
 Anti-Stokes Line will appear at higher energy value
by amount of 6B cm-1 as that required for Rayleigh
Line.
 Energy gap between Stokes Line is 4B cm-1 and
same for Anti-Stokes.
 Energy gap between Stokes an Anti-Stokes gp. of
line is 12B cm-1.

20. PURE VIBRATIONAL RAMAN SPECTRA

21. DEPOLARIZATION OF RAMAN LINES
 In Raman spectroscopy, scattered light is observed after interaction of incident
light with matter.
 The scattered radiations have some fraction perpendicular to the incident light
called perpendicular component. However some fraction has polarisation parallel
to the incident light called parallel components.
 The depolarization ratio is the intensity ratio between the perpendicular component
and the parallel component of the Raman scattered light and can be obtained as
 Where Iperpedicular is intensity of the scattered radiations perpendicular w.r.t. incident
 Iparallel is intensity of the scattered radiations perpendicular w.r.t. incident
III
I

parallel
larperpendicu
I
I

polarised
ddepolarise
I
I


22.  A Raman band whose depolarization ratio is less than
0.75 is called a polarized band, and a band with a 0.75
depolarization ratio is called a depolarized band/spectra.
 Taking example of CH4
:
Case (i): If CH4 is spherically symmetrical
Taking a case that all C-H bond are stretched or
compressed simultaneously then CH4 will remain a
sphere. If a beam of light of light is made to indent over
CH4 then scattered light vibrations will be parallel to
incident.
C
H H
H
H
or C
H H
H
H

23.  Case (ii): If one C-H is stretched and other C-H
bond are compressed:
In such type of vibrartions, the scatted rad. Will
have oscillations in all directions. Hence
Iperpendicular > Iparallel . Hence the spectra
will be depolarised
C
H H
H
H
C
H H
H
H

24. ADVANTAGES OR
WHY RAMAN SPECTRA IS BETTER THAN OTHER?
 Rotational spectroscopy is observed for the substance is
Gaseous state, Vibrational spectra can be observed for gaseous
& Liquid state however Raman Spectra can be used with
solids, liquids or gases.
 Raman Spectra can be used obtained even for O2, N2, Cl2 etc
which have no permanent dipole moment. Such a study has
not been possible by IR spectroscopy.
 Raman spectra is independent of incident frequency. It can be
obtained for visible spectrum range which is easy to adjust
rather than IR or radiowaves.
 No sample preparation needed
 Not interfered by water
 Non-destructive
 Highly specific like a chemical fingerprint of a material
 Raman spectra are acquired quickly within seconds
 Samples can be analyzed through glass or a polymer
packaging.