Presented Presentation on college level about Raman spectroscopy where I describe about Principle and phenomena and their instrumentation and applications to chemistry.

1. RAMAN SPECTROSCOPY
KANCHAN DEVI COLLEGE OF COMPUTER SCIENCE
BHILWARA (RAJASTHAN)
CHEMISTRY DEPARTMENT
M.Sc. Sem. II – 2023-24
President Presenter
Anjali Thadani Piyush Mahatma
Date of submission – 22/02/23
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2. Acknowledgment
In the accomplishment of this project successfully, many people have
best owned upon me their blessings and the heart pledge support ,
this time I am utilizing to thank all the people who have been
concerned with project.
Primarily I would thank god for being to complete this project with
success. Then I would like to thank college Directors Shikha Bhadada
and NaveenBhadada and H.O.D. of Science department Anupama
Sharma, whose valuable guidance has been the ones that helped me
patch this project and make it full proof success his suggestionsand
his instructions has served as the major contributor towards the
completion of the project.
Then I would like to thank my parents and friends who have helped me
with their valuable suggestionsand guidance has been helpful in
various phases of the completion of the project.
Last but not the least I would like to thank my classmates who have
helped me a lot.
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3. OUTLINES OF PRESENTATION
SR.NO. TOPIC PAGE NO.
1. General information 5.
2. Classical Theories of Raman Effect 6.
3. Quantum Theories of Raman Effect 9.
4. Mutual exclusion principle resonance 11.
5. Pure Rotational Raman Spectra 14.
6. Vibrational Raman Spectra 16.
7. Vibrational – Rotational Raman Spectra 17.
8. Resonance Raman Spectroscopy 19.
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4. INTRODUCTION
In 1928, the Indian physicist C.V. Raman discovered that when a beam of coloured light
entered a liquid , a fraction of the light scattered by that liquid was of a different colour.
Raman showed that the nature of this scattered light was dependent on the type of
sample present . Other scientists quickly understood the significance of this
phenomenon as an analytical and research tool and called it the Raman Effect. This
method became even more valuable with the advent of modern computers and lasers.
Its current uses range from the non-destructive identification of minerals to the early
detection of life-threatening diseases . For his discovery Raman was awarded the Nobel
prize in physics in 1930.
OBJECTIVE
 To know about Classical theories of Raman effect.
 To know about Quantum theories of Raman effect.
 To know about Pure Rotational.
 To know about Vibrational Rotational Raman spectra.
 To know about Mutual Exclusion Principle Resonance.
 To know about Resonance Raman spectra.
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5. GENERAL INFORMATION
• When a substance is irradiated with a
monochromatic light of definite frequency
(v) , the light scattered at right angles to the
incident light contains lines of
(i) incident frequency and
(ii) also of lower frequency .
Thus , certain discrete frequencies above and
below that of the incident beam will be
scattered . It is called Raman Scattering.
• The lines with lower frequency are called
Stoke’s lines.
• The lines with higher frequency are called
Anti-Stoke’s lines.
• The lines with the same frequency as that of
the incident light is called Rayleigh lines.
• If vo is the frequency of the incident light and
vr is the frequency of a particular scattered
line then , the difference is called Raman
frequency or Raman shift.
• The various observations made by Raman are
called Raman effect. Also the spectrum
obtained is called Raman Spectrum.
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6. Classical Theories Of Raman Effect
 The classical electromagnetic wave theory gives only an elementary and imperfect
explanation of Raman effect.
 However, it leads to an understanding of a basic concept of Raman spectroscopy.
 This theory said that when a molecule is placed in under E.F.it feel some distortion.
 This is due to the attraction of positively charged nuclei towards negative pole of the field
and of electrons towards positive pole.
 Such seperation of charge centres causes induced electric dipole moment and the molecule
becomes polarized.
 The magnitude of the induced dipole moment (µ) depends on the magnitude of the applied
electric field (E) and cause the polarizibility (α) of the molecule. Thus,
E α µ E = α µ µ = α E eq.........1
 According to the electromagnetic theory of light the variation of the electric field of strength
E, associated with a radiation of frequency v, with time t is given –
E = E ̥Sin 2πνt eq...........2
 Where E ̥ is a constant is the equillibrium value of the field strength,
µ = E ̥Sin 2πνt eq............3
 i.e. Scatters radiation of its own oscillation frequency v. This explains Rayleigh scattering for
which the incident and scattered radiations have the same frequency.
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7. (a) EFFECT OF VIBRATION
 Let us suppose that the molecule is diatomic. Then , as the two nuclei vibrate along the line
joining them, the polarizibility of the molecule will vary. The variation in polarizibility a with
small displacement x from the equillibrium positions is given –
α = α ̥+ β x/A eq........4
Where α ̥ is the euillibrium polarizibility
a) β the rate of variation of the polarizibility with displacement
b) A the vibrational amplitude.
 If the molecule undergoes simple harmonic oscillations , the variation of x with time is given
by- x = A sin 2πν ̥t eq.........5
Where ν ̥is the vibrational frequency of the diatomic molecules for small displacements.
 Therefore,from eq. (4) and (5),we get
α = α ̥ + β sin 2πν ̥t eq..........6
 From eq.3 and eq.6, the variation of the induced dipole moment is given as –
µ = α ̥E ̥sin 2πνt + β E ̥sin 2πνt . sin 2πν ̥t
 Using trigonometric relation , we get –
µ = α ̥E ̥ sin 2πν ̥t + ½ β E ̥[ cos 2π(ν – ν ̥)t – cos 2π(ν + ν ̥) eq........7
 i.e. v – Vv and v + Vv give rise to raman scattering the scattered radiations having these
frequencies result in the Stokes and Anti – stokes lines , respectively.
 It is clear from above discussion that raman effect is due to the variation of polarizibility with
time during the course of vibration.
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8. (b) EFFECT OF ROTATION
 If the molecule is orientation of the molecule with the respect to the electric field Thus, if the
molecule has different polarizabilities in different direction, i.e. If it is optically anisotropic ,
then its polarization will vary with time.
 Analogous to eq. (6) the variation of polarizability in the course of a rotation may be
represented by –
α = α ̥ + β̕ sin 2π(2ν)t eq..........8
Where Vr is the frequency of rotation .
 ‘the rate at which the polarizibility changes is twice as great as the rotation’
 Therefore instead of Vr, 2Vr has been used in eq.8 substituting the value from eq.8 in eq.3 we
get -
µ = α ̥E ̥sin 2πνt + ½ β̕E ̥[cos 2π(ν – 2Vr)t - cos 2π(ν + 2Vr)t] eq..........9
 Eq.9 , shows that raman effect is due to the variation of polarizibility with time during the
course of rotation and that the frequencies of raman lines should be V- 2v , for the stokes
lines and V+2v, for the anti – stokes lines .
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9. QUANTUM THEORY OF RAMAN SPECTROSCOPY
 According to quantum theory , Raman effect is considered to be the outcome of the
collisions between the light photons and molecules. Thus, applying the principle of
conservation of energy
we can write – Ep + ½ mv² + hvi = Eq + ½ mv² + hvs eq............1
a) Where Ep and Eq are the intrinsic energies of the molecules before and after collision
respectively.
b) m is the mass of the molecule
c) v and v’ are the velocities of the molecules before and after collision respectively
d) vi and vs are the frequencies of the incident and scattered photon respectively.
 As the collision does not cause any appreciable change of temperature , it may be assumed
that the kinetic energy of the molecule remains practically unchanged in the process.
Hence , Eq.can be simplified as –
Ep + hvi = Eq + hvs
h (vs – vi ) = Ep - Eq
(vs – vi ) = (Ep - Eq )/h eq.............2
 The three possibilities which arise from eq. are -
(i) If Ep = Eq then vs = vi which represents unmodified lines called Rayleigh scattering.
(ii) If Ep > Eq then vs > vi which refers to anti - Stokes lines.
( In this case , the molecule transfers some of its energy to the incident photon and thus the
scattered photon has higher energy than the incident photon).
(iii) If Ep < Eq then vs < vi which refers to Stokes lines.
( In this case, the molecule absorbs some energy from the incident photon and thus the
scattered photon has lower energy than the incident photon).
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10. INTENSITIES OF RAMAN LINES
 The molecules of a material medium are distributed among a series of quantum states of energies E1 , E2, E3 etc.
 Assuming that the statistical distribution among a series of quantum states is governed by Boltzmann’s law, the number of
molecules Np in the particular state Ep is given by –
Np = CNGp
Where C is a constant ,N is the total number of molecules, Gp is the statistical weight of the state and k is the Boltzmann’s
constant.
(A) The smaller the value of Ep, the larger the value of e-EptkT and thus the greater the number Np
With increase of temperature, the kinetic energy of molecules increases and more molecules will be raised to higher
energy states by inelastic collision .
Energy order = Anti-stokes > Rayleigh >Stokes
Wavelength order = Stoke > Rayleigh > Anti – stokes
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11. RULES OF MUTUAL EXCLUSION PRINCIPAL
o This rule states that if a molecule has a center of symmetry , then Raman active vibrations are IR
inactive and IR active vibrations are Raman inactive.
o Or we can say that if a molecule having a Centre of symmetry (CO2 , C2H2 ) , the IR active vibrations
in these molecules becomes Raman inactive , vice – versa .
o It is used in structural determination in a molecule.
o Homo nuclear diatomic compound are Raman active but IR inactive.
o Hetero nuclear diatomic compound are Raman as well as IR active.
Example :CO2
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Modes of vibration Raman
active/inactive
IR active /IR inactive
1. Symmetrical stretching Active Inactive
1. Asymmetrical stretching Inactive Active
1.In a plane bending Inactive Active
1. Out of plane bending Active Inactive

12. (a) Symmetrical stretching vibration -
- + -
O C O µ = 0 , α ǂ 0
No change in dipole moment therefore , IR inactive.
There is change in polarizibility therefore Raman active.
(b) Asymmetrical stretching vibration –
- + -
O C O µ ǂ 0 , α = 0
There is change in dipole moment therefore , IR active .
No change in polarizibility therefore Raman inactive.
(c) In a plane bonding vibration –
- + - - -
O C O O O µ ǂ 0 , α = 0
+
C
No change in polarizibility therefore , Raman inactive.
There is change in dipole moment therefore , IR active.
(d) Out of plane bonding –
- + -
O C O
No change in polarizibility Raman inactive.
There is change in dipole moment therefore , IR active. 12

13. o But if there is no centre of symmetry in a molecule then some (but not necessarily all )
vibrations may be both Raman and IR active.
Example : H2O
DEGREE OF FREEDOM – For linear structure = 3n – 5
For non – linear structure = 3n – 6
Three vibrations are present in H2O molecule .
VIBRATION CALCULATION
Symmetrical Bonding
(n – 1) n – stretching
(3 – 1)= 2 (3 – 2) = 1
(i) Symmetrical (ii) Asymmetrical
Ō Ō +
+ + + H
H H µ ǂ 0 , α ǂ 0 H µ ǂ 0 , α ǂ 0
Raman and IR both active
(iii) Bonding vibration
Ō
+ +
H H µ ǂ 0 , α ǂ 0 Raman and IR both active
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14. PURE ROTATIONAL RAMAN SPECTRA
 Selection rule for pure rotational spectra for diatomic molecule –
(i) for transition probability of Raman effect , Δα = 0. where Δα = change of polarizibility along the
internuclear axis.
(ii) The Selection rules for the pure rotational Raman spectra of diatomic molecules are : Δ J = 0 , ± 2
where , Δ J = rotational quantum no.
(iii) The Selection rule Δ J = 0 is corresponding to the Rayleigh line .
 Δ J = ± 2 is corresponding to the Raman effect , Stoke’s line Δ J = +2 and Anti – Stoke’s line = -2 .
 Now we know the energy of each rotational level ν with quantum no . J is given
By - ν ̄= BJ (J+1) eq.............1
Where , B = rotational constant
ν̄ = wave no.
J = quantum no. Where J = 0,1,2.........
 When the transition take place from lower rotational level with quantum no. J to higher rotational level
with quantum no. J’ the energy absorb in term of wave no. Is given by –
Δv̄ = BJ’(J’+1) – BJ(J+1) eq...............2
A/Q to Selection rule ,
Δ J = ± 2 when we take Δ J = + 2 so, J’ = J + 2
Δ J = - 2 so, J = J’ + 2
(I) Δ J = + 2 so, J’ = J + 2 apply in formula (II) Δ J = - 2 so, J = J’ + 2 apply in formula
Δv̄ = BJ’(J’+1) – BJ(J+1) Δv̄ = BJ’(J’+1) – BJ(J+1)
= B(J + 2)(J + 2+1) – BJ (J +1) = BJ’(J’ + 1) – B(J’ + 2)(J’ + 2 + 1)
= B(4J + 6) eq. .........3 = - B (4J’ + 6) eq............. 4
compare both 3 and 4 eq. We get ,
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15. compare both 3 and 4 eq. We get ,
Δv̄ = ±B(4J + 6) eq.........5
We know that , v̄ = v̄i + Δv̄ eq..............6
So, v̄ = v̄i ± B(4J + 6) eq.............7
Where J = 0 , 1 , 2................
v̄ = wave no. of Rayleigh line
 Plus (+) sign gives lines with higher wave no. which is called Anti – stock line.
 Minus (-) sign gives lines with lower wave no. which is called Stock line.
Now , if J = 0 then , J = 1 then , J = 2 then ,
v̄ = v̄i ± 6B v̄ = v̄i ± 10B v̄ = v̄i ± 14B
Stock lines Anti – stock lines
14B 14B
6B 6B
10B 10B
4B 4B 4B 4B 4B 4B 4B 4B
0
Rayleigh line 15

16. VIBRATIONAL RAMAN SPECTRA
 The structure of Vibrational Raman spectra can be discussed easily. For every Vibrational
mode we can write as expression for energy as-
ΔE = (v + ½) ω̄e - (v + ½)2 ω̄e xe where, v = 0, 1, 2............
ω̄e = equilibrium vibrational /oscillation frequency
xe = anhormonicity constant
 Selection rule for differene in energy levels are –
v = 0 1 ΔẼfundamental = ω̄e ( 1- 2xe)
v = 0 2 ΔẼovertone = 2 ω̄e ( 1- 3xe)
v = 1 2 ΔẼhot = 3 ω̄e ( 1- 4xe)
 Restricting to fundamental transitions only Raman lines will be obtained at the following
wave numbers-
vR = vO + ω̄e ( 1- 2xe) RAYLEIGH LINE
STOKE LINE ANTI – STOCK LINE
-3 ω̄e -2 ω̄e -1 ω̄e 0 +1 ω̄e +2 ω̄e +3 ω̄e 16

17. VIBRATIONAL – ROTATIONAL RAMAN SPECTRA
o The Vibrational- Rotational Raman spectra for diatomic molecule are given below –
Selection rule for Vibration - Δv = ± 1 ,± 2 ............
Selection rule for Rotation - ΔJ = 0 , ± 2
o Vibration energy expression is –
ΔE = (v + ½) ω̄e – (v + ½) 2ω̄e xe eq..........1
o Rotational energy expression is –
ΔE = BJ (J + 1) eq...........2
o Vibrational – Rotational energy expression is –
Ẽvj = (v + ½) ω̄e – (v + ½) 2ω̄e xe + BJ (J + 1) (on condition)
o Restricting to fundamental vibration transition we have,
For Q – branch : ΔJ = 0 , ΔEq = ω̄e - 2 ω̄e xe
For S – branch : ΔJ = +2 , ΔEs = (ω̄e - 2 ω̄e xe )+ B (4J + 6)
For O – branch: ΔJ = -2 , ΔEo = (ω̄e - 2 ω̄e xe )- B (4J + 6)
Where , ω̄e (Vibrational quantum no.) = 1/2πc(K/µ)1/2 µ-1 .
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18. ΔJ = -2 ΔJ = 0 ΔJ = +2
v = 0 J = 5
high J = 4
energy J = 3
J = 2
J = 1
J = 0
v=0
low J = 5
energy J = 4
J = 3
J = 2
J = 1
J = 0
Rayleigh line
Stock line Anti – stock line
O – branch Q – branch S - branch
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19. RESONANCE RAMAN SPECTROSCOPY(RRS)
 Resonance Raman spectroscopy is an advance technique used to study Vibrational bonds in
the group frequency region and the information obtained is similar to that obtained by
Fourier transform infrared and Raman studies.
 The enhanced of RRS over spontaneous Raman is about 10-6 to 10-8 fold.
 Resonance Raman scattering refers to a phenomenon in which Raman line intensities are
generally enhanced by excitation with wavelength that closely approach of an electronic
absorption peak of an analyte.
 Under this circumstances , the magnitudes of Raman peaks associated with the most
symmetric vibration are enhanced by a factor of 102 to 106.
 As a consequence , Resonance Raman spectrum have been obtained at analyte
concentrations as low as 10-8 M. Resonance Raman
Excited state
 The Resonance effect in RRS arises from
the coupling of Vibrational transition
of the molecule with the motion of electron Virtual state - Raman
associated with the electronic excitation.
 Fig. Schematic diagram of infrared , spontaneous
Raman and Resonance Raman transition .
Ground state
IR 19

20.  Depending of RRS – It depends upon the bond present in the sample , RRS is classified into
(i) Visible RRS (VRRS)
(ii) Ultraviolet RRS (UVRRS)
 APPLICATION OF RRS –
1. RRS is powerful technique for monitoring the structure and dynamics of protein and peptide
in solution.
2. The UVRRS are employed for the confirmation analysis of protein and amyloid fibres.
3. RRS is employed in biomedical applications including single – blood cell detection by the
trapping method.
4. The anti-oxidative capacity of the skin in vivo has been studying using RRS.
 DISADVANTAGES OF RRS –
This technique is that the laser light used for excitation can damage the sample , which can be
solved by either agitation of the sample or using new method.
 ADVANTAGES OF RRS –
1. The state of the sample required for RRS can be solid , fibre , gel or solution.
2. No additional probes are required for analysis.
3. The time scale is on the order of 10-5 to 10-4 S , which is too short and is on the order of
molecular Vibrations.
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21. CONCLUSION
Raman Spectroscopy is one type of vibrational spectroscopy
which acquires good understanding of the properties of light.
It provides a chemical “fingerprint” of the substance
measured and is therefore frequently used whenever
unknown materials need to be identified. Raman spectra also
uses visible and ultra-violet radiations rather than IR
radiations. It can be utilised for the investigation of biological
systems such as proteins is aqueous solutions.
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22. REFFERENCE
• Y.R. Sharma. Organic spectroscopy, S. Chand publishing, 2007.
• H. Kaur. Organic spectrosopy , Pragati Prakashan Merrut , 2017.
• Donald L. Pavia, Gary M. Lampman, George A. Kriz and James R. Vyvyan.
Introduction to Spectroscopy , cengage India Private Limited;5th edition
2015.
• Carey , P.R. Biochemical application of Raman and Resonance Raman
Spectroscopies. Academic press 1982.
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