Introduction to Basics of Raman Spectroscopy

Introduction to Basics of Raman Spectroscopy
Chandrabhas Narayana Chemistry and Physics of Materials Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur P.O., Bangalore , India Lecture at MASTANI Summer School, IISER, Pune, June 30, 2014 to July 12, 2014 July 11, 2014

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

1 in 107 photons is scattered inelastically
Raman Spectroscopy 1 in 107 photons is scattered inelastically Infrared (absorption) Raman (scattering) v” = 0 v” = 1 virtual state Excitation Scattered Rotational Raman Vibrational Raman Electronic Raman

Raman visible through unaided eye

Raman, Fluorescence and IR
Absorption and emission Scattering 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 Molecules
Sym. Stretching 8086 cm-1 = 1 eV HCl n = 2991 cm-1 HF n = 4139 cm-1 Asym. Stretching Sym. Bending H2O n1 = 3835 cm-1 n3 = 3939 cm-1 n2 = 1648 cm-1 Asym. Bending NH3 n1 = cm-1 n3 = cm-1 n2 = 1022 cm-1 n4 = cm-1 SF6 n5 = cm-1 n2 = cm-1 n6 = cm-1 n1 = cm-1 n3 = cm-1 n4 = cm-1

Vibrational Spectroscopy
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: DAB 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: rAB re re = equilibrium distance between A and B DAB = energy required to dissociate into A and B atoms

can be rearranged to solve for k (in N/m): 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 Cl2 557 320 17.5 Br2 321 246 39.5 CO 2143 1855 6.9 NO 1876 1548 7.5 N2 2331 2240 7 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 Cl2, Br2, and N2 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 Raman
Induced Polarization Polarizability Stokes Raman Anti-Stokes Raman

Mutually exclusive principle

Raman Scattering Selection rule: Dv = ±1 Overtones: Dv = ±2, ±3, …
Must also have a change in polarizability Classical Description does not suggest any difference between Stokes and Anti-Stokes intensities

Are you getting the concept?
Calculate the ratio of Anti-Stokes to Stokes scattering intensity when T = 300 K and the vibrational frequency is 1440 cm-1. h = 6.63 x Js k = 1.38 x J/K ~ 0.5

Energy diagram and Quantum picture
Virtual states Raman cross section <eg,p2|Her|p2,eb> <eb,p2|Hep|p1,ea> <ea,p1|Her|p1,eg> |Es-Eb|x|Ei-Ea| S a,b photon ex g Electronic states Vibrational states If Ei = Ea or Es = Eb 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.

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.

Polarized band: p = < 0.76 for totally symmetric modes (A1g) Depolarized band: p = 0.76 for B1g and B2g nonsymmetrical vibrational modes Anomalously polarized band: p = > 0.76 for A2g vibrational modes

Raman spectra of CCl4 Isotope effect Cl has two isotopes 35Cl and 37Cl
Relative abundance is 3:1

CCl4 Spectra 461.5 cm-1 is due to 35Cl4C
458.4 cm-1 is due to 35Cl337ClC 455.1 cm-1 is due to 35Cl237Cl2C What are the two question marks? Why are these bands weak?

Raman Spectra of Methanol and Ethanol
CH stretching CCO stretching Raman Intensity (arbitrary unit) CH3 and CH2 deformation OH stretching CO stretching CH3 deformation Raman Shift (cm-1) CASR Significant identification of alcohols which differ just in one CH2-group

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

The Mass Effect on Raman Spectra
Significant identification of salts (SO42-) which differ just in the metal ion employed Mg - SO4 Na2 - SO4 CASR

Peak positions – Chemical identity Diasteromers
CASR Ephedrine Pseudoephedrine Raman Intensity (arbitrary unit) 500 1000 1500 2000 2500 3000 3500 Raman Shift (cm-1)

Peak Position – Crystal Phases – Polymorphs
Both Anatase and Rutile are TiO2 but of different polymorphic forms - identical chemical composition, different crystalline structures. Rutile Anatase

Peak Shift – Stress and Strain
CASR Larnite (b – Ca2SiO4) inclusion in Diamond 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)

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

Intensity – Concentration
CASR 4-Nitrophenol dissolved in CH2Cl2

Raman technique – what requirements are needed?
CASR 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

What do we need to make a Raman measurement ? CASR
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.) Detector Grating Filter Laser Sample

Rotation-Vibration Spectrum of O2
Demonstration of the very high spectral resolution obtained in the triple additive mode CASR Rotation-Vibration Spectrum of O2 Triple additive mode Slit widths= 30 mm Triple subtractive mode . Slit=30 mm

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

Laser excitation – Laser selection to avoid fluorescence
CASR 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

Laser excitation – Laser selection to avoid fluorescence
CASR Commercial Hand Cream 785 nm – 633 nm – 473 nm Reduction of Fluorescence

Laser excitation – laser radiation power
CASR Laser wavelength: 473 nm Laser power at sample: 25.5 mW Objective N.A. Laser spot size (µm) Radiation power (kW/cm2) 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 Objective N.A. Laser spot size (µm) Radiation power (kW/cm2) 100× 0.90 0.85 ~2200 50× 0.75 1.03 ~1500 10× 0.25 3.09 ~200

Laser excitation – laser radiation power
CASR 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

Collecting Optics CASR θ θ NA = n · sin (Q) n: refraction index
Collection solid angle Large for high N.A. lens Small for low N.A. lens Working distance High N.A. lens Low N.A. lens Sampling volume Small for high N.A. lens Large for low N.A. lens Working distance θ θ Laser spot size Small for high N.A. lens Large for low N.A. lens NA = n · sin (Q) n: refraction index Q: aperture angle

Collecting Optics – Overview on common objectives
CASR 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

Collecting Optics – what objective should be used?
CASR 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/m3)  increases sensitivity Wide collection solid angle  increases sensitivity 100 % Example for an opaque sample: Silicon wafer Silicon x100 – NA = 0.9 – C/s x50 – NA = C/s x10 – NA = C/s 70 % 5 %

Collecting Optics – what objective should be used?
CASR 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 Sample: Cyclohexane Instrument: ARAMIS Red: x100LWD, 7,000 cts/s Blue: Macro lens, 14,500 cts/s 100 % 48 %

Spectral resolution and spectral coverage
CASR Schematic diagram of a Czerny-Turner spectrograph Slit Collimating mirror Grating Detector Focusing mirror Focal Length

Spectral resolution and spectral coverage
CASR 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.

Dispersion as a function of the focal length
CASR Short focal length CCD Detector Same grating Same excitation wavelength Long focal length CCD Detector

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

Dispersion as a function of excitation wavelength
CASR Same focal length Same grating CCD Detector CCD Detector Short wavelength Long wavelength

Spectral coverage - dependence from excitation wavelength
CASR Length of CCD Chip Relative Raman shift of 3100 cm-1 corresponds to 81 nm corresponds to 154 nm corresponds to 252 nm 473 nm – 633 nm – 785 nm Same focal length Same grating Length of CCD Chip Length of CCD Chip

Dispersion as a function of groove density
CASR CCD Detector CCD Detector Low density groove grating High density groove grating Same focal length Same excitation wavelength

Spectral resolution as a function slit width
CASR 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.

Spectral resolution as a function of pixel size
CASR 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) Detector Detector

Choice of Laser for Raman
CASR The choice of laser will influence different parameters: Signal Intensity: (1/l)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. Silicium

Spatial resolution: penetration depth
CASR Penetration depth in Silicon Different wavelength can be used General: The larger the excitation wavelength, the deeper the penetration. The exact values depend on material.

CASR 325 nm 488 nm 785 nm Strained Si of top layer Uniform layer of SiGe Gradient SiGe layer Pure Si substrate Strained silicon layer 325 nm ~nm Si of SiGe layer ~nm ~µm 488 nm Si of silicon substrate Different wavelength can be used 785 nm  The higher the excitation wavelength, the deeper the penetration.

CASR EXAMPLE 325nm laser results Different wavelength can be used The strain is not homogenous. A characteristic, cross-hatch pattern is observed.

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

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

Axial resolution of a Confocal Raman Microprobe
CASR

Confocality and spatial Resolution
Wide 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 Laser focus waist diameter Depth of laser focus (d.o.f) Sampling Volume Narrow Hole Confocal z-scan against silicon with different hole apertures exc = 633 nm 4. Confocal z-scan – attainable axial resolution CASR

Example of application using the confocality principle : depth profile on a multilayer polymer sample x 75 m Polyethylene nylon z CASR

Thank you for your attention!

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 (Cn)
Rotations (Cn) These are rotations about the axes of symmetry. n denotes 360º divided by the number for the rotation. C2 = 180º C3 = 120º C4 = 90º C5 = 72º C6 = 60º

Symmetry – Reflections (s)
Reflections (sh, sd and sv) These are reflections in a symmetry planes (x, y and z). sh - Horizontal Plane (y) perpendicular to the highest rotation axis sv - Vertical Plane (z) parallel to the highest rotation axis sd - Diagonal (dihedral) Plane (x) the Diagonal Plane that bisects two axes

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 (Sn)
Improper rotations (Sn) These are rotations about the axes of symmetry followed by reflections.

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). 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. Translational modes Rotational modes 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 Symmetry
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. Determine the point group of the molecule. Determine the Reducible Representation, tot, for all possible motions of the atoms in the molecule. Identify the Irreducible Representation that provides the Reducible Representation. 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. Determine which of the vibrational modes will be visible in an IR or Raman experiment.

Finding the Point Group

Example, the vibrational modes in water. The point group is C2v 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. 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. top view

Example, the vibrational modes in water. z y The E operation leaves everything where it is so all nine vectors stay in the same place and the character is 9. The C2 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 C2 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. x C2 v (xz) ’v (yz) C2V E C2 v (xz) ’v (yz) tot 9 -1 3 1

C2V E C2 v (xz) ’v (yz) tot 9 -1 3 1 C2V E C2 v (xz) ’v (yz) A1 1 z x2,y2,z2 A2 -1 Rz xy B1 x, Ry xz B2 y, Rx 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: This gives: Which gives: 3 A1’s, 1 A2, 3 B1’s and 2 B2’s or a total of 9 modes, which is what we needed to find because water has three atoms so 3N = 3(3) =9.

Now that we have found that the irreducible representation for tot is (3A1 + A2 + 3B1+ 2B2), 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 (B1, B2, A1) and the three rotational modes have the symmetry of the functions Rx, Ry and Rz (B2, B1, A2). C2V E C2 v (xz) ’v (yz) A1 1 z x2,y2,z2 A2 -1 Rz xy B1 x, Ry xz B2 y, Rx yz The other three modes (3(3)-6 = 3) that are left over for water (2A1 + B1) 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. Translational modes Rotational modes

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 (2A1 + B1) 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. 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”. C2V E C2 v (xz) ’v (yz) A1 1 z x2,y2,z2 A2 -1 Rz xy B1 x, Ry xz B2 y, Rx yz

The Geometry of the Sulfur Dioxide Molecule

Cs structure: 3 normal modes, all having A' symmetry
The Cs 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).

C2v structure: 3 normal modes, two with A1 symmetry, one with B2
The C2v 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 (A1 symmetry) and would be polarized. One Raman line would be depolarized.

The Dooh structure should have two IR active fundamental transitions
The Dooh 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 Observation
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. Fundamental 2 1 3 IR (cm-1) 519 1151 1336 Raman (cm-1) 524

Conclusion The existence of three experimental bands in the IR and Raman corresponding to fundamental transitions weighs strongly against the symmetrical linear (Dooh) 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 (Cs symmetry). The fact that one Raman line is depolarized indicates that the structure must be bent and symmetrical (C2v symmetry). The sulfur dioxide molecule has C2v symmetry.

Problems with Raman: Very Weak – for every 106 photons only 1 photon Raman Resonant Raman not feasible with every sample. Absorption a better process than scattering

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

Introduction to Basics of Raman Spectroscopy

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