18.1 Raman

• an energy level diagram shows four types of scattering processes • Rayleigh scatter is due to microscopic refractive index changes, while Raman scatter is due to time-dependent changes in polarizability • scattering cross-sections are used to compare signal strengths from Rayleigh, Raman and • several Raman spectra are examined that show interesting features

18.1 : 1/12 Energy Level Diagram for Scattering

S * The two horizontal dashed lines are called 1 virtual energy levels. They are called virtual resonance because it does not correspond to any Raman quantum state of the molecule. The energy of a virtual transition is equal to the energy of the . The scatter photon appears Rayleigh within 10-15 seconds of excitation. Raman anti- Stokes There are two general classes of scattering Raman events - elastic and inelastic. Elastic scatter, v = 2 such as Rayleigh, emits the same energy (or v = 1 S v = 0 ) photon as the exciting photon. 0 Inelastic scatter, such as Raman, absorbs sufficient energy to promote a rotational or vibrational transition. The remaining energy is scattered as a lower energy photon.

Scattered light has a directional probability with the shape of a torus. The torus axis is in the direction of the electric vector.

18.1 : 2/12 Polarization of Bound Charge

When the average charge separation in a material is changed from its equilibrium value, the electrons are said

to be polarized. The magnitude of the polarization electron vector, P, is given by cloud

+ - P = NqX * *

! + where N is the number of electrons, q is the electron charge and X is a spatial vector. Polarization can be caused by the electric vector of light, E,

P = ε0χE

where ε0 is the permittivity of free space and χ is the electric susceptibility. χ has values from 0 to ∞, and varies with density. A vacuum has χ = 0, while a metal will have a very large value.

Because the electric vector of light is oscillatory, the polarized electrons will oscillate at the optical frequency. The optically induced polarization will relax within a few cycles and emit a photon due to the moving charge. 18.1 : 3/12 Rayleigh Scatter

When light travels through a vacuum it cannot be seen unless it is impinging on the detector (eye). This is because the waves destructively interfere in all directions except the direction of travel.

When light travels through a transparent material the electric vector polarizes the electron cloud. The extent of polarization depends upon the susceptibility. In turn, the susceptibility depends upon the number of the polarizable electrons contained with a cube of volume λ3. For materials where the number of electrons varies within that volume due to molecular motion, the susceptibility varies. This variation allows a small amount of constructive interference in other directions and is the origin of Rayleigh scatter.

Gases have the largest amount of Rayleigh scatter because their number densities vary ~0.05% at 500 nm and STP. Because liquids are more dense, the variation in the number of molecules per λ3 is only ~0.0016%, and scatter is less. Pure crystalline solids have very little Rayleigh scatter, which arises only because of lattice vibrations.

18.1 : 4/12 Raman Scatter

When the polarizable electrons are within a molecule, vibrations add time dependent components to the polarization.

P =+Δε011⎣⎦⎡⎤χχcos2 πν( ttEt) ++Δ χ… πν 36360NN πν−− cos2() cos2( L)

In the above equation, each numerical subscript refers to one of νπ π ν π ν πν the πν ε χ 3N - 6 normal modes. The Δχ are much smaller than χ.

Consider the polarization due to one of the vibrations, ν1. ε Δχ E P =Δcos() 2tE cos 2 ( t =010 )⎡ cos() 2[] + t + cos() 2 [] − t⎤ 101 10LLL2 ⎣ 1 1⎦

The cos(2π[νL - ν1]t) term is the Stoke's Raman band, while the cos(2π[νL + ν1]t) term is the anti-Stoke's Raman band. The anti- Stoke's band will be temperature dependent, just like an infrared hot band. In order to be Raman active, a vibration must change the polarizability of the electron cloud.

All Raman calculations should be done in cm-1, not nanometers!

18.1 : 5/12 Pre-Resonance Raman Scatter

When the frequency does not quite reach the electronic

energy level, νL < νe, the intensity of Raman scatter is enhanced. The frequency dependence of the enhancement is given by the Albrecht-Hutley equation,

2 ⎡ 2 22⎤ ⎢()ννLvibe−+ νν() L⎥ IR ∝ ()⎢ ⎥ 222 ⎢ eL− ⎥ ⎣ νν ⎦

where νL is the laser frequency, νvib is the vibrational frequency, and νe is the electronic transition frequency. As νL approaches νe, the denominator goes to zero and the intensity increases rapidly.

When νL << νe, the equation can be simplified to a form showing that Raman intensities are proportional to the fourth power of the laser. 4 ()νν− I ∝ Lvib R 4 νe 18.1 : 6/12 Emission Cross-Sections

Both fluorescence and scattering processes have cross-sections which include the solid angle. A solid angle has units of steradians, where a sphere of unit radius is defined as having 4π steradians.

-16 2 A strong absorption band has a cross-section, σa = ~10 cm . A corresponding fluorescence cross-section takes into account the quantum yield and emission into 4π steradians. σ φ σ = f a f 4π

For φf = 1, the fluorescence cross-section corresponding to the above absorption value would be 8×10-18 cm2 sr-1.

A typical cross-section for Rayleigh scatter is 10-28 cm2 sr-1. Because Rayleigh depends upon the fourth power of the frequency, the value will vary a bit. A typical cross-section for Raman scatter is 10-30 cm2 sr-1. Again the exact value will depend upon the excitation frequency and distance from the absorption band.

18.1 : 7/12 Scatter Signal Strength

An equation can be written to estimate and compare signal strengths for fluorescence, Rayleigh, and Raman scatter,

IIlNFEDemission= laserσ ΩΔ(λλλ)() λ( ) where σ is the cross-section, l the pathlength, N the number density, Ω is the lens solid angle, F(λ)Δλ the fraction of the spectrum examined, E(λ) the efficiency of the detection excluding the lens, and D(λ) is the detector efficiency.

The term F(λ)Δλ varies for the three signals. Rayleigh scatter has a width of ~5 cm-1, Raman ~50 cm-1, and fluorescence ~5,000 cm-1. If F(λ)Δλ = 1 for Rayleigh it is ~0.1 for Raman and ~0.001 for fluorescence. The number density for water is 3.3×1022 cm-3. Using the appropriate cross-section and leaving all other terms constant gives the following relative signal strengths.

−6 IIlEDrayleigh=×3.3 10 laser Ω()λ (λ ) −9 () () IIlEDraman=×3.3 10 laser Ωλ λ

18.1 : 8/12 Comparison of Scatter and Fluorescence

What concentration of fluorophore will give a signal strength equal to that for solvent Raman? Use the fluorescence cross-section and F(λ)Δλ information from the last two slides.

−21 INIlEDfluor=×810 laser Ω(λλ) ( ) −−921() () () () 3.3×Ω 10IlEDlaserλ λλλ =×Ω 8 10 NIlED laser 3.3× 10−9 N ==×4.1 1011 cm -3 810× −21 This number density corresponds to 6.8×10-10 M.

Conclusions: (1) In Raman Rayleigh scatter will be a thousand times more intense and needs to be rejected instrumentally. (2) is difficult in the presence of a fluorophore concentration greater than nanomolar. (3) Rayleigh and Raman scatter from the solvent will make fluorescence measurements difficult below nanomolar fluorophore.

18.1: 9/12 Example Spectra (1)

The large peak at 0 cm-1 is Rayleigh scatter. The peak at 1,060 cm-1 is the characteristic ring vibration. The peak above 3,000 cm-1 is the C-H stretch.

Compared to benzene a couple new ring vibrations have been added. The new band just under 3,000 cm-1 is the methyl C-H stretch. The low frequency band

is the CH3-φ stretch.

18.1 : 10/12 Example Spectra (2)

The small peak above 3,000 cm-1 is due to the alkene C-H stretch, while the stronger peak below 3,000 cm-1 is due to the alkane C-H stretch. The peak just below 1,600 cm-1 is the double bond.

All of the frequencies are low because chlorine is so heavy. Note the strength of the ~200 cm-1 anti- Stokes band compared to the Stokes band.

18.1 : 11/12 Example Spectra (3)

The O-H stretch is typically about 3,600 cm-1. Water has most of its vibrational band at lower energies because of hydrogen bonding. The bandwidth is also due to hydrogen bonding. Water Raman is often mis- assigned as fluorescence. Note the real fluorescence interference at lower wave numbers.

The Raman spectrum of motor oil

dissolved in CCl4 is overwhelmed by fluorescent components of the oil. Note the nice progression of anti- Stokes lines.

Ordinarily the instrument is not allowed to scan over the Rayleigh line!!! It wrecks the detector.

18.1 : 12/12