Download 18.1 Raman Scattering

18.1 Raman Scattering
• 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 fluorescence
• several Raman spectra are examined that
show interesting features
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Energy Level Diagram for Scattering
The two horizontal dashed lines are called
virtual energy levels. They are called virtual
because it does not correspond to any
quantum state of the molecule. The energy
of a virtual transition is equal to the energy
of the photon. The scatter photon appears
within 10-15 seconds of excitation.
S1*
resonance
Raman
Rayleigh
Raman
antiStokes
Raman
There are two general classes of scattering
events - elastic and inelastic. Elastic scatter,
such as Rayleigh, emits the same energy (or
S0
wavelength) photon as the exciting photon.
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.
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v=2
v=1
v=0
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
vector, P, is given by
P = NqX
electron
cloud
*+
!
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.
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*+
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.
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Raman Scatter
When the polarizable electrons are within a molecule, vibrations
add time dependent components to the polarization.
P = ε 0 ⎡⎣ χ + Δχ1 cos ( 2πν1t ) + … + Δχ 3 N − 6 cos ( 2πν 3 N − 6t ) ⎤⎦ E0 cos ( 2πν Lt )
In the above equation, each numerical subscript refers to one of
the 3N - 6 normal modes. The Δχ are much smaller than χ.
Consider the polarization due to one of the vibrations, ν1.
P1 = ε 0Δχ1 cos ( 2πν1t ) E0 cos ( 2πν Lt ) =
ε 0Δχ1E0
2
⎡ cos ( 2π [ν L + ν 1 ] t ) + cos ( 2π [ν L −ν 1 ] t ) ⎤
⎣
⎦
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 antiStoke'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!
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Pre-Resonance Raman Scatter
When the laser 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
2
−
+
ν
ν
ν
ν
(
)
L
vib
e
L
⎢
IR ∝ ⎢
2
⎢
ν e2 − ν L2
⎣
(
)
)
⎤
⎥
⎥
⎥
⎦
2
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
ν L − ν vib )
(
IR ∝
4
νe
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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.
A strong absorption band has a cross-section, σa = ~10-16 cm2. A
corresponding fluorescence cross-section takes into account the
quantum yield and emission into 4π steradians.
σf =
φ f σa
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.
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Scatter Signal Strength
An equation can be written to estimate and compare signal
strengths for fluorescence, Rayleigh, and Raman scatter,
I emission = I laserσ lN ΩF ( λ ) Δλ E ( λ ) D ( 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 optics
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.
I rayleigh = 3.3 × 10−6 I laser l ΩE ( λ ) D ( λ )
I raman = 3.3 × 10−9 I laser l ΩE ( λ ) D ( λ )
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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.
I fluor = 8 × 10−21 NI laser lΩE ( λ ) D ( λ )
3.3 × 10−9 I laser lΩE ( λ ) D ( λ ) = 8 × 10−21 NIlaser lΩE ( λ ) D ( λ )
N=
3.3 × 10−9
8 × 10−21
= 4.1 × 1011 cm-3
This number density corresponds to 6.8×10-10 M.
Conclusions:
(1) In Raman spectroscopy Rayleigh scatter will be a thousand
times more intense and needs to be rejected instrumentally.
(2) Raman spectroscopy 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.
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Example Spectra (1)
The large peak at 0 cm-1
is Rayleigh scatter. The
peak at 1,060 cm-1 is the
characteristic benzene
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.
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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 antiStokes band compared to
the Stokes band.
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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 misassigned 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 antiStokes lines.
Ordinarily the instrument is not
allowed to scan over the Rayleigh
line!!! It wrecks the detector.
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