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CHEM 305

Light scattering

In the last chapter, we introduced the concept of scattering and resonance.We will now look at scattering in more detail.

Electric field is E = E0 cos 2πν[t – (x/c)]Magnetic field is H = H0 cos 2πν[t – (x/c)]

where c is the speed of light.

x

z

ysingle molecule

CHEM 305

We will now focus on the electric field. At the particle, x=0, and thus

E = E0 cos 2πνt

Molecules are not hard spheres but rather, are polarizable.

H H

H

H H

H

An oscillating dipole moment is induced: µ = αE0 cos 2πνt

where α is the molecular polarizability.

µµµ

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If the molecule is isotropic, the dipolar moment will be along the z-axis, i.e. thesame direction as the electric vector,and will act as a source of radiation.

In other words, the oscillating dipolewill produce an electric field, whichis given by

Er = αE04π2 sinφ cos 2πν t – r4λ2 c

The term in the round brackets is theamplitude of the scattered wave.

We need, once again, to describe thisin terms of a measurable quantity. How can we do that?

x

z

y

x

y

zφ

r

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The intensity of the scattered radiation, i, or energy flow per meter squared, depends on the square of the amplitude. We can compare it to the intensityof the incident radiation I0 and write

i = [αE04π2 sinφ /rλ2]2 = 16 π4 α2 sin2φI0 E02 r2 λ4

What does this equation tell us?

• intensity falls off with r-2• intensity is zero in the direction in which the dipole vibrates (along z) since

φ = 0• intensity is strongly dependent on the wavelength

Why is the sky blue?

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Sun

White light

Rayleigh scattering

What about red?

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Relationship between scattering intensity and the solution to the driven damped harmonic oscillator

Scattering limit:m2(ω02 – ω2)2 >> f 2 ω2

x’ = F0 m (ω02 – ω2)m2 (ω02 – ω2)2 + f 2ω2

x’’ = F0 f ωm2 (ω02 – ω2)2 + f 2ω2

x’ = F0m (ω02 – ω2)

x’’ = 0

i α constant 2ω0 2 – 1 ω

i α 1λ4

Rayleigh limitω

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The equation for the radiation intensity that we have written is for the case wherethe incident radiation is polarized, i.e. E vectors are only in z-direction.

For more information on polarization, see http://www.colorado.edu/physics/2000/

polarization/polarizationI.html

and http://www.colorado.edu/physics/2000/polarization/blocking_light.html

Normally, unpolarized radiation is used in scattering experiments.

CHEM 305

For unpolarized incidentradiation, the scatteringsurface will look like adumbbell (resulting fromthe sum of all surfaces foreach component of the radiation). The equation for the scattering intensityis given by:

i = 8π4α2 (1+ cos2θ)I0 r2λ4

where θ is the angle betweenthe direction of the incident beam and the direction ofobservation.

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Scattering from more than one molecule

Let us consider two molecules:

1

2 ≈

≈

phase difference

P

Electric field at P:

E = A cos(2πνt + φ1) + A cos(2πνt + φ2)

1 2

common source

CHEM 305

We can use the relationship

cos x + cos y = 2 cos x-y cos x+y2 2

to write

E = 2A cos φ1 – φ2 cos 2πνt + φ1 + φ22 2

amplitude

destructive interference

constructive interference

amplitude=0

amplitude2=i1,2

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Thus we can define the intensity of the scattered radiation to be

i1,2 = 4A2 cos2(∆φ/2)= 2A2 [1 + cos(∆φ) ]= 2i [1 + cos(∆φ)]

where i is the intensity scattered by one molecule. This shows that the intensityfor two scatterers ranges from 0 to 4i and depends on the direction from which we observe them.

1

2

All values of ∆φ areequally probably overa period of time.

Therefore, on average = 0 and = 2i.

CHEM 305

Generalising this to n scatterers, which move randomly with respect to each other, we can write

itot = Σ ij1n

Scattering in a perfect crystal

Destructive interference willoccur in all directions, exceptthat of the incident beam.Here the scattered waves willalways be in phase.

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CHEM 305

Scattering in a pure liquid

In the forward direction, the scatterers interfere constructivelyas in the crystal.

Consider each cell as a scattering center. If they were all equal density thenwe would have the same result as for a crystal. Butin a fluid, density fluctuatesfrom instant to instant. Asa result, the interferenceis not completely destructiveand some scattering occursin all directions.

CHEM 305

Forward scattering and the refractive index

Let us consider the effect of an incident wave (I) going through a certainmaterial of thickness ∆x. We know from the previous two examples thatthere will be scattering in the forward direction (S).

The emergent wave will be the sum of the incident wave (cosine) and a90 degree out of phase scattered wave (sine), allowing us to write:

E = E0 [cos 2πν(t – x/c) – Γ∆x sin 2πν(t – x/c)]

incident wave scattered wave

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where Γ = 4π2Nα/λ and N is the number of scatterers per cubic meter. Theresultant wave is a wave retarded in phase (R). We can define the time bywhich the wave is retarded by τ

Γ ∆x = tan(2πντ)

Thus we can write the equation for electric field as

E = E0 [cos 2πν(t – x/c) – tan(2πντ) sin 2πν(t – x/c)]

= E0 cos 2πν(t – x/c) cos 2πντ – sin 2πντ sin 2πν(t – x/c)cos 2πντ

= E0 cos 2πν(t + τ – x/c)cos 2πντ

where 2πντ will be small.

CHEM 305

We can define a new apparent velocity c’, and so the retardation time τ can also be written as,

τ = ∆x – ∆xc’ c

The refractive index is defined as n = c/c’, so

τ = (n – 1)∆xc

As mentioned before 2πντ is small, therefore tan(2πντ) ≈ 2πντ. Putting this intothe equation above,

Γ∆x ≈ (n – 1)∆x2πν c

or (n – 1) = Γc = 4π2Nα c = 2πNα2πν λ 2πν

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CHEM 305

A more rigorous derivation of the refractive index, which takes into accountthe fact that the electric field in the medium is also altered, yields

n2 – 1 = 4πNα

Now that we have defined the refractive index, we can now look at scatteringin a solution of macromolecules (until now we have only looked at perfectcrystals and pure liquids).

One can use thermodynamics to describe scattering from a solution – wewould need to analyze concentration fluctuations, etc.

Luckily (!) we will use the simpler approach: as on previous occasions, wewill consider that our solute molecules are in a dilute ideal solution. In otherwords, the scatterers are independent particles and we have Rayleighscattering.

CHEM 305

We can use the excess refractive index over the solution over that of thesolvent to tell us something about the scattering, i.e.

solvent

solvent + solute

n0

n

n2 – n02 = 4πNα

where N is the number of solute particles per cubic meter.

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CHEM 305

We can rewrite this last equation to have the polarizability on the left-hand side:

α = (n + n0) (n – n0) C4π C N

where we have multiplied and divided by C, concentration, such that C/N = M/NA.The quantity (n – n0)/C is the specific refractive index increment of the solute andcan be written as dn/dC. Putting this information into the equation above,

α = n0 dn M2π dC NA

where we have also taken into account that for dilute solutions, n + n0 ≈ 2n0.

We can now put this expression for a into the equation of the intensity:

i = 8π4α2 (1+ cos2θ)I0 r2λ4

= 2π2 n02 dn 2 M 2 (1 + cos2θ)r2 λ4 dC NA

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The previous equation is for one particle. For N particles (recall that theintensity is additive), with N = CNA/M:

iθ = 2π2 n02 dn 2 CM (1 + cos2θ)I0 r2 λ4 dC NA

Defining Rθ, the Rayleigh ratio as

Rθ= iθ r2I0 (1 + cos2θ)

= 2π2 n02 (dn/dC)2 CMNAλ4

= KCM.

This indicates that scattering measurements can be used to determinemolecular weight.

Assumptions made?

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CHEM 305

• Ideal solution KC = 1Rθ M

For a non-ideal solution: KC = 1 + 2BC + ….Rθ M

where B is the second virial coefficient.

• Homogeneous solute

If not homogeneous, we will measure an average molecular weight.

Rθ = Σ Rθi = Σ KCiMi = K Σ CiMi C = KMwCΣ Ci

• Particles are small compared with the wavelength.

i i

i

CHEM 305

Scattering by large particles

observer

θ

θ

A

B

A and B are not independent –we must consider interferenceeffects between the differentscattering centers.

We must also consider all orientations that this moleculecan adopt and Brownianmotion.

Complicated!!

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CHEM 305

Instead we simplify the problem, by defining a parameter P(θ), which is the ratioof the scattered intensity at an angle θ to that which would be observed if theparticle had the same molecular weight but was small relative to the wavelengthλ.

P(θ) = 1 ΣΣ sin hRijN2 hRij

where N is the number of scattering centers in the molecule, Rij is the distancebetween a pair of centers i and j, and h is

h = 4π sin(θ/2)λ

and l is the wavelength of the light in the solution, i.e.

λ = λ0/n λ0 = wavelength of light in vacuumn = refractive index.

i=1 j=1

N N

CHEM 305

To infer something about P(θ), we can expand sin(hRij)/hRij in terms of a Taylorseries:

sin x ≈ 1 – x2 + x4 + …x 6 120

Therefore, if h 0 or Rij 0,

P(θ) 1 ΣΣ 1 = N2 = 1N2 N2

i.e. for small θ, for long λ, or forsmall particles, P(θ) 1. Rayleigh scattering isapproached.

i=1 j=1

N N

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For θ = 0, we will have forward scattering as before and constructive interference.

P(θ) = 1

P(θ) < 1

P(θ) < 1

i = 8π4α2 (1+ cos2θ) P(θ)I0 r2λ4

CHEM 305

Apart from the limiting case of small particles relative to the wavelength of theincident radiation, what can P(θ) tell us about a biomolecule?

Let us go back to our Taylor expansion of sin(x)/x and consider the first twoterms (1 – x2/6):

P(θ) = 1 ΣΣ (1) – h2 ΣΣ Rij2N2 6i=1

N N

j=1 i=1

N N

j=1

= 1

contains the definition for the radius of gyration,RG

RG2 = 1 ΣΣ Rij22N2 i=1

N N

j=1

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CHEM 305

unfolded

Radius of gyration: a measure of how much a biomolecule spreads out from itscentre. E.g. protein folding of villin headpiece subdomain

Y. Duan & P.A. Kollman, Science, 282, 740-744 (2001).

CHEM 305

Getting back to P(θ), we can write it in terms of RG:

P(θ) = 1 – h2 RG2+ …. = 1 – 16π2 RG2 sin2 (θ/2) + ….3 3 λ2

There has been no assumption about the shape of the particle. So scatteringcan unambiguously give us information on particle shape via the radius ofgyration.

The equation above is only useful for large macromolecules (relative to wavelength). Why?

Let’s take a molecule with RG = λ/100.

P(θ) = 1 – 16π2 RG2 sin2 (θ/2) + …. = 1 - 16π2 sin2 (θ/2) + ….. 3 λ2 30000

= 1 – 0.005 sin2 (θ/2) ≈ 1

RG = 40 Å

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CHEM 305

We can write an equation relating the Rayleigh ratio Rθ to concentration andmolecular weight as we did for an ideal (or non-ideal) solution. For this weneed a relationship between the Rayleigh ratio for the real particle. That isnothing else but

Rθ(real particle) = P(θ) Rθ(Rayleigh scatterer)

Therefore we can write

KC = 1 1 + 2BC Rθ P(θ) M

or KC = 1 1 + 2BC Rθ 1 – (16π2 RG2/3λ2) sin2 (θ/2) M

≈ 1 + 16π2 RG2 sin2 (θ/2) 1 + 2BC 3λ2 M

for a macromolecule. For heterogeneous samples, M isreplaced by Mw.

CHEM 305

To obtain the molecular weight, we need to extrapolate KC/Rθ to zero angleand zero concentration, i.e.

To achieve this, we can use a Zimm plot:

KC vs sin2 (θ/2) + kCRθ

where k is an arbitrary constant chosen to give a convenient scale.

KC ≈ 1 + 16π2 RG2 sin2 (θ/2) 1 + 2BC Rθ 3λ2 M

0 0

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KC = 1 + 2BC Rθ M

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concentration 1

concentration 2concentration 3 concentration 4

θ = 0

θ1θ2

θ3

…etc.

θ=0

KC = 1 = 1 1 + 16π2 RG2 sin2 (θ/2) Rθ P(θ) M M 3λ2θ=0

CHEM 305

Mw = 89.6 x 106 g.mol-1RG = 38.8 nm

The data shown on the previous slide was for unilamellar vesicles:

J.H. van Zanten, Langmuir, 10, 4391-4393 (1994).

Why characterize these systems?

- drug delivery- drug targeting- medical imaging (deliver contrast

agents)

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CHEM 305

Relationship between RG and particle shape

Shape RG

(0.6)½ a a = radius of the sphere

axes are 2a, 2a and γ2a

L is the length of the rod

h2 is the mean-square end-to-enddistance

2 + (1/γ2)5a

½

L/√12

(h2)½√6