Static Light Scattering

Outline of Static Light Scattering

Measurement system

Rayleigh scattering

Static structure factor

Form factors

Practical problems

Light Scattering Measurement System

Scattering Wavevector

top view

k =k =4πnλ

sinθ2

scattering wavevector

wavevector

ki =ks =2πλ

(in vacuum)

=2πλ /n

(in solution)

Lengths Probed by Light Scattering

Light scattering probes the length of ~1/k.

~ 33 nm

~ 100 nm

Scattering Volume

depends on the focusing of the laser.

specified by the two pinholes.

The scattering volume is an open system.

Rayleigh Scattering by a Small Particle

Why is the sky blue?Why is the sunset reddish?

Polarization in the particle changes in phase with the incoming light.

The particle is now a broad-casting station, emanating radiation in all directions.

Rayleigh Scattering

II0

=π2

λ4α2

ε02

sin2 ′ θ r2

Rayleigh scatteringby a particle in vacuum

: polarizability of the particle particle volume

I maximizes at ´ = 90°.Usually, LS is detected in the horizontal plane.

Scattering by a Chain Molecule (in Vacuum)

The beams scattered by the two particles interfere.Two parts of a large molecule interfere more or less constructively.Therefore, a large molecule scatters the light more strongly than many small particles do.

II0

=π2

λ4α2

ε02

1r2 exp[ik⋅(ri −rj )]

i, j=1

N

∑

Static Structure Factors

S(k) =1nP

exp[ik⋅(ri −rj )]i, j =1

nP

∑ =nP exp[ik⋅(ri −rj )]

suspension of small particles

single large molecule

S1(k) =1N

exp[ik⋅(ri −rj )]i, j=1

N

∑

many large molecules

S(k) =1

nPNexp[ik⋅(rmi−rnj)]

i, j=1

N

∑m,n=1

nP

∑

=S1(k)+nPN

exp[ik⋅(r1i −r2j )]i, j=1

N

∑

Structure Factor of a Polymer Chain

I ∝1

1+k2Rg2 /3

low-angle scattering

Rg

radius of gyration

high-anglescattering

Form Factors P(k)=I(k)I(0)

Angular dependence of P(k) allows us to determine the shape of the molecule.

Form Factor of a Sphere

Rayleigh-Gans formula

EXCEL problems

1. Plot P as a function of kR.2. Plot P as a function of for R = 10, 30, 100, 300, and 1000 nm. Assume specific values of n and .

Psphere(k) =1

Vsp2 dr

Vsp∫ d ′ r

Vsp∫ exp[ik⋅(r − ′ r )]=

1Vsp

drVsp∫ exp(ik⋅r)

2

Psphere(x) =[3x−3(sinx−xcosx)]2 withx =kR

Light Scattering of a Solution

The formula derived for a molecule in vacuum can be used just by replacing with ex.

αex =αmolecule−αsolvent

II0

=π2

(λ / n)4αex

2

(ε0n2)2

1r2 =

π2

λ4αex

2

ε02

1r2 ′ θ =90°

αex

ε0

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= λ ⋅2ndndc

⎛ ⎝

⎞ ⎠

2 cVNA

Iex

I0=

1NA

2πnλ2

dndc

⎛ ⎝

⎞ ⎠

2 cVr2

A more convenient expression

Light Scattering of Polymer Solutions

• Measure I(k) for pure solvent.

• Measure I(k) for solutions of a

given polymer at different

concentrations.

• Calculate Iex(k).

Iex(k)I0

=1

NA

2πnλ2

dndc

⎛ ⎝

⎞ ⎠

2 cVr2 P(k)

Zimm Plot

Iex(k)I0

=1

NA

2πnλ2

dndc

⎛ ⎝

⎞ ⎠

2 cVr2 P(k)

1M

+2A2c+L⎡ ⎣ ⎢

⎤ ⎦ ⎥

−1

Iex

I0≡

RθVr2

H ≡1

NA

2πnλ2

dndc

⎛ ⎝

⎞ ⎠

2

P(k)= 1+k2Rg2 / 3( )

−1

Example of Zimm Plot

Polyguanidine in THF

Differential Refractive Index

dndc

≅(npolymer−nsolvent)vsp

Δn=nsolution−nsolventΔn=

dndc

ΔcAt low concentrations,

Often, we can approximate dn/dc as

Concentration Effect on Scattering Intensity

Iex(k)I0

=1

NA

2πnλ2

dndc

⎛ ⎝

⎞ ⎠

2 cMVr2 P(k) 1−2A2Mc+L[ ]

scattering at low concentrations

Scattering by a Suspension of Spheres

I(kR)= I(0)P(kR)

I(0) ∝cM=ρM2

NA

c =ρMNA

mass/volume

At constant c, I(0) ∝ M ∝ Vsp∝ R3

At constant ρ, I(0) ∝ M2 ∝ Vsp2 ∝ R6 I(kR)∝ R6P(kR)

I(kR)∝ R3P(kR)

number/volume

Scattering by Spheres at Constant c

EXCEL problems

Plot R3P(kR) as a function of for R = 10, 30, 100, 300, and 1000 nm. Assume specific values of n and .

At constant c, I(0) ∝ M ∝ Vsp∝ R3 I(kR)∝ R3P(kR)

Scattering by Spheres at Constant ρ

At constant ρ, I(0) ∝ M2 ∝ Vsp2 ∝ R6 I(kR)∝ R6P(kR)

EXCEL problems

Plot R6P(kR) as a function of for R = 10, 30, 100, 300, and 1000 nm. Assume specific values of n and .

Changes in the Scattering Intensity

I2I1=

R2

R1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

3P(kR2 )P(kR1)

Spheres aggregate into larger spheres:

Porous spheres become nonporous without changing R:

(n porous spheres form 1 nonporous sphere)

Inonporous

Iporous=1n

n2 =n

Nonporous spheres become porous without changing the mass:

I2I1

=R2

R1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

6P(kR2)P(kR1)