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
wavevector
2
ki  ks 

(in vacuum)

2
 /n
(in solution)
scattering wavevector
k  k 
4 n

sin

2
Lengths Probed by Light Scattering
~ 33 nm
~ 100 nm
Light scattering
probes the
length of ~1/k.
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?
The particle is now a broadcasting station, emanating
radiation in all directions.
Polarization in the particle
changes in phase with the
incoming light.
Rayleigh Scattering
I
I0


2
 2 sin 2 
4  0 2
r
2
Rayleigh scattering
by 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)
I
I0

2  2 1
 0 r
4
2
2
N
 exp[ i k  (ri  r j )]
i, j1
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.
Static Structure Factors
suspension of small particles
S (k ) 
nP
1
nP
 exp[ i k  (ri  r j )]  n P
exp[ i k  ( ri  r j )]
i, j 1
single large molecule
S1 (k ) 
1
N
N

exp[ ik  (ri  r j )]
i, j1
many large molecules
S (k ) 
nP
1
nP N
N
 
exp[ ik  (r mi  r nj )]
m, n  1 i , j  1
 S1( k ) 
nP
N
N

i, j  1
exp[ ik  (r1i  r2 j )]
Structure Factor of a Polymer Chain
radius of gyration
high-angle
scattering
Rg
low-angle scattering
I
1
1  k 2 Rg 2 / 3
Form Factors
Angular dependence of P(k)
allows us to determine the
shape of the molecule.
P(k ) 
I( k )
I(0 )
Form Factor of a Sphere
Rayleigh-Gans formula
2
Psp here ( k) 
1
Vsp
2
Vsp  Vsp
dr
Psp here ( x )  [3x
d r  exp[ i k  (r  r )] 
3
(sin x  x cos x )]
2
1
Vsp
Vsp dr exp( ik  r)
with x  kR
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 .
Light Scattering of a Solution
I
I0


 ex 2
2
1
(  / n) (  0 n ) r
4
2 2
2


2

4
 ex 2 1
0
2
r
   90 
2
The formula derived for a molecule in vacuum can be used
just by replacing  with  ex.
 ex   molecule   solvent
A more convenient expression
I ex
I0
2
1  2 n dn  c V

2
2
N A   dc  r
2
 ex 
dn  cV


    2 n
  0 
dc  N A
2
Light Scattering of Polymer Solutions
I ex ( k )
I0
2
1  2  n d n  cV

P( k )
2
2


NA
 dc
r
• Measure I(k) for pure solvent.
• Measure I(k) for solutions of a
given polymer at different
concentrations.
• Calculate Iex(k).
Zimm Plot
2
1  2  n d n  cV
 1

P( k )   2 A 2 c 
2
2


NA
 dc
r
M
I ex ( k )
I0
I ex
I0

R V
r2
2
1  2  n d n 
H 
N A   2 dc 

2
2
1

P(k )  1  k Rg / 3
1



Example of Zimm Plot
Polyguanidine in THF
Differential Refractive Index
 n  n solution  nsolvent
At low concentrations,
n 
dn
c
dc
Often, we can approximate dn/dc as
dn
dc
 ( npolymer  n solvent )v sp
Concentration Effect on Scattering Intensity
scattering at low concentrations
I ex ( k )
I0
2
1  2  n d n  cMV

P( k ) 1  2 A 2 Mc 
2
2


NA
 dc
r

Scattering by a Suspension of Spheres
mass/volume
I( kR )  I(0 )P (kR )
I( 0)  cM  
M
c 
2
NA
NA
number/volume
At constant c,
I( 0)  M  Vsp  R
At constant ,
2
I( 0)  M
M
 Vsp
2
3
3
R
I( kR )  R P( kR )
6
6
I( kR )  R P (kR )
Scattering by Spheres at Constant c
At constant c,
I( 0)  M  Vsp  R
3
3
I( kR )  R P( kR )
EXCEL problems
Plot R3P(kR) as a function of  for R = 10, 30, 100, 300, and
1000 nm. Assume specific values of n and .
Scattering by Spheres at Constant 
At constant ,
I( 0)  M
2
 Vsp 2  R
6
6
I( kR )  R P (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
Spheres aggregate into larger spheres:
3


I2
R2
P(kR 2 )


  
I1
 R1  P (kR1 )
Nonporous spheres become porous without changing the mass:
6
 R 2  P( kR2 )
  
 R1  P(kR1 )
I1
I2
Porous spheres become nonporous without changing R:
(n porous spheres form 1 nonporous sphere)
I nonp orous
I porous

1
n
n
2
 n