today’s outline - march 03, 2015segre/phys570/15s/lecture_13.pdf · today’s outline - march 03,...

Today’s Outline - March 03, 2015

• SAXS review

• Calculating Rg

• Porod regime

• Nucleation mechanism by SAXS

Homework Assignment #04:Chapter 4: 2, 4, 6, 7, 10due Monday, March 10, 2015

C. Segre (IIT) PHYS 570 - Spring 2015 March 03, 2015 1 / 15

• SAXS review

• Calculating Rg

• Porod regime

• SAXS review

• Calculating Rg

• Porod regime

• SAXS review

• Calculating Rg

• Porod regime

• SAXS review

• Calculating Rg

• Porod regime

• SAXS review

• Calculating Rg

• Porod regime

Small Angle Scattering

Recall that there was an additional term in the scattering intensity whichbecomes important at small Q.

I SAXS( ~Q) = f 2∑n

∫Vρate

i ~Q·(~rn−~rm)dVm

= f 2∑n

e i~Q·~rn∫Vρate

−i ~Q·~rmdVm

= f 2∫Vρate

i ~Q·~rndVn

∫Vρate

−i ~Q·~rmdVm

∣∣∣∣∫Vρsle

i ~Q·~rdV

∣∣∣∣2Where we have assumed sufficient averaging and introduced ρsl = f ρat .This final expression looks just like an atomic form factor but the chargedensity that we consider here is on a much longer length scale than anatom.

∫Vρate

= f 2∑n

e i~Q·~rn∫Vρate

−i ~Q·~rmdVm

= f 2∫Vρate

i ~Q·~rndVn

∫Vρate

−i ~Q·~rmdVm

i ~Q·~rdV

∫Vρate

= f 2∑n

e i~Q·~rn∫Vρate

−i ~Q·~rmdVm

= f 2∫Vρate

i ~Q·~rndVn

∫Vρate

−i ~Q·~rmdVm

i ~Q·~rdV

∫Vρate

= f 2∑n

e i~Q·~rn∫Vρate

−i ~Q·~rmdVm

= f 2∫Vρate

i ~Q·~rndVn

∫Vρate

−i ~Q·~rmdVm

i ~Q·~rdV

∫Vρate

= f 2∑n

e i~Q·~rn∫Vρate

−i ~Q·~rmdVm

= f 2∫Vρate

i ~Q·~rndVn

∫Vρate

−i ~Q·~rmdVm

i ~Q·~rdV

∣∣∣∣2

Where we have assumed sufficient averaging and introduced ρsl = f ρat .This final expression looks just like an atomic form factor but the chargedensity that we consider here is on a much longer length scale than anatom.

∫Vρate

= f 2∑n

e i~Q·~rn∫Vρate

−i ~Q·~rmdVm

= f 2∫Vρate

i ~Q·~rndVn

∫Vρate

−i ~Q·~rmdVm

i ~Q·~rdV

∣∣∣∣2Where we have assumed sufficient averaging and introduced ρsl = f ρat .

This final expression looks just like an atomic form factor but the chargedensity that we consider here is on a much longer length scale than anatom.

∫Vρate

= f 2∑n

e i~Q·~rn∫Vρate

−i ~Q·~rmdVm

= f 2∫Vρate

i ~Q·~rndVn

∫Vρate

−i ~Q·~rmdVm

i ~Q·~rdV

The SAXS experiment

Scattering from a dilute solution

The simplest case is for a dilute solution of non-interacting molecules.

Assume that the scattering length density of each identical particle(molecule) is given by ρsl ,p and the scattering length density of the solventis ρsl ,0.

I SAXS( ~Q) =

∣∣∣∣∣∫Vp

ρslei ~Q·~rdVp

∣∣∣∣∣2

= (ρsl ,p − ρsl ,0)2

∣∣∣∣∣∫Vp

e i~Q·~rdVp

∣∣∣∣∣2

If we introduce the single-particleform factor F( ~Q):

F( ~Q) =1

e i~Q·~rdVp

I SAXS( ~Q) = ∆ρ2V 2p |F( ~Q)|2

Where ∆ρ = (ρsl ,p − ρsl ,0), and the form factor depends on themorphology of the particle (size and shape).

I SAXS( ~Q) =

∣∣∣∣∣∫Vp

ρslei ~Q·~rdVp

∣∣∣∣∣2

= (ρsl ,p − ρsl ,0)2

∣∣∣∣∣∫Vp

e i~Q·~rdVp

∣∣∣∣∣2

F( ~Q) =1

e i~Q·~rdVp

I SAXS( ~Q) = ∆ρ2V 2p |F( ~Q)|2

I SAXS( ~Q) =

∣∣∣∣∣∫Vp

ρslei ~Q·~rdVp

∣∣∣∣∣2

= (ρsl ,p − ρsl ,0)2

∣∣∣∣∣∫Vp

e i~Q·~rdVp

∣∣∣∣∣2

F( ~Q) =1

e i~Q·~rdVp

I SAXS( ~Q) = ∆ρ2V 2p |F( ~Q)|2

I SAXS( ~Q) =

∣∣∣∣∣∫Vp

ρslei ~Q·~rdVp

∣∣∣∣∣2

= (ρsl ,p − ρsl ,0)2

∣∣∣∣∣∫Vp

e i~Q·~rdVp

∣∣∣∣∣2

F( ~Q) =1

e i~Q·~rdVp

I SAXS( ~Q) = ∆ρ2V 2p |F( ~Q)|2

I SAXS( ~Q) =

∣∣∣∣∣∫Vp

ρslei ~Q·~rdVp

∣∣∣∣∣2

= (ρsl ,p − ρsl ,0)2

∣∣∣∣∣∫Vp

e i~Q·~rdVp

∣∣∣∣∣2

F( ~Q) =1

e i~Q·~rdVp

I SAXS( ~Q) = ∆ρ2V 2p |F( ~Q)|2

I SAXS( ~Q) =

∣∣∣∣∣∫Vp

ρslei ~Q·~rdVp

∣∣∣∣∣2

= (ρsl ,p − ρsl ,0)2

∣∣∣∣∣∫Vp

e i~Q·~rdVp

∣∣∣∣∣2

F( ~Q) =1

e i~Q·~rdVp

I SAXS( ~Q) = ∆ρ2V 2p |F( ~Q)|2

Scattering from a sphere

There are only a few morphologies which can be computed exactly and thesimplest is a constant density sphere of radius R.

F( ~Q) =1

∫ 2π

∫ π

0e iQr cos θr2 sin θ dθ dφ dr

F( ~Q) =1

sin(Qr)

Qrr2 dr

[sin(QR)− QR cos(QR)

]≡ 3J1(QR)

Where J1(x) is the Bessel functionof the first kind

0 5 10

F( ~Q) =1

∫ 2π

∫ π

F( ~Q) =1

sin(Qr)

Qrr2 dr

]≡ 3J1(QR)

0 5 10

F( ~Q) =1

∫ 2π

∫ π

F( ~Q) =1

sin(Qr)

Qrr2 dr

]≡ 3J1(QR)

0 5 10

F( ~Q) =1

∫ 2π

∫ π

F( ~Q) =1

sin(Qr)

Qrr2 dr

≡ 3J1(QR)

0 5 10

F( ~Q) =1

∫ 2π

∫ π

F( ~Q) =1

sin(Qr)

Qrr2 dr

]≡ 3J1(QR)

0 5 10

F( ~Q) =1

∫ 2π

∫ π

F( ~Q) =1

sin(Qr)

Qrr2 dr

]≡ 3J1(QR)

0 5 10

F( ~Q) =1

∫ 2π

∫ π

F( ~Q) =1

sin(Qr)

Qrr2 dr

]≡ 3J1(QR)

0 5 10

I (vectQ) = ∆ρ2V 2p

∣∣∣∣3J1(QR)

∣∣∣∣2

= ∆ρ2V 2p

∣∣∣∣3sin(QR)− QR cos(QR)

∣∣∣∣2

0 0.05 0.1 0.15 0.2

Q (Å-1

R=100Å

0 0.05 0.1 0.15 0.2

Q (Å-1

R=200Å

∣∣∣∣3J1(QR)

∣∣∣∣2 = ∆ρ2V 2p

∣∣∣∣2

0 0.05 0.1 0.15 0.2

Q (Å-1

R=100Å

0 0.05 0.1 0.15 0.2

Q (Å-1

R=200Å

∣∣∣∣3J1(QR)

∣∣∣∣2 = ∆ρ2V 2p

∣∣∣∣2

0 0.05 0.1 0.15 0.2

Q (Å-1

R=100Å

0 0.05 0.1 0.15 0.2

Q (Å-1

R=200Å

∣∣∣∣3J1(QR)

∣∣∣∣2 = ∆ρ2V 2p

∣∣∣∣2

0 0.05 0.1 0.15 0.2

Q (Å-1

R=100Å

0 0.05 0.1 0.15 0.2

Q (Å-1

R=200Å

Guinier analysis

In the long wavelength limit QR → 0 we can approximate the scatteringfactor with the first terms of the sum

F(Q) ≈ 3

[QR − Q3R3

120− · · ·

− QR

(1− Q2R2

24− · · ·

)]this simplifies to F(Q) ≈ 1− Q2R2

I SAXS(Q) ≈ ∆ρ2V 2p

[1− Q2R2

]2≈ ∆ρ2V 2

[1− Q2R2

]≈ ∆ρ2V 2

p e−Q2R2/5, QR � 1

Guinier analysis

F(Q) ≈ 3

[QR − Q3R3

120− · · ·

− QR

(1− Q2R2

24− · · ·

this simplifies to F(Q) ≈ 1− Q2R2

[1− Q2R2

]2≈ ∆ρ2V 2

[1− Q2R2

]≈ ∆ρ2V 2

p e−Q2R2/5, QR � 1

Guinier analysis

F(Q) ≈ 3

[QR − Q3R3

120− · · ·

− QR

(1− Q2R2

24− · · ·

[1− Q2R2

]2≈ ∆ρ2V 2

[1− Q2R2

]≈ ∆ρ2V 2

p e−Q2R2/5, QR � 1

Guinier analysis

F(Q) ≈ 3

[QR − Q3R3

120− · · ·

− QR

(1− Q2R2

24− · · ·

[1− Q2R2

≈ ∆ρ2V 2p

[1− Q2R2

]≈ ∆ρ2V 2

p e−Q2R2/5, QR � 1

Guinier analysis

F(Q) ≈ 3

[QR − Q3R3

120− · · ·

− QR

(1− Q2R2

24− · · ·

[1− Q2R2

]2≈ ∆ρ2V 2

[1− Q2R2

≈ ∆ρ2V 2p e

−Q2R2/5, QR � 1

Guinier analysis

F(Q) ≈ 3

[QR − Q3R3

120− · · ·

− QR

(1− Q2R2

24− · · ·

[1− Q2R2

]2≈ ∆ρ2V 2

[1− Q2R2

]≈ ∆ρ2V 2

p e−Q2R2/5, QR � 1

Guinier analysis

In the long wavelength limit(QR → 0), the form factor be-comes

F(Q) ≈ 1− Q2R2

I (Q) ≈ ∆ρ2V 2p e

−Q2R2/5

and the initial slope of the log(I ) vsQ2 plot is −R2/5

In terms of the radius of gyration,Rg , which for a sphere is given by√

I (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

0 0.001 0.002

Q2 (Å

R=100Å

R=200Å

Guinier analysis

F(Q) ≈ 1− Q2R2

I (Q) ≈ ∆ρ2V 2p e

−Q2R2/5

I (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

0 0.001 0.002

Q2 (Å

R=100Å

R=200Å

Guinier analysis

F(Q) ≈ 1− Q2R2

I (Q) ≈ ∆ρ2V 2p e

−Q2R2/5

I (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

0 0.001 0.002

Q2 (Å

R=100Å

R=200Å

Guinier analysis

F(Q) ≈ 1− Q2R2

I (Q) ≈ ∆ρ2V 2p e

−Q2R2/5

I (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

0 0.001 0.002

Q2 (Å

R=100Å

R=200Å

Guinier analysis

F(Q) ≈ 1− Q2R2

I (Q) ≈ ∆ρ2V 2p e

−Q2R2/5

I (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

0 0.001 0.002

Q2 (Å

R=100Å

R=200Å

Guinier analysis

F(Q) ≈ 1− Q2R2

I (Q) ≈ ∆ρ2V 2p e

−Q2R2/5

I (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

0 0.001 0.002

Q2 (Å

R=100Å

R=200Å

Guinier analysis

F(Q) ≈ 1− Q2R2

I (Q) ≈ ∆ρ2V 2p e

−Q2R2/5

I (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

0 0.001 0.002

Q2 (Å

R=100Å

R=200Å

Guinier analysis

F(Q) ≈ 1− Q2R2

I (Q) ≈ ∆ρ2V 2p e

−Q2R2/5

I (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

0 0.001 0.002

Q2 (Å

R=100Å

R=200Å

Calculation of Rg

∫Vpρsl ,p(~r)r2dVp∫

Vpρsl ,p(~r)dVp

In terms of the scattering length den-sity, we have

after orientational averaging this ex-pression can be used to extract Rg fromexperimental data using

I SAXS1 (Q) ≈ ∆ρ2V 2p e

−Q2R2g/3

this expression holds for uniform and non-uniform densities

Calculation of Rg

Vpρsl ,p(~r)dVp

−Q2R2g/3

Calculation of Rg

Vpρsl ,p(~r)dVp

−Q2R2g/3

Calculation of Rg

Vpρsl ,p(~r)dVp

−Q2R2g/3

Calculation of Rg

Vpρsl ,p(~r)dVp

−Q2R2g/3

Calculation of Rg

Vpρsl ,p(~r)dVp

−Q2R2g/3

Porod analysis

In the short wavelength limit(QR � 1), the form factor for asphere can be approximated

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

]≈ 3

[−cos(QR)

]I (Q) = 9∆ρ2V 2

[−cos(QR)

]2= 9∆ρ2V 2

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

)I (Q) =

2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100ÅR=200Å

power law drop as Q−4 for spheres

Porod analysis

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

[−cos(QR)

]I (Q) = 9∆ρ2V 2

[−cos(QR)

]2= 9∆ρ2V 2

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

)I (Q) =

2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100ÅR=200Å

Porod analysis

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

]≈ 3

[−cos(QR)

I (Q) = 9∆ρ2V 2p

[−cos(QR)

]2= 9∆ρ2V 2

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

)I (Q) =

2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100ÅR=200Å

Porod analysis

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

]≈ 3

[−cos(QR)

]I (Q) = 9∆ρ2V 2

[−cos(QR)

= 9∆ρ2V 2p

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

)I (Q) =

2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100ÅR=200Å

Porod analysis

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

]≈ 3

[−cos(QR)

]I (Q) = 9∆ρ2V 2

[−cos(QR)

]2= 9∆ρ2V 2

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

)I (Q) =

2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100ÅR=200Å

Porod analysis

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

]≈ 3

[−cos(QR)

]I (Q) = 9∆ρ2V 2

[−cos(QR)

]2= 9∆ρ2V 2

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

I (Q) =2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100ÅR=200Å

Porod analysis

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

]≈ 3

[−cos(QR)

]I (Q) = 9∆ρ2V 2

[−cos(QR)

]2= 9∆ρ2V 2

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

)I (Q) =

2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100ÅR=200Å

Porod analysis

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

]≈ 3

[−cos(QR)

]I (Q) = 9∆ρ2V 2

[−cos(QR)

]2= 9∆ρ2V 2

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

)I (Q) =

2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100Å

Porod analysis

F(Q) = 3

[sin(QR)

Q3R3− cos(QR)

]≈ 3

[−cos(QR)

]I (Q) = 9∆ρ2V 2

[−cos(QR)

]2= 9∆ρ2V 2

⟨cos2(QR)

⟩Q4R4

=9∆ρ2V 2

)I (Q) =

2π∆ρ2

10-1.4

10-1.2

10-1.0

10-0.8

Q (Å-1

R=100ÅR=200Å

Shape effect on scattering

The shape of the particle will have asignificant effect on the SAXS sincethe form factor is derived from anintegral over the particle volume,Vp.

If the particle is not spherical, thenits “dimensionality” is not 3 andthis will affect the form factor andintroduce a different power law inthe Porod regime.

dVp = 4πr2dr sphere

α = 4

dAp = 2πrdr disk

α = 2

dLp = dr rod

α = 1

α = 4

dAp = 2πrdr disk

α = 2

dLp = dr rod

α = 1

α = 4

dAp = 2πrdr disk

α = 2

dLp = dr rod

α = 1

α = 4

dAp = 2πrdr disk

α = 2

dLp = dr rod

α = 1

α = 4

dAp = 2πrdr disk

α = 2

dLp = dr rod

α = 1

dVp = 4πr2dr sphere α = 4

dAp = 2πrdr disk

α = 2

dLp = dr rod

α = 1

dVp = 4πr2dr sphere α = 4

dAp = 2πrdr disk α = 2

dLp = dr rod α = 1

Polydispersivity

Nucleation & growth of glycine

Can SAXS help us understand the nucleation and growth of a simplemolecule which is the prototype for pharmaceutical compounds?

initial studies at 12 keV show change but no crystallization

Glycine nucleation

0.04 0.06 0.08 0.1 0.2

q (Å-1

1.4×10-6

1.6×10-6

1.8×10-6

2.0×10-6

0.04 0.06 0.08 0.1 0.2

q (Å-1

2.2×10-6

2.4×10-6

2.6×10-6

change to 25 keV x-rays

study neutral (top) and acidic (bot-tom) solutions

Glycine nucleation

0.04 0.06 0.08 0.1 0.2

q (Å-1

1.4×10-6

1.6×10-6

1.8×10-6

2.0×10-6

0.04 0.06 0.08 0.1 0.2

q (Å-1

2.2×10-6

2.4×10-6

2.6×10-6

change to 25 keV x-rays

study neutral (top) and acidic (bot-tom) solutions

Glycine Rg

0 20 40 60 80Time (min)

aqueous solution

acidic solution

in aqueous solution, Rg impliesdimerization and increases due toaggregation until crystallization

in acidic solution, Rg remains smalland implies that no dimerizationor aggregation occurs before nucle-ation

“Relationship between Self-Association of Glycine Molecules in Supersaturated Solution and Solid State Outcome”,D. Erdemir et al. Phys. Rev. Lett. 99, 115702 (2007)

today’s outline - march 03, 2015segre/phys570/15s/lecture_13.pdf · today’s outline - march 03,...

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