Today’s Outline - October 03, 2016csrri.iit.edu/~segre/phys570/16F/lecture_12.pdfToday’s Outline - October 03, 2016 Scattering from two electrons Scattering from atoms Scattering

Today’s Outline - October 03, 2016

• Scattering from two electrons

• Scattering from atoms

• Scattering from molecules

• Radial distribution function

• Liquid scattering

APS Visits:10-ID: Friday, October 21, 201610-BM: Friday, October 28, 2016

Homework Assignment #03:Chapter 3: 1, 3, 4, 6, 8due Wednesday, September 05, 2016

C. Segre (IIT) PHYS 570 - Fall 2016 October 03, 2016 1 / 17























































Scattering from two electrons

Consider systems where there is only weak scattering, with no multiplescattering effects. We begin with the scattering of x-rays from twoelectrons.

r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

The scattering from the second electron willhave a phase shift of φ = ~Q ·~r .

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)

= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)C. Segre (IIT) PHYS 570 - Fall 2016 October 03, 2016 2 / 17

Scattering from many electrons

for many electrons

generalizing to a crystal

A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

∑j

e i~Q·~rj

Since experiments measure I ∝ A2, the phase information is lost. This is aproblem if we don’t know the specific orientation of the scattering systemrelative to the x-ray beam.

We will now look at the consequences of this orientation and generalize tomore than two electrons



for many electrons


A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

∑j

e i~Q·~rj





for many electrons


A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

∑j

e i~Q·~rj





for many electrons


A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

∑j

e i~Q·~rj





for many electrons


A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

∑j

e i~Q·~rj





for many electrons


A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

∑j

e i~Q·~rj




Two electrons — fixed orientation

The expression

I (~Q) = 2r20

(1 + cos(~Q ·~r)

)assumes that the two electronshave a specific, fixed orienta-tion. In this case the intensityas a function of Q is.

Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering. 0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



The expression

I (~Q) = 2r20

(1 + cos(~Q ·~r)


Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering.

0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



The expression

I (~Q) = 2r20

(1 + cos(~Q ·~r)


Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering. 0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Orientation averaging

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=

∫e iQr cos θ sin θdθdφ∫

sin θdθdφ

=1

4π2π

∫ π

0e iQr cos θ sin θdθ

=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr

Consider scattering from twoarbitrary electron distribu-tions, f1 and f2. A(~Q), isgiven by

and the intensity, I (~Q), is

if the distance between thescatterers,~r , remains constant(no vibrations) but is allowedto orient randomly in spaceand we take ~Q along the z-axis

substituting x = iQr cos θ anddx = −iQr sin θdθ⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr



if the distance between thescatterers,~r , remains constant(no vibrations) but is allowedto orient randomly in space

and we take ~Q along the z-axis


⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr



if the distance between thescatterers,~r , remains constant(no vibrations) but is allowedto orient randomly in space

and we take ~Q along the z-axis


⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr




substituting x = iQr cos θ anddx = −iQr sin θdθ

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr

=sin(Qr)

Qr





⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

⟨e i~Q·~r⟩

=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr


Randomly oriented electrons

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr

Recall that when we had a fixedorientation of the two electrons,we had and intensity variationI (~Q) = 2r20 (1 + cos(Qr)).

When we now replace the twoarbitrary scattering distributionswith electrons (f1, f2 → −r0),we change the intensity profilesignificantly.⟨I (~Q)

⟩= 2r20

(1 +

sin(Qr)

Qr

)0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



⟩= 2r20

(1 +

sin(Qr)

Qr

)0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



⟩= 2r20

(1 +

sin(Qr)

Qr

)

0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr


When we now replace the twoarbitrary scattering distributionswith electrons (f1, f2 → −r0),we change the intensity profilesignificantly.

⟨I (~Q)

⟩= 2r20

(1 +

sin(Qr)

Qr

)

0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



⟩= 2r20

(1 +

sin(Qr)

Qr

)0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



⟩= 2r20

(1 +

sin(Qr)

Qr

)0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Scattering from atoms

Single electrons are a good first example but a real system involvesscattering from atoms. We can use what we have already used to write anexpression for the scattering from an atom, ignoring the anomalous terms.

with limits in units of r0 of

The second limit can be understoodclassically as loss of phase coher-ence when Q is very large and asmall difference in position resultsin an arbitrary change in phase.

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

where zs is a screening correction

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

=

Z for Q→ 0

0 for Q→∞

However, it is better to usequantum mechanics to calculatethe form factor, starting with ahydrogen-like atom as an example.The wave function of the 1s elec-trons is just






ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs


f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

=

Z for Q→ 0

0 for Q→∞







ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs


f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

=

Z for Q→ 0

0 for Q→∞







ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs


f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

=

Z for Q→ 0

0 for Q→∞







ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs


f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

=

Z for Q→ 0

0 for Q→∞







ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs


f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

=

Z for Q→ 0

0 for Q→∞







ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs


f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

=

Z for Q→ 0

0 for Q→∞







ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs


f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

=

Z for Q→ 0

0 for Q→∞



Hydrogen form factor calculation

Since ρ(r) = |ψ1s(r)|2, the form factor integral becomes

f 01s(~Q) =1

πa3

∫e−2r/ae i

~Q·~r r2 sin θdrdθdφ

the integral in φ gives 2π and if we choose ~Q to be along the z direction,~Q ·~r → Qr cos θ, so

f 01s(~Q) =1

πa3

∫ ∞0

2πr2e−2r/a∫ π

0e iQr cos θ sin θdθdr

=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[−e iQr cos θ

∣∣∣π0dr

=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[e iQr − e−iQr

]dr

=1

πa3

∫ ∞0

2πr2e−2r/a2 sin(Qr)

Qrdr




f 01s(~Q) =1

πa3

∫e−2r/ae i



f 01s(~Q) =1

πa3

∫ ∞0

2πr2e−2r/a∫ π


=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[−e iQr cos θ

∣∣∣π0dr

=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[e iQr − e−iQr

]dr

=1

πa3

∫ ∞0


Qrdr




f 01s(~Q) =1

πa3

∫e−2r/ae i



f 01s(~Q) =1

πa3

∫ ∞0

2πr2e−2r/a∫ π


=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[−e iQr cos θ

∣∣∣π0dr

=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[e iQr − e−iQr

]dr

=1

πa3

∫ ∞0


Qrdr




f 01s(~Q) =1

πa3

∫e−2r/ae i



f 01s(~Q) =1

πa3

∫ ∞0

2πr2e−2r/a∫ π


=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[−e iQr cos θ

∣∣∣π0dr

=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[e iQr − e−iQr

]dr

=1

πa3

∫ ∞0


Qrdr




f 01s(~Q) =1

πa3

∫e−2r/ae i



f 01s(~Q) =1

πa3

∫ ∞0

2πr2e−2r/a∫ π


=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[−e iQr cos θ

∣∣∣π0dr

=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[e iQr − e−iQr

]dr

=1

πa3

∫ ∞0


Qrdr




f 01s(~Q) =1

πa3

∫e−2r/ae i



f 01s(~Q) =1

πa3

∫ ∞0

2πr2e−2r/a∫ π


=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[−e iQr cos θ

∣∣∣π0dr

=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[e iQr − e−iQr

]dr

=1

πa3

∫ ∞0


Qrdr




f 01s(~Q) =1

πa3

∫e−2r/ae i



f 01s(~Q) =1

πa3

∫ ∞0

2πr2e−2r/a∫ π


=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[−e iQr cos θ

∣∣∣π0dr

=1

πa3

∫ ∞0

2πr2e−2r/a1

iQr

[e iQr − e−iQr

]dr

=1

πa3

∫ ∞0


Qrdr


Form factor calculation

f 01s(~Q) =1

πa3

∫ ∞0


Qrdr

If we write sin(Qr) = Im[e iQr

]then the integral becomes

f 01s(~Q) =4

a3Q

∫ ∞0

re−2r/aIm[e iQr

]dr

=4

a3QIm

[∫ ∞0

re−2r/ae iQrdr

]

this can be integrated by parts with

u = r dv = e−r(2/a−iQ)dr

du = dr v = −e−r(2/a−iQ)

(2/a− iQ)

f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]



f 01s(~Q) =1

πa3

∫ ∞0


Qrdr



f 01s(~Q) =4

a3Q

∫ ∞0

re−2r/aIm[e iQr

]dr

=4

a3QIm

[∫ ∞0

re−2r/ae iQrdr

]




(2/a− iQ)

f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]



f 01s(~Q) =1

πa3

∫ ∞0


Qrdr



f 01s(~Q) =4

a3Q

∫ ∞0

re−2r/aIm[e iQr

]dr

=4

a3QIm

[∫ ∞0

re−2r/ae iQrdr

]this can be integrated by parts with



(2/a− iQ)

f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]



f 01s(~Q) =1

πa3

∫ ∞0


Qrdr



f 01s(~Q) =4

a3Q

∫ ∞0

re−2r/aIm[e iQr

]dr =

4

a3QIm

[∫ ∞0

re−2r/ae iQrdr

]




(2/a− iQ)

f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]



f 01s(~Q) =1

πa3

∫ ∞0


Qrdr



f 01s(~Q) =4

a3Q

∫ ∞0

re−2r/aIm[e iQr

]dr =

4

a3QIm

[∫ ∞0

re−2r/ae iQrdr




(2/a− iQ)

f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]



f 01s(~Q) =1

πa3

∫ ∞0


Qrdr



f 01s(~Q) =4

a3Q

∫ ∞0

re−2r/aIm[e iQr

]dr =

4

a3QIm

[∫ ∞0

re−2r/ae iQrdr




(2/a− iQ)

f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]



f 01s(~Q) =1

πa3

∫ ∞0


Qrdr



f 01s(~Q) =4

a3Q

∫ ∞0

re−2r/aIm[e iQr

]dr =

4

a3QIm

[∫ ∞0

re−2r/ae iQrdr




(2/a− iQ)

f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]C. Segre (IIT) PHYS 570 - Fall 2016 October 03, 2016 9 / 17


f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]

the first term becomes zero because re−r(2/a−iQ) → 0 as r → 0, thesecond is easily integrated

f 01s(~Q) =4

a3QIm

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

0

]=

4

a3QIm

[1

(2/a− iQ)2

]=

4

a3QIm

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

]=

4

a3QIm

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]

f 01s(~Q)

=4

a3Q

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

1

[1 + (Qa/2)2]2



f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]the first term becomes zero because re−r(2/a−iQ) → 0 as r → 0, thesecond is easily integrated

f 01s(~Q) =4

a3QIm

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

0

]=

4

a3QIm

[1

(2/a− iQ)2

]=

4

a3QIm

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

]=

4

a3QIm

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]

f 01s(~Q)

=4

a3Q

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

1

[1 + (Qa/2)2]2



f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr


f 01s(~Q) =4

a3QIm

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

0

]

=4

a3QIm

[1

(2/a− iQ)2

]=

4

a3QIm

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

]=

4

a3QIm

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]

f 01s(~Q)

=4

a3Q

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

1

[1 + (Qa/2)2]2



f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr


f 01s(~Q) =4

a3QIm

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

0

]=

4

a3QIm

[1

(2/a− iQ)2

]

=4

a3QIm

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

]=

4

a3QIm

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]

f 01s(~Q)

=4

a3Q

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

1

[1 + (Qa/2)2]2



f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr


f 01s(~Q) =4

a3QIm

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

0

]=

4

a3QIm

[1

(2/a− iQ)2

]=

4

a3QIm

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

]

=4

a3QIm

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]

f 01s(~Q)

=4

a3Q

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

1

[1 + (Qa/2)2]2



f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr


f 01s(~Q) =4

a3QIm

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

0

]=

4

a3QIm

[1

(2/a− iQ)2

]=

4

a3QIm

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

]=

4

a3QIm

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]

f 01s(~Q)

=4

a3Q

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

1

[1 + (Qa/2)2]2



f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr


f 01s(~Q) =4

a3QIm

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

0

]=

4

a3QIm

[1

(2/a− iQ)2

]=

4

a3QIm

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

]=

4

a3QIm

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]

f 01s(~Q)

=4

a3Q

(a/2)4(4Q/a)

[1 + (Qa/2)2]2

=1

[1 + (Qa/2)2]2



f 01s(~Q) =4

a3QIm

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

0

+

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr


f 01s(~Q) =4

a3QIm

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

0

]=

4

a3QIm

[1

(2/a− iQ)2

]=

4

a3QIm

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

]=

4

a3QIm

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]f 01s(~Q) =

4

a3Q

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

1

[1 + (Qa/2)2]2


1s and atomic form factors

f 01s(~Q) =1

[1 + (Qa/2)2]2

This partial form factor will varywith Z due to the Coulomb inter-action

In principle, one can compute thefull atomic form factors, however,it is more useful to tabulate the ex-perimentally measured form factors

0 5

Qao

0

5

10

f0(Q

)

C

O

F

f 0 (Q) =4∑

j=1

aje−bj sin2 θ/λ2 + c =

4∑j=1

aje−bj (Q/4π)2 + c



f 01s(~Q) =1

[1 + (Qa/2)2]2



0 5

Qao

0

5

10

f0(Q

)

C

O

F

f 0 (Q) =4∑

j=1


4∑j=1




f 01s(~Q) =1

[1 + (Qa/2)2]2



0 1 2 3 4 5

Qao

0.0

0.5

1.0

f0 1s(Q

)

Z=1

Z=3

Z=5

f 0 (Q) =4∑

j=1


4∑j=1




f 01s(~Q) =1

[1 + (Qa/2)2]2


In principle, one can compute thefull atomic form factors, however,it is more useful to tabulate the ex-perimentally measured form factors 0 1 2 3 4 5

Qao

0.0

0.5

1.0

f0 1s(Q

)

Z=1

Z=3

Z=5

f 0 (Q) =4∑

j=1


4∑j=1




f 01s(~Q) =1

[1 + (Qa/2)2]2



Qao

0.0

0.5

1.0

f0 1s(Q

)

Z=1

Z=3

Z=5

f 0 (Q) =4∑

j=1

aje−bj sin2 θ/λ2 + c

=4∑

j=1




f 01s(~Q) =1

[1 + (Qa/2)2]2



Qao

0.0

0.5

1.0

f0 1s(Q

)

Z=1

Z=3

Z=5

f 0 (Q) =4∑

j=1


4∑j=1




f 01s(~Q) =1

[1 + (Qa/2)2]2



0 5

Qao

0

5

10

f0(Q

)

C

O

F

f 0 (Q) =4∑

j=1


4∑j=1



Two hydrogen atoms

Previously we derived the scat-tering intensity from two local-ized electrons both fixed andrandomly oriented to the x-rays

when we now replace the two lo-calized electrons with hydrogenatoms, we have, for fixed atoms

and if we allow the hydrogenatoms to be randomly orientedwe have

with no oscillating structure inthe form factor

0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Two hydrogen atoms





0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Two hydrogen atoms





0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Two hydrogen atoms





0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Two hydrogen atoms





0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Two hydrogen atoms





0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Inelastic scattering

The form factors for all atoms dropto zero as Q →∞, however, otherprocesses continue to scatter pho-tons.

In particular, Compton scatteringbecomes dominant.

Compton scattering is an inelasticprocess: |~k| 6= |~k ′| and it is alsoincoherent.

The Compton scattering containsinformation about the momentumdistribution of the electrons in theground state of the atom.

0 5 10

Qao

0

10

20

30

40

f0(Q

)C

Ge

Mo

Cu

Si







0 5 10

Qao

0

10

20

30

40

f0(Q

)C

Ge

Mo

Cu

Si





Compton scattering is an inelasticprocess: |~k| 6= |~k ′|

and it is alsoincoherent.


0 5 10

Qao

0

10

20

30

40

f0(Q

)C

Ge

Mo

Cu

Si







0 5 10

Qao

0

10

20

30

40

f0(Q

)C

Ge

Mo

Cu

Si







0 5 10

Qao

0

10

20

30

40

f0(Q

)C

Ge

Mo

Cu

Si







10500 11000 11500 12000 12500Energy (eV)

X-r

ay Inte

nsity (

arb

units) φ=160

o


Total atomic scattering

We can now write the total scatter-ing from an atom as the sum of twocomponents:

(dσ

dΩ

)el

∼ r20 |f (Q)|2(dσ

dΩ

)in

∼ r20S(Z ,Q)

recall that f (Q)→ Z as Q → 0

so |f (Q)|2 → Z 2 as Q → 0

and for incoherent scattering we ex-pect S(Z ,Q)→ Z as Q →∞

ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

]

1

10

100

Inte

nsity

(in

uni

ts o

f r02 )

Ar

He



We can now write the total scatter-ing from an atom as the sum of twocomponents:(

dσ

dΩ

)el

∼ r20 |f (Q)|2

(dσ

dΩ

)in

∼ r20S(Z ,Q)


so |f (Q)|2 → Z 2 as Q → 0


ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

]

1

10

100

Inte

nsity

(in

uni

ts o

f r02 )

Ar

He




dσ

dΩ

)el

∼ r20 |f (Q)|2(dσ

dΩ

)in

∼ r20S(Z ,Q)


so |f (Q)|2 → Z 2 as Q → 0


ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

]

1

10

100

Inte

nsity

(in

uni

ts o

f r02 )

Ar

He




dσ

dΩ

)el

∼ r20 |f (Q)|2(dσ

dΩ

)in

∼ r20S(Z ,Q)


so |f (Q)|2 → Z 2 as Q → 0


ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

]

1

10

100

Inte

nsity

(in

uni

ts o

f r02 )

Ar

He




dσ

dΩ

)el

∼ r20 |f (Q)|2(dσ

dΩ

)in

∼ r20S(Z ,Q)


so |f (Q)|2 → Z 2 as Q → 0


ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

]

1

10

100

Inte

nsity

(in

uni

ts o

f r02 )

Ar

He




dσ

dΩ

)el

∼ r20 |f (Q)|2(dσ

dΩ

)in

∼ r20S(Z ,Q)


so |f (Q)|2 → Z 2 as Q → 0


ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

]

1

10

100

Inte

nsity

(in

uni

ts o

f r02 )

Ar

He




dσ

dΩ

)el

∼ r20 |f (Q)|2(dσ

dΩ

)in

∼ r20S(Z ,Q)


so |f (Q)|2 → Z 2 as Q → 0


ZHe = 2 ZAr = 18

0 5 10 15 20

Q [Å-1

]

1

10

100

Inte

nsity

(in

uni

ts o

f r02 )

Ar

He




dσ

dΩ

)el

∼ r20 |f (Q)|2(dσ

dΩ

)in

∼ r20S(Z ,Q)


so |f (Q)|2 → Z 2 as Q → 0


ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

]

1

10

100

Inte

nsity

(in

uni

ts o

f r02 )

Ar

He


Scattering from molecules

From the atomic form factor, wewould like to abstract to the nextlevel of complexity, a molecule (wewill leave crystals for Chapter 5).

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

As an example take the CF4

molecule

We have the following relation-ships:

OA

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB

= OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB = OC

= OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB = OC = OD

= 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u

= −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z

= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB




Fmol(~Q) =∑j

fj(~Q)e i~Q·~r


molecule


OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5

but from the triangle OO ′A

(O ′A)2 = 1− z2

Fmol± = f C (Q) + f F (Q)

[3e∓iQR/3 + e±iQR

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B

= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3

u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]

|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f Fsin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]

|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f Fsin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3R



− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

z =1

3u = cos−1(−z) = 109.5


(O ′A)2 = 1− z2



]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)

Q√

8/3RC. Segre (IIT) PHYS 570 - Fall 2016 October 03, 2016 16 / 17

The Radial Distribution Function

Ordered 2D crystal Amorphous solid or liquid




















Documents

Today’s Outline - October 03, 2016csrri.iit.edu/~segre/phys570/16F/lecture_12.pdfToday’s Outline - October 03, 2016 Scattering from two electrons Scattering from atoms Scattering