II. Kinematical Scattering Theory and the Reciprocal...

II. Kinematical Scattering Theory and the Reciprocal Lattice

dΘ

Θ Θ

d sinΘ

Bragg-Gleichung 2d sin = nΘ λ

Netzebenen

Bragg‘s Law (William Lawrence Bragg, 1912)

2d nsin Θ = λ

• Only possible for λ < 2d, i.e., we need radiation with short wavelengths (in the non-visible regime)

• geometrical consideration Diffraction condition results from periodic array of crystal lattice

Kinematical Scattering Theory and the Reciprocal Lattice

‘X-Rays‘:Electromagnetic radiation with wavelengths in the Ångstrøm (10-10 m) regime

[ ][ ]keVE

398.12

E

hcÅ ==λWave vector:

λ

π=

2k

Elastic Scattering of Waves:

Scattering vector

Wave vector transfer

Inelastic Scattering: Compton-Effect

)trk(i0eE)t,r(E ω−⋅=

vvvvv

k'kQv

hv

hr

h −=

k

k’Q

2Θ

Scattering Angle

λ

π2' == kk

Elastic

Inelastic

k ≠ 'k

Basic Elements


II. Kinematical Scattering Theory and the Reciprocal Lattice

dΘ

Θ Θ

d sinΘ

Bragg-Gleichung 2d sin = nΘ λ

Netzebenen

Bragg‘s Law (William Lawrence Bragg, 1912)

2d nsin Θ = λ

• Only possible for λ < 2d, i.e., we need radiation with short wavelengths (in the non-visible regime)

• geometrical consideration Diffraction condition results from periodic array of crystal lattice


Which effects are not described yet by Bragg‘s law?What has been neglected?

• Intensity of diffracted wave• Crystal structure (spatial distribution of atoms within unit cell)• Atomic structure (spatial distribution of electrons within atom)• Thermal vibration of crystal lattice• Shape/Size of scattering object (e.g. crystal)

• Absorption• Refraction• Multiple Scattering Processes•....

On what objects “X-ray waves” will be scattered?Elementary interaction process ?

Bragg’s Law


• Incoming X-ray wave (frequency ω) stimulates (weakly bounded) Electron

• Oscillation of Electron (frequency ω) electron acts as oscillating Hertz dipole

• Re-emission of electromagnetic waves (frequency ω)

• Amplitude of scattered wave in the far field (from Maxwell’s equations)

R

eePErRrE

ikRrki

rad

vvv00),( =

[ ]

−

−

+

=

ertunpolarisi

onPolarisatip

onPolarisatis

)2(cos1

)2cos(

1

22

1 θ

θP

Å1082.2mc4

er 15

20

2

0−×=

πε=

Classical electron radiusThomson scattering length

p-Polarisation s-PolarisationObserver

Observer

2θ

2θincoming x-ray wave

incoming x-ray waveele tronc

ele tronc

Scattering by a Single Electron

Thomson Scattering


For an arbitrary distribution of the electrons (with density ρ(r)) and apart from resonant absorption edges we can assume:

• the electrons can be treated as free electrons

• they equally take part in the scattering process

• the total scattering amplitude is then proportional to the electron density ρ(r)

• the total scattering amplitude at the observer’s position R can be calculated through linear superposition of all partial waves taking into account the phases:

∫−

ρ=

−

Rr

ee)r(dVPEr)R(E

Rrikrki

00rad vvvv

vvvv

Kirchhoff‘s Law

Electron Distribution (r)ρ

dVk

r

Rr-R

Observer

2Θ

0

Scattering by an Ensemble of Electrons


Observer and object are far away: R >> r

r-R is pointing in the same direction. We name this direction as k‘/k‘

( )rQi

Rikrkir'ki

Rikrki

RrRikrki

Rrik

eR

eee

Re

eR

ee

Rr

e vvvvvvvv)v

vvvv

vv⋅−

⋅⋅⋅−

⋅⋅

⋅−⋅⋅

−⋅

=≈≈−

Fraunhofer Approximation

∫⋅−ρ= rQi

ikR

00rad e)r(dVR

ePEr)Q,R(E

vvvvdV

k

r

R

r-R

Observer

2Θ

k’

λ

Θπ=

sin4Qr

k'kQvvr

−=

k

k’Q

2Θ




∫⋅−ρ= rQi

ikR

00rad e)r(dVR

ePEr)Q,R(E

vvvv

Some remarks:

• We “see“ changes in the electron density (Fourieranalysis)

• Expression above does not hold for crystalline materials only

• It is valid also for non-periodic systems (e.g. amorphous materials, liquids)

• Experimentally relevant value is the scattered intensity

pair correlation functionR)dRrρ()Rρ()rg( 3rrrr

+= ∫

∫−

∝∝rQi)erdVg(

2rad

EIrr

r


Home work: Derivation of Eq.(1)

(1)



Differential Scattering Cross Section

Scattered intensity measured by a detector Isc

Isc = photons per second (detectors usually count single photons)

= energy per second (power) flowing through the area of the detector divided by the energy of each photon

I0 = incident beam intensity = .. |E0|2

= energy per second within a cross sectional area A0

AD = area of detector = R2∆Ω

0

2

20

2

0

),(

A

R

E

RQE

I

I radsc ∆Ω⋅=

v


∆Ω: solid angle of detectorA0: cross section of incident beamR: distance of sample to detector


)flux ) ( (Incident into second per scattered photons of Numbers

∆Ω

∆Ω=

Ωd

dσ

( )20

electronPr

dd

=

Ω

σ

( )( )

22

02

02

0

02

00)(

),(

/ ∫−=∆Ω⋅

⋅∆Ω

⋅=

∆Ω=

ΩdVerPrR

AE

ARQE

AI

I

d

d rQiradscvvv

v

ρσ


It is usual to normalize the scattered intensity Isc by

• the incident flux I0/A0

• the solid angle ∆Ω of the detector

2

0

220

0

22

2

220

0

2

20

2

0)()(

),(∫∫

⋅−⋅− ∆Ω=

∆Ω⋅=

∆Ω⋅= rQirQiradsc erdV

A

Pr

A

RerdV

R

Pr

A

R

E

RQE

I

I vvvvvv

v

ρρ


Electron density of a single atom: ρA(r)

)Q(fR

ePEre)r(dV

Re

PEr)Q,R(EikR

00rQi

A

ikR

00radvvvv vv

=ρ= ∫⋅−

atomic scattering factor (amplitude)

atomic form factor

∫⋅−ρ= rQi

A e)r(dV)Q(fvv

vv

Scattering VectorA

tom

ic S

catt

eri

ng

Fa

cto

r0

Z

‘delocalized’density

‘localized’density

experimental/calculated values are listed in:

International Tables of Crystallography

or

http://henke.lbl.gov/optical_constants/asf.html

Scattering by a Single Atom



Properties of atomic scattering factor

• f(0) = Z strong forward scattering• f(∞) = 0

• The stronger the delocalization of the electron density the stronger the Q dependency

• Point like localization ρA(r) = Z δ(r) f(Q) = Z = const

• Spherical symmetry: Qr

QrrrdrQfQf A

sin)(4)()( 2

∫== ρπr

Home Work: Calculate the atomic form factor of a hydrogen atom (Solution: f(Q) = 16 / (4 + Q2a0

2)2) (a0: Bohr radius)

Scattering by a Single Atom

A 3D lattice can be created by a set of vectors Rn :

a1, a2 and a3: Basic lattice vectors defining the unit cell

n1, n2, and n3 : integer numbers

332211n nnn aaaR ++=

Within the unit cell: Atoms with corresponding atomic form factors fj(Q) located at positions inside unit cell r1, r2, ..., rk.

Scattering by a Crystal Lattice

The atoms

• are thus sitting at positions: n1a1 + n2a2 + n3a3 + rj

• exhibit the (complex) atomic scattering factor fj(Q)

The total scattered amplitude is the sum of all partial amplitudes taking into account the relative phases !


Scattering by a Crystal Lattice


For the summation over n1,n2,n3, and j it is very helpful that:

fj(Q) exp [-iQ(n1a1 + n2a2 + n3a3 + rj)] = exp [-iQ(n1a1 + n2a2 + n3a3)] x fj(Q) exp(-iQrj)

S(Q): Structure factor G(Q): Lattice factor

)Q(G)Q(SPRr

E

ee)Q(fRr

PEEE

00

Atoms

G

n,n,n

)ananan(Qi

S

k

1j

rQij

00jcrystal

321

332211j

vv

4444 34444 21444 3444 21

v vvvvvv

××=

=××== ∑ ∑∑ ++−

=

−

We can thus independently perform the summation:

The total scattered amplitude is the sum of all partial amplitudes taking into account the relative phases !

∑ ++−=

321

332211

n,n,n

)ananan(QieGvvvv

• The sum over n1,n2 and n3 depends on the shape and size of the crystal

it might by complicated to perform the summation

• For simplicity we assume that the crystal shape is a parallelepiped (for 90°angle cuboid), with the edges parallel to the basic vectors a1, a2 and a3

)aQinexp()aQinexp()aQinexp(G1N

0n33

1N

0n22

1N

0n11

3

3

2

2

1

1

∑∑∑−

=

−

=

−

=

−⋅−⋅−=vvvvvv

)aQ(sin

)aQN(sin

)aQ(sin

)aQN(sin

)aQ(sin

)aQN(sinG

3212

33212

2212

22212

1212

11212

2vr

vr

vr

vr

vr

vr

⋅⋅=

and thus for the intensity:

Laue Function

Scattering by a Crystal Lattice: The Lattice Factor


• Maxima appear at Q = n ⋅2π/a Bragg‘s Law

• The maxima of the Laue function define the reciprocal lattice

• Width and height of the peaks (Bragg reflections) depend on crystal shape and size (macroscopic symmetry)

• Widths and intensity of Bragg reflections are independent of n: ∆Q = 2π/aN

0

10

20

30In

tens

ity

Qa

2 /Nπ

N2

0−2π−3π 2π 3π−π π

Laue Function: N = 5 N = 10

0

5000

10000

Inte

nsity

Qa0−2π−3π 2π 3π−π π

Laue Function: N = 100

1 1


Scattering by a Crystal Lattice: The Lattice Factor

|G(Q

)|2

|G(Q

)|2

The Reciprocal Lattice


• The Laue function exhibit maxima at

Q1 = h ⋅2π/a1, Q2 = k ⋅2π/a2, Q3 = l ⋅2π/a3

(Laue Equations)

• The Maxima of the Laue function define the reciprocal lattice

• A possible (primitive) basis of reciprocal space is given by

• Every (arbitrary) reciprocal lattice vector Ghkl can be expressed as

ba a

a a ab

a a

a a ab

a a

a a a

2 3

1 2 3

2

3 1

1 2 3

3

1 2

1 2 3

1 2 2 2=×

⋅ ×=

×

⋅ ×=

×

⋅ ×π π π

321 bbbG lkh ++=hkl

hklGQ = Miller Indices

Please note: Diffraction is caused by net-planes of type (hkl)

Indexing of Net-planes


Example: cubic lattice

• Determine intersection of the plane with the axes a1, a2,a3 (in units of the lattice parameters)

• Calculate reciprocal values and search for the smallest integers – h,k,l –, being in the same relationship

• Please note the difference between directions [hkl] and planes (hkl)

• General case: directions and planes are not perpendicular to each other (but is valid for cubic crystal)

Bragg Condition: Reciprocal Space versus Angular Space


In reciprocal space we have found the following relationship for constructive interference:

Since |Ghkl |= 2π/dhkl

and Q = 4π sinθΒ/λ

we can write this as

hklGQ =

λsinθ2d Bhkl =⋅

Homework: Prove the relationship: |Ghkl|= 2π/dhkl

Equivalent!

λ

Θπ=

sin4Qr

k'kQvvr

−=

k

k’Q

2Θ

tQ

πθ

λ

θπ 2cos4=∆=∆

θ

λ=θ∆

cost2

Width of Bragg Reflections – Impact of Crystal Size


∆Q = 2π/aN

Question: Can Bragg reflections become infinitely sharp?

Reciprocal dependence of ∆Q and thickness

Answer: No! Penetration depth inside crystal is finite!

Numerical Values: t = 5 µm (penetration depth into Silicon, CuKα radiation, λ = 0.154 nm, θ = 45° (Bragg angle)

∆θ = 2·10-5 = 0.0011° Sharp reflections !

Consequence: Small Crystals exhibit broad Bragg reflections !

k

k’No refraction

No extinction

Single Scattering

∫⋅−ρ= rQi

ikR

00rad e)r(dVR

ePEr)Q,R(E

vvvvv

• Refraction Re(χ0), Absorption Im(χ0) und Diffraction (χg)

• Re(χ) = 10-5..10-6; Im(χ) = 10-7..10-8 weak interaction

• Bragg Reflection: strong multiple diffraction

• Bragg Reflection: strong extinction t = 1 .. 10 µm

• Intrinsic (Darwin) width of Bragg reflections ∆θD ∝λ-1·χg (1 .. 10 arcsec)

See Lecture Dynamical Theory

Interaction between Material and X-rays is mediated by dielectric polarizability (susceptibility) χ(r)

Solving of Maxwell‘s equations

( ) ∑ ⋅−⋅χ=χg

rgig er

v

vvvv

Width of Bragg Reflections – Dynamical Theory

Darwin Curve

(b) Transmission Geometry(Laue Case)

λ = 1.54 Å, Silicon crystal (thickness = 300 µm), 422 reflection

(a) Reflection Geometry (Bragg Case)


Width of Bragg Reflections – Dynamical Theory

S(Q): Structure factor G(Q): Lattice factor

)()(

)(

00

,,

)(

1

00

321

332211

QGQSPR

rE

eeQfR

rPEEE

Atoms

G

nnn

anananQi

S

k

j

rQi

jjcrystal

j

vv

444 3444 21444 3444 21

v vvvvvv

××=

=××== ∑ ∑∑ ++−

=

−

Scattering by a Crystal Lattice: Structure Factor


• Lattice factor G(Q) Position of Bragg reflections

|G(Q)|2 independent of (hkl)

• Structure factor S(G) Summation over Atoms within unit cell

relative Intensity of Bragg reflections

∑=

−=

k

1j

rQij

je)Q(f)Q(Svvvv

Structure factor (Structure amplitude) S(Q) = S(G)

Crystal structure determines actual value of S(G)

Selection rules forbidden/weak reflections

High symmetry a few allowed reflections only


Defined at Bragg reflections



( )[ ]∑ ++−===j

jyjjhkl lzkyhxifSSS π2exp)( GG

When expressing the coordinates of the atoms within the unit cell as rj = xja1 + yja2 + zja3 = (xj, yj, zj) und using the Miller indices (h,k,l)

we end up with

321 bbbG lkh ++=

Homework: Show that the periodic electron density ρ(r) in a crystal can be expressed by the structure factors S(G) through (V: Volume of primitive unit cell)

∑ ⋅⋅=ρ

G

rGiG eS

V1

)r(v

vvvv

Comment:

In the literature various notations can be found for the term “Structure factor“. Most common are:

Shkl or Fhkl


S = 0 for h+k+l = odd

S = 2f for h+k+l = even

S = 4f if all h, k, l are either even or odd

S = 0 else

Structure factor – Selection Rules


( ) ( )[ ] S hkl f i h k l= + − + +1 exp π

bcc-lattice (body centered cubic): identical atoms at r1 = 0 and r2 = (½ , ½ , ½ )

fcc-Gitter (face centered cubic):identical atoms at r1 = 0, r2 = (0, ½, ½ ), r3 = (½, 0, ½ ), r4 = (½ , ½, 0)

( ) ( )[ ] ( )[ ] ( )[ ] S hkl f i h k i k l i h l= + − + + − + + − +1 exp exp expπ π π

S = 8f for h+k+l = 4n and all h,k,l even (e.g. 220, 400, 444)

S = 8f for all h,k,l odd (e.g. 111, 135)

S = 0 else (z.B. 100, 200, 211)


Diamond structure: identical atoms atr1 = (0,0,0), r2 = (0,½,½), r3 = (½,0,½), r4 = (½ ,½,0), r5 = (¼,¼,¼), r6 = (¼ ,¾ ,¾), r7 = (¾,¼,¾), r8 = (¾ ,¾ ,¼ )

( )

( )[ ] ( )[ ] ( )[ ]( )[ ] ( )[ ]( )[ ] ( )[ ]

S hkl f

i h k i k l i h l

i h k l i h k l

i h k l i h k l

=

+ − + + − + + − + +

+ − + + + − + + +

+ − + + + − + +

1

2 3 3 2

3 3 2 3 3 2

exp exp exp

exp / exp /

exp / exp /

π π π

π π

π π

Homework: Calculate the structure factor for Zinkblende structure and determine the selection rules! Atoms of sort A at: r1 = (0,0,0), r2 = (0,½,½), r3 = (½,0,½), r4 = (½ ,½,0)Atoms of Sort B at: r5 = (¼,¼,¼), r6 = (¼ ,¾ ,¾), r7 = (¾,¼,¾), r8 = (¾ ,¾ ,¼ )

Structure factor – Selection Rules

What happens to the scattered intensity when we introduce a flat surface (with normal vector parallel zu a3 ) with infinite lateral dimension?

• ‘Rod like‘ intensity distribution which is extended along a3

(perpendicular to surface)

• Envelop of Laue function

a1

a3

a2

a *3 a *3

a *1 a *1

G

Flayer(Q): Scattering amplitude of a two-dimensional atomic layer

3z

3zaiQ

layer

0j

jaiQlayerCTR

e1

)Q(Fe)Q(FF

−== ∑

∞

=

−v

v

( )2/aQsin4

)Q(FFI

3z2

2layer2CTRCTR

v

==

The Influence of a Surface – Crystal Truncation Rods

Crystal Truncation Rods (CTR)


|F

| C

TR

2

|F

| C

TR

2

10-1

100

101

102

δz/z = 0.00δz/z = +0.05δz/z = -0.05

Qa0−π π 2π−2π

10-1

100

101

102

Qa0−π π 2π−2π

Perfect atomic layers

( )3z2

2layer2CTRCTR

aQsin4

)Q(FFI

v

==

Vertical relaxation ∆z/z of the top atomic layer

)zz

1(aiQlayeraiQ

layerCTR 3z

3ze)Q(F

e1

)Q(FF

∆+−

⋅+−

=v

v

Qza3 Qza3



Calculated Intensity Profiles for These Surfaces

Two possible ‘ideal‘(111)Si Surfaces

J.K. Robinson, D.J. Tweet, Surface x-ray diffraction, Rep.Prog.Phys. 55, 599 (1992)

Experimental Reflectivity (dots) along with best fitting (line)

Dimer-Adatom-Stacking Fault Model of the r Si(111)7x7 Surface Reconstruction


Applications of CTR-Analysis

• Surface structure:

Vertical lattice relaxation

Surface Reconstruction

• Surface Morphology (Roughness)

• Size and Shape of nano structures

• Identification und Characterization of surface facets

• ...

• ...




Identification und Characterization of surface facets

Laue function of truncated pyramid


(How) Can we determine the crystal structure through measurement of all structure factor?

• The structure factor S(Q) is a complex number

• We measure the diffracted intensity |S(Q)|2

• The complex phases of S(Q) have to be determined

• Things gets easier when the crystal structure are centrosymmetric:

• S(Q) is a real number with unknown sign

∑ ++π−=ρ

l,k,h

)lzkyhx(i2hkleS

V1

)z,y,x(

Structure Factor – Determination of Crystal Structure

II. Kinematical Scattering Theory and the Reciprocal...

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