x-ray scattering

X-ray Scattering

Peak intensitiesPeak widths

Second Annual SSRL Workshop on Synchrotron X-ray Scattering Techniques in Materials and Environmental Sciences: Theory and Application

Tuesday, May 15 - Thursday, May 17, 2007

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Silicon

Peak Intensities and Widths

Basic Principles of Interaction of Waves

Periodic wave characteristics: Frequency : number of waves (cycles) per unit time – =cycles/time.[] = 1/sec = Hz. Period T: time required for one complete cycle – T=1/=time/cycle. [T] =

sec. Amplitude A: maximum value of the wave during cycle. Wavelength : the length of one complete cycle. [] = m, nm, Å.

tx

Atc

xA

tkxAxtE

2exp2exp

exp),(

2

,2 k

A simple wave completes cycle in 360 degrees

For t = 0:

For x = 0:

x

AkxAxEt

2expexp)(0

tAtAtEx

2expexp)(0


Consider two waves with the same wavelength and amplitude but displaced a distance x0.

The phase shift:

tAtE exp)(1

02x

x0

tAtE exp)(1

Waves #1 and #2 are 90o out of phase

Waves #1 and #2 are 180o out of phase

When similar waves combine, the outcome can be constructive or destructive interference


Resulting wave is algebraic sum of the amplitudes at each point

Small difference in phase Large difference in phase

Superposition of Waves

Difference in frequency and phase

Superposition of Waves

If an X-ray beam impinges on a row of atoms, each atom can serve as a source of scattered X-rays.The scattered X-rays will reinforce in certain directions to produce zero-, first-, and higher-order diffracted beams.

The Laue Equations

Consider 1D array of scatterers spaced a apart.Let x-ray be incident with wavelength .

ha 0coscos

lc

kb

ha

0

0

0

coscos

coscos

coscos

2D

3D

The equations must be satisfied simultaneously, it is in general difficult to produce a diffracted beam with a fixed wavelength and a fixed crystal.

In 2D and 3D:

The Laue Equations

Lattice Planes

If the path AB +BC is a multiple of the x-ray wavelength λ, then two waves will give a constructive interference:

nd sin2

n

Bragg’s Law

sin2dBCABn

1sin2

d

nSince sin 1: For n = 1: d2

UV radiation 500 Å

Cu K1 = 1.5406 ÅFor most crystals d ~ 3 Å 6 Å

aa

d

aa

d

a

lkh

d

2

1

002

001

1

2200

2100

2

222

2

sin2sin2 dn

d sin2d

Bragg’s Law

nd sin2

Bragg’s Law

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aSi = 5.43 Å

Silicon lattice constant:

2

2cos1 2

242

4

0

rcm

eII

Elementary scattering unit in an atom is electronClassical scattering by a single free electron – Thomson scattering equation:

If r = few cm:

The polarization factor of an unpolarized primary beam

2

2cm2642

4

1094.7 cm

e

26

0

10I

I

1 mg of matter has ~1020 electrons

Another way for electron to scatter is manifested in Compton effect.

Cullity p.127

Scattering by an Electron

electrononebyscatteredwavetheofamplitude

atomanbyscatteredwavetheofamplitudef

Cu

Atomic Scattering Factor

Scattering by an Atom

Scattering by a group of electrons at positions rn:

Scattering factor per electron:

dVife rss 0/2exp

Assuming spherical symmetry for the charge distribution = (r ) and taking origin at the center of the atom:

0

2 sin4 dr

kr

krrrfe

n

nn

en drkr

krrrff

0

2 sin4

f – atomic scattering factor

Calling Z the number of electrons per atom we get:

Zdrrrn

n

0

24

sin

f

For an atom containing several electrons:


electrononebyscatteredwavetheofamplitude

atomanbyscatteredwavetheofamplitudef

The atomic scattering factor f = Z for any atom in the forward direction (2 = 0): I(2=0) =Z2

As increases f decreases functional dependence of the decrease depends on the details of the distribution of electrons around an atom (sometimes called the form factor)

f is calculated using quantum mechanics


electronsinglebyscatteredamplitude

cellunit in atomsbyscatteredamplitudehklF

h

aACd

dMCN

h

h

00

0012 sin2

a

hx

ha

x

AC

ABMCN

AC

ABRBS

22

2

/

1313

13

For atoms A & C

For atoms A & B

hua

hx 22

13

If atom B position: axu /

For 3D: lwkvhu 2

phase

Scattering by a Unit Cell

We can write:

lwkvhuifAei 2exp

Nlwkvhui

n

Ni

n

iii

nnnn efefF

efefefF

1

2

1

321 ...321

2

hklhklhkl FFFI

22

sincos

sincos

AAeAeAe

AiAAe

xixe

iii

i

ix


Useful expressions

xeeee

e

eeeeee

xixe

ixix

inin

nin

ii

iii

ix

cos2

1 so

11

sincos

420

53

Hint: don’t try to use the trigonometric form – using exponentials is much easier


Examples

Unit cell has one atom at the origin

ffeF i 02

In this case the structure factor is independent of h, k and l ; it will decrease with f as sin/ increases (higher-order reflections)


Examples

Unit cell is base-centered

khikhii effefeF 12/2/202

0

2

F

fF h and k unmixed

h and k mixed

(200), (400), (220)

(100), (121), (300)

224 fFhkl

02 hklF “forbidden”

reflections


Examples

For body-centered cell

lkhilkhilkhi effefeF 12/2/2/20002

0

2

F

fF when (h + k + l ) is even

when (h + k + l ) is odd

(200), (400), (220)

(100), (111), (300)

224 fFhkl

02 hklF “forbidden”

reflections


Examples

For body centered cell with different atoms:

lkhiCsCl

lkhiCs

lkhiCl

eff

efefF

2/2/2/20002

CsCl

CsCl

ffF

ffF

when (h + k + l) is even

when (h + k + l) is odd

(200), (400), (220)

(100), (111), (300)

2ClCs

2ffFhkl

2ClCs

2ffFhkl

Cs+ Cl+


The fcc crystal structure has atoms at (0 0 0), (½ ½ 0), (½ 0 ½) and (0 ½ ½):

lkilhikhiN

lwkvhuinhkl eeefefF nnn 1

1

2

If h, k and l are all even or all odd numbers (“unmixed”), then the exponential terms all equal to +1 F = 4 f If h, k and l are mixed even and odd, then two of the exponential terms will equal -1 while one will equal +1 F = 0

2

hklF16f 2, h, k and l unmixed even and odd

0, h, k and l mixed even and odd


The structure factor contains the information regarding the types ( f ) and locations (u, v, w) of atoms within a unit cell.

A comparison of the observed and calculated structure factors is a common goal of X-ray structural analyses.

The observed intensities must be corrected for experimental and geometric effects before these analyses can be performed.

N

lwkvhuinhkl

nnnefF1

2

electronsingleabyscatteredamplitude

cellunitainatomsallbyscatteredamplitudehklF

The Structure Factor

Diffracted Beam Intensity

Structure factorPolarization factorLorentz factorMultiplicity factorTemperature factorAbsorption factor…..

bC IqFmALpKqI 2)()(

2)( qFqI

The polarization factor p arises from the fact that an electron does not scatter along its direction of vibrationIn other directions electrons radiate with an intensity proportional to (sin )2:

The polarization factor (assuming that the incident beam is unpolarized):

2

2cos1 2 p

The Polarization Factor

The Lorenz factor L depends on the measurement technique used and, for the diffractometer data obtained by the usual θ-2θ or ω-2θ scans, it can be written as

The combination of geometric corrections are lumped together into a single Lorentz-polarization (Lp) factor:

The effect of the Lp factor is to decrease the intensity at intermediate angles and increase the intensity in the forward and backwards directions

2sin

1L

2sin

2cos1 2Lp

The Lorentz - Polarization Factor

As atoms vibrate about their equilibrium positions in a crystal, the electron density is spread out over a larger volume.This causes the atomic scattering factor to decrease with sin/ (or |S| = 4sin/) more rapidly than it would normally.

2

2sinexp

B

where the thermal factor B is related to the mean square displacement of the atomic vibration:

228 uB

MM efeff 220 ~

The temperature factor is given by:

This is incorporated into the atomic scattering factor:

Sca

tteri

ng b

y C

ato

m e

xpre

ssed in e

lect

rons

The Temperature Factor

The multiplicity factor arises from the fact that in general there will be several sets of hkl -planes having different orientations in a crystal but with the same d and F 2 valuesEvaluated by finding the number of variations in position and sign in h, k and l and have planes with the same d and F 2

The value depends on hkl and crystal symmetryFor the highest cubic symmetry we have:

111,111,111,111,111,111,111,111

110,101,110,011,011,101,110,101,011,011,101,110

100,001,010,010,001,100 p100 = 6

p110 = 12

p111=8

The Multiplicity Factor

Angle-dependent absorption within the sample itself will modify the observed intensity

Absorption factor for thin films is given by:

sin

2exp1A

where μ is the absorption coefficient, τ is the total thickness of the film

Absorption factor for infinite thickness specimen is:

2

1A

The Absorption Factor

2

hklhklhkl FFFI

bC IqFKLpApqI 2)()()(

where K is the scaling factor, Ib is the background intensity, q = 4sinθ/λ is the scattering vector for x-rays of wavelength λ

bC IqFKqI

2

2

)(2sin

2cos1

sin

2exp1)(

Diffracted Beam Intensity

Up to this point we have been considering diffraction arising from infinitely large crystals that are strain free and behave like ideally imperfect materials ( x-rays only scattered once within a crystal)

Crystal size and strain affect the diffraction pattern we can learn about them from the diffraction pattern

High quality crystals such as those produced for the semiconductor industry are not ideally imperfect need a different theory to understand how they scatter x-rays

Not all materials are well ordered crystals

Diffraction: Real Samples

As the crystallites in a powder get smaller the diffraction peaks in a powder pattern get wider.Consider diffraction from a crystal of thickness t and how the diffracted intensity varies as we move away from the exact Bragg angle

If thickness was infinite we would only see diffraction at the Bragg angle

nd sin2

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hklhklhkl FFFI

Crystallite Size

Rays A, D, …, M makes angle B

Rays B, …, L makes angle 1

Rays C, …, N makes angle 2

Suppose the crystal of thickness t has (m + 1) planes in the diffraction direction.

Let say is variable with value B that exactly satisfies Bragg’s Law:

Bd sin2

Crystallite Size

For angle B diffracted intensity is maximum

For 1 and 2 – intensity is 0.

For angles 1 > > 2 – intensity is nonzero.

Real case Ideal case

2121 222

1 B

Crystallite Size

.1sin2

,1sin2

2

1

mt

mt

.2

sin2

cos2

,sinsin

2121

21

t

tSubtracting:

1 and 2 are close to B, so:

.22

sin

,2

2121

21

B

Thus:

B

B

Bt

t

cos

,cos2

2 21

The Scherrer Equation

More exact treatment (see Warren) gives:

BBt

cos

94.0 Scherrer’s formula

Suppose = 1.54 Å, d = 1.0 Å, and = 49o:

for crystal size of 1 mm, B = 10-5 deg.

for crystal size of 500 Å, B = 0.2 deg.

The Scherrer Equation

We calculate the diffraction peak at the exact Bragg angle B and at angles that have small deviations from B.

If crystal is infinite then at B intensity = 0.

If crystal is small then at B intensity 0. It varies with angle as a function of the number of unit cells along the diffraction vector (s – s0).

At deviations from B individual unit cells will scatter slightly out of phase.

Vector (s – s0)/ no longer extends to the reciprocal lattice point (RLP).

3210 bbbss

lkh

)(

Interference Function

(1) > B(1) for 001 and (2) > B(2) for 002

If Hhkl = H is reciprocal lattice vector then (s – s0)/ H.

Real space Reciprocal space


We define:

as deviation parameter

HSS 0


In order to calculate the intensity diffracted from the crystal at B, the phase differences from different unit cells must be included.For three unit vectors a1, a2 and a3:

1

0332211

1

0

1

0

3

3

2

2

1

1

2exp

N

n

N

n

N

ntotal nnn

iFA aaass 0

ni – particular unit cellsNi – total number of unit cells along ai

321 bbbHss 0 lkh

From the definition of the reciprocal lattice vector:


1 2 3

1 2 3

332211321

332211

2exp

2exp

n n n

n n ntotal

nnnlkhiF

nnniFA

aaabbb

aaaH

since: ijij ab

321

332211 2exp2exp2expnnn

total anianianiFA

Converting from exp to sines:

332211

3

1

111expsin

sinaaa

a

a

NNNN

FAi i

iitotal


Calculating intensity we lose phase information therefore:

2

totalAI

3

12

22

sin

sin

i i

iiNFI

a

a

interference function

Maximum intensity at Bragg peak is F2N2

Width of the Bragg peak 1/NN is a number of unit cells along (s - s0)


Consider our sample as any form of matter in which there is random orientation.This includes gases, liquids, amorphous solids, and crystalline powders.The scattered intensity from such sample:

rmn

s – s0

s0 s

n

inm

mn

inm

m

n

in

m

im

mnmn

nm

effeffI

efefI

rqrss

rssrss

0

00

2

22 ,

where

nmmn rrr takes allorientations

)(2

0ssq

Small crystallites (nanocrystallites)

Average intensity from an array of atoms which takes all orientations in space:

rmn

s – s0

s0 s

where

Debye scattering equation

sin4

q

It involves only the magnitudes of the distances rmn of each atom from every other atom

,sin

n mn

mnnm

m qr

qrffI


mn

mnmn

iqr

mn

i

qr

qrdre

re mnmn

sinsin2

4

1

0

2cos2

2

rss 0

Consider material consisting of polyatomic molecules.

It is not too dense – there is complete incoherency between the scattering by different molecules.

Intensity per molecule:

Lets take a carbon tetrachloride as example.

It is composed of tetrahedral molecules CCl4.

Then:

Rqr

qrffNI

n mn

mnnm

m

sin correctionfactor

ClCqr

ClCqrf

ClCqr

ClCqrfff

ClCqr

ClCqrfffNI

ClCClCl

ClCC

sin3

sin4

sin4

C Cl4


For tetrahedral CCl4 molecule:

The intensity depends on just one distance r(C – Cl).

Peaks and dips do not require the existence of a crystalline structure.

Certain interatomic distances that are more probable than others are enough to get peaks and dips on the scattering curve.

ClCrClClr 3

8

Intensity (in e.u. per molecule) for a CCl4 gas in which the C – Cl distance is r = 1.82 Å. (Warren)

R

ClCqr

ClCqrf

ClCqr

ClCqrffffNI

Cl

ClCClC

sin12

sin84

2

22

Å82.1 ClCr


Lets treat crystal as a moleculeFCC has 14 atoms: 8 at corners of a cube 6 at face center positions

a is the edge of a cube

Rqr

qrffNI

n mn

mnnm

m

sin

3

3sin8

2

2sin24

5.1

5.1sin48

sin30

2

2sin72142

qa

qa

qa

qa

qa

qa

qa

qa

aq

aqNfI

(111)(200)

(220)

(311)

(222)


321*0 bbbd

sslkhhkl

hklhkl d

ss 1sin2 *0

d

It is equivalent to the Bragg law since :

sin2 hkld

sin20 ss

Reciprocal Lattice and Diffraction

x-ray scattering

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