IX. X-ray diffraction

IX. X-ray diffraction9-1. Production of X-ray

http://www.arpansa.gov.au/radiationprotection/basics/xrays.cfm

Vacuum, thermionic emission, high voltage,and a target

Braking radiation CharacteristicX-ray

Auger electrons

Energetic electron hitting a target

Target

20 /2/2/ hchmvmvE

21 /2/2/ hchmvmvE

(i) Braking radiation:

Vemv 2/20

swhchmvE /2/ max20

V2 > V1

Short wavelength limits

Excitationsource K

CharacteristicsX-Ray photon

Augerelectron

Radiativetransition

NonradiativetransitionM} {

K K2 K1

The figure in Note is wrong!

(ii) characteristic radiation

a typical X-ray spectrum

(iii) Cu K radiationCu K1 =1.54050 ÅCu K2 =1.54434 Å

Cu K =1.5418 Å

High angle line

Low angle line

Why we can resolve K1 and K2 doublelines at high angle?

sin2 hkld dddhkl cos2

tancos

sin22cos2 hkld

tandd dtan

9-2. X-ray diffraction(i) Laue method: variable, white radiation, fixed(ii) Diffractometer method: fixed, characteristic radiation, variable(iii) Powder method: fixed, characteristic radiation, variable(iv) Rotating crystal method: fixed, characteristic radiation, variable

9-2-1. Laue method

(1) Laue conditionVon Laue derived the “ Laue conditions “ in1912 to express the necessary conditions fordiffraction.Assume that are three crystal lattice vectors in a crystal

cba , ,

)()( ****' clbkahaGaSSa hkl

hSSa )( '

Similarly, kSSb )( '

lSSc )( '

Very often , the Laue conditions are expressed as

hSSa )ˆˆ( ' kSSb )ˆˆ( '

lSSc )ˆˆ( '

/ˆ '' SS

Unit vector

The three Laue conditions must be satisfiedsimultaneously for diffraction to occur. The physical meaning of the 3 Laue conditionsare illustrated below.

1st Laue conditions

The path differencebetween the wavesequals to

)ˆˆ( ' SSaBDAC

The criterion for diffraction to occur ishSSa )ˆˆ( ' hSSa )( '

orinteger

For 1-dimensional crystal

Cones of diffracted beams for different h

Stereographic projection representation for 1-D crystalsThe projection of diffracted beams for h = 0 is a great circle if the incident beam direction is perpendicular, i.e.

The projection of diffracted beams for h = 0 is a small circle if the incident beam direction is not perpendicular

2nd Laue condition

kSSb )( '

integer

For a two-dimensional crystal

The 1st and 2nd conditions are simultaneously satisfied only along lines of intersection of cones.

Stereographic projection for 2-D crystals

The lines of intersection of cones are labeled as (h , k)

3rd Laue condition

lSSc )( ' l: integer

For a single crystal and a monochromatic wavelength, it is usually no diffraction to occur

(2) Laue photographLaue photograph is performed at “ single crystal and variable wavelength ”.

(2-1) Ewald sphere construction

*'hklGSS

2 hklGkk

white radiation : wavelength is continuoue;lwl and swl: the longest and shortest wavelength in the white radiation.

Two Ewald spheres of radius OA and OB form

in between two spherical surface meets the diffraction condition since the wavelength is continuous in white radiation.

http://www.xtal.iqfr.csic.es/Cristalografia/archivos_06/laue1.jpg

(2-2) zone axisThe planes belong to a zone

We define several planes belong to a zone [uvw], their plane normal [hi ki li] are perpendicular to the zone axis, In other words,

0][][ iii lkhuvw 0 iii wlvkuh

In the Laue experiments, all the planes meet the diffraction criterion. Each plane meets the 3 Laue conditions.

ihSSa )ˆˆ( '

ikSSb )ˆˆ( '

ilSSc )ˆˆ( '

0)ˆˆ()( ' wlvkuhSScwbvau iii

0 iii wlvkuh

0)ˆˆ()( ' SScwbvau

for all the plane belonged to the zone [uvw].

If we define cwbvauOA

The path difference between the waves equals to

0)ˆˆ( ' SSOA

(A) for the condition < 45o

For all the planes (hikili) in the same zone [uvw]0 wlvkuh iii

0)ˆˆ()( ' wlvkuhSScwbvau iii

Therefore, all the diffracted beam directions Ŝ1', Ŝ2', Ŝ3', …Ŝi' are on the same cone surface.(B) for the condition > 45o

The same as well

9-2-2 Diffractometer method

Instrumentation at MSE, NTHU

Shimatsu XRD 6000

Rigaku TTR

Bruker D2 phaser

9-2-2-1 -2 scan

If a new material is deposited on Si(100) and put on the substrate holder in the diffractometer set-up, is always parallel to the Si(100) surface normal.

Any sample

2'2 kk

Rotate the incident beam at until meets the diffraction condition at

Ewald sphere construction

*111 lkhG

Si (100) substrate

*111 lkhG

This is equivalent to at least one crystal with(h1k1l1) plane normal in parallel with theSi(100) surface normal.

When the incident beam is rotated (rotatingthe Ewald sphere or rotating the reciprocallattice), will meet the diffractioncondition at

*222 lkhG

*111 lkhG

Si (100) substrate

*222 lkhG

This is equivalent to at least one crystal with(h2k2l2) plane normal in parallel with theSi(100) surface normal.

If only one plane of the film is observed in the diffractometer measurement, the film is epitaxial or textured on Si(100).

(b) XRD spectrum of AlN deposited on Si(100)

20 30 40 50 600

AlN(101)AlN(100)

AlN(002)

RFpower=180wI=1.2(A)AC pluse=350(v)

2 £c(degree)

A large number of grains with AlN(002) surface normal in parallel to Si(100) surface normal, but small amount of grains of AlN(100) and AlN(101) are also in parallel to Si(100) surface normal.

9-2-2-2 X-ray rocking curveX-ray rocking is usually used to characterize the crystal quality of an epitaxial or textured film.At a certain diffraction condition

2 lkhGkk

the incident and outgoing directions of X-ray are fixed and the sample is rotated by scan.

Careful scan around a specific diffractionpeak!

When the crystal is rotated by d, this is equivalentto the detector being moved by the same d. (?) The FWHM of the diffracted peak truly reflects thewidth of each reciprocal lattice point, i.e. theshape effect of a crystal. (by high resolution!)

*111 lkhG

9-2-2-3. ScanThe scan is used to characterize the quality of the epitaxial film around its plane normal by rotating (0-2)

Example: the crystal quality of highly oriented grains each of which is a columnar structure shown below

Each lattice point in the reciprocal lattice will be expanded into an orthorhombic volume according to the shape effect. The scan can be used to confirm the orientation of the columnar structure relative to the substrate.

Shape effect

( )Reciprocal

lattice points

9-2-4 Powder method (Debye-Scherrer Camera)

http://www.adias-uae.com/plaster.html

http://www.stanford.edu/group/glam/xlab/MatSci162_172/LectureNotes/06_Geometry,%20Detectors.pdf

(a) Ewald sphere constructionReciprocal lattice of a polycrystallinesample

B.D. Cullity

Reciprocal lattice become a series of concentricspherical shells with as their radius*

Powder can be treated as a very larger number of polycrystals with grain orientation in a random distribution

Intersection of Ewaldsphere with sphericalreciprocal lattice!

forms a cone resulting from the intersection between Ewald sphere and the spherical reciprocal lattice.

The intersection between the concentric spherical shells of in radius and the Ewald sphere of radius are a series of circles (recorded as lines in films)

/12/ k

ExampleDebye-Scherrer powder pattern of Cu made with Cu K radiation

1 2 3 4 5 6 7 8line hkl h2+k2+l2 sin2 sin (o) sin/(Å-1) fCu

12345678

111200220311222400331420

3481112161920

0.13650.18200.3640.5000.5460.7280.8650.910

0.3690.4270.6030.7070.7390.8530.9300.954

21.725.337.145.047.658.568.472.6

0.240.270.390.460.480.550.600.62

22.120.916.814.814.212.511.511.1

1 9 10 11 12 13 14

line |F|2 P Relative integrated intensityCalc.(x105) Calc. Obs.

12345678

78106990452035003230250021201970

122486

12.038.503.702.832.743.184.816.15

7.52 3.562.012.380.710.482.452.91

10.04.72.73.20.90.63.33.9

VsSssmwss

111200220311222400331420

cossin2cos1

(a) Structure factor (Cu: Fm-3m, a = 3.615 Å)

GNSF s

Four atoms at [000], [½ ½ 0], [½ 0 ½],[0 ½ ½]

lwkvhuijG

jjjefS )(2

21(2)000(2 lkhilkhi fefe

210(2)

21(2 lkhilkhi

)1( )()()( lkilhikhi eeef

fSG 40GS

If h, k, l are unmixedIf h, k, l are mixed

222 16|| fNF

(b) sin2 hkld

sin 2222

22 lkh

Small is associated with small h2+k2+l2.

lkhdhkl

3111ad

542.1sin3

615.32542.1sin2 111111111 d

24.0542.13694.0sin 111

o111111 68.213694.0sin

136.0sin 1112

Example: 111

0 0.1 0.2 0.3 0.429 27.19 23.63 19.90 16.48

1.2263.2390.19

2.03.090.19

24.03.0 111111

7814)4( 21112111 CufF

}111{ p = 8 (23 = 8)

05.12cossin2cos1

1111112

753370I

(c) Multiplicity P is the one counted in the point group stereogram.

In Cubic (h k l){hkl}{hhl}{0kl}{0kk}{hhh}{00l}

P = 48P = 24P = 24P = 12P = 8P = 6

(3x2x23)(3x23)(3x2x22)(3x22)(23)(3x2)

How about tetragonal?

(d) Lorentz-polarization factor

cossin2cos1

(i) polarization factorThe scattered beam depends on the angle ofscattering. Thomson equation

0 sinsin4

I0: intensity of the incident beam

0 CmKg104 K: constant

: angle between the scattering direction and the direction of acceleration of the electron

incident beam travel in directionxRandom polarized beam yEy ˆ zEz ˆ

zEyEE zy ˆˆ

222 IIIEEE zyzy

The intensity at point Py component = yOP = /2

20 rKII yPy

z component

2cos220 r

KII zPz

= zOP = /2 -2

2cos22020 r

KIrKIIII zyPzPyP

Polarization factor

22cos1 2

(ii) Lorentz factor

1cos2sin

(ii-1) factor due to grain orientation or crystal rotation2sin

The factor is counted in powder method and in rotating crystal method.

First, the integratedintensity )2( Id

BIId max)2(

(a) sin

1max I

Crystals with reflection planes make an angleB with the incident beam Bragg conditionmet intensity diffracted in the direction 2B.But, some other crystals are still diffracted inthis direction when the angle of incidentdiffers slightly from B!

Diffraction Angle 2

Imax/2

IntegratedIntensity

path difference for 11-22= AD – CB = acos2 - acos1

= a[cos(B-) - cos (B+)]= 2asin()sinB ~ 2a sinB.

2Na sinB = completelycancellation (1- N/2, 2- (N/2+1) …)

BNa sin2

Maximum angular range of the peak

(b) cos

the width B increases as the thickness of the crystal decreases , as shown in the figure in the next page

… …012

Constructive Interference

sin)3(23sin)2(22

);sin(2 Bhkld );sin()2(2)(2 Bhkld

);sin()3(2)(3 Bhkld …

Destructive interference:extra path difference (plane 0 and plane m/2): /2

);sin()2/(2))(2/( Bhkldmm 2/)2/( m

BBB cossincossin)sin( << 1s cos ~ 1 and sin ~ .

);cos)(sin2/(22/)2/( BBhkldmm

BhklBhkl mddmm cos2/sin)2/(2)2/( = t

Broadening (B): FWHMThickness dependent (just like slits)

max BI

(iii-1) :factor due to the number of crystal counted in the powder method

The number of crystal is not constant for aparticular B even through the crystal are oriented completely at random.

For the hkl reflection, the range of angle near the Bragg angle, over which reflection is appreciable, is . Assuming that N is the number of crystals located in the circular band of width r.

2cos B

# of grains satisfying the diffraction condition cosB. more grains satisfying the diffraction condition at higher B.

(iii-2) factor due to the segment factor in the Debye-Scherrer film

The film receives a greater proportion of a diffraction cone when the reflection is in the forward or backward direction than it does near 2/2

The relative intensity per unit length is proportional to

BR 2sin21

i.e. proportional toB2sin

Therefore, the Lorentz factor for the powder method is

cossin41

2sin1cos

and for the rotating crystal method and the diffractometer method is

(iv) Plot of polarization and Lorentz factors

0 20 40 60 80 100 120 140 160 180 2000

2 (Degrees)

cossin2cos1

IX. X-ray diffraction

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