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X-ray Scattering. Peak intensities Peak widths. Peak Intensities and Widths. Silicon. Basic Principles of Interaction of Waves. Periodic wave characteristics: Frequency : number of waves (cycles) per unit time – =cycles/time. [] = 1/sec = Hz. - PowerPoint PPT Presentation
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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
Basic Principles of Interaction of Waves
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
Basic Principles of Interaction of Waves
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:
Scattering by an Atom
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
Scattering by an Atom
Scattering by an Atom
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
Scattering by a Unit Cell
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
Scattering by a Unit Cell
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)
Scattering by a Unit Cell
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
Scattering by a Unit Cell
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
Scattering by a Unit Cell
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+
Scattering by a Unit Cell
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
Scattering by a Unit Cell
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
Interference Function
We define:
as deviation parameter
HSS 0
Interference Function
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:
Interference Function
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
Interference Function
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)
Interference Function
Interference Function
Interference Function
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
Small crystallites (nanocrystallites)
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
Small crystallites (nanocrystallites)
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
Small crystallites (nanocrystallites)
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)
Small crystallites (nanocrystallites)
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