X-ray Scattering
Peak intensities
Peak 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
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.

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.
A simple wave completes
cycle in 360 degrees
E (t , x)  A exp kx  t 
x 
x

 A exp 2   t   A exp 2  t 
c 


  2 , k 
2

Basic Principles of Interaction of Waves
For t = 0:
x

E ( x) t 0  A exp kx  A exp  2 
 
For x = 0:
E(t ) x0  A exp t   A exp 2t 
Basic Principles of Interaction of Waves
Consider two waves with the same wavelength and amplitude but displaced a distance x0.
The phase shift:
 x0 


  2 
E1 (t )  A exp t 
Waves #1 and #2 are 90o out of phase
E1 (t )  A exp t   
x0
Waves #1 and #2 are 180o out of phase
When similar waves combine, the outcome can be constructive or destructive interference
Superposition 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
The Laue Equations
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 .
acos   cos  0   h
In 2D and 3D:
acos   cos  0   h
2D
bcos   cos  0   k
3D
ccos   cos  0   l
The equations must be satisfied simultaneously, it is in general difficult to produce a diffracted
beam with a fixed wavelength and a fixed crystal.
Lattice Planes
Bragg’s Law
If the path AB +BC is a multiple of the x-ray wavelength λ, then two waves will give a
constructive interference:
n
n  AB  BC  2d sin 
2d sin   n
Bragg’s Law
The incident beam and diffracted beam are always coplanar.
The angle between the diffracted beam and the transmitted beam is always 2.
Since sin  1:
n
 sin   1
2d
For most crystals d ~ 3 Å
For n = 1:
  2d
6Å
Cu K1  = 1.5406 Å
Rewrite Bragg’s law:
d
n
  2 sin     2d  sin 
1 h2  k 2  l 2

2
d
a2
a
d100 
a
2
1 00
a
1
d 200 
 a
22  0  0 2
UV radiation   500 Å
  2d sin 
Bragg’s Law
Silicon lattice constant:
aSi = 5.43 Å
2d sin   n
Scattering by an Electron
Elementary scattering unit in an atom is
electron
Classical scattering by a single free electron
– Thomson scattering equation:
  2
e 4  1  cos 2 2 

I  I 0 2 4 2 
mcr 
2

The polarization factor of an
unpolarized primary beam
e4
 26
2

7
.
94

10
cm
m 2c 4
If r = few cm:
I
 10  26
I0
1 mg of matter has ~1020 electrons
Another way for electron to scatter is
manifested in Compton effect.
Cullity p.127
Scattering by an Atom
Atomic
Scattering
Factor
f 
Cu
amplitude of the wave scattered by an atom
amplitude of the wave scattered by one electron
Scattering by an Atom
Scattering by a group of electrons at positions rn:
Scattering factor per electron:
f e   exp 2i /  s  s 0   r  dV
Assuming spherical symmetry for the charge
distribution  = (r ) and taking origin at the center of
the atom:

f e   4r 2  r 
0
sin kr
dr
kr
For an atom containing several electrons:

f   f en    4r 2  n r 
n
n
0
f – atomic scattering factor
f 
sin kr
dr
kr
sin 

Calling Z the number of electrons per atom we get:

n

0
4r 2  n r dr  Z
Scattering by an Atom
f 
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
amplitude of the wave scattered by an atom
amplitude of the wave scattered by one electron
Scattering by an Atom
Scattering by a Unit Cell
Fhkl 
For atoms A & C
 21  MCN  2d h 00 sin   
d h 00  AC 
a
h
For atoms A & B
 31  RBS 
AB
AB
x
MCN 


AC
AC
a/h


2hx
phase
  2
31  2
 31


a
If atom B position: u  x / a
31 
For 3D:
2hx
 2hu
a
  2 hu  kv  lw
amplitude scattered by atoms in unit cell
amplitude scattered by single electron
e ix  cos x  i sin x
Scattering by a Unit Cell
Aei  A cos   Ai sin 
2
We can write:
Aei  Aei Ae i  A2
Aei  f exp 2ihu  kv  lw
F  f1ei1  f 2 ei2  f 3ei3  ...
N
F   f ne
1
i n
N
  f ne
1
2i  hun  kvn  lwn 

I  Fhkl Fhkl
 Fhkl
2
Scattering by a Unit Cell
Useful expressions
eix
e i
e0
so e ni
e ni
eix  e ix
 cos x  i sin x
 e3i  e5 i    1
 e 2 i  e 4 i    1
n
  1
 e  ni
 2 cos x
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
F  fe2i 0  f
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

F  fe2i 0  fe2i h / 2 k / 2   f 1  ei h k 
F 2f
h and k unmixed
F 0
h and k mixed

2
(200), (400), (220)   Fhkl  4 f
2
(100), (121), (300)   F
hkl
2
0
“forbidden”
reflections
Scattering by a Unit Cell
Examples
For body-centered cell

F  fe2i h0 k 0l 0   fe2i h / 2 k / 2l / 2   f 1  ei h k l 
F 2f
when (h + k + l ) is even
F 0
when (h + k + l ) is odd

2
F

4
f
(200), (400), (220)   hkl
2
(100), (111), (300)  
Fhkl  0
2
“forbidden”
reflections
Scattering by a Unit Cell
Examples
For body centered cell with different atoms:
F  f Cl e 2i h 0 k 0l 0   f Cs e 2i h / 2 k / 2l / 2 
 f Cl  f Cs ei h k l 
F  f Cl  f Cs
when (h + k + l) is even
F  f Cl  f Cs
when (h + k + l) is odd
Cs+
(200), (400), (220)  
Fhkl   f Cs  f Cl 
(100), (111), (300)  
Fhkl   f Cs  f Cl 
2
2
2
2
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):
Fhkl   f n e 2i hun  kvn lwn   f 1  ei h  k   ei h l   ei k l  
N
1
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
Fhkl 
2
16f 2,
h, k and l unmixed even and odd
0,
h, k and l mixed even and odd
The Structure Factor
N
Fhkl   f n e 2 i  hun  kvn lwn 
1
Fhkl 
amplitude scattered by all atoms in a unit cell
amplitude scattered by a single electron
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.
Diffracted Beam Intensity
Structure factor
Polarization factor
Lorentz factor
Multiplicity factor
Temperature factor
Absorption factor
…..
I (q )  F q 
I C (q)  mALpK F (q)  I b
2
2
The Polarization Factor
The polarization factor p arises from the fact that an electron does not scatter
along its direction of vibration
In other directions electrons radiate with an intensity proportional to (sin )2:
The polarization factor (assuming that the incident beam is unpolarized):
1  cos 2 2
p
2
The Lorentz - 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
L
1
sin 2
The combination of geometric corrections are lumped together into a single
Lorentz-polarization (Lp) factor:
1  cos 2 2
Lp 
sin 2
The effect of the Lp factor is to decrease the intensity at intermediate angles
and increase the intensity in the forward and backwards directions
The Temperature Factor
The temperature factor is given
by:

sin 2  
exp  B 2 
 

where the thermal factor B is
related to the mean square
displacement of the atomic
vibration:
B  82  u 2
This is incorporated into the
atomic scattering factor:
f  f 0e  M  f 2 ~ e 2 M
Scattering by C atom expressed in electrons
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| =
4sin/) more rapidly than it would normally.
The Multiplicity 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
values
Evaluated 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 symmetry
For the highest cubic symmetry we have:
100, 1 00, 010, 0 1 0, 001, 00 1
p100 = 6
110, 1 10, 1 1 0, 1 1 0, 101, 10 1, 1 0 1, 1 01, 011, 0 1 1, 01 1, 0 1 1
p110 = 12
111, 11 1, 1 1 1, 1 11, 1 1 1, 1 1 1, 1 1 1, 1 1 1
p111=8
The Absorption Factor
Angle-dependent absorption within the sample itself will modify the
observed intensity
Absorption factor for infinite thickness specimen is:
A
1
2
Absorption factor for thin films is given by:
 2  
A  1  exp 

 sin 
where μ is the absorption coefficient, τ is the total thickness of the film
Diffracted Beam Intensity

I  Fhkl Fhkl
 Fhkl
2
I C (q)  Ap( Lp ) K F (q)  I b
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
2

2
 2   1  cos 2
I C ( q)  1  exp 
K
F
(
q
)
 Ib

 sin  sin 2

Diffraction: Real Samples
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
2d sin   n
Crystallite Size

I  Fhkl Fhkl
 Fhkl
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
2
Crystallite Size
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:
  2d sin  B
Rays A, D, …, M makes angle B
Rays B, …, L makes angle 1
Rays C, …, N makes angle 2
Crystallite Size
For angle B diffracted intensity is maximum
For 1 and 2 – intensity is 0.
For angles 1 >  > 2 – intensity is nonzero.
B
Real case
Ideal case
1
21  2 2   1   2
2
The Scherrer Equation
2t sin 1  m  1 ,
2t sin  2  m  1.
t sin 1  sin  2    ,
Subtracting:
       
2t cos 1 2  sin  1 2   .
 2   2 
1 and 2 are close to B, so:
1   2  2 B ,
       
sin  1 2    1 2 .
 2   2 
Thus:
   
2t  1 2  cos  B   ,
 2 
t

B cos  B
The Scherrer Equation
More exact treatment (see Warren) gives:
t
0.94
B cos  B
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.
Scherrer’s formula
Interference Function
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).
(s  s 0 )

 hb1  kb 2  lb 3
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:
S  S0   H  
as deviation parameter
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:
Atotal 
N1 1 N 2 1 N 3 1
 
n1  0
 F exp
n2  0 n3  0
2i

s  s 0  n1a1  n2a 2  n3a3 
ni – particular unit cells
Ni – total number of unit cells along ai
From the definition of the reciprocal lattice vector:
s  s0    H    hb1  kb 2  lb3   
Interference Function
Atotal  F  exp 2i H     n1a1  n2a 2  n3a 3 
n1
n2
n3
n1
n2
n3
 F  exp 2i hb1  kb 2  lb 3     n1a1  n2a 2  n3a 3 
since:
b j a i   ij
Atotal  F  exp 2i  n1a1  exp 2i  n2 a2  exp 2i  n3a3 
n1
n2
n3
Converting from exp to sines:
 3 sin   Ni ai 
Atotal  F 
 exp N1  1a1  N 2  1a 2  N3  1a3 
 i 1 sin   ai 
Interference Function
Calculating intensity we lose phase information therefore:
I  Atotal
2
sin 2   N i a i 
I F 
sin 2   a i 
i 1
3
2
interference function
Maximum intensity at Bragg peak is F2N2
Width of the Bragg peak  1/N
N is a number of unit cells along (s - s0)
Interference Function
Interference Function
Small crystallites (nanocrystallites)
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:
I   f m e 2i  s s0  rm  f n e  2i  s s0  rn ,
m
n
I 
f
m
fe
2i  s s 0  rmn
m n
n

m
f
fe
iq  rmn
m n
n
s0
where
rmn  rm  rn
q
2

(s  s 0 )
s – s0
takes all
orientations

rmn
s
Small crystallites (nanocrystallites)
Average intensity from an array of atoms which takes all orientations in space:
e
2i  s s 0 rmn
I 

m

sin qrmn
1
iqrmn cos
2

e
2rmn sin d 
2 
4rmn  0
qrmn
fm fn
n
sin qrmn
,
qrmn
Debye scattering equation
where
q
s – s0
4 sin 

It involves only the magnitudes of the distances
rmn of each atom from every other atom
s0

rmn
s
Small crystallites (nanocrystallites)
Consider material consisting of polyatomic molecules.
It is not too dense – there is complete incoherency between the scattering by different
molecules.
Intensity per molecule:
I N 
m

n
fm fn
sin qrmn
R
qrmn
correction
factor
Lets take a carbon tetrachloride as example.
It is composed of tetrahedral molecules CCl4.
Then:

sin qr C  Cl 
I N  f C  f C  4 f Cl



qr
C

Cl



sin qr C  Cl 
sin qr C  Cl 
 4 f Cl  f Cl  f C
 3 f Cl





qr
C

Cl
qr
C

Cl


C Cl4
Small crystallites (nanocrystallites)
For tetrahedral CCl4 molecule:
r Cl  Cl  
8
r C  Cl 
3
sin qr C  Cl 
qr C  Cl 
sin qr C  Cl 
 12 f Cl2
R
qr C  Cl 
I N  f C2  4 f Cl2  8 f C f Cl
r C  Cl   1.82 Å
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.
Intensity (in e.u. per molecule) for a CCl4 gas in which the
C – Cl distance is r = 1.82 Å. (Warren)
Small crystallites (nanocrystallites)
Lets treat crystal as a molecule
FCC has 14 atoms:


8 at corners of a cube
6 at face center positions
I N 
m

fm fn
n
sin qrmn
R
qrmn
a is the edge of a cube
I Nf 2  14  72

sin q a 2 30 sin qa

qa
qa 2
(311)
48 sin qa 1.5 24 sin qa 2

qa 1.5
qa 2
8 sin qa 3

qa 3
(220)
(111)
(200)
(222)
Reciprocal Lattice and Diffraction
s  s0

 d*hkl  hb1  kb 2  lb 3
It is equivalent to the Bragg law since s  s 0  2 sin  :
s  s0


2 sin 

 d*hkl 
  2d hkl sin 
1
d hkl