1. Peak intensities Peak widths
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

2. Peak Intensities and Widths
Silicon

3. 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

4. Basic Principles of Interaction of Waves
For t = 0:
For x = 0:

5. Basic Principles of Interaction of Waves
Consider two waves with the same wavelength and amplitude but displaced a distance x0.
The phase shift:
Waves #1 and #2 are 90o out of phase x0
Waves #1 and #2 are 180o out of phase
When similar waves combine, the outcome can be constructive or destructive interference

6. Superposition of Waves
Resulting wave is algebraic sum of the amplitudes at each point
Small difference in phase
Large difference in phase

7. Superposition of Waves
Difference in frequency and phase

8. 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.

9. The Laue Equations Consider 1D array of scatterers spaced a apart.
Let x-ray be incident with wavelength . 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:

10. Lattice Planes

11. 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

12. 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.
Rewrite Bragg’s law:
Since sin  1:
For n = 1:
UV radiation   500 Å
Cu K1  = Å
For most crystals d ~ 3 Å
  6 Å

13. Bragg’s Law
Silicon lattice constant:
aSi = 5.43 Å

14. Scattering by an Electron
Elementary scattering unit in an atom is electron
Classical scattering by a single free electron – Thomson scattering equation:
The polarization factor of an unpolarized primary beam
If r = few cm:
Another way for electron to scatter is manifested in Compton effect.
Cullity p.127
1 mg of matter has ~1020 electrons

15. Scattering by an Atom Cu
Atomic Scattering Factor

16. Scattering by an Atom
Scattering by a group of electrons at positions rn:
Scattering factor per electron:
Assuming spherical symmetry for the charge distribution  = (r ) and taking origin at the center of the atom:
For an atom containing several electrons:
f – atomic scattering factor
Calling Z the number of electrons per atom we get:

17. Scattering by an Atom
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

18. Scattering by an Atom

19. Scattering by a Unit Cell
For atoms A & C
For atoms A & B
phase
If atom B position:
For 3D:

20. Scattering by a Unit Cell
We can write:

21. Scattering by a Unit Cell
Useful expressions
Hint: don’t try to use the trigonometric form – using exponentials is much easier

22. Scattering by a Unit Cell
Examples
Unit cell has one atom at the origin
In this case the structure factor is independent of h, k and l ; it will decrease with f as sin/ increases (higher-order reflections)

23. Scattering by a Unit Cell
Examples
Unit cell is base-centered
h and k unmixed
h and k mixed
(200), (400), (220)  
(100), (121), (300)  
“forbidden” reflections

24. Scattering by a Unit Cell
Examples
For body-centered cell
when (h + k + l ) is even
when (h + k + l ) is odd
(200), (400), (220)  
(100), (111), (300)  
“forbidden” reflections

25. Scattering by a Unit Cell
Examples
For body centered cell with different atoms:
when (h + k + l) is even
Cs+
Cl+
when (h + k + l) is odd
(200), (400), (220)  
(100), (111), (300)  

26. 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):
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
16f 2, h, k and l unmixed even and odd
0, h, k and l mixed even and odd

27. The Structure Factor
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.

28. Diffracted Beam Intensity
Structure factor
Polarization factor
Lorentz factor
Multiplicity factor
Temperature factor
Absorption factor
…..

29. 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 a)2:
The polarization factor (assuming that the incident beam is unpolarized):

30. 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
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

31. The Temperature 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 sinq/l (or |S| = 4psinq/l) more rapidly than it would normally.
The temperature factor is given by:
where the thermal factor B is related to the mean square displacement of the atomic vibration:
Scattering by C atom expressed in electrons
This is incorporated into the atomic scattering factor:

32. 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:
p100 = 6
p110 = 12
p111=8

33. The Absorption Factor
Angle-dependent absorption within the sample itself will modify the observed intensity
Absorption factor for infinite thickness specimen is:
Absorption factor for thin films is given by:
where μ is the absorption coefficient, τ is the total thickness of the film

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

35. 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

36. Crystallite Size
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

37. 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:
Rays A, D, …, M makes angle B
Rays B, …, L makes angle 1
Rays C, …, N makes angle 2

38. 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

39. The Scherrer Equation Subtracting: 1 and 2 are close to B, so:
Thus:

40. The Scherrer Equation More exact treatment (see Warren) gives:
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.

41. 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).

42. 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

43. Interference Function
We define:
as deviation parameter

44. 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:
ni – particular unit cells
Ni – total number of unit cells along ai
From the definition of the reciprocal lattice vector:

45. Interference Function
since:
Converting from exp to sines:

46. Interference Function
Calculating intensity we lose phase information therefore:
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)

47. Interference Function

48. Interference Function

49. 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:
rmn f
s – s0 s0 s
where
takes all
orientations

50. Debye scattering equation
Small crystallites (nanocrystallites)
Average intensity from an array of atoms which takes all orientations in space:
rmn f
s – s0 s0 s
Debye scattering equation
where
It involves only the magnitudes of the distances rmn of each atom from every other atom

51. 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:
Lets take a carbon tetrachloride as example.
It is composed of tetrahedral molecules CCl4.
Then:
correction
factor
C Cl4

52. 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.
Intensity (in e.u. per molecule) for a CCl4 gas in which the C – Cl distance is r = 1.82 Å. (Warren)

53. Small crystallites (nanocrystallites)
Lets treat crystal as a molecule
FCC has 14 atoms:
8 at corners of a cube
6 at face center positions
a is the edge of a cube
(111)
(200)
(220)
(311)
(222)

54. Reciprocal Lattice and Diffraction
It is equivalent to the Bragg law since :