1. Last Time Brillouin Zones and Intro to Scattering
Homework due today and next Thursday
XRD is a nondestructive and cheap technique providing information on: crystal structure, orientation, crystallinity, texture, thickness, strain and electron distribution

2. Learning Objectives for Diffraction
After our diffraction topic you should be able to:
Understand/ apply Bragg’s law
Discuss a few different diffraction techniques and their purposes
(Next time) Calculate the structure factor for simple cubic, bcc, fcc, diamond, rock salt, cesium chloride
Alternative reference: Ch. 2 Kittel
Utilize Bragg’s law to determine angles of diffraction

3. Diffraction In a Crystal
Incoming radiation amplitude:
To calculate amplitude of scattered waves at detector position, sum over contributions of all scattering centers Pi with scattering amplitude (form factor) f: Pi ri ko R
R’-ri R’
source
R, R’ >> ri
Generic for any point in the sample. (ignoring the time dependence, will disappear when we multiply by the complex conjugate)
If R is much greater than ri, then can pull out the exponents involving R from both si incident and si outgoing
Detector
The intensity that is measured (can’t measure amplitude) is
Scattering vector
The book calls K, but G is another common notation.

4. The Bottom Line
If you do the math you can prove that the peaks only occur when (a1, a2, a3 = lattice vectors):
n1, n2, n3 integers
This requires a bunch of math to show that could take a good portion of the class today if I wanted to. However, that’s not really the point of this class. Better just to trust me here and learn how to us it.
Compare these relations to the properties of reciprocal lattice vectors:

5. The Laue Condition
Replacing n1 n2 n3 with the familiar h k l, we see that these three conditions are equivalently expressed as: K
(Max von Laue, 1911)
So, the condition for nonzero intensity is that the scattering vector K is a translation vector of the reciprocal lattice.

6. Show vector subtraction on the board
From Laue to Bragg
The magnitude of the scattering vector K depends on the angle between the incident wave vector and the scattered wave vector:
Show vector subtraction on the board
Notice this angle is 2!
Elastic scattering requires:
Draw vector k-ko on the board to illustrate correct direction of K.
Elastic scattering requires no energy lose, so k’s magnitude is the same
Use right triangles to show 2ksin theta
You can use a similar argument to show how d and lambda are related from the diagram on the left. a=dsin theat
So from the wave vector triangle and the Laue condition we see:
Leaving Bragg’s law:
If the Bragg condition is not met, the incoming wave just moves through the lattice and emerges on the other side of the crystal (neglecting absorption)

7. How does this limit ?
where, d is the spacing of the planes and n is the order of diffraction.
Bragg reflection can only occur for wavelength
This is why we cannot use visible light. No diffraction occurs when the above condition is not satisfied.
What jumps out at you as a necessary condition for this to be true?

8. Bragg Equation:
The diffracted beams (reflections) from any set of lattice planes can only occur at particular angles pradicted by the Bragg law.
Bragg-Brentano diffractometer (θin=θout)
When atoms are specific phases apart. Otherwise they destructively interfere over the many layers of the crystal. Also shows why higher order diffraction occurs (just off by a multiple of the wavelength).
Note: While peaks should not technically be labeled with parentheses, it is very common to see them labelled that way. Likely even my examples will sometimes show them that way.
Above are 1st, 2nd, 3rd and 4th order “reflections” from the (111) face of NaCl. Orders of reflections are given as 111, 222, 333, 444, etc. (without parentheses!)

9. Why might you use this technique?
A single crystal specimen in a Bragg-Brentano diffractometer (θin=θout) would produce only one family of peaks in the diffraction pattern.
Why might you use this technique? 2q
Good for determining lattice parameters. It’s also simple. It’s not too hard to figure out what all of your peaks are.
The (110) planes would diffract at 29.3 °2q; however, the detector is not at that position (the perpendicular to those planes does not bisect the incident and diffracted beams). Only background is observed.
At 20.6° (2q), Bragg’s law fulfilled for the (100) planes, producing a diffraction peak.
The (200) planes are parallel to the (100) planes. Therefore, they also diffract for this crystal. Since d200 is ½ d100, they appear at 42° (2q).

10. THE EWALD SPHERE (Will show a few ways)
Consider an arbitrary sphere
passing through the reciprocal lattice,
with the crystal arranged in the center of the sphere.
We specify two conditions:
the sphere radius is 2 / - the inverse wavelength of X-ray radiation
the origin of the reciprocal lattice lies on the surface of the sphere
X-rays are ON
diffracted ray
Sometimes you’ll see this as just one over lambda (dropping the 2 pi)
At this point, the Bragg condition is said to be satisfied 2 O
2/
The diffraction spot will be observed when a reciprocal lattice point crosses the Ewald sphere

11. The Ewald Sphere Reciprocal Space K K D K Only a few angles
A sphere of radius k
Surface intersects a point in reciprocal space and its origin is at the tip of the incident wavevector.
Sphere rotated around point (0,0) in reciprocal lattice space.
Any points which intersect the surface of the sphere indicate where diffraction peaks will be observed if the structure factor is nonzero (later).
Only a few angles
Reciprocal Space 2 q K
So, we can arbitrarily rotate this sphere of radius k about the 00 point. When it intersects with another point, there is a chance of a diffraction peak. In a simple cubic lattice, there will always (barring effects from potential defects) be a peak at these intersections. K D i 02
The Ewald Sphere touches the reciprocal lattice (for point 41)
 Bragg’s equation is satisfied for 41 01 D K
(41) 00 10 20

12. 1. Longitudinal or θ-2θ scan
Sample moves as θ, Detector follows as 2θ k0 k’

13. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ K k0 k’
Reciprocal lattice rotates by θ during scan

14. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ K k0 k’ 2q

15. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ K k0 k’ 2q

16. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ K k0 2q k’

17. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ K k0 2q k’

18. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ k’ K k0 2q
Provides information about relative arrangements, angles, and spacings between crystal planes.

19. Higher order diffraction peaks

20. 3 COMMON X-RAY DIFFRACTION METHODS
Laue
Rotating Crystal
Powder
Orientation
Single Crystal
Polychromatic Beam
Fixed Angle
Lattice constant
Single Crystal
Monochromatic Beam
Variable Angle
Lattice Parameters
Polycrystal/Powder
Monochromatic Beam
Fixed Angle

21. Back-reflection vs. Transmission Laue Methods
In the back-reflection method, the film is placed between the x-ray source and the crystal. The beams which are diffracted backward are recorded.
Which is this?
X-rays have wide wavelength range (called white beam).
Single
Crystal
X-Ray
Film
Single
Crystal
Film
X-Ray
The diffraction spots generally lay on:
an ellipse
a hyperbola

22. LAUE METHOD
The diffracted beams form arrays of spots, that lie on curves on the film.
Each set of planes in the crystal picks out and diffracts a particular wavelength from the white radiation that satisfies the Bragg law for the values of d and θ involved.
Laue on one of our thin films.

23. Great for symmetry and orientation determination
Laue Pattern
The symmetry of the spot pattern reflects the symmetry of the crystal when viewed along the direction of the incident beam.
Great for symmetry and orientation determination
It’s a bit of an art to interpret these things. Takes some practice.

24. Crystal structure determination by Laue method?
Although the Laue method can be used, several wavelengths can reflect in different orders from the same set of planes, making structure determination difficult (use when structure known for orientation or strain).
Rotating crystal method overcomes this problem. How?
You would interpret from the intensity of the spots, but if multiple spots are at the same location, that’s a problem.
Materials scientists typically already know the crystal structure for a material they’ve worked on a lot and then this method can be used to understand orientation or strain tensors.

25. ROTATING CRYSTAL METHOD
A single crystal is mounted with a rotation axis perpendicular to a monochromatic x-ray beam.
A cylindrical film is placed around it and the crystal is rotated. 
Sets of lattice planes will at some point make the correct Bragg angle, and at that point a diffracted beam will be formed.  

26. Rotating Crystal Method
Reflected beams are located on imaginary cones. 
By recording the diffraction patterns (both angles and intensities), one can determine the shape and size of unit cell as well as arrangement of atoms inside the cell.
But around what axis should you rotate?
Requires various orientations measured and you may not know where those orientations are, so hard if that is the case.
Film

27. THE POWDER METHOD Least crystal information needed ahead of time
If a powder is used, instead of a single crystal, then there is no need to rotate the sample, because there will always be some crystals at an orientation for which diffraction is permitted.
A monochromatic X-ray beam is incident on a powdered or polycrystalline sample.
Common method if you don’t know much about your material (or if you already have it in powder form).

28. The Powder Method
A sample of some hundreds of crystals (i.e. a powdered sample) show that the diffracted beams form continuous cones.
A circle of film is used to record the diffraction pattern as shown.
Each cone intersects the film giving diffraction arcs.
If a monochromatic x-ray beam is directed at a single crystal, then only one or two diffracted beams may result.
If the sample consists of some tens of randomly orientated single crystals, the diffracted beams are seen to lie on the surface of several cones.
The cones may point both forwards and backwards.

29. Powder diffraction film
When the film is removed from the camera, flattened and processed, it shows the diffraction lines and the holes for the incident and transmitted beams.

30. K

31. Useful for Phase Identification
The diffraction pattern for every phase is as unique as your fingerprint
Phases with the same element composition can have drastically different diffraction patterns.
Use the position and relative intensity of a series of peaks to match experimental data to the reference patterns in the database

32. Databases such as the Powder Diffraction File (PDF) contain dI lists for thousands of crystalline phases.
The PDF contains over 200,000 diffraction patterns.
Modern computer programs can help you determine what phases are present in your sample by quickly comparing your diffraction data to all of the patterns in the database.

33. Quantitative Phase Analysis
With high quality data, you can determine how much of each phase is present
The ratio of peak intensities varies linearly as a function of weight fractions for any two phases in a mixture
RIR method is fast and gives semi-quantitative results
Whole pattern fitting/Rietveld refinement is a more accurate but more complicated analysis
Reference Intensity Ratio Method
RIR method: need to know the constant of proportionality

34. Applications of Powder Diffractometry
phase analysis (comparison to known patterns)
unit cell determination (dhkl′s depend on lattice parameters)
particle size estimation (line width)
crystal structure determination (line intensities and profiles)

35. Extra slides
There is a lot of useful information on diffraction. Following are some related slides that I have used or considered using in the past.
A whole course could be taught focusing on diffraction so I can’t cover everything here.

36. XRD: “Rocking” Curve Scan
Sample normal
“Rock” Sample
Vary ORIENTATION of K relative to sample normal while maintaining its magnitude. How? “Rock” sample over a very small angular range.
Resulting data of Intensity vs. Omega (w, sample angle) shows detailed structure of diffraction peak being investigated. Can inform about quality of sample.
Draw a flat line versus a line with small tilts in either direction

37. XRD: Rocking Curve Example
GaN Thin Film
How do you know if this is good?
Compare to literature to see how good (some materials naturally easier than others)
(002) Reflection
16000
Intensity (Counts/s)
8000
Generally limited by quality of substrate
Could give a whole course on diffraction, but lots more to cover
I’m not talking politics, I just think having Obama cropped in the picture is funny.
16.995
17.195
17.395
17.595
17.795
Omega (deg)
Rocking curve of single crystal GaN around (002) diffraction peak showing its detailed structure.

38. X-ray reflectivity (XRR) measurement
A glancing, but varying, incident angle, combined with a matching detector angle collects the X rays reflected from the samples surface
Calculation of the density, composition, thickness and interface roughness for each particular layer
r t [Å] s [Å]
Edge of TER
Kiessig oscillations (fringes) Mo Mo Mo W Si
The surface must be smooth (mirror-like)

39. XRD: Reciprocal-Space Map
GaN(002)
AlN 
/2
Vary Orientation and Magnitude of k.
Diffraction-Space map of GaN film on AlN buffer shows peaks of each film.

40. The X-ray Shutter is the most important safety device on a diffractometer
X-rays exit the tube through X-ray transparent Be windows.
X-Ray safety shutters contain the beam so that you may work in the diffractometer without being exposed to the X-rays.
Being aware of the status of the shutters is the most important factor in working safely with X rays.

41. Non-xray Diffraction Methods (more in later chapters)
Any particle will scatter and create diffraction pattern
Beams are selected by experimentalists depending on sensitivity
X-rays not sensitive to low Z elements, but neutrons are
Electrons sensitive to surface structure if energy is low
Atoms (e.g., helium) sensitive to surface only
For inelastic scattering, momentum conservation is important
X-Ray
Neutron
Electron
Replace top figure (not so informative, probably more confusing)
λ = 1A°
E ~ 104 eV
interact with electron
Penetrating
λ = 1A°
E ~ 0.08 eV
interact with nuclei
Highly Penetrating
λ = 2A°
E ~ 150 eV
interact with electron
Less Penetrating

42. Group: Consider Neutron Diffraction
Qualitatively discuss the atomic scattering factor (e.g., as a function of scattering angle) for neutron diffraction (compared to x-ray) by a crystalline solid.
For x-rays, we saw that f is related to Z and has a strong angular component. For neutrons?
The same equation applies, but since the neutron scatters off a tiny nucleus, scattering is more point-like, and f is ~ independent of .
3.9

43. Preferred Orientation (texture)
Diffracting crystallites
Preferred Orientation (texture)
Preferred orientation of crystallites can create a systematic variation in diffraction peak intensities
can qualitatively analyze using a 1D diffraction pattern
a pole figure maps the intensity of a single peak as a function of tilt and rotation of the sample
this can be used to quantify the texture
(111)
(311)
(200)
(220)
(222)
(400) 40 50 60 70 80 90
100
Two-Theta (deg)
x10 3
2.0
4.0
6.0
8.0
10.0
Intensity(Counts)
> Gold - Au