nomenclature directions are chosen that lead aong special symmetry points. these points are labeled...
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Nomenclature
Directions are chosen that lead aong special symmetry points. These points are labeled according to the following rules:
• Points (and lines) inside the Brillouin zone are denoted with Greek letters.
• Points on the surface of the Brillouin zone with Roman letters.
• The center of the Wigner-Seitz cell is always denoted by a G
Usually, it is sufficient to know the energy En(k) curves - the dispersion relations - along the major directions.
Ener
gy o
r Fre
quen
cy
Direction along BZ
Brillouin Zones in 3Dfcc
hcp
•The BZ reflects lattice symmetry•Construction leads to primitive unit cell in rec. space
bcc
Note: fcc lattice in reciprocal space is a bcc lattice
Note: bcc lattice in reciprocal space is a fcc lattice
What kind of crystal structure is
Si?
Brillouin Zone of Silicon
Points of symmetry on the BZ are important (e.g. determining
bandstructure). Electrons in semiconductors are
perturbed by the potential of the crystal, which varies across unit cell.
Symbol DescriptionΓ Center of the Brillouin zone
Simple CubicM Center of an edgeR Corner pointX Center of a face
FCC
K Middle of an edge joining two hexagonal faces
L Center of a hexagonal face C6
U Middle of an edge joining a hexagonal and a square face
W Corner pointX Center of a square face C4
BCC
H Corner point joining 4 edges
N Center of a faceP Corner point joining 3 edges
Learning Objectives for Diffraction
After our diffraction class you should be able to:• Explain why diffraction occurs• Utilize Bragg’s law to determine angles of
diffraction• Discuss some different diffraction techniques• (Next time) Determine the lattice type and lattice
parameters of a material given an XRD pattern and the x-ray energy
• Alternative reference: Ch. 2 Kittel
Continuum limit:Where the wavelength is bigger than the spacing between
atoms. Otherwise diffraction effects dominate.
Application of XRD
1. Determination of the structure of crystalline materials2. Measurement of strain and small grain size3. Determination of the orientation of single crystals4. Measurement of layer thickness5. Differentiation between crystalline and amorphous
materials6. Determination of electron distribution within the
atoms, and throughout the unit cell7. Determination of the texture of polygrained materials
XRD is a nondestructive and cheap technique. Some of the uses of x-ray diffraction are:
DIFFRACTION• Diffraction is a wave phenomenon in which the apparent
bending and spreading of waves when they meet an obstruction is measured.
• Diffraction occurs with electromagnetic waves, such as light and radio waves, and also in sound waves and water waves.
• X-ray diffraction is optimally sensitive to the periodic nature of the solid’s atomic structure.
When X-rays interact with atoms, you get scattering
Scattering is the emission of X-rays of
the same frequency/energy as the incident X-rays in all
directions (but with much lower intensity)
Similar to the double slit experiment, this scattering will
sometimes be constructive
Incident beam
Zeroth Order
Second order
Will
look
at t
his
agai
n sh
ortly
Physical Model for X-ray Scattering
Consider a plane wave scattering on an atom.
ok
'kAtom
)( tRkioincident
oeA
)'( tRkioscattered eA
R
'R
Diffraction Theory
ko
Detector
Pi
ri
To calculate amplitude of scattered waves at detector position, sum over contributions of all scattering centers Pi with scattering amplitude (form factor) f:
R’)()()( ii
iiInDet ef rRkrr R’-ri
Generic incoming radiation amplitude is:)(
00 ii
In eA rRk
R
)'()'(0 )( kkrRkRk 00 r ii
i
iDet efeA
The intensity that is measured (can’t measure amplitude) is
2
)()( rrK Kr defI i0' kkK
source
R, R’ >> ri
The book calls K, but G is another common notation.Scattering vector
Diffraction Theory
ko
Detector
Pi
ri
R’ R’-ri
)(0
0 iiIn eA rRk
R
The intensity that is measured (can’t measure amplitude) is
2
)()( rrK Kr defI i0' kkK
source
The book calls K, but G is another common notation.Scattering vector
ko
k’K=k’-ko
The Bottom Line
If you do a whole bunch of math you can prove that the peaks only occur when (a1, a2, a3=lattice vectors):
n1, n2, n3 integers 11 2 nKa
22 2 nKa
33 2 nKa
Compare these relations to the properties of reciprocal lattice vectors:
laK
kaK
haK
hkl
hkl
hkl
2
2
2
3
2
1
2
)()( rrK Kr defI i0' kkK
The Laue Condition
Replacing n1 n2 n3 with the familiar h k l, we see that these three conditions are equivalently expressed as:
321 blbkbhK
The Laue condition (Max von Laue, 1911)
So, the condition for nonzero intensity is that the scattering vector K is a translation vector of the
reciprocal lattice.
n1, n2, n3 integers 11 2 nKa
22 2 nKa
33 2 nKa
From Laue to Bragg
Notice this angle is 2!
The magnitude of the scattering vector K depends on the angle between the incident wave vector and the scattered wave vector:
ok
'k
K
2
hkld
ok
'k
Elastic scattering requires: 2
' kkko
So from the wave vector triangle and the Laue condition we see:
sin
4sin2 kK
sin2 hkldLeaving Bragg’s law:
hkld
2
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)
0kkK Show vector
subtraction on the board
Bragg Equation
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.
ndhkl sin2
dn 2
X-ray Diffraction
/hcE Typical interatomic distances in solid are of the order of 0.4 nm.
Upon substituting this value for the wavelength into the energy equation, we find that E is of the order of 3000 eV, which is a
typical x-ray energy. Thus x-ray diffraction of crystals is a standard diffraction probe.
dn 2
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!)
Bragg Equation: nd sin2The diffracted beams (reflections) from any set
of lattice planes can only occur at particular angles pradicted by the Bragg law.
A single crystal specimen in a Bragg-Brentano diffractometer (θin=θout) would produce only one family of peaks in the diffraction pattern.
At 20.6 °2 , 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
°2 .
2q
The (110) planes would diffract at 29.3 °2 ; however, they are not properly aligned to produce a
diffraction peak (the perpendicular to those planes does not bisect the
incident and diffracted beams). Only background is observed.
20
THE EWALD SPHERE
Consider an arbitrary spherepassing through the reciprocal lattice,with the crystal arranged in the center of the sphere.
We specify two conditions:
(1)the sphere radius is 2 / - the inverse wavelength of X-ray radiation
(2)the origin of the reciprocal lattice lies on the surface of the sphere
X-rays are ON
O2/
2
diffracted ray
The diffraction spot will be observed when a reciprocal lattice point crosses the Ewald sphere
01
10
02
00 20
2
(41)
Ki
KD
DK
Reciprocal Space
The Ewald Sphere
The Ewald Sphere touches the reciprocal lattice (for point 41)
Bragg’s equation is satisfied for 41
A sphere of radius k Surface intersects a point in reciprocal space and its origin is at the tip of the incident wavevector.Sphere can be moved in reciprocal lattice space arbitrarily.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
1. Longitudinal or θ-2θ scanSample moves as θ, Detector follows as 2θ
k0 k’
0 10 20 30 40
1. Longitudinal or θ-2θ scanSample moves on θ, Detector follows on 2θ
K
0 10 20 30 40
Reciprocal lattice rotates by θ during scan
k0 k’
1. Longitudinal or θ-2θ scanSample moves on θ, Detector follows on 2θ
2
0 10 20 30 40
Kk0 k’
0 10 20 30 40
1. Longitudinal or θ-2θ scanSample moves on θ, Detector follows on 2θ
2K
k0 k’
1. Longitudinal or θ-2θ scanSample moves on θ, Detector follows on 2θ
2
0 10 20 30 40
Kk0 k’
1. Longitudinal or θ-2θ scanSample moves on θ, Detector follows on 2θ
0 10 20 30 40
2
0 10 20 30 40
Kk0 k’
1. Longitudinal or θ-2θ scanSample moves on θ, Detector follows on 2θ
0 10 20 30 400 10 20 30 40
•Provides information about relative arrangements, angles, and spacings between crystal planes.
2
0 10 20 30 40
Kk0
k’
Higher order diffraction peaks
http://www.doitpoms.ac.uk/tlplib/reciprocal_lattice/ewald.phphttp://www.physics.byu.edu/faculty/campbell/animations/x-ray_diffraction.html
3 COMMON X-RAY DIFFRACTION METHODS
X-Ray Diffraction Method
Laue
OrientationSingle Crystal
Polychromatic BeamFixed Angle
Rotating Crystal
Lattice constantSingle Crystal
Monochromatic BeamVariable Angle
Powder
Lattice ParametersPolycrystal/Powder
Monochromatic BeamVariable Angle
X-rays have wide wavelength range
(called white beam).
Back-reflection vs. TransmissionLaue Methods
The diffraction spots generally lay on: an ellipse
X-RayFilmSingle
Crystal
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?
a hyperbola
X-Ray Film
SingleCrystal
32
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 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
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?
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.
Rotating Crystal Method
Film
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.
Reflected beams are located on imaginary cones.
THE POWDER METHODLeast 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.
38
The Powder Method
• 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.
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.
39
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.
Useful for Phase IdentificationThe 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
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.
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
0
10
20
30
40
50
60
0 0.2 0.4 0.6 0.8 1
X(phase a)/X(phase b)I(p
hase
a)/I(p
hase
b) ..
Reference Intensity Ratio Method
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)
XRD: “Rocking” Curve Scan
• 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.
ikfk
“Rock” Sample
Sample normalK K
XRD: Rocking Curve Example
• Rocking curve of single crystal GaN around (002) diffraction peak showing its detailed structure.
16.995 17.195 17.395 17.595 17.7950
8000
16000
GaN Thin Film(002) Reflection
Inte
nsity
(C
ount
s/s)
Omega (deg)
Compare to literature to see how good (some
materials naturally easier than others)
Generally limited by quality of substrate
X-Ray Reflectivity (XRR)
• A glancing, but varying, incident angle, combined with a matching detector angle collects the X rays reflected from the samples surface
• Interference fringes in the reflected signal can be used to determine:– thickness of thin film layers– density and composition of thin
film layers– roughness of films and interfaces
X-ray reflectivity measurement
Si
Mo
Mo
Mo
r t [Å] s[Å]0.68 19.6 5.8
0.93 236.5 34.0
1.09 14.1 2.71.00 5.0 2.7
1.00 2.8
Calculation of the electron density, thickness and interface roughness for each particular layer
W
The surface must be smooth (mirror-like)
0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,010
0
101
102
103
104
105
106
Inte
nsity
(a.
u.)
Diffraction angle (o2)
Edge of TER
Kiessig oscillations (fringes)
Lots of 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 tough focusing on diffraction so I can’t cover everything here.
XRD: Reciprocal-Space Map
• Vary Orientation and Magnitude of k.• Diffraction-Space map of GaN film on AlN buffer
shows peaks of each film.
/2
GaN(002) AlN
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 100Two-Theta (deg)
x103
2.0
4.0
6.0
8.0
10.0
Inte
nsity
(Cou
nts)
00-004-0784> Gold - Au
Diffracting crystallites
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.
Cu
H2O In H2O Out
e-
Be
XRAYS
windowBe
XRAYS
FILAMENT
ANODE
(cathode)
AC CURRENT
window
metal
glass
(vacuum) (vacuum)
Primary
Shutter
Secondary
Shutter
Solenoid
SAFETY SHUTTERS
Neutron
λ = 1A°
E ~ 0.08 eV
interact with nucleiHighly Penetrating
Electron
λ = 2A°
E ~ 150 eV
interact with electronLess Penetrating
Diffraction Methods
• 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
λ = 1A°
E ~ 104 eV
interact with electronPenetrating
54
Electron Diffraction(Covered in Chapter
18)If low electron energies are used, the penetration depth will be very small (only about 50 A°), and the beam will be reflected from the surface. Consequently, electron diffraction is a useful technique for surface structure studies.
Electrons are scattered strongly in air, so diffraction experiment must be carried out in a high vacuum. This brings complication and it is expensive as well.
55
Electron Diffraction
Electron diffraction has also been used in the analysis of crystal structure. The electron, like the neutron, possesses wave properties;
Electrons are charged particles and interact strongly with all atoms. So electrons with an energy of a few eV would be completely absorbed by the specimen. In order that an electron beam can penetrate into a specimen , it necessitas a beam of very high energy (50 keV to 1MeV) as well as the specimen must be thin (100-1000 nm)
02AeVm
h
m
kE
ee
4022 2
222
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