x-ray diffraction (xrd)
DESCRIPTION
X-Ray Diffraction (XRD). Principle. In XRD, an incident wave is directed into a material and a detector is typically moved about to record the directions and intensities of the outgoing diffracted waves. X-ray diffraction has acted as the corner stone of the twentieth century science. - PowerPoint PPT PresentationTRANSCRIPT
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School of Materials and Mineral Resources EngineeringAzizan Aziz
X-Ray Diffraction (XRD)
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In XRD, an incident wave is directed into a material and a detector is typically movedabout to record the directions and intensities of the outgoing diffracted waves.
Principle
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X-ray diffraction has acted as the corner stone of the twentieth century science. Its development has
catalyzed the develpments of all of the rest of solid state science and much of understanding of chemical bonding
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X-rays are high energy electromagnetic radiation having energies from ~200eV to 1 MeV
Between the γ-rays and ultraviolet (UV) in the electromagnetic spectrum.
Gamma rays and x-rays are essentially identical with γ –rays being more energetic and shorter in wavelength
What is X-rays
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Each quantum of of electromagnetic radiation or photon has an energy, E which is proportional to its frequency, :
E=h h=Plank’s constant=4.136 x 10-15eV.s
λ=hc/E λ= wavelength ; c=2.998 x 108 m/s
The useful range of wavelength for x-ray diffraction studies is between 0.05 and 0.25 nm.
* Interatomic spacings in crystals are typically about 0.2 nm (2Å)
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Production of X-rays
X-rays are produced in an x-ray-tube consisting of two metals electrodes enclosed in a vacuum chamber
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Electrons are produced by heating a tungsten filament cathode which is at a high negative potential
The e are accelerated at high velocity towards the anode (water-cooled)
Loss of energy due to the collision with the metal anode produced X-rays
Only 1% of the e beam converted to x-rays – the rest as heat
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A typical x-ray spectrum
- A continuous radiation due to electrons losing their energy in a series collisions with atoms of target anode.
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Illustration of the origin of continuous radiation
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Short Wavelength Limit (λSWL)
When e loses all its energy in a single collision with a target atom x-rays photon with maximum energy or shortest wavelength is produced - short wavelength limit
Characteristic lines
When an e has sufficient energy to eject an inner shell e →atom will be in the xcited state with vacancy in the inner shell
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An e from an outer shell will fill up the vacancy
Energy equals to the difference in the e energy levels will b released in d form of x-ray photon. This is characteristic of the
target metal producing a sharp peaks in the spectrum –known as characteristic lines
It is this characteristic that are most useful in XRD
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The accelerating potentials necessary to produce x-rays having comparable to interatomic spacings are about 10kV.
Usually high accelerating potentials are used to produce higher intensity line spectrum of target metal
Higher accelerating potentials changes λSWL but not the characteristic wavelengths.
Intensity of characteristic line depends both on the applied potential and the tube current I
I = Bi(V – VK)n B=proportional constant I = current VK = potential req. to eject an e from K shell
V= applied potential n =a cosntant for a particular value of V
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The fig indicate that there are more than 1 characteristic line
This line correspond to electron transitions line between different energy levels
Characteristic line classified as K, L, M (Bohr model)
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Transition L → K Kά M → L Lα
M → K Kβ
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Due to the presence of subshells Kά or Kβ can be further resolved into Kά1 and Kά2
LIII → K Kά1
LII → K Kά2
Level Energy(keV)
K -20.00
LII -2.63
LIII 2.52
Energies of the K,LII ,and LIII Levels of Molybdenum
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Wide choice of characteristic Kά lines obtained by using different target metals as shown on Table2 but Cu Kά is the most common radiation used.
The most important radiations in diffraction work are those corresponding to the filling of the inner most K shell from adjacent shells giving Kά1,Kά2 dan Kβ lines
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The Kάlines are used bcos they are more energetic than Lά, therefore less strongly absorbed by the material we want to examine.
The wavelength spread of each line is extremely narrow and each wavelength is known with high precision
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Interaction of X-rays with matter
Any mechanism which causes a photon, in the collimated incident X-ray beam to miss the detector is called absorption.
Most mechanism – conversion of photon energy to another form, while some simply change the direction- diffraction
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There are various processes taking place when x-ray interact w matter:
a. No interaction
b. Conversion to heat
c. Photoelectric effect – flourescence and auger electron
d. Compton Scattering
e. Coherent Scattering – leads to the phenomena of diffraction
**Note: please do ur own reading on d above topics
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X-ray absorption
• Just like an incident electron, X-ray photons can initiate electronic transitions
• Decrease in intensity distance traversed by the X-ray beam
xoeIxIdx
I
dI )(
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x
oeIxI
)(
and for a mixture (or alloy):
Beer’s law where is the linear absorption coefficient
• Problem: depends on the density of the absorbing material, but the ratio does not (mass absorption coefficient)
2
2
1
1
ww
weight fractions
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The way μ varies with wavelength λ gives the indication to the interaction of X-rays and atom
Properties of the absorption coefficient
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Properties of the absorption coefficient
33 mZk m
• There is a sharp discontinuity in the dependence of the absorption coefficient on energy (wavelength) at the energy corresponding to the energy required to eject an inner-shell electron
• The discontinuity is known as an absorption edge
• Away from an absorption edge, each “branch” of the absorption curve is given by:
E
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Tabulated values of mass absorption coefficients
• Values of / are tabulated in the International Tables for Crystallography as well as in most X-ray diffraction textbooks
• Note the discontinuities in the tabulated data at the absorption edges
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We see many absorption edges…
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An interesting application of absorption edge is when the edge of one element A is located between the Kά and Kβ lines of another element B
When this occurs the Kβ from the B atoms will be very strongly absorbed while the longer wavelength Kά will be slightly absorbed.
With suitable thickness ,element A can act as a beta-filter for characteristic radiation from element B
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Absorption and X-ray filters
• Using an absorber as an X-ray filter can reduce undesirable wavelength contamination in a diffraction experiment
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X-ray fluorescence• Fluorescence is the opposite of absorption -- when
energy is absorbed, a vacancy is produced in an electron shell
• Other electrons fill that vacancy, producing radiation• Absorption at an edge generates high fluorescence• Fluorescence can be a source of background in a
diffraction experimentCu K -- = 1.54Å Cu-radiation
fluorescesFe K-edge -- = 1.74Å iron, but Cr-radiationCr K -- = 1.79Å does not
• Fluorescent radiation is characteristic to specific elements and is widely used for chemical analysis
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DIFFRACTION
When a beam of x-rays incident on an atom, ē’s in the atom oscillate about their mean positions
The process of absorption and reemission of electromagnetic radiation by ē is known as scattering
Diffraction is a general characteristic of all waves can be define as modification of the behaviour of light or other waves by its interaction with an object.
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When there is no change in energy bet. Incident and emitted photon- radiation is elastically scattered and is known as coherent scattering.
When there is a photon energy loss – inelastic scattering , also known as Compton scattering
The photon changes direction after colliding with the electron but transfer none of its energy to the electron.Thus this scattered photon leaves in a new direction but with the same phase and energy as the incident photon. ------ phenomenon of diffraction
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The above Fig. shows an atom containing several electrons arranged as points around the nuclues
Consider two waves that are incident on the atom
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The upper wave is scattered by electron A in the forward direction
The two waves scattered in the forward direction are said to be in phase across wavefront XX’ bcos the waves have traveled the same distance before and after.
Similarly, the lower wave is scattered in the forward direction by electron
No path difference → in phase
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The other scattered waves in the above Fig will not be in phase across the wavefront YY’ when the path difference (CB-AD) is not an integral number of wavelength
If we add these two waves across the wave front the resultant amplitude of the scattered waves is less than the wave scattered by the same electron in the forward direction.
Hence adding these two waves will give twice the amplitude but same wavelength
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The efficiency of how an atom scatter a beam of X-rays is define by the atomic scattering factor, f
electron oneby scattered waveof Amplitude
atom aby scattered waveof Amplitudef
When scattering is in the forward direction (scattering angle =0o ) f =Z ,since the waves scattered by all the electrons in the atom are in phase and the amplitude sum up
As θ increases, the waves become more and more out of phase bcos they travel different path length and therefore the amplitude,or f, decreases.
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The atomic scattering factor also depends on the wavelength of the incident rays
Take note that most of the scattering occurs in the forward direction θ=0o
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DIFFRACTION FROM CRYSTALLINE MATERIALS-BRAGG’S LAW
Atoms scatter x-rays and these scattered waves from all atoms can interfere.
If scattered waves are in phase(coherent) , constructive interference and the beams are diffracted in specific directions.
These directions are governed by the wavelength λ, and the nature of the crystalline sample – Braggs Law
nλ = 2dsinθ
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In deriving Bragg’s Law, it is convenient to think the x-ray as being reflected from plane of atoms (mind you x-rays are really not being reflected)
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Consider diffracted waves in the above Fig and assumed that it make the same angle,θ with atomic plane as does the incident wave.
Criterion for wave to be diffracted,the reflected x-rays should all be in phase across the wavefront such as BB’
To be in phase, the path lengths between wfs AA’ and BB’ must differ by exactly an integral number n of wavelength λ
The path difference, δ=nλ , n is an integer
Since lines CC’ and CD are also wavefronts,
δ = DE +EC’ = 2EC’
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But,
δ = 2d’ Sinθ
nλ = 2d’Sinθ Bragg’s Law
Bcos CE is the interplanar spacing d’
But,
δ = 2CE Sinθ
This is extremely important equation in indexing x-ray diffraction pattern and hence crystal structure of the materials.
n known as order of reflection, is the path difference,in terms of the number of wavelengths between waves scattered by adjacent planes.
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Although Bragg's law was used to explain the interference pattern of X-rays scattered by crystals, diffraction has been developed to study the structure of all states of matter with any beam, e.g., ions, electrons, neutrons, and protons, with a wavelength similar to the distance between the atomic or molecular structures of interest.
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What is X-ray Diffraction ?
The atomic planes of a crystal cause an incident beam of X-rays to interfere with one another as they leave the crystal. The phenomenon is called X-ray diffraction.
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Why XRD?
Measure the average spacings between layers or rows of atoms
Determine the orientation of a single crystal or grain
Find the crystal structure of an unknown material
Measure the size, shape and internal stress of small crystalline regions
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Bragg’s Law and Diffraction:
How waves reveal the atomic structure of crystals
nλ = 2dsinθ n = integerDiffraction occurs only when Bragg’s Law is satisfied Condition for constructive interference (X-rays 1 & 2) from planes with spacing d
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Basics of Crystallography
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Seven Crystal Systems - Review
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Miller Indices: hkl - Review
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Several Atomic Planes and Their d spacings in a Simple Cubic - Review
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Planes and Spacings - Review
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Indexing of Planes and Directions - Review
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Structure Factor
The structure factor F, describe the effect of the crystal structure on the intensity of the diffracted beam.
Consider 2 orthorombic unit cells below
Body-centeredBased-centered
Both cells have 2 atoms per unit cell
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Diffraction from the the (001) planes of the two cells is shown below
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• For the based-centered orthorhombic lattice,if Bragg Law is satisfied path difference ABC bet. Wave 1’ and 2’ is λ and diffraction will occur
• Similarly, for the body-centered lattice wave 1’ and 2’ will have
constructive interference, but in this case there is another plane of atoms midway btween (001) plane
• The path difference bet wave 1’ and 3’ is λ/2 and therefore will b
completely out of phase, destructive interference hence no diffracted beam.
• Wave 2’ will also be cancel by a wave reflected by a next plane below and so on.
• Net result – NO 001 reflection for body-centered lattice.
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• For certain structures some reflections will be absent from the diffraction pattern – they are known as forbidden reflection
• The intensity of the beam diffracted by all the atoms in the unit cell in a given direction predicted by Bragg Law is proportional to F2.
• F2 = 0 , no reflection
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• The equation for calculating the structure factor F is given by
i
lwkvhuii
iiiefF )(2
• f = atomic scattering factor; u,v,and w coordinates of atom in unit cell; h,k and l are the Miller indices.
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• The above eqn tells us what reflection hkl to expect in a diffraction pattern from a given crystal structure with atom located at u,v,w – these are known as selection rules
• F = 0 will have zero intensity and will not appear in the diffraction pattern – forbidden reflection
• A completely general eqn. Applies to all crystal lattice, structure (cubic,triclinic or hexagonal) and whether there are 1 or 100 atoms per unit cell.
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Examples
Simple cubic crystal structure
• There is 1 atom/unit cell located at (0,0,0)
ffeF i )0(2
• For simple cubic structure F is independent of h,k,l meaning all reflections are allowed
• We will observe reflections from (100),(110),(111),(200),(210),etc
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School of Materials and Mineral Resources EngineeringAzizan Aziz
FCC structure
• There is 4 atom/unit cell located at (0,0,0), (½,½,0), (½,0,½), (0,½,½)
222
222
222
)0(2lk
ilh
ikh
ii fffefeF
)()(1 lkilhikhi eeefF
• If h,k,and l are all even or all odd (unmixed),then the sums h+k, h+l, and k+l even integers and each term has a value of 1. Hence for fcc structure
fF 4 for unmixed indices
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School of Materials and Mineral Resources EngineeringAzizan Aziz
• If h,k, and l are mixed (i.e., some are odd and some are even),then the sum of the three exponentials is -1.
• Say h=0,k=1 and l=2 the plane has indices (012)
F = f(1-1+1-1)=0 there is no reflection
• Thus F = 0 for mixed indices.
• For materials with FCC structure, we will see reflections corresponding to planes such as (111),(200) and (220),but not for planes (100),(110),(210),(211) etc.
![Page 66: X-Ray Diffraction (XRD)](https://reader035.vdocuments.site/reader035/viewer/2022081501/56814f31550346895dbcc28d/html5/thumbnails/66.jpg)
School of Materials and Mineral Resources EngineeringAzizan Aziz
• The structure factor is independent of shape and size of the unit cell
• Any primitive cell will show reflections corresponding to diffraction from all the lattice planes
• fcc cell will show reflections only when h,k,l are unmixed. However not all unmixed reflection may be seen in all face-centered cells;
some may be absent depending on the number of atoms present in the unit cell.
What can be certain of is that no reflection will be observed when h,k,and l are mixed.
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School of Materials and Mineral Resources EngineeringAzizan Aziz
Consider unit cell containing a basis of two atoms/lattice point eg NaCl structure.
One Na and one Cl per lattice pt.
To calculate the structure factor for NaCl we use result for fcc plus contribution of the basis.
In NaCl structure, 8 atoms/unit cell (4 Na and 4 Cl)
Normally, for NaCl, we take the basis , 1 Na located at 0,0,0 and one Cl located at ½,0,0. But we can also take Na(0,0,0) and Cl(½,½,½)
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School of Materials and Mineral Resources EngineeringAzizan Aziz
Using the formula, the structure factor for NaCl structure as
lkilhikhilkh
i
Cli
Na eeeefefF
)(222
2)0(2 1
Simplifying the above equation will give
lkilhikhilkhiClNa eeeeffF 1
The first term represent the basis of the unit cell, Na ion at 0,0,0 and Cl ion at ½,½,½
The second term is similar to fcc structure
![Page 69: X-Ray Diffraction (XRD)](https://reader035.vdocuments.site/reader035/viewer/2022081501/56814f31550346895dbcc28d/html5/thumbnails/69.jpg)
School of Materials and Mineral Resources EngineeringAzizan Aziz
F=4(fNa + fCl) if h,k,l are even
F=4(fNa - fCl) if h,k,l are odd
F= 0 if h,k,l are mixed
The first two conditions are new and depend on fNa± fCl .
The third condition is the same for fcc structure.
The selection rules for the NaCl structure are:
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School of Materials and Mineral Resources EngineeringAzizan Aziz
NaCl structure contain more than 4 atoms/unit cell has not led elimination of any reflections seen for fcc structure but change in the intensity
E.g 111 reflection for NaCl weaker than 200 reflection bcos the former involves the difference rather than sum of the atomic scattering factor
Whenever a lattice contains a basis, the equation for the structure factor will contain the first term which is the atom positions in the basis, and the second term is for the Bravais lattice crystal structure.
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School of Materials and Mineral Resources EngineeringAzizan Aziz
Another example to illustrate the point diamond cubic structure.
Diamond cubic structure is face-centered cubic Bravais with a basis of two atoms, giving a total of eight atoms/unit cell.
The atoms in the basis are located at 0,0,0 and ¼,¼,¼
Both atoms are the same type (in diamond they are all carbon atoms)
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School of Materials and Mineral Resources EngineeringAzizan Aziz
lkilhikhilkhi
CC eeeeffF
12
Therefore, the structure factor is given by
first term represents the basis
this term represents face-centered cubic Bravais lattice.
![Page 73: X-Ray Diffraction (XRD)](https://reader035.vdocuments.site/reader035/viewer/2022081501/56814f31550346895dbcc28d/html5/thumbnails/73.jpg)
School of Materials and Mineral Resources EngineeringAzizan Aziz
The selection rules for the diamond structure cubic are a litle more complicated:
F = 0 if h,k,l mixed (same condition as for the fcc structure)
F = 4fc(1 + e(π/2)i(h+k+l)) if h,k,l are odd
F = 0 if h,k,l are even and h+k+l =2N where N is odd (eg.,the 200 reflection)
F = 8fc if h,k,l are even and h+k+l = 2N where N is even (e.g., the 400 reflection)
![Page 74: X-Ray Diffraction (XRD)](https://reader035.vdocuments.site/reader035/viewer/2022081501/56814f31550346895dbcc28d/html5/thumbnails/74.jpg)
School of Materials and Mineral Resources EngineeringAzizan Aziz
In the diffraction pattern that we see the intensities of the reflections we consider F2 not F
F=0 , same condition as fcc,but it applies to the diamond cubic structure bcos it has the same Bravais lattice as the fcc
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School of Materials and Mineral Resources EngineeringAzizan Aziz
NaCl structures and diamond cubic structures
• both structures are based on face – centered cubic Bravais lattice and have eight atoms per unit cell
• bcos atom positions are different in the two structure the x-ray diffraction patterns are different: different reflections and different intensities.
![Page 76: X-Ray Diffraction (XRD)](https://reader035.vdocuments.site/reader035/viewer/2022081501/56814f31550346895dbcc28d/html5/thumbnails/76.jpg)
School of Materials and Mineral Resources EngineeringAzizan Aziz
Table 8
![Page 77: X-Ray Diffraction (XRD)](https://reader035.vdocuments.site/reader035/viewer/2022081501/56814f31550346895dbcc28d/html5/thumbnails/77.jpg)
School of Materials and Mineral Resources EngineeringAzizan Aziz
Diffraction from Amorphous Materials
The crystalline materials shows a series of sharp peaks,or reflections due to diffracted beams arising from different lattice planes.
The glass shows one broad peak centered in the range in which the strong occurs