# Anandh Subramaniam & Kantesh Balani

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Anandh Subramaniam & Kantesh Balani
X-RAY DIFFRACTION MATERIALS SCIENCE & ENGINEERING Anandh Subramaniam & Kantesh Balani Materials Science and Engineering (MSE) Indian Institute of Technology, Kanpur URL: home.iitk.ac.in/~anandh AN INTRODUCTORY E-BOOK Part of A Learner’s Guide X- Ray Sources Diffraction: Bragg’s Law Crystal Structure Determination Elements of X-Ray Diffraction B.D. Cullity & S.R. Stock Prentice Hall, Upper Saddle River (2001) Recommended websites:

What will you learn in this ‘sub-chapter’?
How to produce monochromatic X-rays? How does a crystal scatter these X-rays to give a diffraction pattern?  Bragg’s equation What determines the position of the XRD peaks?  Answer) the lattice. What determines the intensity of the XRD peaks?  Answer) the motif. How to analyze a powder pattern to get information about the lattice type? (Cubic crystal types). What other uses can XRD be put to apart from crystal structure determination?  Grain size determination  Strain in the material…

Some Basics For electromagnetic radiation to be diffracted* the spacing in the grating (~a series of obstacles or a series of scatterers) should be of the same order as the wavelength. In crystals the typical interatomic spacing ~ 2-3 Å**  so the suitable radiation for the diffraction study of crystals is X-rays. Hence, X-rays are used for the investigation of crystal structures. Neutrons and Electrons are also used for diffraction studies from materials. Neutron diffraction is especially useful for studying the magnetic ordering in materials. ** If the wavelength is of the order of the lattice spacing, then diffraction effects will be prominent. Click here to know more about this ** Lattice parameter of Cu (aCu) = 3.61 Å  dhkl is equal to aCu or less than that (e.g. d111 = aCu/3 = 2.08 Å)

Generation of X-rays X-rays can be generated by decelerating electrons. Hence, X-rays are generated by bombarding a target (say Cu) with an electron beam. The resultant spectrum of X-rays generated (i.e. X-rays versus Intensity plot) is shown in the next slide. The pattern shows intense peaks on a ‘broad’ background. The intense peaks can be ‘thought of’ as monochromatic radiation and be used for X-ray diffraction studies. Target X-rays Beam of electrons An accelerating (or decelerating) charge radiates electromagnetic radiation

Mo Target impacted by electrons accelerated by a 35 kV potential shows the emission spectrum as in the figure below (schematic) X-ray sources with different  for doing XRD studies Target Metal  Of K radiation (Å) Mo 0.71 Cu 1.54 Co 1.79 Fe 1.94 Cr 2.29 The high intensity nearly monochromatic K x-rays can be used as a radiation source for X-ray diffraction (XRD) studies  a monochromator can be used to further decrease the spread of wavelengths in the X-ray

X-ray sources with different  for doing XRD studies Elements (KV)
 Of K1 radiation (Å)  Of K2 radiation (Å)  Of Kβ radiation (Å) Kβ-Filter (mm) Ag 25.52 0.5638 Pd 0.0461 Mo 20 0.7093 Zr 0.0678 Cu 8.98 Ni 0.017 8.33 Co 0.0158 7.71 Fe 0.0166 7.11 Mn 0.0168 Cr 5.99 2.2897 V 0.169 C.Gordon Darwin, Grandson of C. Robert Darwin developed the dynamic theory of scattering of x-rays (a tough theory!) in 1912

Incident X-rays Fluorescent X-rays Electrons Scattered X-rays
When X-rays hit a specimen, the interaction can result in various signals/emissions/effects. The coherently scattered X-rays are the ones important from a XRD perspective. Incident X-rays SPECIMEN Absorption (Heat) Fluorescent X-rays Electrons Scattered X-rays Compton recoil Photoelectrons Coherent From bound charges Incoherent (Compton modified) From loosely bound charges Transmitted beam Click here to know more X-rays can also be refracted (refractive index slightly less than 1) and reflected (at very small angles)

Diffraction Click here to “Understand Diffraction” Now we shall consider the important topic as to how X-rays interact with a crystalline array (of atoms, ions etc.) to give rise to the phenomenon known as X-ray diffraction (XRD). Let us consider a special case of diffraction → a case where we get ‘sharp[1] diffraction peaks’. Diffraction (with sharp peaks) (with XRD being a specific case) requires three important conditions to be satisfied: Radiation related Coherent, monochromatic, parallel waves& (with wavelength ). Sample related Crystalline array of scatterers* with spacing of the order of (~) . Diffraction geometry related Fraunhofer diffraction geometry (& this is actually part of the Fraunhofer geometry) Coherent, monochromatic, parallel wave Aspects related to the wave Diffraction pattern with sharp peaks Aspects related to the material Crystalline*,** Aspects related to the diffraction set-up (diffraction geometry) Fraunhofer geometry [1] The intensity- plot looks like a ‘’ function (in an ideal situation). * A quasicrystalline array will also lead to diffraction with sharp peaks (which we shall not consider in this text). ** Amorphous material will give broadened (diffuse) peak (additional factors related to the sample can also give a diffuse peak).

Not all objects act like scatterers for all kinds of radiation.
Some comments and notes The waves could be:  electromagnetic waves (light, X-rays…),  matter waves** (electrons, neutrons…) or  mechanical waves (sound, waves on water surface…). Not all objects act like scatterers for all kinds of radiation. If wavelength is not of the order of the spacing of the scatterers, then the number of peaks obtained may be highly restricted (i.e. we may even not even get a single diffraction peak!). In short diffraction is coherent reinforced scattering (or reinforced scattering of coherent waves). In a sense diffraction is nothing but a special case of constructive (& destructive) interference. To give an analogy  the results of Young’s double slit experiment is interpreted as interference, while the result of multiple slits (large number) is categorized under diffraction. Fraunhofer diffraction geometry implies that parallel waves are impinging on the scatteres (the object), and the screen (to capture the diffraction pattern) is placed far away from the object. Click here to know more about Fraunhofer and Fresnel diffraction geometries ** With a de Broglie wavelength

XRD  the first step A beam of X-rays directed at a crystal interacts with the electrons of the atoms in the crystal. The electrons oscillate under the influence of the incoming X-Rays and become secondary sources of EM radiation. The secondary radiation is in all directions. The waves emitted by the electrons have the same frequency as the incoming X-rays  coherent. The emission can undergo constructive or destructive interference. Schematics

Incident and scattered waves are in phase if
Some points to recon with We can get a better physical picture of diffraction by using Laue’s formalism (leading to the Laue’s equations). However, a parallel approach to diffraction is via the method of Bragg, wherein diffraction can be visualized as ‘reflections’ from a set of planes. As the approach of Bragg is easier to grasp we shall use that in this elementary text. We shall do some intriguing mental experiments to utilize the Bragg’s equation (Bragg’s model) with caution. Let us consider a coherent wave of X-rays impinging on a crystal with atomic planes at an angle  to the rays. Incident and scattered waves are in phase if the: i) in-plane scattering is in phase and ii) scattering from across the planes is in phase. In plane scattering is in phase Incident and scattered waves are in phase if Scattering from across planes is in phase

Let us consider in-plane scattering
There is more to this Click here to know more and get introduced to Laue equations describing diffraction Extra path traveled by incoming waves  AY These can be in phase if  incident = scattered Extra path traveled by scattered waves  XB But this is still reinforced scattering and NOT reflection

BRAGG’s EQUATION Let us consider scattering across planes Click here to visualize constructive and destructive interference See Note Ӂ later A portion of the crystal is shown for clarity- actually, for destructive interference to occur many planes are required (and the interaction volume of x-rays is large as compared to that shown in the schematic). The scattering planes have a spacing ‘d’. Ray-2 travels an extra path as compared to Ray-1 (= ABC). The path difference between Ray-1 and Ray-2 = ABC = (d Sin + d Sin) = (2d.Sin). For constructive interference, this path difference should be an integral multiple of : n = 2d Sin  the Bragg’s equation. (More about this sooner). The path difference between Ray-1 and Ray-3 is = 2(2d.Sin) = 2n = 2n. This implies that if Ray-1 and Ray-2 constructively interfere Ray-1 and Ray-3 will also constructively interfere. (And so forth).

Which remains same thereafter (like in the BB’ plane)
The previous page explained how constructive interference occurs. How about the rays just of Bragg angle? Obviously the path difference would be just off  as in the figure below. How come these rays ‘go missing’? Click here to understand how destructive interference of just ‘of-Bragg rays’ occur Interference of Ray-1 with Ray-2 Which remains same thereafter (like in the BB’ plane) Note that they ‘almost’ constructively interfere! Funda Check How to ‘see’ that path difference increases with angle? Clearly A’BC’ > ABC

Reflection versus Diffraction
Laue versus Bragg In Laue’s picture constructive and destructive interference at various points in space is computed using path differences (and hence phase differences) given a crystalline array of scatterers. Bragg simplified this picture by considering this process as ‘reflections from atomic planes’. Click here to know more about the Laue Picture Reflection versus Diffraction Though diffraction (according to Bragg’s picture) has been visualized as a reflection from a set of planes with interplanar spacing ‘d’  diffraction should not be confused with reflection (specular reflection). Reflection Diffraction Occurs from surface Occurs throughout the bulk (though often the penetration of x-rays in only of the order of 10s of microns in a material) Takes place at any angle Takes place only at Bragg angles ~100 % of the intensity may be reflected Small fraction of intensity is diffracted Planes are imaginary constructs Note: X-rays can ALSO be reflected at very small angles of incidence

Understanding the Bragg’s equation
n = 2d Sin The equation is written better with some descriptive subscripts: n is an integer and is the order of the reflection (i.e. how many wavelengths of the X-ray go on to make the path difference between planes). Note: if hkl reflection (corresponding to n=1) occurs at hkl then 2h 2k 2l reflection (n=2) will occur at a higher angle 2h 2k 2l. Bragg’s equation is a negative statement  If Bragg’s eq. is NOT satisfied  NO ‘reflection’ can occur  If Bragg’s eq. is satisfied  ‘reflection’ MAY occur (How?- we shall see this a little later). The interplanar spacing appears in the Bragg’s equation, but not the interatomic spacing ‘a’ along the plane (which had forced incident = scattered); but we are not free to move the atoms along the plane ‘randomly’  click here to know more.  For large interplanar spacing the angle of reflection tends towards zero → as d increases, Sin decreases (and so does ).  The smallest interplanar spacing from which Bragg diffraction can be obtained is /2 → maximum value of  is 90, Sin is 1  from Bragg equation d = /2. If this equation is satisfied, then  is Bragg Note: Ӂ

Order of the reflection (n)
For Cu K radiation ( = 1.54 Å) and d110= 2.22 Å n Sin = n/2d 1 0.34 20.7º First order reflection from (110)  110 2 0.69 43.92º Second order reflection from (110) planes  110 Also considered as first order reflection from (220) planes  220 Relation between dnh nk nl and dhkl e.g.

In XRD nth order reflection from (h k l) is considered as 1st order reflection from (nh nk nl)
All these form the (200) set Hence, (100) planes are a subset of (200) planes Important point to note: In a simple cubic crystal, 100, 200, 300… are all allowed ‘reflections’. But, there are no atoms in the planes lying within the unit cell! Though, first order reflection from 200 planes is equivalent (mathematically) to the second order reflection from 100 planes; for visualization purposes of scattering, this is better thought of as the later process (i.e. second order reflection from (100) planes). Note: Technically, in Miller indices we factor out the common factors. Hence, (220)  2(110)  (110). In XRD we extend the usual concept of Miller indices to include planes, which do not pass through lattice points (e.g. every alternate plane belonging to the (002) set does not pass through lattice points) and we allow the common factors to remain in the indices.

Funda Check I have seen diagrams like in Fig.1 where rays seem to be scattered from nothing! What does this mean? Few points are to be noted in this context. The ray ‘picture’ is only valid in the realm of geometrical optics, where the wave nature of light is not considered. In diffraction we are in the domain of physical optics. The wave impinges on the entire volume of material and plane of atoms (the effect of which can be quantified using the atomic scattering power* and the density of atoms in the plane). Due to the ‘incoming’ wave the atomic dipoles are set into oscillation, which further act like emitter of waves In Bragg’s viewpoint, the atomic planes are to be kept in focus and the wave (not just a ray) impinges on the entire plane (some planes have atoms in contact and most have atoms which are not in contact along the plane  see Fig.2). Wave impinging on a crystal (parallel wave-front) (note there are no ‘rays’) Fig.2 ?? Fig.1 * To be considered later A plane in Bragg’s viewpoint can be characterized by two factors: (a) atomic density (atoms/unit area on the plane), (b) atomic scattering factor of the atoms.

“It is difficult to give an explanation of the nature of the semi-transparent layers or planes that is immediately convincing, as they are a concept rather than a physical reality. Crystal structures, with their regularly repeating patterns, may be referred to a 3D grid and the repeating unit of the grid, the unit cell, can be found. The grid may be divided up into sets of planes in various orientations and it is these planes which are considered in the derivation of Bragg’s law. In some cases, with simple crystal structures, the planes also correspond to layers of atoms, but this is not generally the case. See Section 1.5 for further information. Some of the assumptions upon which Bragg’s law is based may seem to be rather dubious. For instance, it is known that diffraction occurs as a result of interaction between X-rays and atoms. Further, the atoms do not reflect X-rays but scatter or diffract them in all directions. Nevertheless, the highly simplified treatment that is used in deriving Bragg’s law gives exactly the same answers as are obtained by a rigorous mathematical treatment. We therefore happily use terms such as reflexion (often deliberately with this alternative, but incorrect, spelling!) and bear in mind that we are fortunate to have such a simple and picturesque, albeit inaccurate, way to describe what in reality is a very complicated process.” [1] [1] Anthony R West, Solid State Chemistry and its Applications, Second Edition, John Wiley & Sons Ltd., Chichester, 2014.

Funda Check How is it that we are able to get information about lattice parameters of the order of Angstroms (atoms which are so closely spaced) using XRD? Diffraction is a process in which ‘linear information’ (the d-spacing of the planes) is converted to ‘angular information’ (the angle of diffraction, Bragg). If the detector is placed ‘far away’ from the sample (i.e. ‘R’ in the figure below is large) the distances along the arc of a circle (the detection circle) get amplified and hence we can make ‘easy’ measurements. This also implies that in XRD we are concerned with angular resolution instead of linear resolution. Later we will see that in powder diffraction this angle of deviation (2) is plotted instead of .

Forward and Back Diffraction
Here a guide for quick visualization of forward and backward scattering (diffraction) is presented

Funda Check What is  (theta) in the Bragg’s equation?
 is the angle between the incident x-rays and the set of parallel atomic planes (which have a spacing dhkl). Which is 10 in the above figure. Usually,  in this context implies Bragg (i.e. the angle at which Bragg’s equation is satisfied). It is NOT the angle between the x-rays and the sample surface (note: specimens could be spherical or could have a rough surface).

The missing ‘reflections’
We had mentioned that Bragg’s equation is a negative statement: i.e. just because Bragg’s equation is satisfied a ‘reflection’ may not be observed. Let us consider the case of Cu K radiation ( = 1.54 Å) being diffracted from (100) planes of Mo (BCC, a = 3.15 Å = d100). But this reflection is absent in BCC Mo The missing reflection is due to the presence of additional atoms in the unit cell (which are positions at lattice points)  which we shall consider next The wave scattered from the middle plane is out of phase with the ones scattered from top and bottom planes. I.e. if the green rays are in phase (path difference of ) then the red ray will be exactly out of phase with the green rays (path difference of /2).

Why does the 110 reflection not go missing? (Why is it present?)
Continuing with the case of BCC Mo… However, the second order reflection from (100) planes (which is equivalent to the first order reflection from the (200) planes is observed This is because if the green rays have a path difference of 2 then the red ray will have path difference of → which will still lead to constructive interference! Funda Check Why does the 110 reflection not go missing? (Why is it present?) Let us look at the (110) planes in projection. Note that (110)blue coloured planes existed before and after introducing an atom at unit cell centre at (½, ½ ½)grey coloured. Thus lattice centering does not lead to any waves being scattered out of phase.

Some of the reflections may even go missing
Important points Presence of additional atoms/ions/molecules in the UC  at lattice points (as we may chose a non-primitive unit cell)  or as a part of the motif can alter the intensities of some of the reflections Some of the reflections may even go missing Position of the ‘reflections’/‘peaks’ tells us about the lattice type. The Intensities tells us about the motif.

A B C Intensity of the Scattered Waves Electron Atom Unit cell (uc)
Bragg’s equation tells us about the position of the diffraction peaks (in terms of )  but tells us nothing about the intensities. The intensities of the peaks depend on many factors as considered here. Scattering by a crystal can be understood in three steps To understand the scattering from a crystal leading to the ‘intensity of reflections’ (and why some reflections go missing), three levels of scattering have to be considered: 1) scattering from electrons 2) scattering from an atom 3) scattering from a unit cell Click here to know the details A Electron Polarization factor B Structure Factor (F): The resultant wave scattered by all atoms of the unit cell The Structure Factor is independent of the shape and size of the unit cell; but is dependent on the position of the atoms/ions etc. within the cell Atom Atomic scattering factor (f) C Unit cell (uc) Structure factor (F) Structure factor calculations & Intensity in powder patterns Click here to know more about

The concept of a Reciprocal lattice and the Ewald Sphere construction: Reciprocal lattice and Ewald sphere constructions are important tools towards understanding diffraction. (especially diffraction in a Transmission Electron Microscope (TEM)) A lattice in which planes in the real lattice become points in the reciprocal lattice is a very useful one in understanding diffraction.  click here to go to a detailed description of these topics. Click here to know more about Reciprocal Lattice & Ewald Sphere construction

Reflections which may be present Reflections necessarily absent
Selection / Extinction Rules As we have noted before even if Bragg’s equation is satisfied, ‘reflections may go missing’  this is due to the presence of additional atoms in the unit cell. The reflections present and the missing reflections due to additional atoms in the unit cell are listed in the table below. Click here to see the derivations Structure factor calculations Bravais Lattice Reflections which may be present Reflections necessarily absent Simple all None Body centred (h + k + l) even (h + k + l) odd Face centred h, k and l unmixed h, k and l mixed End centred (C centred) h and k unmixed h and k mixed Bravais Lattice Allowed Reflections SC All BCC (h + k + l) even FCC h, k and l unmixed DC Either,  h, k and l are all odd or  all are even & (h + k + l) divisible by 4

Allowed reflections in SC*, FCC*, BCC* & DC crystals
h2 + k2 + l2 SC FCC BCC DC 1 100 2 110 3 111 4 200 5 210 6 211 7 8 220 9 300, 221 10 310 11 311 12 222 13 320 14 321 15 16 400 17 410, 322 18 411, 330 19 331 Allowed reflections in SC*, FCC*, BCC* & DC crystals Cannot be expressed as (h2+k2+l2) * lattice decorated with monoatomic/monoionic motif

Crystal structure determination
As diffraction occurs only at specific Bragg angles, the chance that a reflection is observed when a crystal is irradiated with monochromatic X-rays at a particular angle is small (added to this the diffracted intensity is a small fraction of the beam used for irradiation). The probability to get a diffracted beam (with sufficient intensity) is increased by either varying the wavelength () or having many orientations (rotating the crystal or having multiple crystallites in many orientations). The three methods used to achieve high probability of diffraction are shown below. Many s (orientations) Powder specimen POWDER METHOD Monochromatic X-rays Single  LAUE TECHNIQUE Panchromatic X-rays ROTATING CRYSTAL METHOD  Varied by rotation Monochromatic X-rays Only the powder method (which is commonly used in materials science) will be considered in this text.

THE POWDER METHOD In the powder method the specimen has crystallites (or grains) in many orientations (usually random). Monochromatic* X-rays are irradiated on the specimen and the intensity of the diffracted beams is measured as a function of the diffracted angle. In this elementary text we shall consider cubic crystals. Cubic crystal (1) (2) (2) in (1) * In reality this is true only to an extent

The ratio of (h2 + k2 + l2) derived from extinction rules (earlier page)
As we shall see soon the ratios of (h2 + k2 + l2) is proportional to Sin2  which can be used in the determination of the lattice type SC 1 2 3 4 5 6 8 BCC 7 FCC 11 12 DC 16 Note that we have to consider the ratio of only two lines to distinguish FCC and DC. I.e. if the ratios are 3:4 then the lattice is FCC. But, to distinguish between SC and BCC we have to go to 7 lines!

Usually the source is fixed and the detector and sample are rotated
POWDER METHOD In the powder sample there are crystallites in different ‘random’ orientations (a polycrystalline sample too has grains in different orientations) The coherent x-ray beam is diffracted by these crystallites at various angles to the incident direction All the diffracted beams (called ‘reflections’) from a single plane, but from different crystallites lie on a cone. Depending on the angle there are forward and back reflection cones. A diffractometer can record the angle of these reflections along with the intensities of the reflection The X-ray source and diffractometer move in arcs of a circle- maintaining the Bragg ‘reflection’ geometry as in the figure (right) Usually the source is fixed and the detector and sample are rotated Different cones for different reflections Also called Debye ring

Random assemblage of crystallites in a material
How to visualize the occurrence of peaks at various angles It is ‘somewhat difficult’ to actually visualize a random assembly of crystallites giving peaks at various angels in a XRD scan. The figures below are expected to give a ‘visual feel’ for the same. [Hypothetical crystal with a = 4Å is assumed with =1.54Å. Only planes of the type xx0 (like (100,110)are considered]. For convenience the source may be stationary (and the sample and detector may rotate– but the effect is equivalent) The sample is not rotating only the source and detector move in arcs of a circle Random assemblage of crystallites in a material As the scan takes place at increasing angles, planes with suitable ‘d’, which diffract are ‘picked out’ from favourably oriented crystallites h2 hkl d Sin(q) q 1 100 4.00 0.19 11.10 2 110 2.83 0.27 15.80 3 111 2.31 0.33 19.48 4 200 2.00 0.39 22.64 5 210 1.79 0.43 25.50 6 211 1.63 0.47 28.13 8 220 1.41 0.54 32.99 9 300 1.33 0.58 35.27 10 310 1.26 0.61 37.50

Determination of Crystal Structure from 2 versus Intensity Data in Powder Method
In the power diffraction method a 2 versus intensity (I) plot is obtained from the diffractometer (and associated instrumentation). The ‘intensity’ is the area under the peak in such a plot (NOT the height of the peak).  The information of importance obtained from such a pattern is the ‘relative intensities’ and the absolute value of the intensities is of little importance (the longer we irradiate the sample the more will be the intensity under the peak) (for now).  I is really diffracted energy (as Intensity is Energy/area/time). A table is prepared as in the next slide to tabulate the data and make calculations to find the crystal structure (restricting ourselves to cubic crystals for the present). Powder diffraction pattern from Al Intensity (I) has units of [Energy/area/time] → but here it is plotted as arbitrary units. Radiation: Cu K,  = 1.54 Å Increasing d This is peak (sometimes called a line- a hangover from Debye Scherrer camera usage) Usually in degrees () Increasing 

n 2→  Intensity Sin Sin2  ratio
Determination of Crystal Structure (lattice type) from 2 versus Intensity Data The following table is made from the 2 versus Intensity data (obtained from a XRD experiment on a powder sample (empty starting table of columns is shown below- completed table shown later). n 2→  Intensity Sin Sin2  ratio

Powder diffraction pattern from Al Radiation: Cu K,  = 1.54 Å
Note: This is a schematic pattern In real patterns peaks or not idealized  peaks  broadened Increasing splitting of peaks with g  (1 & 2 peaks get resolved in the high angle peaks) Peaks are all not of same intensity No brackets are used around the indexed numbers (the peaks correspond to planes in the real space) Note that there are no brackets around the indices!

Powder diffraction pattern from Al
Radiation: Cu K,  = 1.54 Å Powder diffraction pattern from Al Note: Peaks or not idealized  peaks  broadened. Increasing splitting of peaks with g . Peaks are all not of same intensity. There is a ‘noisy’ background. 111 200 220 311 222 400 In low angle peaks K1 & K2 peaks merged K1 & K2 peaks resolved in high angle peaks (in 222 and 400 peaks this can be seen)

What is the maximum value of  possible (experimentally)? Ans: 90
Funda Check How are real diffraction patterns different from the ‘ideal computed ones? We have seen real and ideal diffraction patterns. In ideal patterns the peaks are ‘’ functions. Real diffraction patterns are different from ideal ones in the following ways:  Peaks are broadened Could be due to instrumental, residual ‘non-uniform’ strain (microstrain), grain size etc. broadening.  Peaks could be shifted from their ideal positions Could be due to uniform strain→ macrostrain.  Relative intensities of the peaks could be altered Could be due to texture in the sample. Note peak splitting has not been included here as this comes from ‘symmetry lowering’ (i.e. crystal with lower symmetry) Instrumental broadening Peak Broadening Small crystallite size Crystal defects (‘bent’ planes) Including those coming from strain fields associated with these defects Click here to know more Funda Check What is the maximum value of  possible (experimentally)? Ans: 90 At  = 90 the ‘reflected ray’ is opposite in direction to the incident ray. Beyond this angle, it is as if the source and detector positions are switched.  2max is 180.

Funda Check What will determine how many peaks I will get? 1)   smaller the wavelength of the X-rays, more will be the number of peaks possible.  From Bragg’s equation: [=2dSin], (Sin)max will correspond to dmin. (Sin)max=1. Hence, dmin=/2. Hence, if  is small then planes with smaller d spacing (i.e. those which occur at higher 2 values) will also show up in a XRD patter (powder pattern). Given that experimentally  cannot be greater than 90. 2) Lattice type  in SC we will get more peaks as compared to (say) FCC/DC. Other things being equal. 3) Lower the symmetry of the crystal, more the number of peaks (e.g., in tetragonal crystal the 100 peak will lie at a different 2 as compared to the 001 peak).

From the ratios in column 6 we conclude that
Solved example Determination of Crystal Structure (lattice type) from 2 versus Intensity Data 1 Let us assume that we have the 2 versus intensity plot from a diffractometer  To know the lattice type we need only the position of the peaks (as tabulated below) # 2 Sin Sin2  ratio Index d 1 38.52 19.26 0.33 0.11 3 111 2.34 2 44.76 22.38 0.38 0.14 4 200 2.03 65.14 32.57 0.54 0.29 8 220 1.43 78.26 39.13 0.63 0.40 11 311 1.22 5 82.47 41.235 0.66 0.43 12 222 1.17 6 99.11 49.555 0.76 0.58 16 400 1.01 7 112.03 56.015 0.83 0.69 19 331 0.93 116.60 58.3 0.85 0.72 20 420 0.91 9 137.47 68.735 0.87 24 422 10 163.78 81.89 0.99 0.98 27 333 0.78 Note that Sin cannot be > 1 From the ratios in column 6 we conclude that FCC Note Using We can get the lattice parameter  which correspond to that for Al Note: Error in d spacing decreases with  → so we should use high angle lines for lattice parameter calculation Click here to know more XRD_lattice_parameter_calculation.ppt

Solved example 2 Another example Given the positions of the Bragg peaks we find the lattice type 2→  Sin Sin2  Ratios of Sin2 Dividing Sin2 by 0.134/3 = Whole number ratios 1 21.5 0.366 0.134 3 2 25 0.422 0.178 1.33 3.99 4 37 0.60 0.362 2.70 8.10 8 45 0.707 0.500 3.73 11.19 11 5 47 0.731 0.535 11.98 12 6 58 0.848 0.719 5.37 16.10 16 7 68 0.927 0.859 6.41 19.23 19 FCC

More Solved Examples on XRD

Funda Check What happens when we increase or decrease ?
We had pointed out that  ~ a is preferred for diffraction. Let us see what happens if we ‘drastically’ increase or decrease . (This is only a thought experiment!!) If we make  small→ all the peaks get crowded to small angles If we ~double  → we get too few peaks With CuK  = 1.54 Å And the detector may not be able to resolve these peaks if they come too close!

Applications of XRD Bravais lattice determination
We have already seen these applications Lattice parameter determination Determination of solvus line in phase diagrams Long range order Crystallite size and Strain Click here to know more Determine if the material is amorphous or crystalline Next slide And More….

Diffraction angle (2) →
Crystal Schematics Schematic of difference between the diffraction patterns of various phases Intensity → Sharp peaks 90 180 Diffraction angle (2) → Monoatomic gas No peak Intensity → Diffraction angle (2) → Diffraction angle (2) → Liquid / Amorphous solid Intensity → 180 90 Diffuse Peak 90 180

Diffuse peak from Cu-Zr-Ni-Al-Si Metallic glass
Amorphous solid Actual diffraction pattern from an amorphous solid Diffuse peak from Cu-Zr-Ni-Al-Si Metallic glass A amorphous solid which shows glass transition in a Differential Scanning Calorimetry (DSC) plot is also called a glass. In ‘general usage’ a glass may be considered equivalent to a amorphous solid (at least loosely in the structural sense). Sharp peaks are missing. Broad diffuse peak survives → the peak corresponds to the average spacing between atoms which the diffraction experiment ‘picks out’ (XRD patterns) courtesy: Dr. Kallol Mondal, MSE, IITK

With increasing indices the interplanar spacing decreases
Funda Check What is the minimum spacing between planes possible in a crystal? How many diffraction peaks can we get from a powder pattern? Let us consider a cubic crystal (without loss in generality) As h,k, l increases, ‘d’ decreases  we could have planes with infinitesimal spacing The number of peaks we obtain in a powder diffraction pattern depends on the wavelength of x-ray we are using. Planes with ‘d’ < /2 are not captured in the diffraction pattern. These peaks with small ‘d’ occur at high angles in diffraction pattern. With increasing indices the interplanar spacing decreases

Q & A How to increase the number of peaks in a XRD pattern? We have noted that (e.g. for DC crystal) the number of available peaks in the 2 regime could be insufficient for a given analysis. The number of peaks can be increased in two ways: 1) using Mo Kα instead of Cu Kα 2) first obtain pattern with β filter and then remove the filter to get more lines.