Presentation on theme: "Anandh Subramaniam & Kantesh Balani"— Presentation transcript:
1Anandh Subramaniam & Kantesh Balani X-RAY DIFFRACTIONMATERIALS SCIENCE&ENGINEERINGAnandh Subramaniam & Kantesh BalaniMaterials Science and Engineering (MSE)Indian Institute of Technology, KanpurURL: home.iitk.ac.in/~anandhAN INTRODUCTORY E-BOOKPart ofA Learner’s GuideX- Ray SourcesDiffraction: Bragg’s LawCrystal Structure DeterminationElements of X-Ray DiffractionB.D. Cullity & S.R. StockPrentice Hall, Upper Saddle River (2001)Recommended websites:
2What 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 equationWhat 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…
3Click here to know more about this Some BasicsFor 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)
4Generation of X-raysX-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.TargetX-raysBeam of electronsAn accelerating (or decelerating) charge radiates electromagnetic radiation
5Mo 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 studiesTarget Metal Of K radiation (Å)Mo0.71Cu1.54Co1.79Fe1.94Cr2.29The high intensity nearly monochromatic K x-rays can be used as a radiation source forX-ray diffraction (XRD) studies a monochromator can be used to further decrease the spread of wavelengths in the X-ray
6X-ray sources with different for doing XRD studies Elements (KV) Of K1 radiation (Å) Of K2 radiation (Å) Of Kβ radiation (Å)Kβ-Filter(mm)Ag25.520.5638Pd0.0461Mo200.7093Zr0.0678Cu8.98Ni0.0178.33Co0.01587.71Fe0.01667.11Mn0.0168Cr5.992.2897V0.169C.Gordon Darwin, Grandson of C. Robert Darwin developed the dynamic theory of scattering of x-rays (a tough theory!) in 1912
7Incident 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-raysSPECIMENAbsorption (Heat)Fluorescent X-raysElectronsScattered X-raysCompton recoilPhotoelectronsCoherentFrom bound chargesIncoherent (Compton modified)From loosely bound chargesTransmitted beamClick here to know moreX-rays can also be refracted (refractive index slightly less than 1) and reflected (at very small angles)
8DiffractionClick 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 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 waveAspects related to the waveDiffraction pattern with sharp peaksAspects related to the materialCrystalline*,**Aspects related to the diffraction set-up(diffraction geometry)Fraunhofer geometry 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).
9Not all objects act like scatterers for all kinds of radiation. Some comments and notesThe 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
10XRD the first stepA 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
11Incident and scattered waves are in phase if Some points to recon withWe 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 phaseIncident and scattered waves are in phase ifScattering from across planes is in phase
12Let us consider in-plane scattering There is more to this Click here to know more and get introduced to Laue equations describing diffractionExtra path traveled by incoming waves AYThese can be in phase if incident = scatteredExtra path traveled by scattered waves XBBut this is still reinforced scattering and NOT reflection
13Click here to visualize constructive and destructive interference BRAGG’s EQUATIONLet us consider scattering across planesClick here to visualize constructive and destructive interferenceSee Note Ӂ laterA 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) = 2n = 2n. This implies that if Ray-1 and Ray-2 constructively interfere Ray-1 and Ray-3 will also constructively interfere. (And so forth).
14Which 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’ occurInterference of Ray-1 with Ray-2Which remains same thereafter (like in the BB’ plane)Note that they ‘almost’ constructively interfere!Funda CheckHow to ‘see’ that path difference increases with angle?Clearly A’BC’ > ABC
15Reflection versus Diffraction Laue versus BraggIn 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 PictureReflection versus DiffractionThough 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).ReflectionDiffractionOccurs from surfaceOccurs 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 angleTakes place only at Bragg angles~100 % of the intensity may be reflectedSmall fraction of intensity is diffractedPlanes are imaginary constructsNote: X-rays can ALSO be reflected at very small angles of incidence
16Understanding 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 BraggNote: Ӂ
17Order of the reflection (n) For Cu K radiation ( = 1.54 Å) and d110= 2.22 ÅnSin = n/2d10.3420.7ºFirst order reflection from (110) 11020.6943.92ºSecond order reflection from (110) planes 110Also considered as first order reflection from (220) planes 220Relation between dnh nk nl and dhkle.g.
18In XRD nth order reflection from (h k l) is considered as 1st order reflection from (nh nk nl) All these form the (200) setHence, (100) planes are a subset of (200) planesImportant 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.
19Funda CheckI 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 wavesIn 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 laterA 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.
20More about the Bragg’s viewpoint “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.”  Anthony R West, Solid State Chemistry and its Applications, Second Edition, John Wiley & Sons Ltd., Chichester, 2014.
21Funda CheckHow 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 .
22Forward and Back Diffraction Here a guide for quick visualization of forward and backward scattering (diffraction) is presented
23Funda 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).
24The 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 MoThe missing reflection is due to the presence of additional atoms in the unit cell(which are positions at lattice points) which we shall consider nextThe 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).
25Why 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 observedThis 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 CheckWhy 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.
26Some of the reflections may even go missing Important pointsPresence 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 reflectionsSome of the reflections may even go missingPosition of the ‘reflections’/‘peaks’ tells us about the lattice type.The Intensities tells us about the motif.
27A 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 stepsTo 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 electrons2) scattering from an atom3) scattering from a unit cell Click here to know the detailsAElectronPolarization factorBStructure Factor (F): The resultant wave scattered by all atoms of the unit cellThe 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 cellAtomAtomic scattering factor (f)CUnit cell (uc)Structure factor (F)Structure factor calculations&Intensity in powder patternsClick here to know more about
28Click 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 aboutReciprocal Lattice & Ewald Sphere construction
29Reflections which may be present Reflections necessarily absent Selection / Extinction RulesAs 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 derivationsStructure factor calculationsBravais LatticeReflections which may be presentReflections necessarily absentSimpleallNoneBody centred(h + k + l) even(h + k + l) oddFace centredh, k and l unmixedh, k and l mixedEnd centred (C centred)h and k unmixedh and k mixedBravais LatticeAllowed ReflectionsSCAllBCC(h + k + l) evenFCCh, k and l unmixedDCEither, h, k and l are all odd or all are even & (h + k + l) divisible by 4
30Allowed reflections in SC*, FCC*, BCC* & DC crystals h2 + k2 + l2SCFCCBCCDC110021103111420052106211782209300, 2211031011311122221332014321151640017410, 32218411, 33019331Allowed reflections in SC*, FCC*, BCC* & DC crystalsCannot be expressed as (h2+k2+l2)* lattice decorated with monoatomic/monoionic motif
31Crystal 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 specimenPOWDER METHODMonochromatic X-raysSingle LAUE TECHNIQUEPanchromatic X-raysROTATING CRYSTAL METHOD Varied by rotationMonochromatic X-raysOnly the powder method (which is commonly used in materials science) will be considered in this text.
32THE POWDER METHODIn 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
33The 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 typeSC1234568…BCC7FCC1112DC16Note 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!
34Usually the source is fixed and the detector and sample are rotated POWDER METHODIn 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 directionAll 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 reflectionThe 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 rotatedDifferent cones for different reflectionsAlso called Debye ring
35Random assemblage of crystallites in a material How to visualize the occurrence of peaks at various anglesIt 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 circleRandom assemblage of crystallites in a materialAs the scan takes place at increasing angles, planes with suitable ‘d’, which diffract are ‘picked out’ from favourably oriented crystallitesh2hkldSin(q)q11004.000.1911.1021102.830.2715.8031112.310.3319.4842002.000.3922.6452101.790.4325.5062111.630.4728.1382201.410.5432.9993001.330.5835.27103101.260.6137.50
36Determination 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 AlIntensity (I) has units of [Energy/area/time] → but here it is plotted as arbitrary units.Radiation: Cu K, = 1.54 ÅIncreasing dThis is peak (sometimes called a line- a hangover from Debye Scherrer camera usage)Usually in degrees ()Increasing
37n 2→ Intensity Sin Sin2 ratio Determination of Crystal Structure (lattice type) from 2 versus Intensity DataThe 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).n2→ IntensitySinSin2 ratio
38Powder diffraction pattern from Al Radiation: Cu K, = 1.54 Å Note:This is a schematic patternIn real patterns peaks or not idealized peaks broadenedIncreasing splitting of peaks with g (1 & 2 peaks get resolved in the high angle peaks)Peaks are all not of same intensityNo 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!
39Powder diffraction pattern from Al Radiation: Cu K, = 1.54 ÅPowder diffraction pattern from AlNote:Peaks or not idealized peaks broadened.Increasing splitting of peaks with g .Peaks are all not of same intensity.There is a ‘noisy’ background.111200220311222400In low angle peaks K1 & K2 peaks mergedK1 & K2 peaks resolved in high angle peaks (in 222 and 400 peaks this can be seen)
40What is the maximum value of possible (experimentally)? Ans: 90 Funda CheckHow 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 broadeningPeak BroadeningSmall crystallite sizeCrystal defects (‘bent’ planes)Including those coming from strain fields associated with these defectsClick here to know moreFunda CheckWhat is the maximum value of possible (experimentally)?Ans: 90At = 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. 2max is 180.
41Funda CheckWhat 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).
42From the ratios in column 6 we conclude that Solved exampleDetermination of Crystal Structure (lattice type) from 2 versus Intensity Data1Let 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)#2SinSin2 ratioIndexd138.5219.260.330.1131112.34244.7622.380.380.1442002.0365.1432.570.540.2982201.4378.2639.130.630.40113111.22582.4741.2350.660.43122221.17699.1149.5550.760.58164001.017112.0356.0150.830.69193310.93116.6058.30.850.72204200.919137.4768.7350.872442210163.7881.890.990.98273330.78Note that Sin cannot be > 1From the ratios in column 6 we conclude thatFCCNoteUsingWe can get the lattice parameter which correspond to that for AlNote: Error in d spacing decreases with → so we should use high angle lines for lattice parameter calculationClick here to know moreXRD_lattice_parameter_calculation.ppt
43Solved example2Another example Given the positions of the Bragg peaks we find the lattice type2→ SinSin2 Ratios of Sin2Dividing Sin2 by0.134/3 =Whole number ratios121.50.3660.13432250.4220.1781.333.994370.600.3622.708.108450.7070.5003.7311.19115470.7310.53511.98126580.8480.7195.3716.10167680.9270.8596.4119.2319FCC
44More Solved Examples on XRD Comparison of diffraction patterns of SC, BCC & B2 structuresClick hereMore SolvedExamples on XRDClick here
45Funda 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 anglesIf we ~double → we get too few peaksWith CuK = 1.54 ÅAnd the detector may not be able to resolve these peaks if they come too close!
46Applications of XRD Bravais lattice determination We have already seen these applicationsLattice parameter determinationDetermination of solvus line in phase diagramsLong range orderCrystallite size and StrainClick here to know moreDetermine if the material is amorphous or crystallineNext slideAnd More….
47Diffraction angle (2) → CrystalSchematicsSchematic of difference between the diffraction patterns of various phasesIntensity →Sharp peaks90180Diffraction angle (2) →Monoatomic gasNo peakIntensity →Diffraction angle (2) →Diffraction angle (2) →Liquid / Amorphous solidIntensity →18090Diffuse Peak90180
48Diffuse peak from Cu-Zr-Ni-Al-Si Metallic glass Amorphous solidActual diffraction pattern from an amorphous solidDiffuse peak from Cu-Zr-Ni-Al-Si Metallic glassA 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
49With increasing indices the interplanar spacing decreases Funda CheckWhat 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 spacingThe 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
50Q & AHow 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.