X-RAY DIFFRACTION X- Ray Sources Diffraction: Bragg‟s Law Crystal Structure Determination Part of MATERIALS SCIENCE & A Learner’s Guide ENGINEERING AN INTRODUCTORY E-BOOK Anandh Subramaniam & Kantesh Balani Materials Science and Engineering (MSE) Indian Institute of Technology, Kanpur- 208016 Email: [email protected], URL: home.iitk.ac.in/~anandh http://home.iitk.ac.in/~anandh/E-book.htm Elements of X-Ray Diffraction B.D. Cullity & S.R. Stock Prentice Hall, Upper Saddle River (2001) Recommended websites: http://www.matter.org.uk/diffraction/ http://www.ngsir.netfirms.com/englishhtm/Diffraction.htm 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. ** 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) Click here to know more about this 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. Beam of electrons Target X-rays An accelerating (or decelerating) charge radiates electromagnetic radiation 4 Wavelength () 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 .71 Cu 1.54 Co 1.79 Fe 1.2 0.29 1.0 X-ray sources with different for doing XRD studies Target Metal Of K radiation (Å) Mo 0.94 Cr 2. nearly monochromatic Intensity K White radiation K Characteristic radiation → due to energy transitions in the atom 0.6 1.Mo Target impacted by electrons accelerated by a 35 kV potential shows the emission spectrum as in the figure below (schematic) Intense peak. . the interaction can result in various signals/emissions/effects. • The coherently scattered X-rays are the ones important from a XRD perspective.• When X-rays hit a specimen. Incident X-rays SPECIMEN Fluorescent X-rays Electrons Scattered X-rays Coherent From bound charges Absorption (Heat) Compton recoil Photoelectrons 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) . parallel wave Aspects related to the wave Diffraction pattern with sharp peaks Aspects related to the material Crystalline*. Fraunhofer diffraction geometry Coherent. Let us consider a special case of diffraction → a case where we get „sharp[1] diffraction peaks’. parallel waves (with wavelength ). monochromatic. ions etc. monochromatic.) to give rise to the phenomenon known as Xray diffraction (XRD). . Crystalline array of scatterers* with spacing of the order of (~) . Diffraction (with sharp peaks) (with XRD being a specific case) requires three important conditions to be satisfied: Coherent.** Aspects related to the diffraction set-up (diffraction geometry) Fraunhofer geometry [1] The intensity- plot looks like a ‘’ function. * A quasicrystalline array will also lead to diffraction with sharp peaks ** Amorphous material will give diffuse peak.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. (which we shall not consider in this text). Not all objects act like scatterers for all kinds of radiation. then the number of peaks obtained may be highly restricted (i. Click here to know more about Fraunhofer and Fresnel diffraction geometries ** With a de Broglie wavelength . In short diffraction is coherent reinforced scattering (or reinforced scattering of coherent waves). To give an analogy the results of Young‟s double slit experiment is interpreted as interference. we may even not even get a single diffraction peak!). waves on water surface…). In a sense diffraction is nothing but a special case of constructive (& destructive) interference. neutrons…) or mechanical waves (sound.e. Fraunhofer diffraction geometry implies that parallel waves are impinging on the scatteres (the object). while the result of multiple slits (large number) is categorized under diffraction. X-rays…). If wavelength is not of the order of the spacing of the scatterers.Some comments and notes The waves could be: electromagnetic waves (light. and the screen (to capture the diffraction pattern) is placed far away from the object. matter waves** (electrons. The waves emitted by the electrons have the same frequency as the incoming X-rays coherent. The electrons oscillate under the influence of the incoming X-Rays and become secondary sources of EM radiation. Secondary emission Incoming X-rays Oscillating charge re-radiates In phase with the incoming x-rays Schematics Sets nucleus into oscillation Sets Electron cloud into oscillation Small effect neglected . The secondary radiation is in all directions.XRD the first step A beam of X-rays directed at a crystal interacts with the electrons of the atoms in the crystal. The emission can undergo constructive or destructive interference. wherein diffraction can be visualized as „reflections‟ from a set of planes. However. 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.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). a parallel approach to diffraction is via the method of Bragg. 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. 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 A B X Y Atomic Planes 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 Extra path traveled by scattered waves XB But this is still reinforced scattering and NOT reflection A B X These can be in phase if incident = scattered Y . (More about this sooner). Ray-2 travels an extra path as compared to Ray-1 (= ABC). For constructive interference.Sin). The scattering planes have a spacing „d‟. 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). this path difference should be an integral multiple of : n = 2d Sin the Bragg‟s equation. This implies that if Ray-1 and Ray-2 constructively interfere Ray-1 and Ray-3 will also constructively interfere.actually.Sin) = 2n = 2n.BRAGG‟s EQUATION Let us consider scattering across planes Click here to visualize constructive and destructive interference A portion of the crystal is shown for clarity. The path difference between Ray-1 and Ray-2 = ABC = (d Sin + d Sin) = (2d. The path difference between Ray-1 and Ray-3 is = 2(2d. . (And so forth). How about the rays just of Bragg angle? Obviously the path difference would be just off as in the figure below. The previous page explained how constructive interference occurs. 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 Note that they ‘almost’ constructively interfere! . 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 Takes place at any angle Takes place only at Bragg angles ~100 % of the intensity may be reflected Small fraction of intensity is diffracted Note: X-rays can ALSO be reflected at very small angles of incidence . how many wavelengths of the X-ray go on to make the path difference between planes).we shall see this a little later). is NOT satisfied NO „reflection‟ can occur If Bragg‟s eq. is satisfied „reflection‟ MAY occur (How?. The interplanar spacing appears in the Bragg‟s equation.e. Sin decreases (and so does ). Bragg‟s equation is a negative statement If Bragg‟s eq.Understanding the Bragg‟s equation n = 2d Sin The equation is written better with some descriptive subscripts: n Cu K 2 d hkl Sin hkl n is an integer and is the order of the reflection (i. 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. . 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. but not the interatomic spacing „a‟ along the plane (which had forced incident = scattered). 7º 0.g. a d 220 8 a d110 2 d 220 1 d110 2 .34 20.22 Å n Sin = n/2d 1 0.69 • Second order reflection from (110) planes 110 43.92º • Also considered as first order reflection from (220) planes 220 2 • First order reflection from (110) 110 Relation between dnh nk nl and dhkl d Cubic crystal hkl d nh nk nl d nh nk nl a h2 k 2 l 2 a (nh)2 (nk )2 (nl )2 d hkl 2 2 2 n n h k l a e.54 Å) and d110= 2.Order of the reflection (n) For Cu K radiation ( = 1. this is better thought of as the later process (i. first order reflection from 200 planes is equivalent (mathematically) to the second order reflection from 100 planes. But. there are no atoms in the planes lying within the unit cell! Though.e. (100) planes are a subset of (200) planes d 200 1 d100 2 d 300 1 d100 3 Important point to note: In a simple cubic crystal. 300… are all allowed „reflections‟. 100. for visualization purposes of scattering. 200. . second order reflection from (100) planes).In XRD nth order reflection from (h k l) is considered as 1st order reflection from (nh nk nl) n 2d hkl sin d hkl 2 sin n 2d nh nk nl sin d nhnk nl d hkl 1 n Hence. Bragg).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. If the detector is placed „far away‟ from the sample (i.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. . 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. It is NOT the angle between the x-rays and the sample surface (note: specimens could be spherical or could have a rough surface). Let us consider the case of Cu K radiation ( = 1.e.149 2(3.54 Å) being diffracted from (100) planes of Mo (BCC.15) 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.15 Å = d100). 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). I. a = 3.e.54 0. just because Bragg‟s equation is satisfied a „reflection‟ may not be observed. 2 d100 Sin100 Sin100 2 d100 1. .244 100 14.The missing „reflections‟ We had mentioned that Bragg‟s equation is a negative statement: i. 15 2nd order 100 ~ 1nd order 200 29.267 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! .54 0.Continuing with the case of BCC Mo… However.488 2 d100 3. the second order reflection from (100) planes (which is equivalent to the first order reflection from the (200) planes is observed Sin100 2 1. .Important points Presence of additional atoms/ions/molecules in the UC at lattice points 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. within the cell C Unit cell (uc) Structure factor (F) Click here to know more about Structure factor calculations & Intensity in powder patterns . but is dependent on the position of the atoms/ions etc. The intensities of the peaks depend on many factors as considered here.Intensity of the Scattered waves Bragg‟s equation tells us about the position of the intensity peaks (in terms of ) but tells us nothing about the intensities. 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 B Atom Atomic scattering factor (f) 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. Scattering by a crystal can be understood in three steps A Electron Polarization factor To understand the scattering from a crystal leading to the „intensity of reflections‟ (and why some reflections go missing). Click here to know more about Reciprocal Lattice & Ewald Sphere construction . (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.The concept of a Reciprocal lattice and the Ewald Sphere construction: Reciprocal lattice and Ewald sphere constructions are important tools towards understanding diffraction. 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. ‘reflections may go missing’ this is due to the presence of additional atoms in the unit cell. 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. The reflections present and the missing reflections due to additional atoms in the unit cell are listed in the table below. k and l are all odd or all are even & (h + k + l) divisible by 4 . k and l unmixed h.Selection / Extinction Rules As we have noted before even if Bragg’s equation is satisfied. h. k and l unmixed DC Either. 221 10 310 11 311 311 12 222 222 13 320 14 321 220 220 310 311 222 321 15 * lattice decorated with monoatomic/monoionic motif 16 400 17 410. BCC* & DC crystals h2 + k2 + l2 SC 1 100 2 110 3 111 111 4 200 200 5 210 6 211 FCC BCC DC 110 111 200 211 7 Cannot be expressed as (h2+k2+l2) 8 220 220 9 300. 330 19 331 400 400 400 411. 330 331 331 .Allowed reflections in SC*. 322 18 411. FCC*. But.e. I.The ratio of (h2 + k2 + l2) derived from extinction rules (previous 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 1 2 3 4 5 6 7 … FCC 3 4 8 11 12 … DC 3 8 11 16 … Note that we have to consider the ratio of only two lines to distinguish FCC and DC. if the ratios are 3:4 then the lattice is FCC. to distinguish between SC and BCC we have to go to 7 lines! . 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. . Monochromatic X-rays Panchromatic X-rays Monochromatic X-rays Many s (orientations) Powder specimen Single Varied by rotation POWDER METHOD λ fixed θ variable LAUE TECHNIQUE λ variable θ fixed ROTATING CRYSTAL METHOD λ fixed θ rotated Only the powder method (which is commonly used in materials science) will be considered in this text. 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).Crystal structure determination As diffraction occurs only at specific Bragg angles. 2 d Sin Cubic crystal (1) d Cubic a 2 2 2 h k l (2) in (1) 2 2 4 a sin 2 2 h k2 l2 a 2 2 4 sin 2 (h 2 k 2 l 2 ) * In reality this is true only to an extent (h 2 k 2 l 2 ) 4a 2 2 sin 2 (h 2 k 2 l 2 ) sin 2 (2) . In this elementary text we shall consider cubic crystals. Monochromatic* X-rays are irradiated on the specimen and the intensity of the diffracted beams is measured as a function of the diffracted angle.THE POWDER METHOD In the powder method the specimen has crystallites (or grains) in many orientations (usually random). Depending on the angle there are forward and back reflection cones.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.maintaining the Bragg „reflection‟ geometry as in the figure (right) Different cones for different reflections . 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. but from different crystallites lie on a cone. 26 0.80 3 111 2.54Å.50 .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.99 9 300 1.00 0.43 25.33 19.00 0. Random assemblage of crystallites in a material As the scan takes place at increasing angles. Only planes of the type xx0 (like (100.39 22.47 28.50 6 211 1.110)are considered].27 15.54 32.83 0.13 8 220 1.61 37.27 10 310 1.48 4 200 2.31 0.63 0.33 0.79 0. [Hypothetical crystal with a = 4Å is assumed with =1.64 5 210 1.41 0.10 2 110 2.58 35. which diffract are „picked out‟ from favourably oriented crystallites h2 hkl d Sin() 1 100 4. planes with suitable „d‟. The figures below are expected to give a ‘visual feel’ for the same.19 11. Powder diffraction pattern from Al Radiation: Cu K. The information of importance obtained from such a pattern is the „relative intensities‟ and the absolute value of the intensities is of little importance (for now).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). = 1. 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). The „intensity‟ is the area under the peak in such a plot (NOT the height of the peak).54 Å Increasing d Increasing . I is really diffracted energy (as Intensity is Energy/area/time). completed table shown later). n 2→ Intensity Sin Sin2 ratio .Determination of Crystal Structure 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. 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) .Powder diffraction pattern from Al Radiation: Cu K. = 1. Radiation: Cu K.54 Å Powder diffraction pattern from Al In low angle peaks K1 & K2 peaks merged 400 222 311 220 200 111 Note: Peaks or not idealized peaks broadened Increasing splitting of peaks with g Peaks are all not of same intensity K1 & K2 peaks resolved in high angle peaks (in 222 and 400 peaks this can be seen) . = 1. In ideal patterns the peaks are „‟ functions. it is as if the source and detector positions are switched. Real diffraction patterns are different from ideal ones in the following ways: Peaks are broadened Could be due to instrumental. Funda Check What is the maximum value of possible (experimentally)? At = 90 the „reflected ray‟ is opposite in direction to the incident ray. residual ‘non-uniform’ strain (microstrain).Funda Check How are real diffraction patterns different from the ‘ideal computed ones? We have seen real and ideal diffraction patterns. grain size etc. Peaks could be shifted from their ideal positions Could be due to uniform strain→ macrostrain. Ans: 90 . broadening. Relative intensities of the peaks could be altered Could be due to texture in the sample. 2max is 180. Beyond this angle. 2) Lattice type in SC we will get more peaks as compared to (say) FCC/DC.g. Other things being equal. From Bragg‟s equation: [=2dSin]. 2dSin 2dmin Sin max 2 d min .. dmin=/2. (Sin)max=1. if is small then planes with smaller d spacing (i. (Sin)max will correspond to dmin.e. more will be the number of peaks possible. Hence.Funda Check What will determine how many peaks I will get? 1) smaller the wavelength of the X-rays. more the number of peaks (e. Hence. 3) Lower the symmetry of the crystal. in tetragonal crystal the 100 peak will lie at a different 2 as compared to the 001 peak). 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. 83 10 163.14 4 200 2.63 0.015 0.54 2 d111 Sin111 2 a 0.47 41.14 32.43 4 78.01 7 112.26 39.60 58.03 3 65.38 0.17 6 99.72 20 420 0.11 49.76 0.13 0.83 0.555 0.03 56.735 0.33 0.58 16 400 1.93 8 116.98 27 333 0.66 0.78 Note that Sin cannot be > 1 From the ratios in column 6 we conclude that Using 2 d Sin o a 4.3 0.22 5 82.38 0.43 12 222 1.04 A Al FCC 1.87 24 422 0.33 3 Note (h 2 k 2 l 2 ) sin 2 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.26 0.47 68.69 19 331 0.11 3 111 2.ppt .91 9 137.85 0.57 0.29 8 220 1.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 38.78 81.54 0.40 11 311 1.93 0.52 19.235 0.34 2 44.76 22.89 0.99 0. 41 19.70 8.134 1 2 25 0.73 11.Solved example 2 Another example Given the positions of the Bragg peaks we find the lattice type Dividing Sin2 by 0.98 12 6 58 0.19 11 5 47 0.33 3.23 19 3 FCC .10 8 4 45 0.99 4 3 37 0.37 16.719 5.859 6.500 3.366 0.927 0.5 0.178 1.362 2.535 4 11.134/3 = 0.044667 Whole number ratios 2→ Sin Sin2 Ratios of Sin2 1 21.848 0.731 0.422 0.10 16 7 68 0.60 0.707 0. Comparison of diffraction patterns of SC. BCC & B2 structures Click here More Solved Examples on XRD Click here . 38 44.47 1.22 1.99 81.91 0.17 1.57 65.56 99.05 - - 0.90 0.23 - - 0.What happens when we increase or decrease ? Funda Check We had pointed out that ~ a is preferred for diffraction.00 4.68 0.68 7.04 2.34 2.85 58.13 0.49 - - 0.91 0.82 1.02 1.16 0.14 0.05 3.15 6.01 78.47 1.63 39.26 1.45 47.84 5.01 0.45 4.45 6.22 2.28 - - 0.60 1.78 = 1.41 2.04 2.89 163.06 3.74 0.02 112.1 Å Sin() 2 79.93 0.11 1.92 - - 0.38 =3Å Sin() 38.03 1.66 2 2 = 0.81 - - 0. (This is only a thought experiment!!) Aluminium hkl d 111 200 220 311 222 400 331 420 422 333 2.76 49.05 3.08 6.03 1.54 32.64 39.05 2.24 82.76 0.26 0.61 - - 0.28 0.93 68.83 0.52 0.06 3.30 116.74 137.70 41. Let us see what happens if we ‘drastically’ increase or decrease .74 0.54 Å If we ~double → we get too few peaks If we make small→ all the peaks get crowded to small angles And the detector may not be able to resolve these peaks if they come too close! .64 95.54 Å Sin() 0.03 2.02 1.78 1.30 0.87 22.43 1.35 With CuK = 1.35 4.83 56.65 - - 0.33 19. 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 . Intensity → 0 Crystal Schematic of difference between the diffraction patterns of various phases Sharp peaks 90 180 Diffraction angle (2) → Intensity → Monoatomic gas No peak Intensity → Diffraction angle (2) → Liquid / Amorphous solid 90 0 Diffuse Peak 0 90 180 Diffraction angle (2) → 180 . Kallol Mondal. IITK . MSE.Actual diffraction pattern from an amorphous solid Diffuse peak from Cu-Zr-Ni-Al-Si Metallic glass Note 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. These peaks with small „d‟ occur at high angles in diffraction pattern. l increases. What is the minimum spacing between planes possible in a crystal? How many diffraction peaks can we get from a powder pattern? Funda Check Let us consider a cubic crystal (without loss in generality) Cubic crystal d hkl a h2 k 2 l 2 As h. Planes with „d‟ < /2 are not captured in the diffraction pattern. „d‟ decreases we could have planes with infinitesimal spacing d13 a 10 d 34 a a 25 5 With increasing indices the interplanar spacing decreases d10 a d12 5 a a 1 d11 a 2 The number of peaks we obtain in a powder diffraction pattern depends on the wavelength of x-ray we are using.k. .