X-RAY DIFFRACTION X- Ray Sources Diffraction: Bragg s Law Crystal Structure Determination Part of MATERIALS SCIENCE & ENGINEERING A Learner s Guide AN INTRODUCTORY E-BOOK Anandh Subramaniam & Kantesh Balani Materials Science and Engineering (MSE) Indian Institute of Technology, Kanpur- 08016 Email: anandh@iitk.ac.in, 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 (001) 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 ~ -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 (a Cu ) = 3.61 Å d hkl is equal to a Cu or less than that (e.g. d 111 = a Cu /3 =.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
Mo Target impacted by electrons accelerated by a 35 kv potential shows the emission spectrum as in the figure below (schematic) K Intense peak, nearly monochromatic X-ray sources with different for doing XRD studies Intensity White radiation K Characteristic radiation due to energy transitions in the atom Target Metal Of K radiation (Å) Mo 0.71 Cu 1.54 Co 1.79 Fe 1.94 Cr.9 0. 0.6 1.0 1.4 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
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: Coherent, monochromatic, parallel waves (with wavelength ). Crystalline array of scatterers* with spacing of the order of (~). Fraunhofer diffraction geometry Aspects related to the wave Coherent, monochromatic, parallel 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. * A quasicrystalline array will also lead to diffraction with sharp peaks (which we shall not consider in this text). ** Amorphous material will give diffuse peak.
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. Incoming X-rays Secondary emission Schematics Oscillating charge re-radiates In phase with the incoming x-rays Sets Electron cloud into oscillation Sets nucleus into oscillation Small effect neglected
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. Incident and scattered waves are in phase if In plane scattering is in phase Scattering from across planes is in phase
Let us consider in-plane scattering A B Atomic Planes X Y 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 These can be in phase if incident = scattered A B But this is still reinforced scattering and NOT reflection X Y
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- 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- travels an extra path as compared to Ray-1 (= ABC). The path difference between Ray-1 and Ray- = ABC = (d Sin + d Sin) = (d.sin). For constructive interference, this path difference should be an integral multiple of : n = d Sin the Bragg s equation. (More about this sooner). The path difference between Ray-1 and Ray-3 is = (d.sin) = n = n. This implies that if Ray-1 and Ray- constructively interfere Ray-1 and Ray-3 will also constructively interfere. (And so forth).
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- 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 Occurs from surface Takes place at any angle Diffraction Occurs throughout the bulk 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
Understanding the Bragg s equation n = d Sin The equation is written better with some descriptive subscripts: n d Sin Cu K hkl hkl 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). 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 / maximum value of is 90, Sin is 1 from Bragg equation d = /.
Order of the reflection (n) For Cu K radiation ( = 1.54 Å) and d 110 =. Å n Sin = n/d 1 0.34 0.7º First order reflection from (110) 110 0.69 43.9º Second order reflection from (110) planes 110 Also considered as first order reflection from (0) planes 0 Relation between d nh nk nl and d hkl d d d Cubic crystal hkl nhnk nl nhnk nl a h k l a ( nh) ( nk) ( nl) a dhkl n h k l n e.g. d d 0 110 a 8 a d d 0 110 1
In XRD n th order reflection from (h k l) is considered as 1 st order reflection from (nh nk nl) n d sin hkl d hkl sin n d nhnk nl d hkl 1 n d sin nh nk nl Hence, (100) planes are a subset of (00) planes d d 00 100 1 d d 300 100 1 3 Important point to note: In a simple cubic crystal, 100, 00, 300 are all allowed reflections. But, there are no atoms in the planes lying within the unit cell! Though, first order reflection from 00 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).
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.
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 d hkl ). 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).
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 = d 100 ). d100 Sin100 Sin 100 1.54 0.44 100 14.149 d100 (3.15) 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. 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 /).
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 (00) planes is observed 1.54 nd order 1 Sin 100 0.488 d 3.15 100 nd ~ 9.67 order 100 00 This is because if the green rays have a path difference of then the red ray will have path difference of which will still lead to constructive interference!
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.
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. The intensities of the peaks depend on many factors as considered here. Scattering by a crystal can be understood in three steps A Electron Polarization factor B Atom 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 ) scattering from an atom 3) scattering from a unit cell Click here to know the details 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; but is dependent on the position of the atoms/ions etc. within the cell C Unit cell (uc) Structure factor (F) Click here to know more about Structure factor calculations & Intensity in powder patterns
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
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 SC BCC FCC DC Allowed Reflections All (h + k + l) even h, k and l unmixed 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 Cannot be expressed as (h +k +l ) * lattice decorated with monoatomic/monoionic motif h + k + l SC FCC BCC DC 1 100 110 110 3 111 111 111 4 00 00 00 5 10 6 11 11 7 8 0 0 0 0 9 300, 1 10 310 310 11 311 311 311 1 13 30 14 31 31 15 16 400 400 400 400 17 410, 3 18 411, 330 411, 330 19 331 331 331
The ratio of (h + k + l ) derived from extinction rules (previous page) As we shall see soon the ratios of (h + k + l ) is proportional to Sin which can be used in the determination of the lattice type SC 1 3 4 5 6 8 BCC 1 3 4 5 6 7 FCC 3 4 8 11 1 DC 3 8 11 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!
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. Monochromatic X-rays Many s (orientations) Powder specimen POWDER METHOD λ fixed θ variable Panchromatic X-rays Single LAUE TECHNIQUE λ variable θ fixed Monochromatic X-rays Varied by rotation 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 POWDER METHOD sin ) ( l k h sin 4 ) ( a l k h ) ( 4sin l k h a l k h a d Cubic d Sin sin 4 l k h a Cubic crystal 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. (1) () () in (1) * In reality this is true only to an extent
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) Different cones for different reflections
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]. 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 h hkl d Sin() 1 100 4.00 0.19 11.10 110.83 0.7 15.80 3 111.31 0.33 19.48 4 00.00 0.39.64 5 10 1.79 0.43 5.50 6 11 1.63 0.47 8.13 8 0 1.41 0.54 3.99 9 300 1.33 0.58 35.7 10 310 1.6 0.61 37.50
Determination of Crystal Structure from versus Intensity Data in Powder Method In the power diffraction method a 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 (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 Radiation: Cu K, = 1.54 Å Increasing d Increasing
Determination of Crystal Structure from versus Intensity Data The following table is made from the 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 Intensity Sin Sin 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 & 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)
400 311 0 00 111 Powder diffraction pattern from Al Radiation: Cu K, = 1.54 Å Note: Peaks or not idealized peaks broadened Increasing splitting of peaks with g Peaks are all not of same intensity In low angle peaks K 1 & K peaks merged K 1 & K peaks resolved in high angle peaks (in and 400 peaks this can be seen)
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. 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. 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: [=dsin], (Sin) max will correspond to d min. (Sin) max =1. Hence, d min =/. Hence, if is small then planes with smaller d spacing (i.e. those which occur at higher values) will also show up in a XRD patter (powder pattern). Given that experimentally cannot be greater than 90. ) 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 as compared to the 001 peak). dsin Sin max d min d min
Determination of Crystal Structure (lattice type) from versus Intensity Data Let us assume that we have the versus intensity plot from a diffractometer To know the lattice type we need only the position of the peaks (as tabulated below) Solved example 1 # Sin Sin ratio Index d 1 38.5 19.6 0.33 0.11 3 111.34 44.76.38 0.38 0.14 4 00.03 3 65.14 3.57 0.54 0.9 8 0 1.43 4 78.6 39.13 0.63 0.40 11 311 1. 5 8.47 41.35 0.66 0.43 1 1.17 6 99.11 49.555 0.76 0.58 16 400 1.01 7 11.03 56.015 0.83 0.69 19 331 0.93 8 116.60 58.3 0.85 0.7 0 40 0.91 9 137.47 68.735 0.93 0.87 4 4 0.83 10 163.78 81.89 0.99 0.98 7 333 0.78 From the ratios in column 6 we conclude that FCC a Using d Sin 1.54 d111 Sin 111 0.33 3 o a 4.04A Al Note that Sin cannot be > 1 We can get the lattice parameter which correspond to that for Al Note ( h k l ) sin 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 Another example Given the positions of the Bragg peaks we find the lattice type Sin Sin Ratios of Sin Dividing Sin by 0.134/3 = 0.044667 Whole number ratios 1 1.5 0.366 0.134 1 3 5 0.4 0.178 1.33 3.99 4 3 37 0.60 0.36.70 8.10 8 4 45 0.707 0.500 3.73 11.19 11 5 47 0.731 0.535 4 11.98 1 6 58 0.848 0.719 5.37 16.10 16 7 68 0.97 0.859 6.41 19.3 19 FCC
Comparison of diffraction patterns of SC, BCC & B structures Click here More Solved Examples on XRD Click here
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!!) Aluminium = 1.54 Å = 3 Å = 0.1 Å hkl d Sin() Sin() Sin() 111.34 0.33 19.6 38.5 0.64 39.87 79.74 0.0 1..45 00.03 0.38.38 44.76 0.74 47.64 95.8 0.0 1.41.8 0 1.43 0.54 3.57 65.14 1.05 - - 0.03.00 4.01 311 1. 0.63 39.13 78.6 1.3 - - 0.04.35 4.70 1.17 0.66 41.4 8.47 1.8 - - 0.04.45 4.90 400 1.01 0.76 49.56 99.11 1.49 - - 0.05.84 5.68 331 0.93 0.83 56.0 11.03 1.61 - - 0.05 3.08 6.16 40 0.91 0.85 58.30 116.60 1.65 - - 0.05 3.15 6.30 4 0.83 0.93 68.74 137.47 1.81 - - 0.06 3.45 6.91 333 0.78 0.99 81.89 163.78 1.9 - - 0.06 3.68 7.35 With Cu K = 1.54 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!
Applications of XRD Bravais lattice determination Lattice parameter determination We have already seen these applications 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 Intensity 0 90 180 Intensity Crystal Schematic of difference between the diffraction patterns of various phases Sharp peaks 0 90 180 Diffraction angle () Monoatomic gas No peak Diffraction angle () Liquid / Amorphous solid 0 Diffraction angle () 90 180 Diffuse Peak
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. Kallol Mondal, MSE, IITK
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) d Cubic crystal hkl a h k l As h,k, l increases, d decreases we could have planes with infinitesimal spacing d 13 a 10 With increasing indices the interplanar spacing decreases a a d34 5 5 d 10 a 1 a The number of peaks we obtain in a powder diffraction pattern depends on the wavelength of x-ray we are using. Planes with d < / are not captured in the diffraction pattern. These peaks with small d occur at high angles in diffraction pattern. d 1 a 5 d 11 a