X-RAY DIFFRACTION

1. MATERIALS SCIENCE & ENGINEERING Part of A Learner’s Guide AN INTRODUCTORY E-BOOK Anandh Subramaniam & Kantesh Balani Materials Science and Engineering (MSE) Indian Institute of Technology, Kanpur- 208016 Email:anandh@iitk.ac.in, URL:home.iitk.ac.in/~anandh http://home.iitk.ac.in/~anandh/E-book.htm X-RAY DIFFRACTION • 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: • http://www.matter.org.uk/diffraction/ • http://www.ngsir.netfirms.com/englishhtm/Diffraction.htm

2. 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…

3. 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)

4. 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

5. 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 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

7. 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 Click here to know more Transmitted beam X-rays can also be refracted (refractive index slightly less than 1) and reflected (at very small angles)

8. 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 Coherent, monochromatic, parallel wave Aspects related to the wave Diffraction pattern with sharp peaks Crystalline*,** Aspects related to the material 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.

9. 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

10. 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

11. 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

12. 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 scatteringand NOT reflection

13. 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-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).

14. 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 Note that they ‘almost’ constructively interfere!

15. 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). Note: X-rays can ALSO be reflected at very small angles of incidence

16. 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). • 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.

17. Order of the reflection (n) • For Cu K radiation ( = 1.54 Å) and d110= 2.22 Å Relation between dnh nk nl and dhkl e.g.

18. In XRD nth order reflection from (h k l) is considered as 1st order reflection from (nh nk nl) 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).

19. 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.

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

21. 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).

22. 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).

23. 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!

24. Important points • Presence of additional atoms/ions/molecules in the UC  at lattice points or as a part of the motifcan 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.

25. 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 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 cellClick 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 Structure factor calculations & Intensity in powder patterns Unit cell (uc) Structure factor (F) Click here to know more about

26. The concept of a Reciprocal latticeand 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. Reciprocal Lattice & Ewald Sphere construction Click here to know more about

27. 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

28. Allowed reflections in SC*, FCC*, BCC* & DC crystals Cannot be expressed as (h2+k2+l2) * lattice decorated with monoatomic/monoionic motif

29. 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 • 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!

30. 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  LAUETECHNIQUE Panchromatic X-rays ROTATINGCRYSTALMETHOD  Varied by rotation Monochromatic X-rays Only the powder method (which is commonly used in materials science) will be considered in this text.

31. 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 (2) (1) (2) in (1)   * In reality this is true only to an extent

32. 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

33. 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]. 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

34. 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 (for now). Iis 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 

35. 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- completed table shown later).

36. 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 • Nobrackets are used around the indexed numbers(the peaks correspond to planes in the real space)

37. 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 111 200 220 311 222 400 K1 & K2 peaks resolved in high angle peaks(in 222 and 400 peaks this can be seen) In low angle peaks K1 & K2 peaks merged

38. 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 broadenedCould be due to instrumental, residual ‘non-uniform’ strain (microstrain), grain size etc. broadening. Peaks could be shifted from their ideal positionsCould be due to uniform strain→ macrostrain. Relative intensities of the peaks could be alteredCould be due to texture in the sample. Funda Check Ans: 90 What is the maximum value of  possible (experimentally)? • 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.

39. 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).

40. 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) Note that Sin cannot be > 1 Note FCC From the ratios in column 6 we conclude that 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

41. Solved example 2 Another example Given the positions of the Bragg peaks we find the lattice type FCC

42. Comparison of diffraction patterns of SC, BCC & B2 structures Click here More Solved Examples on XRD Click here

43. What happens when we increase or decrease ? Funda Check 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!

44. 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 Next slide Determine if the material is amorphous or crystalline

45. Crystal Schematic of difference between the diffraction patterns of various phases Intensity→ Sharp peaks 0 90 180 Diffraction angle (2)→ Monoatomic gas No peak Intensity→ Diffraction angle (2)→ Diffraction angle (2)→ Liquid / Amorphous solid 180 Intensity→ 90 0 Diffuse Peak 0 90 180

46. 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

47. 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