5 X-ray beam sourceBruker D8 ADVANCE uses an x-ray tube with a Cu anode as the primary x-ray beam source. In this component x-rays are generated when a focused electron beam accelerated across a high voltage field bombards a stationary solid Cu target. As electrons collide with atoms in the target and slow down, a continuous spectrum of x-rays is emitted, which is termed Bremsstrahlung radiation.The high energy electrons also eject inner shell electrons in atoms through the ionization process. When a free electron fills the shell, an x-ray photon with energy characteristic of the target material is emitted.Common targets used in x-ray tubes include Cu and Mo, that emit 8 keV and 14 keV x-rays with corresponding wavelengths of 1.54 Å and 0.8 Å, respectively.
6 Wavelengths for X-Ray source CopperAnodesBearden(1967)Holzer et al.(1997)CobaltCu Ka1ÅÅCo Ka1ÅÅCu Ka2ÅÅCo Ka2ÅÅCu KbÅÅCo KbÅÅMolybdenumChromiumMo Ka1ÅÅCr Ka1ÅÅMo Ka2ÅÅCr Ka2ÅÅMo KbÅÅCr KbÅÅOften quoted values from Cullity (1956) and Bearden, Rev. Mod. Phys. 39 (1967) are incorrect.Values from Bearden (1967) are reprinted in international Tables for X-Ray Crystallography and most XRD textbooks.Most recent values are from Hölzer et al. Phys. Rev. A 56 (1997)
7 d dSin The path difference between ray 1 and ray 2 = 2d Sin BRAGG’s EQUATIONDeviation = 2Ray 1Ray 2ddSinThe path difference between ray 1 and ray 2 = 2d SinFor constructive interference: n = 2d Sin
8 θ - 2θ ScanThe θ - 2θ scan maintains these angles with the sample, detector and X-ray sourceNormal to surfaceOnly planes of atoms that share this normal will be seen in the θ - 2θ ScanNanoLab/NSF NUE/Bumm
9 Occurs throughout the bulk Takes place at any angle Powder diffraction data can be collected using either transmission or reflection geometry, as shown below. Because the particles in the powder sample are randomly oriented, these two methods will yield the same dataReflectionDiffractionOccurs from surfaceOccurs throughout the bulkTakes place at any angleTakes place only at Bragg angles~100 % of the intensity may be reflectedSmall fraction of intensity is diffracted
10 Incident X-rays Fluorescent X-rays Electrons Scattered X-rays SPECIMENHeatFluorescent X-raysElectronsScattered X-raysCompton recoilPhotoelectronsCoherentFrom bound chargesIncoherent (Compton modified)From loosely bound chargesTransmitted beamX-rays can also be refracted (refractive index slightly less than 1) and reflected (at very small angles)Refraction of X-rays is neglected for now.
11 How does it work?In powder XRD method, a sample is ground to a powder (±10µm) in order to expose all possible orientations to the X-ray beam of the crystal values of , d and for diffraction are achieved as follows:1. is kept constant by using filtered X- radiation that isapproximately monochromatic.2. d may have value consistent with the crystal structure3. is the variable parameters, in terms of which thediffraction peaks are measured.
12 How does XRD Works???Every crystalline substance produce its own XRD pattern, which because it is dependent on the internal structure, is characteristic of that substance.The XRD pattern is often spoken as the “FINGERPRINT” of a mineral or a crystalline substance, because it differs from pattern of every other mineral or crystalline substances.
13 Basic Component Of XRD Machine Therefore any XRD machine will consist of three basic component.Monochromatic X-ray source ()Sample-holder (goniometer).Data collector- such as film, strip chart or magnetic medium/storage.By varying the angle , the Bragg’s Law conditions are satisfied by different d-spacing in polycrystalline materials. Plotting the angular positions and intensities of the resultant diffraction peaks produces a pattern which is characterised of the sample
14 X-ray ComponentsA typical X-ray instrument is built by combining high performance components such as X-ray tubes, X-ray optics, X-ray detectors, sample handling device etc. to meet the analytical requirements. A consequent modular design is the key to configure the best instrumentation..
15 Diffraction Pattern Collected Where A Ni Filter Is Used To Remove Kβ K alpha 1 and K alpha 2 overlap heavily at low angles and are easier to discriminate at high angles.Kb
16 Typical experimental data from Bruker XRD TiO22-thetaintensitas20405357381371376356370395373335397I101 Anatase110 Rutile2
18 SC Lattice = SC Reciprocal Crystal = SC Examples of 3D Reciprocal Lattices weighed in with scattering power (|F|2)SC001011101111Lattice = SC000010100110No missing reflectionsReciprocal Crystal = SCFigures NOT to Scale
19 BCC Lattice = BCC Reciprocal Crystal = FCC 002 022 202 222 011 101 020 000Lattice = BCC110200100 missing reflection (F = 0)220Reciprocal Crystal = FCCWeighing factor for each point “motif”Figures NOT to Scale
20 FCC Lattice = FCC Reciprocal Crystal = BCC 002 022 202 222 111 020 000 200220100 missing reflection (F = 0)110 missing reflection (F = 0)Weighing factor for each point “motif”Reciprocal Crystal = BCCFigures NOT to Scale
36 Rock Salt Why are peaks missing? The sample is made from Morton’s Salt JCPDF#111200220311222The sample is made from Morton’s SaltJCPDF# is supposed to fit it (Sodium Chloride Halite)
37 It’s a single crystal1112002203112222qThe (200) planes would diffract at °2q; however, they are not properly aligned to produce a diffraction peakThe (222) planes are parallel to the (111) planes.At °2q, Bragg’s law fulfilled for the (111) planes, producing a diffraction peak.
38 A random polycrystalline sample that contains thousands of crystallites should exhibit all possible diffraction peaks2002201112223112q2q2qFor every set of planes, there will be a small percentage of crystallites that are properly oriented to diffract (the plane perpendicular bisects the incident and diffracted beams).Basic assumptions of powder diffraction are that for every set of planes there is an equal number of crystallites that will diffract and that there is a statistically relevant number of crystallites, not just one or two.
39 Hint: Why are the intensities different? Which of these diffraction patterns comes from a nanocrystalline material?6667686970717273742q(deg.)Intensity (a.u.)Hint: Why are the intensities different?1o0.0015oThese two diffraction patterns come from the exact same sample (silicon).The apparent difference in peak broadening is due to the instrument optics, not due to specimen broadeningThese diffraction patterns were produced from the exact same sampleThe apparent peak broadening is due solely to the instrumentation0.0015° slits vs. 1° slits optical cofigurationsScan speed ( stepsize)
40 Crystallite Size Broadening Scherrer’s FormulaPeak Width B(2q) varies inversely with crystallite sizeThe constant of proportionality, K (the Scherrer constant) depends on the how the width is determined, the shape of the crystal, and the size distributionthe most common values for K are 0.94 (for FWHM of spherical crystals with cubic symmetry), 0.89 (for integral breadth of spherical crystals with cubic symmetry, and 1 (because 0.94 and 0.89 both round up to 1).K actually varies from 0.62 to 2.08For an excellent discussion of K, refer to JI Langford and AJC Wilson, “Scherrer after sixty years: A survey and some new results in the determination of crystallite size,” J. Appl. Cryst. 11 (1978) pRemember:Instrument contributions must be subtracted
41 Scherrer’s Formula t = thickness of crystallite / crystallite size K = constant dependent on crystallite shape (0.89)l = x-ray wavelengthB = FWHM (full width at half max) or integral breadthqB = Bragg Angle
42 Scherrer’s Formula What is B? B = (2θ High) – (2θ Low) B is the difference in angles at half maxPeak2θ low2θ highNoise
43 When to Use Scherrer’s Formula Crystallite size <1000 ÅPeak broadening by other factorsCauses of broadeningSizeStrainInstrumentIf breadth consistent for each peak then assured broadening due to crystallite sizeK depends on definition of t and BWithin 20%-30% accuracy at bestSherrer’s Formula ReferencesCorman, D. Scherrer’s Formula: Using XRD to Determine Average Diameter of Nanocrystals.
45 Scherrer’s Example = 0.89*1.54 Ǻ / ( 0.00174 * Cos (98.25/ 2 ) ) t = 0.89*λ / (B Cos θB) λ = 1.54 Ǻ= 0.89*1.54 Ǻ / ( * Cos (98.25/ 2 ) )= 1200 ǺB = ( )*π/180 =Simple Right!Target Metal Of K radiation (Å)Mo0.71Cu1.54Co1.79Fe1.94Cr2.29
46 Methods used to Define Peak Width 46.746.846.947.047.147.247.347.447.547.647.747.847.92q(deg.)Intensity (a.u.)Full Width at Half Maximum (FWHM)the width of the diffraction peak, in radians, at a height half-way between background and the peak maximumIntegral Breadththe total area under the peak divided by the peak heightthe width of a rectangle having the same area and the same height as the peakrequires very careful evaluation of the tails of the peak and the backgroundFWHM46.746.846.947.047.147.247.347.447.547.647.747.847.92q(deg.)Intensity (a.u.)
47 Remember, Crystallite Size is Different than Particle Size A particle may be made up of several different crystallitesCrystallite size often matches grain size, but there are exceptionsTEM images collected by Jane Howe at Oak Ridge National Laboratory.
48 Anistropic Size Broadening The broadening of a single diffraction peak is the product of the crystallite dimensions in the direction perpendicular to the planes that produced the diffraction peak.
49 Instrumental Peak Profile A large crystallite size, defect-free powder specimen will still produce diffraction peaks with a finite widthThe peak widths from the instrument peak profile are a convolution of:X-ray Source ProfileWavelength widths of Ka1 and Ka2 linesSize of the X-ray sourceSuperposition of Ka1 and Ka2 peaksGoniometer OpticsDivergence and Receiving Slit widthsImperfect focusingBeam sizePenetration into the sample47.047.247.447.647.82q(deg.)Intensity (a.u.)Patterns collected from the same sample with different instruments and configurations at MIT
51 What Instrument to Use?The instrumental profile determines the upper limit of crystallite size that can be evaluatedif the Instrumental peak width is much larger than the broadening due to crystallite size, then we cannot accurately determine crystallite sizeFor analyzing larger nanocrystallites, it is important to use the instrument with the smallest instrumental peak widthVery small nanocrystallites produce weak signalsthe specimen broadening will be significantly larger than the instrumental broadeningthe signal:noise ratio is more important than the instrumental profile
53 Comparison of Peak Widths at Crystallite Sizes FWHM (deg)100 nm0.09950 nm0.18210 nm0.8715 nm1.745Rigaku XRPD is better for very small nanocrystallites, <80 nm (upper limit 100 nm)PANalytical X’Pert Pro is better for larger nanocrystallites, <150 nm
54 Decreasecrystallite sizeA = anatase, R = rutile, B = brokite, (B)=TiO2(B)Wahyuningsih, S., 2009
55 Polycrystalline films on Silicon Why do the peaks broaden toward each other?Solid Solution InhomogeneityVariation in the composition of a solid solution can create a distribution of d-spacing for a crystallographic planeCeO219 nm45464748495051522q(deg.)Intensity (a.u.)ZrO246nmCexZr1-xO20<x<1
56 Many factors may contribute to the observed peak profile Instrumental Peak ProfileCrystallite SizeMicrostrainNon-uniform Lattice DistortionsFaultingDislocationsSolid Solution InhomogeneityThe peak profile is a convolution of the profiles from all of these contributions
57 Thank you for your attending! Workshop & Analysis Informations:Dr. Sayekti Wahyuningsih, M.SiDr. Yoventina Iriani, M.SiLaboratorium MIPA TerpaduFMIPA Universitas Sebelas MaretPhone / fax : (0271)