XRD Intensity Calculations -Example FCC Cu (centric)
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1 Ihkl F hkl F hkl hkl f f Cu Cu ( e (1 e XRD Intensity Calculations -Example FCC Cu (centric) Consider Copper which is F 3 with a=3.615å; atoms in positions [0,0,0] m m [½,½,0][½,0,½][0,½,½] and l=1.54å 1 cos e e e ) I F p [3] Know sin cos this k) i( hl ) i( k ) e e ) i (0) i ( h/ k / ) i ( h/ l / ) i ( k / l / ) i( h l S (hkl) F hkl I hkl f Cu 16f Cu f Cu 16f Cu f Cu 16f Cu All (hkl) reflections will have different intensities! Let s determine why. Agreement between calculated and observed intensities is satisfactory. Note how the p + LP values have a strong influence on line (hkl) intensity. Column 4: sin l 4a Column 9: F 4 where I=relative integrated intensity (relative area under the curve of intensity vs. ), F=structure factor, p= multiplicity factor, see handout appendix 13 (table); and =Bragg angle. The trig. terms in parentheses are the Lorentz-polarization (LP) factor (in degrees). This is an approximate solution, since other factors contribute slightly such as temperature and absorption factors. 16 f ( h Cu k Column 1: Eqn. [3] l Column 13: values from column 1 are recalculated to normalize the intensity to 10 Column 14: observed intensities (with visual scale: vs=very strong, s=strong, m=medium, w=weak) ) I hkl e ix cos x
2 XRD Powder Diffraction File (PDF) -Example: Cubic ZrO fluorite structure (centric) PDF# (RDB): QM=Blank(B); d=calculated; I=Calculated Zirconia, (Zirconium Oxide) ZrO Radiation=CuKa Lambda= Filter= Calibration= Theta= Ref: Bouvier, P., Djurado, E., Lucazeau, G., Le-Bihan, T. Phys. Rev. B: Condens. Matter, v6 p8731 (000) Cubic - Profile Analysis, Fm-3m (space group 5) CELL: x x <90.0 x 90.0 x 90.0> P.S.=cF1.00 Density(c)=6.889 Mwt=13. Ref: Ibid. Where 1/(d) = sin/l and n^ = S (=h^+k^+l^) -Theta d(å) I(f) ( h k l) Theta() 1/(d) pi/d n^ ( 1 1 1) ( 0 0) ( 0) (111) ( 3 1 1) ( ) allowed ( 4 0 0) ( 3 3 1) ( 4 0) ( 4 ) ( 5 1 1) ( 4 4 0) (110) ( 5 3 1) extinct ( 4 4 ) class6/
3 The effect of changes in symmetry on the multiplicity and peak intensity of diffraction peaks XRD Intensity Calculations -Example Rutile (TiO ) (centric) e ix cos x (next slide) class6/3
4 S. G. #136 e ix cos x class6/4
5 XRD Intensity Calculations -Example Rutile (TiO ) (centric) class6/5
6 Atomic Scattering Factors for TiO (Rutile ) atomic scattering factor (f) Ti +4 (f values from appendix 1) O - (f values from appendix 1) Ti +4 (f values interpolated for I 110 & I 001 ) O - (f values interpolated for I 110 & I 001 ) sin/l class6/6
7 (Previous H.W. Question): XRD Intensity Calculations -Example Zincblende (ZnTe) (non-centric) 1. For ZnTe (Zincblende structure), SG # 16: F43m Zn in 4a (0,0,0) + F = (0,½,½) (½,½,0) (½,0,½) Te in 4c (¼,¼,¼) + F = (¼,¾,¾) (¾,¼,¾) (¾,¾,¼) a. Draw the unit cell, a=6.106å b. Calculate the structure factor, F c. Determine the selection rules for hkl reflections. You should be able to determine the selection rules going from S=1 to S=16. Show all your work. d. Based on your answers in (b) and (c), which set of reflections will have the highest diffraction intensity and the lowest diffraction intensity? Do not calculate the Intensities, just look at your answers. You may want to draw the (hkl) planes to help you. e. Now calculate I 111 and I 00 and I 0 with CuKa (x-ray wavelength is 1.54Å). How do these values compare with your answer in (d). class6/7
8 XRD Intensity Calculations -Example Zincblende (ZnTe) (non-centric) Te atoms in ZnTe at (¼,¼,¼) (¼,¾,¾) (¾,¼,¾) (¾,¾,¼) (¾,¾,¾) (¾,¼,¼) (¼,¾,¼) (¼,¼,¾) Non-Equivalent points thus non-centric e ix cos x i sinx + class6/8
9 XRD Intensity Calculations -Example Zincblende (ZnTe) (non-centric) class6/9
10 XRD Intensity Calculations -Example Zincblende (ZnTe) (non-centric) class6/10
11 XRD Powder Diffraction File (PDF) ZnTe zincblende structure (non-centric) Intensity values calculated in (e) agree well with planar packing in (d) as well as the intensities from the PDF (on right) for ZnTe. class6/11
12 Simple Problems for XRD from Callister with Solutions given out Problem 1: For Aluminum compute the interplanar spacing for the (110) set of planes. Problem : Determine the expected diffraction angle for the 1 st order reflection from the (310) set of planes for BCC Chromium when monochromatic radiation of wavelength l= nm is used. Problem 3: The metal Rhodium is FCC crystal structure. If the angle of diffraction for the (311) plane occurs at 36.1º (first order reflection) when monochromatic radiation of wavelength nm is used, compute (a) the interplanar spacing for this set of planes and (b) the atomic radius for a rhodium atom. Problem 4: For which set of crystallographic planes will a first-order diffraction peak occur at a diffraction angle of 44.53º for FCC Nickel when monochromatic radiation having a wavelength of nm is used? Problem 5: Below shows first 5 peaks of the XRD pattern for Tungsten (BCC crystal structure). Monochromatic radiation of wavelength nm was used. (a) Index (h,k,l) for each of these peaks. (b) Determine the interplanar spacing for each of these peaks. (c) For each peak, determine the atomic radius for W and compare these with tabulated value of class6/1
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