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Determination of Crystal Structure (From Chapter 10 of Textbook 2)

Determination of Crystal Structure (From Chapter 10 of Textbook 2). Unit cell  line positions Atom position  line intensity (known chemistry) Three steps to determine an unknown structure: (1) Angular position of diffracted lines  shape and size of the unit cell.

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Determination of Crystal Structure (From Chapter 10 of Textbook 2)

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  1. Determination of Crystal Structure (From Chapter 10 of Textbook 2) Unit cell  line positions Atom position  line intensity (known chemistry) Three steps to determine an unknown structure: (1) Angular position of diffracted lines  shape and size of the unit cell. (2) sizes of unit cell, chemical composition, density  # atoms/unit cell (3) Relative intensities of the peaks  positions of the atoms within the unit cell

  2. Preliminary treatment of data:  Ensure true random orientation of the particles of the sample  Remove extraneous lines from (1) K or other wavelength: In the analyzing step, the sin2 is used  For most radiations  filament contamination  W KLradiation  Diffraction by other substance  Calibration curve using known crystal

  3. Effect of sample height displacement + R s: sample height displacement R: diffractometer radius   s Length of a larger error for low angle peaks  for the most accurate unit cell parameters it is generally better to use the high angle peaks for this calculation.

  4. Example for sample height displacement Assume a crystal; cubic structure; a = 0.6 nm. Consider the error that can be introduced if the sample was displaced by 100 microns (0.1 mm) for (100) and (400) diffraction peaks? Assume λ = 0.154 nm and R = 225 mm. The displacement cause these peaks to shift by (100):  = 7.395o (400):  = 30.892o

  5. Pattern Indexing →assign hkl values to each peak Simplest example: indexing cubic pattern Constant for a given crystal Define

  6. Values for h2 + k2 + l2 for cubic system SC: 1, 2, 3, 4, 5, 6, 8, … BCC: 2, 4, 6, 8, 10, 12, 14, .. FCC: 3, 4, 8, 11, 12, 16, 19, 20, … Diamond: 3, 8, 11, 16, 19, …

  7. s is doubled in BCC, No s = 7 in SC SC: 1, 2, 3, 4, 5, 6, 8, … BCC: 2, 4, 6, 8, 10, 12, 14, ..

  8. Indexing Tetragonal system When l = 0 (hk0 lines), Possible values for h2 + k2: 1, 2, 4, 5, 8, 9, 10, … Possible values for l2 are 1, 4, 9, 16, …

  9. Indexing Hexagonal system Possible values for h2 + hk + k2 are 1, 3, 4, 7, 9, 12, … The rest of lines are Possible values for l2 are 1, 4, 9, 16, …

  10. Example: sin2 sin2/3 sin2/4 sin2/7 0.097 0.032 0.112 0.037 0.136 0.045 0.209 0.070 0.332 0.111 0.390 0.130 0.434 0.145 0.472 0.157 0.547 0.182 0.668 0.223 0.722 0.241 1 2 3 4 5 6 7 8 9 10 11 0.024 0.028 0.034 0.052 0.083 0.098 0.109 0.118 0.137 0.167 0.180 0.014 0.016 0.019 0.030 0.047 0.056 0.062 0.067 0.078 0.095 0.103 100 110 Let’s say A = 0.112

  11. sin2 sin2-A sin2-3A 0.097 belongs to Cl2. What is the l? There are two lines between 100 and 110. Probably, 10l1 and 10l2  0.024 and 0.097 are different ls. 0.097/0.024 ~ 4 0.220/0.024 ~ 9 0.390/0.024 ~ 16  C = 0.02441 002 0 0.024 0.097 0.220 0.278 0.322 0.360 0.435 0.556 0.610 0.694 0.767 0.054 0.098 0.136 0.211 0.332 0.386 0.470 0.543 0.097 0.112 0.136 0.209 0.332 0.390 0.434 0.472 0.547 0.668 0.722 0.806 0.879 1 2 3 4 5 6 7 8 9 10 11 12 13 100 101 102 ,103 110 004 112

  12. Indexing orthorhombic system: More difficult! Consider any two lines having indices hk0 and hkl  Cl2 put it back get A and B.  guess right (consistent)  not right, try another guesses C Indexing Monoclinic and Triclinic system Even more complex, 6 variables must have enough diffraction lines for the computer to indexing.

  13. Effect of Cell distortion on the powder Pattern

  14. Autoindexing http://www.ccp14.ac.uk/solution/indexing/index.html

  15. Determination of the number of atoms in a unit cell Indexing the power pattern  shape and size of unit cell (volume) number of atoms in that unit cell. VC: unit cell volume; : density N0: Avogodro’s number; M: molecular weight; n: number of molecules in a unit cell in Å3 in g/cm3

  16. Determination of Atom positions Relative intensities determine atomic positions (a trial and error process) (unvnwn): position of nth atom in a unit cell. Trial and error: known composition, known number of molecules, known structure  eliminate some trial structure. Space groups and Patterson Function (selection of trial structures)

  17. Example: CdTe • Chemical analysis which revealed: • 49.8 atomic percent as Cd (46.6 weight percent) • 50.2 atomic percent as Te (53.4 weight percent) * Make powder diffraction and list sin2: index the pattern! Assume it is cubic

  18. SC s BCC s FCC s Diamond s sin2 close 0.0462 0.1194 0.1615 0.179 0.234 0.275 0.346 0.391 0.461 0.504 0.575 0.616 0.688 0.729 0.799 1 2 3 4 5 6 8 9 10 11 12 13 14 16 17 0.0462 0.0597 0.05383 0.04475 0.0468 0.04583 0.04325 0.04344 0.0461 0.04582 0.04792 0.04738 0.04914 0.04556 0.047 2 4 6 8 10 12 14 16 18 20 22 24 26 30 32 0.0231 0.02985 0.02692 0.02237 0.0234 0.02292 0.02471 0.02444 0.02561 0.0252 0.02614 0.02567 0.02646 0.0243 0.02497 3 4 8 11 12 16 18 20 24 27 32 35 36 40 43 0.0154 0.02985 0.02019 0.01627 0.0195 0.01719 0.01922 0.01955 0.01921 0.01867 0.01797 0.0176 0.01911 0.01822 0.01858 3 8 11 16 19 24 27 32 35 40 43 49 51 56 59 0.0154 0.01493 0.01468 0.01119 0.01232 0.01146 0.01281 0.01222 0.01317 0.0126 0.01337 0.01257 0.01349 0.01302 0.01354 Very weak

  19. Diamond structure and Zinc blend structure: forbidden peaks! Which one has more peaks?  removing line 4 first 0.0462 0.1194 0.1615 0.234 0.275 0.346 0.391 0.461 0.504 0.575 0.616 0.688 0.729 0.799 0.84 3 8 11 16 19 24 27 32 35 40 43 49 51 56 59 0.0154 0.01493 0.01468 0.01463 0.01447 0.01442 0.01448 0.01441 0.0144 0.01437 0.01433 0.01404 0.01429 0.01427 0.01424 very close

  20. Alternative: Assume the first line is from s = 1, s = 2 and s = 3, … 3.000 7.779 10.487 11.623 15.195 17.857 22.468 25.390 29.935 32.727 37.338 40.000 44.675 47.338 51.883 1.000 2.593 3.496 3.874 5.065 5.952 7.489 8.463 9.978 10.909 12.446 13.333 14.892 15.779 17.294 1 2 3 4 5 6 8 9 10 11 12 13 14 16 17 2.000 5.189 6.991 7.749 10.130 11.905 14.978 16.926 19.957 21.818 24.892 26.667 29.784 31.558 34.589 2 4 6 8 10 12 14 16 18 20 22 24 26 30 32 3 4 8 11 12 16 18 20 24 27 32 35 36 40 43 3 8 11 16 19 24 27 32 35 40 43 49 51 56 59 0.0462 0.1198 0.1615 0.179 0.234 0.275 0.346 0.391 0.461 0.504 0.575 0.616 0.688 0.729 0.799

  21. Larger error smaller , use the first three lines to fit a more correct A.  0.01444, Use this number to divide sin2 ! 3.235 8.389 11.310 12.535 16.387 19.258 24.230 27.381 32.283 35.294 40.266 43.137 48.179 51.050 55.952 3.199 8.296 11.184 12.396 16.205 19.044 23.961 27.078 31.925 34.903 39.820 42.659 47.645 50.485 55.332 3 8 11 16 19 24 27 32 35 40 43 49 51 56 59 0.0462 0.1198 0.1615 0.179 0.234 0.275 0.346 0.391 0.461 0.504 0.575 0.616 0.688 0.729 0.799 . . . Larger Deviation New A is 0.01428

  22. 0.1790/0.01413 = 12.66; 222 s = 12 (forbidden diffraction lines for diamond structure, OK for zinc-blend structure!) *  =5.82 g/cm3 then M for CdTe is 112.4+127.6 = 240  n = 3.97 ~ 4 There are 4 Cd and 4 Te atoms in a unit cell. * FCC based structure with 4 molecules in a unit cell – NaCl and ZnS; CdTe: NaCl or ZnS

  23. NaCl –Cd at 000 & Te at ½½½ + fcc translations FCC when h, k, l all even when h, k, l all odd ZnS –Cd at 000 & Te at ¼ ¼ ¼ + fcc translations Unmixed hkl only when h + k + l is odd when h +k + l = odd multiple of 2 = even multiple of 2

  24. 0.0 0.2 0.4 0.6 0.8 1.0 1.2 48 37.7 27.5 21.8 17.6 14.3 12.0 Cd Te 52 41.3 30.3 24.0 19.5 16.0 13.3 fCd + fTe = 100 and |fCdfTe| = 4 at sin / = 0 to fCd + fTe = 30.3 and |fCdfTe| = 1.7 at sin / ~ 1.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 625 482 426 415 381 318 379 Several hundreds NaCl –Cd at 000 & Te at ½½½ + fcc translations h, k, l all even: strong diffracted lines h, k, lall odd: week diffracted lines

  25. ZnS fit better!

  26. Order-Disorder Determination Substitutionalsolid solution  A, B elements  AB atoms’ arrangement order disorder temperature TC low high Example – Cu-Au system (AuCu3), TC = 390 oC Cu Au Cu-Au average ordered disordered

  27. Complete Disordered structure: the probability of each site being occupied: ¼ Au, ¾ Cu  simple FCC with fav disordered For mixed h, k, l Fhkl = 0 For unmixed h, k, l Fhkl = (fAu + 3fCu)

  28. Complete Ordered structure: 1 Au atom, at 000, three Cu atoms at ½ ½ 0, ½ 0 ½, 0 ½ ½. ordered For mixed h, k, l Fhkl = (fAu  fCu) Peaks show up For unmixed h, k, l Fhkl = (fAu + 3fCu)

  29. Define a long range order parameter S: rA: fraction of A sites occupied by the right atoms; FA: fraction of A atoms in the alloy complete order: rA = 1  S = 1; complete disorder: rA = FA  S = 0 0 S 1

  30. AuCu3 : order parameter S Average atomic factor for A-site A-site is the 000 equipoint rAu :fraction of Au atoms in 000 site  the average atomic form factor = fav

  31. Average atomic factor for B-site (1  rA): is the fraction of Au occupying the Bsite  in Bsite (1  rA)/3 Au and 1 (1rA)/3 Cu = (2 + rA)/3 Cu 

  32. The structure factor is For mixed h, k, l For unmixed h, k, l Intensity  |F|2  superlattice lines  S2

  33. Using different S definition What would you get? Homework!

  34. Intensity  weak diffuse background  If atoms A and B completely random in a solid solution  diffuse scattering k: a constant for any one composition f decreases as sin/ increases  ID  as sin/ Weak signal, very difficult to measure

  35. Example – Cu-Zn system (CuZn), TC = 460 oC fZn fav fCu disordered ordered

  36. * Completely random: a BCC structure For h + k + l even  Fhkl = fCu + fZn For h + k + l odd  Fhkl = 0 * Completely order: a CsCl For h + k + l even  Fhkl = fCu + fZn For h + k + l odd  Fhkl = fCu  fZn

  37. Define a long range order parameter S: (practice yourself) For h + k + l even  Fhkl = fCu + fZn For h + k + l odd  Fhkl = S(fCu  fZn)

  38. Different system 1 0 T TC

  39. Relative intensity from the superlatticeline and the fundamental line: * Case AuCu3: ignoring the multiplication factor and Lorentz-polarization factor, just look at the |F|2. Assume sin/ = 0  f = z Assume sin/ = 0.2 About 10%, can be measured without difficulty.

  40. * Case CuZn: the atomic number Z of Cu and Zn is close  atomic form factor is close!! Assume sin/ = 0  f = z Assume sin/ = 0.2 About 0.03%, very difficult to measure choose a proper wavelength to resolve the case!

  41. Resonance between the radiation and the K shell electrons  larger absorption  f Produce extra difference in fCufZn Using Zn K radiation. f for Cu is -3.6 and for Zn is -2.7  About 0.13%, possible to be detected.

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