The document discusses crystal indexing and diffraction. It introduces Miller indices for labeling crystal planes, how to calculate d-spacings, and Bragg's law for deriving x-ray diffraction from crystal planes. The objectives are to understand plane identification in crystals, calculate Miller indices and d-spacings, and derive and use Bragg's law to solve diffraction problems. Examples are provided to illustrate key concepts and calculations.

1. Pertemuan 4CRYSTAL INDEXING IwanSugihartono, M.Si JurusanFisika, FMIPA UniversitasNegeri Jakarta 1

3. unit cells

4. symmetry

5. lattices

6. Diffraction

7. how and why - derivation

8. Some important crystal structures and properties

9. close packed structures

10. octahedral and tetrahedral holes

11. basic structures

13. know that planes are identified by their Miller Index and their separation, d

14. be able to calculate Miller Indices for planes

15. know and be able to use the d-spacing equation for orthogonal crystals

16. understand the concept of diffraction in crystals

17. be able to derive and use Bragg’s law06/01/2011 3 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

18. Lattice Planes and Miller Indices Imagine representing a crystal structure on a grid (lattice) which is a 3D array of points (lattice points). Can imagine dividing the grid into sets of “planes” in different orientations 06/01/2011 4 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

19. All planes in a set are identical The planes are “imaginary” The perpendicular distance between pairs of adjacent planes is the d-spacing Need to label planes to be able to identify them Find intercepts on a,b,c: 1/4, 2/3, 1/2 Take reciprocals 4, 3/2, 2 Multiply up to integers: (8 3 4) [if necessary] 06/01/2011 5 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

20. Exercise - What is the Miller index of the plane below? Find intercepts on a,b,c: Take reciprocals Multiply up to integers: 06/01/2011 6 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

21. General label is(h k l)which intersects at a/h, b/k, c/l (hkl) is the MILLER INDEX of that plane (round brackets, no commas). Plane perpendicular to y cuts at , 1,   (0 1 0) plane This diagonal cuts at 1, 1,   (1 1 0) plane NB an index 0 means that the plane is parallel to that axis 06/01/2011 7 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

22. Using the same set of axes draw the planes with the following Miller indices: (0 0 1) (1 1 1) 06/01/2011 8 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

23. Using the same set of axes draw the planes with the following Miller indices: (0 0 2) (2 2 2) NOW THINK!! What does this mean? 06/01/2011 9 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

24. Planes - conclusions 1 Miller indices define the orientation of the plane within the unit cell The Miller Index defines a set of planes parallel to one another (remember the unit cell is a subset of the “infinite” crystal (002) planes are parallel to (001) planes, and so on 06/01/2011 10 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

25. d-spacing formula For orthogonal crystal systems (i.e. ===90) :- For cubic crystals (special case of orthogonal) a=b=c :- e.g. for (1 0 0) d = a (2 0 0) d = a/2 (1 1 0) d = a/2 etc. 06/01/2011 11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

26. A cubic crystal has a=5.2 Å (=0.52nm). Calculate the d-spacing of the (1 1 0) plane A tetragonal crystal has a=4.7 Å, c=3.4 Å. Calculate the separation of the: (1 0 0) (0 0 1) (1 1 1) planes 06/01/2011 12 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

27. Question 2 in handout: If a = b = c = 8 Å, find d-spacings for planes with Miller indices (1 2 3) Calculate the d-spacings for the same planes in a crystal with unit cell a = b = 7 Å, c = 9 Å. Calculate the d-spacings for the same planes in a crystal with unit cell a = 7 Å, b = 8 Å, c = 9 Å. 06/01/2011 13 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

28. X-ray Diffraction 06/01/2011 14 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

29. Diffraction - an optical grating Path difference XY between diffracted beams 1 and 2: sin = XY/a  XY = a sin  For 1 and 2 to be in phase and give constructive interference, XY = , 2, 3, 4…..n so a sin  = n where n is the order of diffraction 06/01/2011 15 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

30. Consequences: maximum value of  for diffraction sin = 1  a =  Realistically, sin <1  a >  So separation must be same order as, but greater than, wavelength of light. Thus for diffraction from crystals: Interatomic distances 0.1 - 2 Å so  = 0.1 - 2 Å X-rays, electrons, neutrons suitable 06/01/2011 16 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

31. Diffraction from crystals X-ray Tube Detector ? 06/01/2011 17 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

32. Beam 2 lags beam 1 by XYZ = 2d sin  so 2d sin  = nBragg’s Law 06/01/2011 18 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

33. e.g. X-rays with wavelength 1.54Å are reflected from planes with d=1.2Å. Calculate the Bragg angle, , for constructive interference.  = 1.54 x 10-10 m, d = 1.2 x 10-10 m, =? n=1 :  = 39.9° n=2 : X (n/2d)>1 2d sin  = n We normally set n=1 and adjust Miller indices, to give 2dhkl sin  =  06/01/2011 19 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

34. Example of equivalence of the two forms of Bragg’s law: Calculate  for =1.54 Å, cubic crystal, a=5Å 2d sin  = n (1 0 0) reflection, d=5 Å n=1, =8.86o n=2, =17.93o n=3, =27.52o n=4, =38.02o n=5, =50.35o n=6, =67.52o no reflection for n7 (2 0 0) reflection, d=2.5 Å n=1, =17.93o n=2, =38.02o n=3, =67.52o no reflection for n4 06/01/2011 20 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

35. Use Bragg’s law and the d-spacing equation to solve a wide variety of problems 2d sin  = n or 2dhkl sin  =  06/01/2011 21 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

36. Combining Bragg and d-spacing equation X-rays with wavelength 1.54 Å are “reflected” from the (1 1 0) planes of a cubic crystal with unit cell a = 6 Å. Calculate the Bragg angle, , for all orders of reflection, n. d = 4.24 Å 06/01/2011 22 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

37. d = 4.24 Å n = 1 :  = 10.46° n = 2 :  = 21.30° n = 3 :  = 33.01° n = 4 :  = 46.59° n = 5 :  = 65.23° = (1 1 0) = (2 2 0) = (3 3 0) = (4 4 0) = (5 5 0) 2dhkl sin  =  06/01/2011 23 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

39. Each set of planes is uniquely identified by its Miller index (h k l)

40. We can calculate the separation, d, for each set of planes (h k l)

41. Crystals diffract radiation of a similar order of wavelength to the interatomicspacings

42. We model this diffraction by considering the “reflection” of radiation from planes - Bragg’s Law06/01/2011 24 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

43. THANK YOU 06/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 25