Indexing a Diffraction Pattern Indexing a diffraction pattern means assigning Miller indices hkl to each value of d If we know the unit cell, we can assign hkl 2dsin0=λ values to each d value using 1h2k212 Similarly. if we know hkl values, we can calculate the unit cell The split of the XRD lines However, often we don' t know hklor the unit 0o1100 he symmetry of structure decreased. the lines increased How Many Lines Are Possible? Observable diffraction n a cubic material, the largest d-spacing that can be observed is 100=010=001. For a primitive cell, we count h2+k2+P2 +k:+1 +上+ g e nd oes of M Danetng Planes tor acc and ke :1,2,3,4 1,5,6,8,9, Scumble -i Bcc: z4.6.8.10, 12. FCC:3,4,8,11,12,16,24 1 15 impossible Note we start with the largest d-spacing and work down Note: not all lines are present in every case 5i 3+1: 1 19. 3to 2d sin 0=n7. Ex: An element, BCC or FCC, shows diffraction peaks at Systematic Abser d centering Determine (a) Crystal structure?(b) Lattice constant? The presence of a centered lattice leads to the (e) What is the element. systematic absence of certain types of peak in For i centered lattices +k+l=2n for a line to be For an f centered lattice 43404721 h+k=2n. k+l=2n and h+I For a C centered lattice h+k=2 a=318A,BCC,→10 Indexing a diffraction pattern means assigning Miller indices hkl to each value of d If we know the unit cell, we can assign hkl values to each d value using: Similarly, if we know hkl values, we can calculate the unit cell However, often we don’t know hkl or the unit cell… . 2 2 2 2 2 2 2 c l b k a h d 1 = + + Indexing a Diffraction Pattern 2dsinq = l The split of the XRD lines: The symmetry of structure decreased, the lines increased. h k l h2 + k 2 + l 2 h k l h2 + k2 + l 2 1 0 0 1 2 2 1, 3 0 0 9 1 1 0 2 3 1 0 10 1 1 1 3 3 1 1 11 2 0 0 4 2 2 2 12 2 1 0 5 3 2 0 13 2 1 1 6 3 2 1 14 2 2 0 8 4 0 0 16 In a cubic material, the largest d-spacing that can be observed is 100=010=001. For a primitive cell, we count according to h2+k2+l2 Note: 7 and 15 impossible Note: we start with the largest d-spacing and work down Note: not all lines are present in every case How Many Lines Are Possible? Observable diffraction peaks 2 2 2 h + k + l Ratio Simple cubic SC: 1,2,3,4,5,6,8,9,10,11,12.. BCC: 2,4,6,8,10, 12… . FCC: 3,4,8,11,12,16,24…. 2 2 2 hkl h k l a d + + = 2dsin q = nl Systematic Absences and Centering The presence of a centered lattice leads to the systematic absence of certain types of peak in the diffraction pattern For I centered lattices: h + k + l = 2n for a line to be present For an F centered lattice: h + k =2n, k + l = 2n and h + l = 2n For a C centered lattice: h + k =2n Ex: An element, BCC or FCC, shows diffraction peaks at 2q: 40, 58, 73, 86.8,100.4 and 114.7. Determine:(a) Crystal structure? (b) Lattice constant? (c) What is the element? 114.7 57.35 0.7090 6 (222) 100.4 50.2 0.5903 5 (310) 86.8 43.4 0.4721 4 (220) 73 36.5 0.3538 3 (211) 58 29 0.235 2 (200) 40 20 0.117 1 (110) 2theta theta (hkl) q 2 sin 2 2 2 h + k + l a =3.18 Å , BCC, ‡ W