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Indexing (cubic) powder patterns Learning Outcomes By the end of this section you should: • know the reflection conditions for the different Bravais lattices • understand the reason for systematic absences • be able to index a simple cubic powder pattern and identify the lattice type • be able to outline the limitations of this technique! First, some revision Key equations/concepts: Miller indices Bragg’s Law 2dhkl sin = d-spacing equation for orthogonal crystals 1 h2 k 2 l 2 2 2 2 d hkl a 2 b c Now, go further 1 We can rewrite: d hkl h2 k 2 l 2 2 2 2 a b c a and for cubic this simplifies: d hkl h 2 k2 l2 a Now put it together with Bragg: 2 sin h 2 k2 l2 Finally h 2 k l 2 2 2a sin h 2 k2 l2 2a sin How many lines? Lowest angle means lowest (h2 + k2 + l2). hkl are all integers, so lowest value is 1 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 Quick question – why does 100=010=001 in cubic systems? How many lines? hkl h2 + k 2 + l 2 hkl h2 + k2 + l2 100 1 2 2 1, 3 0 0 9 110 2 310 10 111 3 311 11 200 4 222 12 210 5 320 13 211 6 321 14 220 8 400 16 Note: 7 and 15 impossible Note: we start with the largest d-spacing and work down Largest d-spacing = smallest 2 This is for PRIMITIVE only. Some consequences • Note: not all lines are present in every case so beware • What are the limiting (h2 + k2 + l2) values of the last reflection? h 2 k2 l2 2a sin or 4a 2 h 2 k 2 l 2 2 sin 2 sin2 has a limiting value of 1, so for this limit: 4a 2 h 2 k 2 l2 2 Wavelength SPINEL: LAMBDA = 1 .5 4 4a 2 Lamb d a: 1 .5 4 1 78 Mag n if: 1 .0 FW HM : 0. 20 0 Space g rp: F d -3 m:2 Direct cell: 8 . 08 0 0 8 .0 80 0 8 .0 8 0 0 9 0 .0 0 9 0. 00 9 0 .0 0 h 2 k 2 l2 2 113 = 1.54 Å 004 This is obviously 111 115 022 wavelength dependent 224 333 222 133 Hence in principle using a 10 20 30 40 50 60 smaller wavelength will SPINEL: LAMBDA = 1 .2 2 Lamb d a: 1 .2 2 0 00 Mag n if: 1 .0 FW HM : 0. 20 0 Space g rp: F d -3 m:2 Direct cell: 8 . 08 0 0 8 .0 80 0 8 .0 8 0 0 9 0 .0 0 9 0. 00 9 0 .0 0 access higher hkl values 113 = 1.22 Å 044 004 115 111 022 224 333 335 135 026 222 133 244 10 20 30 40 50 60 Indexing Powder Patterns • Indexing a powder pattern means correctly assigning the Miller index (hkl) to the peak in the pattern. • If we know the unit cell parameters, then it is easy to do this, even by hand. high QUARTZ Lambda: 1.54178 Magnif: 1.0 FWHM: 0.200 Space grp: P 62 2 2 Direct cell: 5.0800 5.0800 5.5807 90.00 90.00 120.00 011 1 h2 k 2 l 2 2 2 2 d hkl a 2 b c 010 0 0 31 1 2 022 3 121 021 110 020 01 012 120 111 10 20 30 40 50 60 Indexing Powder Patterns • The reverse process, i.e. finding the unit cell from the powder pattern, is not trivial. • It could seem straightforward – i.e. the first peak must be (100), etc., but there are other factors to consider • Let’s take an example: The unit cell of copper is 3.613 Å. What is the Bragg angle for the (100) reflection with Cu K radiation ( = 1.5418 Å)? Question sin 1 = 12.32o, so 2 = 24.64o BUT…. 2dhkl Copper, [W. L. Bragg (Philosophical Magazine, Serie 6 (1914) 28, 255-360] Lambda: 1.54180 M agnif: 1.0 FWHM: 0.200 Space grp: F m -3 m Direct cell: 3.6130 3.6130 3.6130 90.00 90.00 90.00 10 20 30 40 50 60 70 80 Systematic Absences • Due to symmetry, certain reflections cancel each other out. • These are non-random – hence “systematic absences” • For each Bravais lattice, there are thus rules for allowed reflections: P: no restrictions (all allowed) I: h+k+l =2n allowed F: h,k,l all odd or all even Reflection Conditions So for each Bravais lattice: PRIMITIVE BODY FACE h2 + k2 + l2 All possible h+k+l=2n h,k,l all odd/even 1 100 2 110 110 3 111 111 4 200 200 200 5 210 6 211 211 8 220 220 220 9 2 2 1, 3 0 0 10 310 310 11 311 311 12 222 222 222 13 320 14 321 321 16 400 400 400 General rule Characteristic of every cubic pattern is that all 1/d2 values have a common factor. 1 h2 k 2 l 2 2 d a2 The highest common factor is equivalent to 1/d2 when (hkl) = (100) and hence = 1/a2. The multiple (m) of the hcf = (h2 + k2 + l2) We can see how this works with an example Indexing example 2 2 d (Å) 1/d m hkl = 1.5418 Å 21.76 4.08 0.06 3 111 Highest common factor = 0.02 25.20 3.53 0.08 4 200 So 0.02 = 1/a2 35.88 2.50 0.16 8 220 11 311 a = 7.07Å 42.38 2.13 0.22 12 222 44.35 2.04 0.24 Lattice type? 51.57 1.77 0.32 16 400 (h k l) all odd or all even F-centred Try another… 2 m Highest common d (Å) 1/d hkl factor = 3.892 2.752 So a= Å 2.247 1.946 1.741 Lattice type? 1.589 1.376 1.297 In real life, the numbers are rarely so “nice”! …and another 2 m Highest common d (Å) 1/d hkl factor = 3.953 2.795 So a= Å 2.282 1.976 1.768 Lattice type? 1.614 1.494 1.398 Watch out! You may have to revise your hcf… So if the numbers are “nasty”? Remember the expression we derived previously: h 2 k l 2 2 2a sin So a plot of (h2 +k2 + l2) against sin has slope 2a/ 0.5 y = 0.1089x Very quickly (with the 0.4 2 R =1 aid of a computer!) we sin theta 0.3 can try the different 0.2 options. Primitive 0.1 Body Centred (Example from above) Face-centred 0 0 1 2 3 Linear (Face- 5 4 centred) sqrt(h2+k2+l2) Caveat Indexer • Other symmetry elements can cause additional systematic absences in, e.g. (h00), (hk0) reflections. • Thus even for cubic symmetry indexing is not a trivial task • Have to beware of preferred orientation (see previous) • Often a major task requiring trial and error computer packages • Much easier with single crystal data – but still needs computer power! Ba2 La0.667 V2 O8 Lambda: 1.54178 Magnif: 1.0 FWHM: 0.200 Space grp: RB3M Direct cell: 5.7527 5.7527 21.0473 90.0 0 90.00 120 .00 012 202 113 110 003 009 006 104 101 018 107 015 021 10 20 30 40