# Indexing (cubic) powder patterns by 816Igq4

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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
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
• 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

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