Indexing cubic powder patterns

• systematic absences for Bravais lattices (Simple cubic, body centered cubic and face centered cubic)

• to index a simple cubic powder pattern and identify the lattice type

(Use the program given to you. Input can be either two theta or d values).

Bragg’s Law 2dhkl sin = n

(n is generally ignored. Why??)

d-spacing equation for orthogonal crystals – cubic, tetragonal or orthorhombic)

2

2

2

2

2

2

2

1

c

l

b

k

a

h

d hkl

2

2

2

2

2

2

1

c

l

b

k

a

hdhkl

222 lkh

adhkl

for cubic this simplifies as,

sin2222 lkh

aHence,

sin2222 a

lkh i.e.,

How many lines?

Bragg peak at lowest angle means, the indices will be the lowest i.e., (h2 + k2 + l2) will be minimum.

Remember, h,k,l are all integers, so the lowest value is 1.

For a cubic material, the largest d-spacing may be assigned the reflection, 100 or 010 or 001 (Multiplicity or equivalent reflections).

sin2222 a

lkh

This is true only for a primitive lattice (P). For BCC and FCC, the first Bragg peak will be 110 and 111.

How many lines?

h k l h2 + k2 + l2 h k l h2 + k2 + l2

1 0 0 1 2 2 1, 3 0 0 91 1 0 2 3 1 0 101 1 1 3 3 1 1 112 0 0 4 2 2 2 122 1 0 5 3 2 0 132 1 1 6 3 2 1 142 2 0 8 4 0 0 16

Note: 7 and 15 are not possible.Note: we start with the largest d-spacing and work down.Largest d-spacing = smallest 2This is for PRIMITIVE only.

Remember …..

• Not all reflections are present in every substance.• What are the limiting (h2 + k2 + l2) values or where do

you expect the last reflection?

22

2222 4

sina

lkh

sin2222 a

lkh or

sin2 has a limiting value of 1, so for this limit:

2

2222 4

alkh

Wavelength of the X-ray source

The number of observable reflections are wavelength dependent

(Relative intensities remain the same but the position changes)

A smaller wavelength will access higher hkl values

2

2222 4

alkh

1 1

1

0 2

2

1 1

3

2 2

2

0 0

4

1 3

3

2 2

4

1 1

53

3 3

10 20 30 40 50 60

SPINEL: LAMBDA = 1.54Lambda: 1.54178 Magnif: 1.0 FWHM: 0.200Space grp: F d -3 m:2 Direct cell: 8.0800 8.0800 8.0800 90.00 90.00 90.00

1 1

1

0 2

2

1 1

32

2 2

0 0

4

1 3

3 2 2

4

1 1

53

3 3

0 4

4

1 3

52

4 4

0 2

6 3 3

5

10 20 30 40 50 60

SPINEL: LAMBDA = 1.22Lambda: 1.22000 Magnif: 1.0 FWHM: 0.200Space grp: F d -3 m:2 Direct cell: 8.0800 8.0800 8.0800 90.00 90.00 90.00

= 1.54 Å

= 1.22 Å

Indexing Powder Patterns

• Indexing a powder pattern means correctly assigning the Miller index (hkl) to the peak in the pattern, i.e., to estimate the unit cell dimensions.

• If we know the unit cell parameters (known compound or known structure), You may generate ‘d’ values by using the program HKLGEN.

0 1

0

0 1

1

1 1

0

0 1

21

1 1

0 2

0

0 2

1

1 1

20

0 3

0 2

20

1 3

1 2

0

1 2

1

10 20 30 40 50 60

high QUARTZLambda: 1.54178 Magnif: 1.0 FWHM: 0.200Space grp: P 62 2 2 Direct cell: 5.0800 5.0800 5.5807 90.00 90.00 120.00

2

2

2

2

2

2

2

1

c

l

b

k

a

h

d hkl

Indexing Powder Patterns

• Note finding the unit cell from the powder pattern, is not trivial even for cubic systems.

The unit cell of copper is 3.613 Å. What is the Bragg angle for the lowest observable reflection with CuK radiation ( = 1.5418 Å)?

Question

10 20 30 40 50 60 70 80

Copper, [W. L. Bragg (Philosophical Magazine, Serie 6 (1914) 28, 255-360]Lambda: 1.54180 Magnif: 1.0 FWHM: 0.200Space grp: F m -3 m Direct cell: 3.6130 3.6130 3.6130 90.00 90.00 90.00

= 12.32o, so 2 = 24.64o BUT….

hkld2sin 1

Systematic Absences

• Due to symmetry, certain reflections cancel each other out (out of phase).

• These are non-random – hence “systematic absences”• For each Bravais lattice, there are thus rules for

allowed reflections:

P (primitive): no restrictions (all allowed)

I (Body centered): h+k+l =2n allowed

F (face centered): h,k,l all are odd or all even

Reflection Conditions (Extinctions)For each Cubic Bravais lattice:

  PRIMITIVE B.C.C F.C.C.h2 + k2 + l2 All possible h+k+l=2n h,k,l all odd/even

1 1 0 0    

2 1 1 0 1 1 0  

3 1 1 1   1 1 14 2 0 0 2 0 0 2 0 05 2 1 0    

6 2 1 1 2 1 1  

8 2 2 0 2 2 0 2 2 09 2 2 1, 3 0 0    

10 3 1 0 3 1 0  

11 3 1 1   3 1 112 2 2 2 2 2 2 2 2 213 3 2 0    

14 3 2 1 3 2 1  

16 4 0 0 4 0 0 4 0 0

General rule

Characteristic of every cubic pattern is that all 1/d2 values have a common factor.

2

222

2

1

a

lkh

d

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 d (Å) 1/d2

m h k l

21.76 4.08 0.06

25.20 3.53 0.08

35.88 2.50 0.16

42.38 2.13 0.22

44.35 2.04 0.24

51.57 1.77 0.32

= 1.5418 Å

3

4

8

11

12

16

1 1 1

2 0 0

2 2 0

3 1 1

2 2 2

4 0 0

Lattice type?

(h k l) all odd or all even

F-centred

Highest common factor = 0.02

So 0.02 = 1/a2

a = 7.07Å

Try another…

In real life, the numbers are rarely so “nice”!

d (Å) 1/d2

m h k l

3.892

2.752

2.247

1.946

1.741

1.589

1.376

1.297

Lattice type?

Highest common factor =

So a = Å

…and another

Watch out! You may have to revise your hcf…

d (Å) 1/d2

m h k l

3.953

2.795

2.282

1.976

1.768

1.614

1.494

1.398

Lattice type?

Highest common factor =

So a = Å

Remember, the error is more when you calculate cell dimension with larger ‘d’ values. Hence, unit cell dimensionas should be obtained only by curve fitting.

sin2222 a

lkh

So a plot of (h2 +k2 + l2) against sin has slope 2a/

y = 0.1089x

R2 = 1

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5sqrt(h2+k2+l2)

sin

th

eta

Primitive

Body Centred

Face-centred

Linear (Face-centred)

KCl has rock salt (NaCl) structure. Explain why the patterns are different. Notice the reflections (111) and (131) are barely visible in the case of KCl.

Shape of the crystals do not change the position of the peaks.