Linear and Cubic Box Splines for the Body Centered Cubic Lattice
Indexing cubic powder patterns systematic absences for Bravais lattices (Simple cubic, body centered...
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Transcript of Indexing cubic powder patterns systematic absences for Bravais lattices (Simple cubic, body centered...
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.