solid structure

1. University of Hyderabad
School of Chemistry
CY551 : Chemistry of Materials
Part A
Basic Concepts
T. P. Radhakrishnan
This manual is provided as an aid for learning the subject by the students attending the lectures in
the above course; any typos/errors found may please be brought to the notice of the teacher.
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

2. 2
Section I: Solid State Structure
1. Types of solids 3
2. Order – spatial, dimensional 4
3. Symmetry in crystals 5
1. Translation symmetry 6
2. Crystal systems in 2-dimension 7
3. Crystal systems in 3-dimension 8
4. Bravais lattices in 2-dimension 9
5. Bravais lattices in 3-dimension 10
6. Lattice + Basis = Crystal structure 13
7. Miller planes 14
4. X-ray diffraction 17
1. Interference of waves 18
2. Bragg’s law for diffraction 18
3. Powder X-ray diffraction 19
4. Systematic absence 22
5. Structure factor 23
6. von Laue condition for X-ray diffraction 26
7. Reciprocal lattice 28
8. Ewald construction 31
9. Structure factor in terms of the reciprocal lattice vector 32
10. Basic concepts of X-ray structure solution and refinement 33
5. Common crystal structure motifs 34
6. Quasicrystals : A brief note 35

3. 3
I-1. Types of solids
 Solids can be classified based on several different criteria, for example,
o elemental composition
o nature of interactions between the constituent atoms/molecules
o properties or applications etc.
 Common classification based on the nature of interactions are (Fig. I-1):
o ionic
 NaCl, KCl, MgO
o Covalent
 Si, diamond
o Metallic
 Cu, Fe
o molecular (van der Waals)
 Ar, I2, C60
Figure I-1. Classes of solids based on the nature of interactions between constituent atoms/molecules

4. 4
I-2. Order – spatial and dimensional
 Order implies predictability of atom/molecule position and molecule orientation;
translational periodicity (see the schematic in Fig. I-2): if moving along a specific
direction, an atom/molecule is repeated at a distance r, then it will be found to repeat
along that direction at integral multiples of the same distance (2r, 3r, 4r ….)
 Order can be visualized at different levels:
o Spatial : depending on the extent in space (length scale) up to which the 3-
dimensional order persists :
 crystal (millimeters or above)
 microcrystal (micrometers)
 nanocrystal (nanometers)
 amorphous solid (none)
o Dimensional : depending on the number of dimensions in which long range order
persists :
 crystal (typically 3-dimension)
 liquid-crystal (usually 1- or 2-dimension)
 nematic, smectic, cholesteric etc.
 liquid (none, isotropic)
Figure I-2. Schematic showing translation symmetry in 2-dimension, along two non-orthogonal directions.

5. 5
I-3. Symmetry in crystals
 Recall point group symmetry operations
o identity (E)
o rotation (Cn)
o reflection (σ)
o rotation-reflection (Sn)
o inversion (i)
 In addition to the above symmetries, crystals possess translation symmetry
 Consider a 2-D square lattice; note that the lattice extends to infinity along x, y directions.
Fig. I-3 shows the lattice; the rotation (by 90o
) and translation (by vectors connecting the
nearest points, called unit cell vectors) operations leave the lattice invariant (unchanged);
the operations can be visualized by coloring some lattice points.
 Combination of rotation and translation (along the rotation axis) gives a new symmetry
operation, ‘screw rotation’ in crystals. Similarly, combination of reflection and
translation (along an axis parallel to the plane of reflection) gives ‘glide reflection’.
Figure I-3. (a) 2-D square lattice; (b) C4 rotation of the lattice; (c) unit vector (along x) translation of the lattice.
(a)
90o
(b)
(c)
Unit vector translation 

6. 6
I-3.1 Translation symmetry imposes restrictions on the possible rotation symmetries. This
can be geometrically illustrated as shown in Fig. I-4; a is the unit cell length (n has to be
an integer, positive, negative or zero, due to the translational symmetry) and θ is the
rotation angle allowed by symmetry. Table I-1 shows that rotations of order 1, 2, 3, 4
and 6 only are compatible with the translational symmetry of the crystal lattice.
Table I-1. Possible values of n with the condition, -1  cosθ  +1, and the
corresponding values of θ and the order of the allowed rotation operations.
n 3 2 1 0 -1
θ (o
) 180 120 90 60 0
Order of rotation 2 3 4 6 1
 
a
Figure I-4. Geometric construction showing lattice points with a unit cell length of a
along the x-axis, and the relation derived between the angle (θ) of rotational symmetry and
integers, n.
cos 180-θ = - cos θ = n-1 /2
a
na
θ
n-1 a/2
180 - θ

7. 7
I-3.2 Crystal systems in 2-dimension: Considering all the possible point group symmetries,
the only crystal systems that can be visualized in 2-dimension are the following [a, b are
the unit cell lengths, and γ is the angle between the unit cell axes] shown in Fig. I-5.
o square (a = b, γ = 90o
)
o rectangular (a  b, γ = 90o
)
o hexagonal (a = b, γ = 120o
)
o oblique (a  b, γ  90o
)
 Try to list the symmetry elements in each case above.
 Try to see if any periodic lattices in 2-D are possible with a set of point group
symmetries that are different from the four above.
Figure I-5. Crystal systems in 2-dimension; unit cell vectors are indicated.
square rectangular
hexagonal oblique
a
b γ

8. 8
I-3.3 Crystal systems in 3-dimension: Extending the logic as earlier, we get seven crystal
systems (Fig. I-6) [a, b, c are the unit cell lengths, and α, β, γ are the angles between the
unit cell axes b and c, a and c, and a and b].
o cubic (a = b = c, α = β = γ = 90o
)
o tetragonal (a = b  c, α = β = γ = 90o
)
o orthorhobmic (a  b  c, α = β = γ = 90o
)
o monoclinic (a  b  c, α = γ = 90o
 β)
o triclinic (a  b  c, α  β  γ)
o trigonal (a = b = c, α = β = γ < 120o
,  90o
)
o hexagonal (a = b  c, α = β = 90o
, γ = 120o
)
Figure I-6. Crystal systems in 3-dimension.
Cubic Orthorhombic
Trigonal Hexagonal
Monoclinic Triclinic
Tetragonal

9. 9
I-3.4 Bravais lattices in 2-dimension: Adding no further point group symmetries, but
incorporating any additional translational symmetry operations possible to the crystal
systems, one ends up with 5 Bravais lattices in 2 dimensions. They are the following
(Fig. I-7):
o square
o rectangular
o centered rectangular
o hexagonal
o oblique
 Note that the rectangular and centered rectangular have exactly the
same set of point group symmetries, but the latter has a new
translational symmetry.
 Try to imagine centered lattice of the square system; figure out why it
does not produce a new Bravais lattice; which of the five would it be?
Figure I-7. Bravais lattices in 2-dimension.
square rectangular
hexagonal oblique
centered rectangular

10. 10
I-3.5 Bravais lattices in 3-dimension follow similarly from the crystal systems shown in Fig.
I-6. The 14 Bravais lattices are listed in Table I-2; P (primitive), F (face-centered), I
(body-centered), C (edge-centered).
Table I-2. Bravais lattices in 3-dimension
Crystal System Bravais Lattices Number
Cubic P, F, I 3
Tetragonal P, I 2
Orthorhombic P, C, F, I 4
Monoclinic P, C 2
Triclinic P 1
Trigonal P 1
Hexagonal P 1
Total 14
o As an example, consider the three Bravais lattices under the cubic crystal
system (Fig. I-8); they all belong to the point group, Oh and hence have the
same set of point group symmetries.
o It is an interesting exercise to figure out why the face-centered and body-
centered lattices are the same in the tetragonal system, but not in the cubic
system. This has to do with the fact that c  a = b in tetragonal, and the F and
I lattices have the same symmetry with only a rotation of the a, b axes system.
o Orthorhombic system has a new lattice, C; if you tried to imagine a similar
situation in the tetragonal system, it would simply be another tetragonal P
lattice with a different a = b value.
 Our understanding of the crystal symmetries at this stage can be summarized as:
Point group operations 7 Crystal systems
Point group operation +
Translation Symmetries
14 Bravais lattices
Primitive cube (P) Face-centered cube (F) Body-centered cube (I)
Figure I-8. Bravais lattices in the cubic system..

11. 11
 A little more insight into a Bravais lattice is useful. In 3-D, a Bravais lattice can be
imagined as arising from the collection of all points represented by the set of vectors,
𝑹
⃗⃗ = 𝑛1𝒂
⃗
⃗ 1 + 𝑛2𝒂
⃗
⃗ 2 + 𝑛3𝒂
⃗
⃗ 3
where 𝑛1, 𝑛2, 𝑛3are integers and 𝒂
⃗
⃗ 1, 𝒂
⃗
⃗ 2, 𝒂
⃗
⃗ 3are primitive vectors that determine the
symmetry of the lattice.
o primitive vectors for a cubic lattice are:
𝒂
⃗
⃗ 1 = 𝑎𝒙
̂; 𝒂
⃗
⃗ 2 = 𝑎𝒚
̂; 𝒂
⃗
⃗ 3 = 𝑎𝒛
̂ where a is the unit cell length and 𝒙
̂, 𝒚
̂, 𝒛
̂ are the
unit vectors along the three Cartesian axes.
o similarly, the primitive vectors for the face-centered cubic (fcc) and body-
centered cubic lattices are (these are convenient, but not unique choices):
 fcc: 𝒂
⃗
⃗ 1 =
𝑎
2
𝒚
̂ + 𝒛
̂ ; 𝒂
⃗
⃗ 2 =
𝑎
2
𝒛
̂ + 𝒙
̂ ; 𝒂
⃗
⃗ 3 =
𝑎
2
𝒙
̂ + 𝒚
̂
 bcc : 𝒂
⃗
⃗ 1 =
𝑎
2
𝒚
̂ + 𝒛
̂ − 𝒙
̂ ; 𝒂
⃗
⃗ 2 =
𝑎
2
𝒛
̂ + 𝒙
̂ − 𝒚
̂ ; 𝒂
⃗
⃗ 3 =
𝑎
2
𝒙
̂ + 𝒚
̂ − 𝒛
̂
o the primitive vectors define a primitive unit cell; Fig. I-9a shows the fcc cell.
o primitive vectors for tetragonal and orthorhombic lattices are respectively:
𝒂
⃗
⃗ 1 = 𝑎𝒙
̂; 𝒂
⃗
⃗ 2 = 𝑎𝒚
̂; 𝒂
⃗
⃗ 3 = 𝑏𝒛
̂
𝒂
⃗
⃗ 1 = 𝑎𝒙
̂; 𝒂
⃗
⃗ 2 = 𝑏𝒚
̂; 𝒂
⃗
⃗ 3 = 𝑐𝒛
̂
 As mentioned above, the unit cell can be chosen in multiple ways. A special choice is the
Wigner-Seitz cell, constituted by all points closer to one lattice point than any other; it is
the space enclosed by perpendicular planes (lines in 2-D) bisecting the lines connecting a
lattice point to all its nearest neighbors (Fig. I-9b).
Figure I-9. (a) Primitive vectors and primitive unit cell of an fcc lattice; (b) construction of the Wigner-Seitz
cell for a 2-D hexagonal lattice.
𝒂
⃗
⃗ 1
𝒂
⃗
⃗ 2
𝒂
⃗
⃗ 3
y
x
z
(a) (b)
Blue hexagon : Wigner-Seitz cell

12. 12
 Every point in a Bravais lattice has identical environment; observation from any point is
identical. This aspect can be used to distinguish a non-Bravais lattice from a Bravais
lattice.
o For example, a honeycomb lattice (Fig. I-10) is not a Bravais lattice, as the points
o and o do not have identical environment; however, combination of two of these
points leads to the hexagonal Bravais lattice in 2-D.
o A cubic close-packed (ccp or fcc) lattice is a Bravais lattice, but a hexagonal
close-packed (hcp) lattice is not (Fig. I-11).
Figure I-11. Cubic and hexagonal close packing; the layer sequences are shown for each. The
identical environment of each site in ccp and different environment in hcp are shown schematically.
ccp
C
C C
A
B
B
B
A
A
A
B C
C
C
A C
B C
B
A
A and B have identical environments
hcp
B
A
A
A
A
B
B B
A B A B A B
Eclipsed A
layer below
Eclipsed B
layer below
A and B have different environments
Figure I-10. (a) Honeycomb lattice is not a Bravais lattice; notice the differences in the points
marked with blue and red circles. (b) Combination of these points forms the hexagonal 2-D
Bravais lattice shown by the broken blue lines; the unit cell is also indicated.
o
o
(a) (b)

13. 13
I-3.6 Lattice + Basis = Crystal structure
o We have discussed so far, the symmetries of the lattice; ‘lattice’ is a set of points
in space described by a set of coordinates, two in 2-D or three in 3-D.
o A real crystal is made up of atoms, ions or molecules; the molecule can be as
small as H2 or a large protein.
o The object or set of objects (with atoms at specific locations with respect to each
other) placed on the lattice points, is described technically as the ‘basis’.
o A crystal consists of the basis organized on a lattice with a specified symmetry.
 The basis can be spherical (perfectly symmetric), eg. a single atom, or non-spherical
(with lower symmetry than a sphere) if it has more than one atom, a molecule etc..
o Symmetry of the unit cell remains unchanged when a spherical basis is added.
o However, symmetry of the unit cell is reduced when a non-spherical basis is
added; it goes to one of the sub-groups of the original point group.
o The example below, of the square lattice unit cell (Fig. I-12a) illustrates the point.
Figure I-12a. Square lattice with (a) spherical basis possessing C4 symmetry,
and (b) non-spherical basis not having C4 symmetry.
(a)
C4
C4
(b)

14. 14
 In a similar way, subgroups of the various crystal systems in 3-D (Fig. I-6) can be
identified. The different point groups based on the cubic system are shown in Fig. I-12b.
 The hierarchy of crystal symmetries can now be summarized as:
Lattice +
Spherical Basis
Lattice +
Non-spherical basis
Point group operations 7 Crystal systems 32 Crystallographic point groups
Point group operations +
Translation Symmetries
14 Bravais lattices 230space groups
Figure I-12b. Schematic diagrams showing the symmetry of the five cubic point groups. The point group of each is
indicated at the left top and the international notation is shown at right bottom; the three fold rotation axis is shown in the
undecorated cube. [Source: Solid State Physics, N. W. Ashcroft and N. D. Mermin]

15. 15
I-3.7 Miller planes provide a convenient way of designating planes in the crystal lattice; this
will have a direct bearing on the discussion and analysis of the X-ray diffraction process
later.
 We look at 2-D lattices first, in which they should strictly be called Miller lines.
Examples in Fig. I-13 show the designations and the distance between the lines. Note
that the distances keep decreasing with increasing values of the indices.
 The logical steps involved in naming the Miller planes can be explained using Fig. 1-13c.
o Consider the point at arbitrary origin (0, 0) in the x, y axes framework
o The ‘plane’ intercepts the axes at fractional coordinates 1/2 and 1/3 respectively
o Take the inverse values; convert to the lowest set of integers, if not integers: 2, 3.
d =
𝒂
√𝟐
= 0.71a
Figure I-13. Miller ‘planes’ in a 2-D square lattice, with unit cell vectors along x and y axes and length a. ‘Planes’ (a)
(1 0) and (0 1), (b) (1 1), (c) (2 3) and (d) (2 0). Distances between the ‘planes’(d) in each case is shown.
(d)
d = a/2
(20)
d = a
(01)
(10)
a
a
x
y
(a)
(c)
d =
𝒂
𝟐𝟐+𝟑𝟐
= 0.27a
(23)
(b)
(11)

16. 16
 Distance between the planes can be calculated using simple geometry. In the case of the
square lattice, following the above protocol, a ‘plane’ (n m) will intercept the axes at a/n
and a/m. Distance between the planes, d, can be obtained using the geometry of the right
angled triangle shown below.
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 =
1
2
(
𝑎
𝑛
×
𝑎
𝑚
) =
1
2
ℎ × 𝑑
(
𝑎
𝑛
×
𝑎
𝑚
) = (√(
𝑎
𝑛
)
2
+ (
𝑎
𝑚
)
2
× 𝑑)
𝑑 =
𝑎2
𝑛𝑚√(
𝑎
𝑛
)
2
+ (
𝑎
𝑚
)
2
=
𝑎2
𝑎√(
𝑛𝑚
𝑛
)
2
+ (
𝑛𝑚
𝑚
)
2
=
𝑎
√𝑛2 + 𝑚2
 For a cubic lattice (3-D), the lines in Fig. I-13c would extend parallel to the z axis to form
planes
o intercepts will be a/2, a/3 and 
o the Miller plane will be designated as (2 3 0)
o some common Miller planes are shown in Fig. I-14
o based on the above discussion, the interplanar distance, d for an (h k l) plane of a
cubic lattice of unit cell length a, can be shown to be: 𝑑ℎ𝑘𝑙 =
𝑎
ℎ2
+𝑘2
+𝑙2
Figure I-14. Miller planes in a 3-D cubic lattice, with unit cell of length a. Distances between the planes in
each case is shown.
d010 = a
(0 1 0)
(1 1 0)
d110 =
𝒂
√𝟐
= 0.71 a d111 =
𝒂
√𝟑
= 0.58 a
(1 1 1)
d100 = a
(1 0 0)
a
a/n
a/m
d
h

17. 17
I-4. X-ray Diffraction
 X-rays are electromagnetic waves with wavelengths typically in the range of Ångströms
(10-10
m).
 Waves can be represented mathematically as a function of displacement (x) and time (t):
𝑓 𝑥, 𝑡 = 𝐴 𝑠𝑖𝑛 [2𝜋 (
𝑥
𝜆
− 𝜐𝑡)] where A = amplitude,  = wavelength,  = frequency
At any time t, f(x) can be visualized as:
Superposition of waves can be represented as follows; cases where they are in phase and
out of phase by /2 or  are shown in Fig. I-15:
Figure I-15. Superposition of waves leading to (a) constructive, (b)
partially destructive and (c) destructive interference.
(a)
(c)
(b)
𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑣𝑒 = 𝐴 {𝑠𝑖𝑛 (
2𝜋𝑥
𝜆
) + 𝑠𝑖𝑛 (
2𝜋 (𝑥 +
𝜆
4
)
𝜆
)}
= 𝐴 {𝑠𝑖𝑛 (
2𝜋𝑥
𝜆
) + 𝑠𝑖𝑛 (
2𝜋𝑥
𝜆
+
𝜋
2
)}
= 𝐴 {𝑠𝑖𝑛 (
2𝜋𝑥
𝜆
) + 𝑐𝑜𝑠 (
2𝜋𝑥
𝜆
)}
𝜕
𝜕𝑥
[𝐴 {𝑠𝑖𝑛 (
2𝜋𝑥
𝜆
) + 𝑐𝑜𝑠 (
2𝜋𝑥
𝜆
)} ] = 0
At maximum,
⟹
2𝜋
𝜆
{𝑐𝑜𝑠 (
2𝜋𝑥
𝜆
) − 𝑠𝑖𝑛 (
2𝜋𝑥
𝜆
)} = 0 ⟹
2𝜋𝑥
𝜆
=
𝜋
4
the amplitude is : 𝐴 {𝑠𝑖𝑛 (
𝜋
4
) + 𝑐𝑜𝑠 (
𝜋
4
)} = 1.4𝐴
0  2
0 /2 
Phase
Displacement
sin (0) = sin (n) = 0
sin ([n+1/2] = +1 (n even)
= -1 (n odd)

18. 18
I-4.1 Interference of waves with different path lengths in a medium and hence relative phase
differences lead to enhancement or annihilation as shown below:
I-4.2 Bragg’s law for diffraction is essentially the above phenomenon in action. X-ray (could
equally well be electron on neutron) waves reflected from parallel planes (h k l) of a
crystal undergo interference, with non-zero intensity emerging only when the path
difference between the two is an integral multiple of the wavelength,  (Fig. I-16); the
condition is related to the angle of incidence, θ and the interplanar distance, dhkl.
o It is an interesting question to consider what happens if the beam deviates slightly
from the Bragg angle, θ; will there be partial interference or complete destructive
interference ? (This has relevance to line broadening in finitely small crystals).
o Another interesting point to note: the Bragg’s law equation shows that a reflection
for n = 2 is equivalent to an n = 1 reflection from a Miller plane with spacing of
dhkl/2; therefore in practice one always works with just first order reflections.
Figure I-16. Derivation of Bragg’s law for X-ray diffraction; n is an integer, and is called the order of reflection.
Bragg’s law : 2dhkl.sinθ = n
θ
θ
dhkl.sinθ
dhkl
(h k l) plane
Total path difference
= 2dhkl.sinθ
X-ray with
wavelength, 

19. 19
I-4.3 Powder X-ray diffraction experiment involves scattering a monochromatic X-ray beam
from a collection of randomly oriented microcrystals. The diffracted beams satisfying
the Bragg condition for each Miller plane will form a cone (due to the random orientation
of the microcrystals) at scattering angle, 2θ with respect to the incident beam (Fig. I-17).
Figure I-17. (a) Scattering of an X-ray beam from a Miller plane satisfying the Bragg condition at θ,
deviates from its path by 2θ; (b) X-rays Bragg scattered from different Miller planes in randomly oriented
collection of microcrystals form cones.
(b)
Microcrystals
Scattered X-rays
from different
Miller planes
(a)
A Miller plane
Incident X-ray
Scattered X-ray

20. 20
 Typical example of a powder X-ray diffraction pattern is provided in Fig. I-18. The
pattern corresponds to NaCl; incidentally, the first crystal structure determination by X-
ray diffraction (Bragg and Bragg) was that of NaCl.
o The basic analysis of the powder X-ray diffraction pattern involves fitting the
observed peak positions (in terms of the 2θ values) to Miller planes corresponding
to a unit cell belonging to a specific crystal system. The process is called
indexing.
o The different crystal systems, usually starting with the highest symmetry one
(cubic) are explored. In each case, there is a relationship between the unit cell
parameters, the Miller plane indices (h k l) and the interplanar spacing dhkl; the
equation shown earlier for the cubic system is the simplest.
o Indexing involves the determination of a common set of a, b, c, α, β, γ within a
chosen system that satisfies the observed diffraction peaks (just the 2θ values), so
that at the end of this process, one knows the crystal system and unit cell
dimensions of the crystal under investigation.
Figure I-18. Powder X-ray diffraction pattern of NaCl.
1
1
1
2
0
0
2
2
0
3
1
1
2
2
2
4
0
0
3
3
1
4
2
0
4
2
2
5
1
1
2 (degree)
Counts
s

21. 21
 The essential idea of indexing can be described using the NaCl diffraction data, as shown
in Table I-3.
o Using the 2θ values from the experimental diffraction pattern and the X-ray
wavelength, the corresponding d values are calculated (n in the Bragg equation is
taken as 1; see Sec. I-4.2).
o Various combinations of integral values of h, k and l (starting with, say, (1 0 0))
are tried for each d value so as to get the same value of 𝑎 = 𝑑 ℎ2 + 𝑘2 + 𝑙2
(Sec. I-3.7) within acceptable numerical deviations (this is essentially a statistical
fitting protocol).
o In the NaCl case, this works out fine as the crystal does belong to a cubic system.
o If the crystal did not belong to the cubic system, then this fitting would never be
satisfactory, and one has to move on to the next system, say tetragonal, and try a
similar fitting, now with two unit cell parameters a and c.
o The process can go on until the triclinic system where the six independent unit
cell parameters will need to be fitted.
o A careful observation reveals that in Table I-3, all the (h k l) sets contain either all
odd or all even numbers; cross combinations (for example, (1 0 0)) are completely
absent. This is not accidental, and is determined by the Bravais lattice as
explained below.
Table I-3. Indexing of the NaCl powder X-ray diffraction data; the X-ray source
is CuKα ( = 1.540598 Å).
2 (deg.) d (Å) h k l (h2
+k2
+l2
)½
a (Å)
27.367 3.256 1 1 1 1.732 5.639
31.704 2.820 2 0 0 2.000 5.640
45.448 1.994 2 2 0 2.828 5.639
53.869 1.700 3 1 1 3.317 5.639
56.473 1.628 2 2 2 3.464 5.639
66.227 1.410 4 0 0 4.000 5.640
73.071 1.294 3 3 1 4.359 5.641
75.293 1.261 4 2 0 4.472 5.639
83.992 1.151 4 2 2 4.899 5.639
90.416 1.085 5 1 1 5.196 5.638
90.416 1.085 3 3 3 5.196 5.638

22. 22
I-4.4 Systematic absence
 The X-ray diffraction pattern observed for crystals with primitive cube, body-centered
cube and face-centered cubic lattices (Fig. I-19) demonstrate the idea of systematic
absences.
o The primitive cube with atoms only at the vertices of the cube show the
diffraction from all the possible Miller planes.
o The systematic absences in the bcc and fcc lattices can be visualized as arising
due to the destructive interference between the X-rays scattered from the atoms at
the vertices and atoms at the body center / face center respectively.
o A deeper look at the interference of waves from atoms at specifically related
spatial positions is required to understand this phenomenon better.
o The idea of structure factor provides a general framework to understand this.
Figure I-19. Schematic X-ray diffraction pattern of crystals with primitive cube, bcc and fcc lattices; systematic
absences in the latter two are indicated by the light blue lines. The systematic absence conditions are stated.
fcc (h, k, l all even or all odd present )
bcc (h+k+l = odd absent )

23. 23
I-4.5 Structure factor
 A physical approach to construct the structure factor for the X-ray scattering from a
Miller plane (h k l) involves the path and hence phase difference between the X-rays
(satisfying the Bragg condition for that plane) reflected by the different atoms of the
basis. The atom positions are well-defined with respect to each other. A different
approach will be discussed in Sec. I-4.9.
 As an illustrative case, we first look at the 1-D problem with a unit cell of length a (Fig.
I-20) along the x-axis.
o 𝑑100 = 𝑎; for plane (h 0 0) that is parallel to (1 0 0), 𝑑ℎ00 =
𝑎
ℎ
o The reflection from (h 0 0) is equivalent to the hth
order reflection from (1 0 0)
(Sec. 1-4.2); this follows from the Bragg’s law:
2𝑑. 𝑠𝑖𝑛𝜃 = 𝑛𝜆 ⇒ 2
𝑑
𝑛
. 𝑠𝑖𝑛𝜃 = 𝜆
o Path difference between the waves 1 and 2 (1' 2') satisfying the Bragg condition,
Δ𝑙 = 𝜹 = 𝜆
o Using similar triangles, the path difference between the waves 1 and 3 (1' 3'),
Δ𝑙 = 𝜹 = 𝜹 (
𝑥𝑎
𝑎/ℎ
) = 𝜆ℎ𝑥
o Phase difference for 1' 3', Δ𝜑 = (
2𝜋
𝜆
) 𝜆ℎ𝑥 = 2𝜋ℎ𝑥 [note: Δ𝑙 = 𝜆 ⇒ Δ𝜑 = 2𝜋].
Figure I-20. Schematic diagram showing X-ray scattering from two atoms (red and
blue) of a basis, satisfying the Bragg condition for reflection from (h 0 0) plane.

24. 24
 For the 3-D case, phase difference between the two diffracted X-rays is given by,
Δ𝜑 = 2𝜋 ℎ𝑥 + 𝑘𝑦 + 𝑙𝑧 .
 Two waves with same frequency, scattered from different atomic layers with phases 1
and 2, and amplitudes A1 and A2 (for example, if the atoms are not the same) can add up
to form a wave 3 as shown in Fig. I-21a.
o Scattered X-ray intensity will be due to the sum of the contributions of different
individual waves.
o A wave can be represented as a vector in complex space (Fig. I-21b), written as
𝐴 𝑐𝑜𝑠𝜑 + 𝑖𝑠𝑖𝑛𝜑 = 𝐴𝑒𝑖𝜑
 A measure of the scattering amplitude of the X-ray from an atom is the atomic
scattering factor, defined as : 𝑓 =
𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑤𝑎𝑣𝑒 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑒𝑑 𝑏𝑦 𝑎𝑛 𝑎𝑡𝑜𝑚
𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑤𝑎𝑣𝑒 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑒𝑑 𝑏𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
. This is
clearly related to the electron density on the atom.
 X-ray wave scattered from an atom, n (scattering factor, fn), located at a position with
coordinates xn, yn, zn, will contribute to the structure factor for the Miller plane (h k l)
with phase 2𝜋 ℎ𝑥𝑛 + 𝑘𝑦𝑛 + 𝑙𝑧𝑛 and amplitude fn. The Structure Factor can now be
written as :
𝑆ℎ𝑘𝑙 = ∑ 𝑓𝑛𝑒2𝜋𝑖 ℎ𝑥𝑛+𝑘𝑦𝑛+𝑙𝑧𝑛
𝑛 ; n represents the atoms in the basis. In this
expression, information on the atom type comes from fn and the atom position from the
coordinates xn, yn, zn.
 The physical significance of the structure factor is that the intensity of the scattered X-ray
from the (h k l) plane, 𝐼ℎ𝑘𝑙 ∝ 𝑆ℎ𝑘𝑙
∗
. 𝑆ℎ𝑘𝑙 (Sec. I-4.10).
Figure I-21. (a) Addition of two waves with same frequency, but different amplitude and phase. (b)
Representation of a wave as a vector in complex space.
(b)
(a)

25. 25
 Relevance of the structure factor can be illustrated by using it to understand the
systematic absence in a bcc lattice as follows.
o Imagine the bcc lattice unit cell to be a simple cubic cell with two atoms in the
basis, A at position (0 0 0) and B at (½ ½ ½); the latter are the fractional
coordinates expressing the position with respect to the unit cell axis.
o 𝑆ℎ𝑘𝑙 = 𝑓𝐴 + 𝑓𝐵𝑒2𝜋𝑖
ℎ
2
+
𝑘
2
+
𝑙
2 = 𝑓𝐴 + 𝑓𝐵𝑒𝜋𝑖 ℎ+𝑘+𝑙
o As atoms at A and B are the same in a bcc lattice, fA = fA= f and therefore,
𝑆ℎ𝑘𝑙 = 𝑓 1 + 𝑒𝜋𝑖 ℎ+𝑘+𝑙
o As 𝑒𝑛𝜋𝑖
= ±1 for n even/odd, the equation for the structure factor shows that
Shkl = 2f when h+k+l is even
Shkl = 0 when h+k+l is odd
o This explains the absence of X-ray scattering from those Miller planes which
satisfy h+k+l = odd in a bcc lattice.

26. 26
I-4.6 von Laue condition for X-ray diffraction
 The Bragg’s law for X-ray diffraction (Sec. I-4.2) can be recast in the language of wave
vectors and translations in momentum space (reciprocal lattice space) obtained by the
Fourier transform of the periodic real lattice space. Meaning of the latter will become
clear as we go along.
 A wave can be described by a wave vector, defined as 𝒌
⃗
⃗ =
2𝜋
𝜆
𝑖̂ where  is the
wavelength and 𝑖̂ is the unit vector along the direction of propagation of the wave. Note
that 𝒌
⃗
⃗ represents a momentum vector, since ℏ𝒌
⃗
⃗ =
ℎ
𝜆
𝑖̂ = 𝑝
 Consider the X-ray scattering in a manner very similar to that described in Fig.I-16, but
now represented using the incident wave vector, 𝒌
⃗
⃗ and scattered wave vector, 𝒌′
⃗⃗⃗ (from a
pair of atoms at lattice positions related by a lattice vector, 𝒅
⃗
⃗ ) (Fig. I-22).
o Projection of the vector 𝒅
⃗
⃗ on the vector 𝒌
⃗
⃗ (direction of which is represented by
the unit vector 𝑖̂) is given by 𝒅
⃗
⃗ . 𝑖̂, similarly 𝒅
⃗
⃗ on the vector −𝒌′
⃗⃗⃗ is given by −𝒅
⃗
⃗ . 𝑖′
̂
o The total path difference of the second wave with respect to the first one is given
by Δ𝑙 = 𝒅
⃗
⃗ . 𝑖̂ − 𝒅
⃗
⃗ . 𝑖′
̂ = 𝒅
⃗
⃗ . (𝑖̂ − 𝑖′
̂)
o Since 𝒌
⃗
⃗ =
2𝜋
𝜆
𝑖̂ and 𝒌′
⃗⃗⃗ =
2𝜋
𝜆
𝑖′
̂ , Δ𝑙 =
𝜆
2𝜋
𝒅
⃗
⃗ . (𝒌
⃗
⃗ − 𝒌′
⃗⃗⃗ ) =
𝜆
2𝜋
𝒅
⃗
⃗ . Δ𝒌
⃗
⃗
o For constructive interference, Δ𝑙 = 𝑚𝜆 where m is an integer
o Therefore, 𝑚𝜆 =
𝜆
2𝜋
𝒅
⃗
⃗ . (𝒌
⃗
⃗ − 𝒌′
⃗⃗⃗ )  𝒅
⃗
⃗ . Δ𝒌
⃗
⃗ = 2𝜋𝑚
o If 𝑅
⃗ represents a general translation vector in the Bravais lattice, the condition for
constructive interference can be generalized as : 𝑹
⃗⃗ . Δ𝒌
⃗
⃗ = 2𝜋𝑚
o Equivalently: 𝑒𝑖𝑹
⃗⃗ .Δ𝒌
⃗
⃗
= 1, as 𝑒2𝜋𝑚𝑖
= [cos 2𝑚𝜋 + 𝑖. 𝑠𝑖𝑛 2𝑚𝜋 ] = 1
Figure I-22. Schematic diagram of X-ray scattering from the lattice atoms (red
filled circles) used to derive the von Laue condition for constructive interference.
𝑘
⃗
𝑘′
⃗⃗⃗
𝑑
𝑑. 𝑖̂
−𝑑. 𝑖′
̂
𝑖̂
𝑖′
̂

27. 27
 Introducing the Reciprocal lattice : Similar to the Bravais lattice vector for a 3-D lattice,
𝑹
⃗⃗ = 𝑛1𝒂
⃗
⃗ 1 + 𝑛2𝒂
⃗
⃗ 2 + 𝑛3𝒂
⃗
⃗ 3 (Sec. I-3.5), we now define a reciprocal lattice vector,
𝑲
⃗⃗⃗ = ℎ𝒃
⃗
⃗ 1 + 𝑘𝒃
⃗
⃗ 2 + 𝑙𝒃
⃗
⃗ 3; the primitive vectors in reciprocal space are a set of orthonormal
vectors for the primitive (real) lattice vectors, 𝒂
⃗
⃗ 1, 𝒂
⃗
⃗ 2, 𝒂
⃗
⃗ 3:
𝒃
⃗
⃗ 1 = 2𝜋
𝒂
⃗
⃗ 2×𝒂
⃗
⃗ 3
𝒂
⃗
⃗ 1.𝒂
⃗
⃗ 2×𝒂
⃗
⃗ 3
𝒃
⃗
⃗ 2 = 2𝜋
𝒂
⃗
⃗ 3×𝒂
⃗
⃗ 1
𝒂
⃗
⃗ 1.𝒂
⃗
⃗ 2×𝒂
⃗
⃗ 3
𝒃
⃗
⃗ 3 = 2𝜋
𝒂
⃗
⃗ 1×𝒂
⃗
⃗ 2
𝒂
⃗
⃗ 1.𝒂
⃗
⃗ 2×𝒂
⃗
⃗ 3
o This definition ensures that 𝑹
⃗⃗ . 𝑲
⃗⃗⃗ = 2𝜋𝑚, where m is an integer {this is easily
proved; try}
o Combining the two equations in boxes above, we get the final von Laue condition
for constructive interference:
Δ𝒌
⃗
⃗ = 𝑲
⃗⃗⃗
o Physical meaning of this relation is that constructive interference occurs when the
change in the X-ray wave vector is equal to one of the reciprocal lattice vectors.
o Mathematically, reciprocal lattice is the Fourier transform of the Bravais lattice.
The latter is referred to as the real lattice as it represents the real spatial
arrangement of the crystal; reciprocal lattice can be visualized as the lattice in
momentum space.
 We discuss the meaning and construction of reciprocal lattice briefly before returning to
the X-ray diffraction problem.

28. 28
I-4.7 Reciprocal lattice
 As noted above, the reciprocal lattice vectors form a set of orthonormal vectors for the
real lattice vectors. This can be shown geometrically using first, a 2-D Bravais lattice.
o In the oblique lattice shown in Fig. I-23a, 𝒂
⃗
⃗ 1 and 𝒂
⃗
⃗ 2 are the real lattice vectors;
the unit cell defined by these vectors is shown as the green parallelogram.
o The perpendiculars to 𝒂
⃗
⃗ 2 and 𝒂
⃗
⃗ 1 give the axial directions of 𝒃
⃗
⃗ 1and 𝒃
⃗
⃗ 2
respectively; this ensures the orthogonality criteria, 𝒃
⃗
⃗ 1 ⊥ 𝒂
⃗
⃗ 2 and 𝒃
⃗
⃗ 2 ⊥ 𝒂
⃗
⃗ 1.
o Clearly, 𝒃
⃗
⃗ 1 is perpendicular to the (1 0) plane (this is to be imagined, as the
reciprocal vector is in inverse space whereas the Miller planes are in real space),
and 𝒃
⃗
⃗ 2 is perpendicular to the (0 1) plane. This shows that each reciprocal lattice
vector is associated uniquely to a Miller plane.
o Magnitude of 𝒃
⃗
⃗ 1 is chosen to be
2𝜋
𝑑10
and of 𝒃
⃗
⃗ 2 to be
2𝜋
𝑑01
; this ensures that
𝒂
⃗
⃗ 1. 𝒃
⃗
⃗ 1 = |𝒂
⃗
⃗ 1||𝒃
⃗
⃗ 1|𝑐𝑜𝑠𝜃 = |𝒂
⃗
⃗ 1|𝑐𝑜𝑠𝜃 |𝒃
⃗
⃗ 1| = 𝑑10 ×
2𝜋
𝑑10
= 2𝜋 (Fig. I-23b shows
that |𝒂
⃗
⃗ 1|𝑐𝑜𝑠𝜃 = 𝑑10, the spacing for the (1 0) planes). Similarly, 𝒂
⃗
⃗ 2. 𝒃
⃗
⃗ 2 = 2𝜋.
o Based on the above definitions, it is clear that the reciprocal lattice vector
corresponding to a Miller plane with larger spacing will be shorter and vice versa.
o Using the reciprocal lattice vectors, 𝒃
⃗
⃗ 1 and 𝒃
⃗
⃗ 2 we construct the reciprocal lattice
(blue) shown in Fig. I-23c; each reciprocal lattice point can be labeled by the
coordinates (h, k) corresponding to the Miller plane (h k) associated with it. The
point (1, 1) in the reciprocal lattice shows that the vector (𝒃
⃗
⃗ 1 + 𝒃
⃗
⃗ 2) is
perpendicular to the (1 1) plane in the real lattice.
Figure I-23. (a) A 2-D oblique lattice showing the primitive
lattice vectors, 𝒂
⃗
⃗ 1and 𝒂
⃗
⃗ 2, the Miller planes (1 0) and (0 1) and the
direction of the reciprocal lattice vectors 𝒃
⃗
⃗ 1and 𝒃
⃗
⃗ 2. (b) Geometry
to calculate 𝒂
⃗
⃗ 1. 𝒃
⃗
⃗ 1. (c) Reciprocal lattice of the oblique lattice
defined by the reciprocal lattice vectors 𝒃
⃗
⃗ 1 and 𝒃
⃗
⃗ 2.
𝒂
⃗
⃗ 2
𝒂
⃗
⃗ 1
𝒃
⃗
⃗ 2
𝒃
⃗
⃗ 1
(1 0)
(0 1)
(a)
𝜃
𝒃
⃗
⃗ 1
𝒂
⃗
⃗ 1
𝑑10
(b)
𝒃
⃗
⃗ 2
𝒃
⃗
⃗ 1
(1, 0)
(0, 1)
(0, 0)
(1, 1)
(c)

29. 29
 Similarly one can visualize geometrically, the construction of the reciprocal lattice for 3-
D lattices.
o Fig. I-24 shows a triclinic lattice cell with primitive vectors 𝒂
⃗
⃗ 1, 𝒂
⃗
⃗ 2, 𝒂
⃗
⃗ 3.
o From the definition of the reciprocal vector 𝒃
⃗
⃗ 3, it is clear that it is perpendicular
to the plane defined by 𝒂
⃗
⃗ 1and 𝒂
⃗
⃗ 2; the magnitude of this vector (in reciprocal
length, of course) is given by :
|𝒃
⃗
⃗ 3| = 2𝜋 |
𝒂
⃗
⃗ 1 × 𝒂
⃗
⃗ 2
𝒂
⃗
⃗ 1. 𝒂
⃗
⃗ 2 × 𝒂
⃗
⃗ 3
| = 2𝜋
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
=
2𝜋
ℎ𝑒𝑖𝑔ℎ𝑡
=
2𝜋
𝑑001
o Similarly the reciprocal vectors, 𝒃
⃗
⃗ 1and 𝒃
⃗
⃗ 2 can be visualized.
Figure I-24. Constructing geometrically, the reciprocal lattice vectors of a 3-D
triclinic Bravais lattice .
𝒂
⃗
⃗ 1
𝒂
⃗
⃗ 2
𝒂
⃗
⃗ 3
𝒃
⃗
⃗ 3
𝒃
⃗
⃗ 2
𝒃
⃗
⃗ 1
(0 0 1) plane
d001

30. 30
 The reciprocal lattice vectors can be derived analytically. Using the definition given in
Sec. I-4.6 and shown geometrically in Sec. I-4.7, the primitive reciprocal lattice vectors
of some typical 3-D Bravais lattices (cubic lattices) can be derived as follows.
o Primitive cube: 𝒂
⃗
⃗ 1 = 𝑎𝒙
̂; 𝒂
⃗
⃗ 2 = 𝑎𝒚
̂; 𝒂
⃗
⃗ 3 = 𝑎𝒛
̂
o 𝒃
⃗
⃗ 1 = 2𝜋
𝑎𝒚
̂×𝑎𝒛
̂
𝑎𝒙
̂.𝑎𝒚
̂×𝑎𝒛
̂
= 2𝜋
𝑎2𝒙
̂
𝑎3
=
2𝜋
𝑎
𝒙
̂
o Similarly, 𝒃
⃗
⃗ 2 =
2𝜋
𝑎
𝒚
̂ and 𝒃
⃗
⃗ 3 =
2𝜋
𝑎
𝒛
̂
o Clearly, the primitive reciprocal lattice vectors, 𝒃
⃗
⃗ 1, 𝒃
⃗
⃗ 2, 𝒃
⃗
⃗ 3 of a simple cubic
Bravais lattice form a lattice with cubic symmetry in inverse space
o For a simple cubic lattice of unit cell length, a Å, the reciprocal lattice is cubic
with unit cells of length
2𝜋
𝑎
Å-1
 Reciprocal lattice of the fcc lattice can be shown to be a lattice with bcc symmetry!
o 𝒂
⃗
⃗ 1 =
𝑎
2
𝒚
̂ + 𝒛
̂ ; 𝒂
⃗
⃗ 2 =
𝑎
2
𝒛
̂ + 𝒙
̂ ; 𝒂
⃗
⃗ 3 =
𝑎
2
𝒙
̂ + 𝒚
̂
o 𝒃
⃗
⃗ 1 = 2𝜋
(𝑎
2
⁄ )
2
(𝑎
2
⁄ )
3
𝒛
̂+𝒙
̂ × 𝒙
̂+𝒚
̂
𝒚
̂+𝒛
̂ . 𝒛
̂+𝒙
̂ × 𝒙
̂+𝒚
̂
=
4𝜋
𝑎
𝒚
̂+𝒛
̂−𝒙
̂
𝒚
̂+𝒛
̂ . 𝒚
̂+𝒛
̂−𝒙
̂
=
2𝜋
𝑎
𝒚
̂ + 𝒛
̂ − 𝒙
̂
o Similary, 𝒃
⃗
⃗ 2 =
2𝜋
𝑎
𝒛
̂ + 𝒙
̂ − 𝒚
̂ , and 𝒃
⃗
⃗ 3 =
2𝜋
𝑎
𝒙
̂ + 𝒚
̂ − 𝒛
̂
 The reverse is also true; reciprocal lattice of the bcc lattice has fcc symmetry (Fig. I-25).
o 𝒂
⃗
⃗ 1 =
𝑎
2
𝒚
̂ + 𝒛
̂ − 𝒙
̂ ; 𝒂
⃗
⃗ 2 =
𝑎
2
𝒛
̂ + 𝒙
̂ − 𝒚
̂ ; 𝒂
⃗
⃗ 3 =
𝑎
2
𝒙
̂ + 𝒚
̂ − 𝒛
̂
o 𝒃
⃗
⃗ 1 =
2𝜋
𝑎
𝒚
̂ + 𝒛
̂ ; 𝒃
⃗
⃗ 2 =
2𝜋
𝑎
𝒛
̂ + 𝒙
̂ ; 𝒃
⃗
⃗ 3 =
2𝜋
𝑎
𝒙
̂ + 𝒚
̂
Figure I-25. Real bcc lattice and the corresponding reciprocal fcc lattice.
𝒂
⃗
⃗ 𝟏
𝒂
⃗
⃗ 𝟐
𝒂
⃗
⃗ 𝟑
Real lattice (bcc)
Unit cell length = a
(0, 0, 0)
(1, 1, 0)
(1, 0, 1)
(2, 0, 0)
(0, 1, 1)
(1, 1, 2)
𝒃
⃗
⃗ 𝟏
𝒃
⃗
⃗ 𝟐
𝒃
⃗
⃗ 𝟑
Reciprocal lattice (fcc)
Unit cell length = 2/a

31. 31
I-4.8 Ewald construction
 In Sec. I-4.6 we derived the condition for constructive interference of scattered X-ray
waves, Δ𝒌
⃗
⃗ = 𝑲
⃗⃗⃗ .
o 𝑲
⃗⃗⃗ = ℎ𝒃
⃗
⃗ 1 + 𝑘𝒃
⃗
⃗ 2 + 𝑙𝒃
⃗
⃗ 3 is any reciprocal lattice vector; based on the discussions in
Sec. I-4.7, this vector is uniquely associated with the (h k l) Miller plane in the
real lattice, and |𝑲
⃗⃗⃗ | =
2𝜋
𝑑ℎ𝑘𝑙
o The von Laue condition can be visualized geometrically using the Ewald
construction in reciprocal space (Fig. I-26).
o Placing the incident X-ray wave vector, 𝒌
⃗
⃗ on a lattice point in reciprocal space
(chosen as (0, 0, 0) in the figure), and using it as the radius, a circle (sphere in 3-D
construction) is drawn. If the circle passes through any other reciprocal lattice
point, then it satisfies the condition, 𝒌
⃗
⃗ + 𝑲
⃗⃗⃗ = 𝒌′
⃗⃗⃗ , ie. 𝒌′
⃗⃗⃗ − 𝒌
⃗
⃗ = 𝑲
⃗⃗⃗ (note that the
scattering is elastic; the wavelength and hence the magnitude of the wave vector
remain the same).
o Combining the diffraction picture in the real lattice (Fig. I-26), one can see that
this is consistent with the Bragg’s law.
𝑠𝑖𝑛𝜃ℎ𝑘𝑙 =
1
2
|𝑲
⃗⃗⃗ |
|𝒌′
⃗⃗⃗⃗ |
=
1
2
(
2𝜋
𝑑ℎ𝑘𝑙
)
2𝜋
𝜆
=
𝜆
2𝑑ℎ𝑘𝑙
 2𝑑ℎ𝑘𝑙𝑠𝑖𝑛𝜃ℎ𝑘𝑙 = 𝜆  2𝑑. 𝑠𝑖𝑛𝜃ℎ𝑘𝑙 = 𝑛𝜆
as every observed d can be related to an integral multiple of a dhkl.
Figure I-26. Ewald construction in reciprocal lattice (blue points) showing the von Laue condition for
constructive interference of X-rays; scattering from the real lattice plane is combined to show how the
Bragg’s law can be inferred from the same picture.
𝟏, 𝟏, 𝟎
𝟎, 𝟏, 𝟎
𝟎, 𝟎, 𝟎
𝟏, 𝟎, 𝟎
𝟏, 𝟏, 𝟎
(h k l ) planes
of the crystal
𝒌
⃗
⃗
𝒌
⃗
⃗
𝒌′
⃗⃗⃗
𝑲
⃗⃗⃗
θhkl

32. 32
I-4.9 Structure factor in terms of the reciprocal lattice vector
 Structure factor can be derived based on the fact that there is interference between the X-
rays scattered by the different atoms in the basis (Sec. I-4.5).
 In the derivation of the von Laue condition (Sec. I-4.6) we saw that the path length
difference is given by
𝜆
2𝜋
𝒅
⃗
⃗ . Δ𝒌
⃗
⃗ =
𝜆
2𝜋
𝒅
⃗
⃗ . 𝑲
⃗⃗⃗ . This translates to a phase difference of 𝒅
⃗
⃗ . 𝑲
⃗⃗⃗ (if
path difference is , the phase difference is 2). Amplitude of the wave can be written as
𝑒𝑖𝒅
⃗⃗ .𝑲
⃗⃗⃗
.
 Summing up the contributions from atoms i in the basis, the structure factor can be
written as : 𝑆𝐾 = ∑ 𝑓𝑖𝑒𝑖𝒅𝒊
⃗⃗⃗⃗ .𝑲
⃗⃗⃗
𝑖 ; fi and di are the atomic scattering factor and position
vector of atom i respectively.
 We will apply the above equation for the structure factor to determine the systematic
absence in bcc lattice.
o As in Sec. I-4.5, we consider the bcc lattice unit cell to be a simple cubic cell with
two atoms in the basis, 1 at position (0 0 0) and 2 at (½ ½ ½). Therefore, 𝒅𝟏
⃗⃗⃗⃗ = 𝟎
and 𝒅𝟐
⃗⃗⃗⃗ =
𝒂
𝟐
𝒙
̂ + 𝒚
̂ + 𝒛
̂ ; a is the length of the unit cell.
o The reciprocal lattice vector corresponding to an (h k l) plane of the cubic lattice
considered above, is 𝑲
⃗⃗⃗ =
2𝜋
𝑎
ℎ𝒙
̂ + 𝑘𝒚
̂ + 𝑙𝒛
̂
o 𝑆𝐾 = 𝑆ℎ𝑘𝑙 = 1 + 𝑒𝑖
𝒂
𝟐
𝒙
̂+𝒚
̂+𝒛
̂ .
2𝜋
𝑎
ℎ𝒙
̂+𝑘𝒚
̂+𝑙𝒛
̂
= 1 + 𝑒𝑖𝜋 ℎ+𝑘+𝑙
; this is the same
relation as derived in Sec.I-4.5 and the systematic absence conditions follow as
discussed there.

33. 33
I-4.10 Basic concepts of X-ray structure solution and refinement
 Intensity of X-rays scattered from an (h k l) plane, 𝐼ℎ𝑘𝑙 = 𝑆ℎ𝑘𝑙
∗
𝑆ℎ𝑘𝑙.
 From the intensities measured from different planes in a single crystal X-ray diffraction
experiment (corrected for various experimental factors such as absorption, polarization
etc.), the structure factors are determined (this involves the tricky issue of deciding the
phase factor, the famous ‘phase problem’, discussed in specialized text books on
crystallography).
 As the atomic scattering factor is related to the electron density, one can use the picture
of the spatial electron density distribution in the crystal, and replace the summation in the
equation for the structure factor by the integral so that,
𝑆𝐾 = ∫ 𝑓 𝑟 . 𝑒𝑖𝐾
⃗
⃗ .𝑟
𝑑𝑟
 Fourier transformation leads to an electron density map called the Fourier map:
𝜌 𝑟 ∝ 𝑓 𝑟 = ∫ 𝑆𝐾. 𝑒−𝑖𝐾
⃗
⃗ .𝑟
𝑑𝐾
⃗
⃗
 The Fourier map provides the initial structure solution; the total electron densities at
specific points determine the type of atom present there.
 Using the initial solution, structure factors are calculated for each (h k l) plane; this gives
the calculated structure factor list, Shkl(calc). From the experiment, one already has the
experimental structure factor list, Shkl(expt).
 Least square method is used to carry out regression of Shkl(calc) against Shkl(expt).
Quality of refinement is represented by the so-called ‘r factor’.
 The model is revised iteratively to obtain decreasing r factor values. The final model
used for the best Shkl(calc) that gives the lowest ‘r factor’ is the refined structure.

34. 34
I-5. Common crystal structure motifs
 It is important to be familiar with specific types of crystal structures which occur rather
frequently or have unique features. A few examples selected at random are listed below
with materials that possess them.
o Rock salt : NaCl, FeO
o Fluorite : CaF2, Li2O
o Perovskite (ABO3) : BaTiO3
o Spinel (AB2O4) : MgAl2O4 (inverse spinel : Fe3O4)
o Rutile : TiO2, NbO2
o Diamond : C, Si
o Graphite : C
o Zinc blende : ZnS, GaAs
o Wurtzite : ZnS, AgI
 There are several more; see textbooks on solid state chemistry, crystallography etc. for
more examples.

35. 35
I-6. Quasicrystals: a brief note
 As shown in Sec.I-3.1, translational periodicity ensures that rotational symmetry occurs
only with orders 1, 2, 3, 4 and 6, leading eventually to the 230 space groups.
 Clearly, a crystal cannot have rotational symmetry of order 5, 7, 9, 10 etc.. However, in
1982, Daniel Shechtman (Nobel Prize, 2011) found that electron diffraction of a rapidly
cooled Al-Mn alloy produced a diffraction pattern that had 10-fold symmetry !
 The term ‘quasicrystal’ refers to a structure that is ordered, but not periodic in the
observable dimensions like 1, 2 or 3-D.
o The meaning of ‘ordered’ but not ‘periodic’ can be understood from figures like
the famous Penrose tiling (Fig. I-27a). It is clearly an ordered structure with a 5-
fold symmetry; obviously there cannot be and there is no translational symmetry.
o The meaning of quasi-periodicity in a lower dimension mapped on to a periodic
structure in a higher dimension can be understood from Fig. I-27b.
 The 2-D space has a square lattice (periodic), with the unit cell shaded
deep yellow.
 The points within a projection strip (light yellow) are projected on to an
external space (Ve axis).
 If the Ve axis has an irrational slope, the projection gives a quasi-periodic
sequence [green (G) and red (R) segments in the ratio equal to the slope of
the Ve axis (Fig. I-27c)]; for example, if the slope = √5+1
2
= 1.61803 …, the
sequence is a Fibonacci chain (1-D quasi periodic structure).
o Quasicrystals are such manifestations which are quasi-periodic in 2 or 3-D, but
periodic in higher dimensions.
Figure I-27. (a) Penrose tiling; (b) a 1-D quasi-periodic sequence (Fibonacci chain) embedded in a higher dimensional
(2-D) space in the form of a periodic square lattice [Courtsey: http://www.jcrystal.com/steffenweber/qc.html]; (c)
Derivation of the G/R ratio.
(a) (b)
𝐺
𝑥
= 𝑠𝑖𝑛𝜃;
𝑅
𝑥
= 𝑐𝑜𝑠𝜃

𝐺
𝑅
= 𝑡𝑎𝑛𝜃 = 𝑠𝑙𝑜𝑝𝑒
𝜃
𝜃
𝜃 𝜃
G
R
x
x
G R
(c)