CRYSTALOGRAPHY
CRYSTAL STRUCTURE
A crystal may be defined as a solid composed of atoms arranged in a pattern periodic in three dimensions.
As such, crystals differ in a fundamental way from gases and liquids because the atomic arrangements in
the latter do not possess the essential requirement of periodicity. Not all solids are crystalline, however;
some are amorphous, like glass, and do not have any regular interior arrangement of atoms.
There is, in fact, no essential difference between an amorphous solid and a liquid, and the former is often
referred to as an "undercooled liquid." In thinking about crystals, it is often convenient to ignore the actual
atoms composing the crystal and their periodic arrangement in space, and to think instead of a set of
imaginary points which has a fixed relation in space to the atoms of the crystal and which may be regarded
as a sort of framework or skeleton on which the actual crystal is built.
LATTICE
This set of points can be formed as follows. Imagine space to be divided by three sets of planes, the planes
in each set being parallel and equally spaced. This division of space will produce a set of cells each identical
in size, shape, and orientation to its neighbors. Each cell is a parallelepiped, since its opposite faces are
parallel and each face is a parallelogram. The space-dividing planes will intersect each other in a set of
lines (Fig. 1), and these lines in turn intersect in the set of points referred to above.
Figure 1: (Left) A point latice and (right) projection of front/top/side view.
A set of points so formed has an important property: it constitutes a point lattice, which is defined as an
array of points in space so arranged that each point has identical surroundings. By "identical surroundings"
we mean that the lattice of points, when viewed in a particular direction from one lattice point, would
have exactly the same appearance when viewed in the same direction from any other lattice point.
Since all the cells of the lattice shown (Fig. 2) are identical, we may choose any one as a unit cell. The size
and shape of the unit cell can in turn be described by the three vectors a, b, and c drawn from one corner
of the cell taken as origin (Fig. 2). These vectors define the cell and are called the crystallographic axes of
the cell. They may also be described in terms of their lengths (a, b, c) and the angles between them (α, β,
γ). These lengths and angles are the lattice constants or lattice parameters of the unit cell.
Crystallography Note, Dr. Md. Muktadir Billah
Note that, the vectors a, b, c define, not only the unit cell, but also the whole point lattice through the
translations. In other words, the whole set of points in the lattice can be produced by repeated action of
the vectors a, b, c on one lattice point located at the origin, or, stated alternatively, the vector coordinates
of any point in the lattice are Pa, Qb, and Rc, where P, Q, and R, are whole numbers. It follows that the
arrangement of points in a point lattice is absolutely periodic in three dimensions, points being repeated
at regular intervals along any line one chooses to draw through the lattice.
Figure 2: (Left) Representation of space lattice and (right) unit cell with lattice parameters.
CRYSTAL SYSTEM
In dividing space by three sets of planes, we can of course produce unit cells of various shapes, depending
on how we arrange the planes. For example, if the planes in the three sets are all equally spaced and
mutually perpendicular, then the unit cell is cubic. In this case the vectors a, b, c are all equal and at right
angles to one another, or a = b = c and α = β = λ = 90O. By thus giving special values to the axial lengths
and angles, we can produce unit cells of various shapes and therefore various kinds of point lattices, since
the points of the lattice are located at the cell corners. It turns out that only seven different kinds of cells
are necessary to include all the possible point lattices. These correspond to the seven crystal systems into
which all crystals can be classified. These systems are listed in (Fig. 3) (Some writers consider the
rhombohedral system as a subdivision of the hexagonal, thus reducing the number of crystal systems to
six.)
Seven different point lattices can be obtained simply by putting points at the corners of the unit cells of
the seven crystal systems. However, there are other arrangements of points which fulfill the requirements
of a point lattice, namely, that each point have identical surroundings. The French crystallographer Bravais
worked on this problem and in 1848 demonstrated that there are fourteen possible point lattices and no
more; this important result is commemorated by our use of the terms Bravais lattice and point lattice as
synonymous. For example, if a point is placed at the center of each cell of a cubic point lattice, the new
array of points also forms a point lattice. Similarly, another point lattice can be based on a cubic unit cell
having lattice points at each corner and in the center of each face.
The fourteen Bravais lattices are described and illustrated in (Fig. 3), where the symbols P, F, I, etc., have
the following meanings. We must first distinguish between simple, or primitive, cells (symbol P or R) and
Crystallography Note, Dr. Md. Muktadir Billah
non-primitive cells (any other symbol): primitive cells have only one lattice point per cell while nonprimitive have more than one. A lattice point in the interior of a cell "belongs" to that cell, while one in a
cell face is shared by two cells and one at a corner is shared by eight. The number of lattice points per cell
is therefore given by,
𝑁𝑓
𝑁𝑐
𝑁 = 𝑁𝑖 +
+
2
2
Where, 𝑁𝑖 = Number of interior points
𝑁𝑓 = Number of points on faces
𝑁𝑐 = Number of points on corners
Figure 3: (Left) Seven crystal systems and (right) fourteen Bravais lattices.
Number of atoms per unit cell is shown in Table 1 for different types of lattices. Any cell containing lattice
points on the corners only is therefore primitive, while one containing additional points in the interior or
on faces is non-primitive. The symbols F and I refer to face-centered and body-centered cells, respectively,
while A, B, and C refer to base-centered cells, centered on one pair of opposite faces A, B, or C. (The A
face is the face defined by the b and c axes, etc.) The symbol R is used especially for the rhombohedral
system. At first glance, the list of Bravais lattices appears incomplete. Why not, for example, a basecentered tetragonal lattice? However, these will be discussed in advanced crystallography.
Crystallography Note, Dr. Md. Muktadir Billah
ATOMIC PACKING FACTOR
The filling factor of a crystal structure is defined as the fraction of the total crystal volume filled with atoms
considered hard spheres. Sometimes instead of filling factor the expressions “atomic packing factor” or
“packing fraction” are used. The filling factor gives us an idea how close “the atoms are packed” in the
crystal structure. In conclusion, the filling factor together with the coordination number give us an idea
about the degree of filling the crystal volume with atoms, and, at the same time, tells us how close the
atoms in a crystal are packed. In order to calculate the filling factor, we have to know first the radii for
atoms considered hard spheres. To do the calculations, we can limit ourselves to the conventional unit
cell of the crystal structure. The filling factor is defined as, filling factor cell volume = volume occupied by
atoms within the unit cell/cell volume.
Table 1: Number of atoms per unit cell.
Simple Cubic Structure
Let us assume that the vertices of the cubic unit cell of the simple cubic structure coincide with the centers
of atoms. Fig. 4 shows the plane of one of the faces of the cube with the cross sections of the atoms, being
considered hard spheres, are represented by circles on this plane. Each atom from a cube face has two of
its Nearest Neighbors (NNs) on this face. The NNs are at a distance equal to the lattice constant a. The
atomic radius is equal to half of 𝑎 (𝑎 = 2𝑟), so the volume of the only one atom belonging to the cube is
given by,
4
𝑉𝑆 = ( ) π 𝑎3
3
Atomic packing factor for is given by,
𝐴𝑃𝐹 =
=
=
𝑉𝑆
𝑉𝐶
4
( ) π 𝑟3
3
𝑎3
4
𝑎
( ) π ( )3
3
2
𝑎3
= 0.52
1
Number of atoms per unit cell of simple cubic structure = 8 corner atoms ( ) = 1 atom.
8
Crystallography Note, Dr. Md. Muktadir Billah
Figure 4: The plane that contains a face of a cubic unit cell of the simple cubic structure with the cross
sections of atoms considered hard spheres.
The result for the atomic packing factor shows that in the case of the simple cubic structure about half of
the crystal volume is filled with atoms and the other corresponds to the interstices. Of course this is
reflected also by the coordination number; the number of the NNs of an atom in the case of the structure
is only 6. The next nearest neighbors (NNN) are already 12, but at the distance about 40% higher than the
NNs. In conclusion, the interstitial crystal structure is quite large. Under normal conditions, only one
element, polonium in the α-phase, crystallizes in simple cubic structure. However, there are three
elements, As, Sb and Bi crystallize in the trigonal structure, which is a slightly distorted simple cubic
structure.
Body Centered Cubic Structure
In this case, the atoms of the vertices of the cube are NNs of the atom that is in the center of the cube
(Fig. 5a), and those atoms are in contact with it. The point of contact between two atoms is found in a
plane defined by two body diagonals of the cube. The atoms are in contact with each other only along the
body diagonals of the cube, while, the atoms that are at the vertices are at a distance greater than 2𝑟 (𝑟
– radius of the atom). There are two atoms in the cubic cell of the bcc structure, so the atomic packing
factor is given by,
𝐴𝑃𝐹 =
=
=
𝑉𝑆
𝑉𝐶
4
3
2 ( ) π 𝑟3
𝑎3
4
𝑎√3 3
2 ( )π(
)
3
4
3
𝑎
= 0.68
1
Number of atoms per unit cell of body centered cubic structure = 8 corner atoms ( ) + 1 body center
8
atoms = 2 atoms.
APF for the body centered cubic structure is higher than that one for the simple cubic structure. This is
consistent with the fact that the number of NNs in body centered cubic is also higher than in simple cubic
(8 and 6, respectively). Moreover, the distance of the 6 NNNs of an atom in the body centered cubic
structure differs from the distance of its NNs by less than 15%. Therefore, an atom in this structure has
effectively 14 atoms close to it.
Crystallography Note, Dr. Md. Muktadir Billah
Figure 5: (a) Unit cell of the body centered cubic structure. (b) A plane defined by two body diagonals of
the cube shown in (a). In this plane, there are the points of contact between the central atom and its NNs.
Face Centered Cubic Structure
The atom placed in the center of a face of the cube has 4 of its NNs at the vertices of this face (Fig. 6a).
Fig. 6b shows the plane of the front face of the cube with the cross sections of 5 atoms considered hard
spheres. The points of contact between the atoms are found on the face diagonals.
Figure 6: (a) Unit cell of the face centered cubic structure. (b) Plane of the front face of the cube from (a)
with the cross sections of 5 atoms considered hard spheres. The points of contact between the NNs are
found in this plane.
In the case of the face centered cubic structure, there are four atoms in the cubic cell, therefore the atomic
packing factor is given by,
𝐴𝑃𝐹 =
=
=
𝑉𝑆
𝑉𝐶
4
3
4 ( ) π 𝑟3
𝑎3
4
𝑎√2 3
4 ( )π(
)
3
4
𝑎3
= 0.74
1
Number of atoms per unit cell of face centered cubic structure = 8 corner atoms ( ) + 6 face center atoms
1
8
( ) = 4 atoms.
2
Crystallography Note, Dr. Md. Muktadir Billah
This APF is the largest one among the filling factors for the cubic structures and at the same time the
3
largest one among the filling factors for all structures for the elements. In this case of the crystal volume
4
is filled with atoms considered hard spheres and only
1
4
is empty. The number of the NNs, equal to 12, is
also the largest possible. The face centered cubic structure represents one of the close-packed structures.
We will discuss them below.
POINTS, DIRECTIONS AND PLANES
Miller indices are used to specify directions and planes. These directions and planes could be in lattices or
in crystals. The number of indices will match with the dimension of the lattice or the crystal. Naming
points, directions, and planes in a unit cell can seem overwhelming at first, but will become easy as you
practice and follow the following procedures. Indices of crystallographic points, directions, and planes are
given in terms of the lattice constants of the unit cell. For points and directions, you can consider the
indices to be coefficients of the lattice constants. Remember that you only need to invert the indices for
planes. It is essential to label axes, lattice constants, and identifying information for directions (vector
arrow head) and planes (axes intercepts) in order to receive full credit.
Points
Labeling points in a unit cell follows the same procedure for listing points in any Cartesian coordinate
system. The indices used to refer to points are q, r, and s. They are listed without commas, parentheses,
or brackets. Consider point P, in Figure 10a. If you were standing at the origin of the unit cell, you could
travel q · a in the x-direction, r · v in the y-direction, and s · c in the z-direction to get to point P. Thus we
would say that point P corresponds to the q r s point coordinates. To find q, r, and s when you are shown
a drawing with a point:
Figure 10: Crystallographic point coordinates.
1. Start with your pencil at the origin.
2. Count the number lattice constants you must move in the x-, y-, and z-directions to reach the point.
3. Write the point as q r s without commas, parentheses, or brackets.
Do not convert the coordinates to reduced integers. The
1 1 1
2 2 2
point in the BCC structure is not the same
as the 1 1 1 point. To draw a point given qrs:
Crystallography Note, Dr. Md. Muktadir Billah
1. Start with your pencil at the origin.
2. Count q · a in the x-direction, r · v in the y-direction, and s · c in the z-direction.
3. Draw and label the point.
In the bcc system shown in Figure 10b, the path to point 9 would be 0 · a in the x-direction, 1 · a in the ydirection, and 1 · a in the z-direction. Thus the coordinates of point 9 are 0 1 1. The coordinates for point
5 are
1 1 1
.
2 2 2
Directions
To draw a direction given the [u v w] indices:
1. Choose a point for the origin of your vector. Simple directions, such as those shown in Figure 11a, are
easy to draw using the 0 0 0 crystallographic point as the origin. For more complicated directions (see
Figure 11b), it is more convenient to translate (add or subtract) or scale (multiply or divide) the indices of
the direction. If you translate your origin to the point 0 0 1 for example, the tail of your direction vector
will be at this point.
2. Start with your pencil on your chosen origin. Move your pencil u · a in the x-direction, v · b in the ydirection, and w · c in the z-direction. If u, v, or w, are negative, move in the negative direction on that
axis.
3. Draw and label a point.
4. Draw a line from your chosen origin to the point you just drew. Add an arrow head at the point.
Figure 11: Crystallographic directions.
It is extremely important when drawing directions that you label the point that lies at the end of your
direction vector. Is it also highly preferred that you shift the origin and scale of the direction vector so that
it is contained inside the unit cell. If you shifted the origin, a way to check that you drew your direction
properly is to subtract the point coordinates of the arrow head from those of its tail. The resulting indices
should be [u v w].
To write the [u v w] indices given a drawing of a direction:
1. Place your pencil at the tail of the given vector.
Crystallography Note, Dr. Md. Muktadir Billah
2. Count how many lattice constants you must move in the x-, y-, and z-directions to reach the head of
the vector. Write them down. Be mindful of negatives.
3. If necessary, multiply by a common number to make the indices integers.
4. Rewrite the indices and enclose them in square brackets [u v w].
Family of Directions
Table 2 lists some family of directions. Directions having similar indices are equivalent, e.g. faces of the
cube [1 0 0], [0 1 0] and [0 0 1]. This is termed as a family of planes and denoted as <1 0 0> which includes
all the [1 0 0] combinations including negative indices.
Table 2: It’s a set of directions related by symmetry operations of the lattice.
Planes
To draw a plane given the (h k l) indices:
1. Find the reciprocal of (h k l). If one or more of the indices is zero, consider the reciprocal to be infinity.
The values
1 1
1
ℎ 𝑘
𝑙
, and are the intercepts of the x-, y-, and z-axes, respectively. If one or more intercept is
infinity, this means that the plane will never intercept that axis (it is parallel to that axis). For the (2 2 1)
plane shown in Figure 12a, the reciprocals would be,
1 1
2 2
𝑎
1
𝑏
2. For all non-zero indices, draw a point on the x-axis at , a point on the y-axis at , and a point on the z𝑐
ℎ
𝑘
axis at .
𝑙
3. Connect the dots.
To list the (h k l) indices given a drawing of a plane:
1. Check to see if the plane goes through the origin, as in Figure 13a. If not, proceed to step 2. If it does,
shift the plane 1 lattice constant in an axis such that it no longer passes through the origin (see Figure
13b). This new plane is equivalent to the first.
2. Find the coefficients by which you would multiply the lattice constants to identify where the plane
intercepts the x-, y-, and z-axes. Be mindful of negative intercepts. If the plane doesn’t intercept an axis
(it is parallel to that axis), call the intercept infinity. Write down these values.
3. Take the reciprocal of these intercepts. The value of 1 ∞ is zero.
Crystallography Note, Dr. Md. Muktadir Billah
4. If necessary, multiply or divide by a common number to put the indices in the most reduced integer
form. Sometimes you will be asked to leave the indices unreduced - for example, (2 2 0) instead of (1 1
0) - for use in x-ray diffraction questions.
5. Enclose the indices in parentheses without commas (h k l). It is extremely important when drawing
planes that you label where the plane intersects the axes.
Figure 12: Crystallographic planes.
Figure 13: The method used to determine the indices of a plane that passes through the origin.
Family of Planes/Multiplicity Factor
Table 3 lists some family of planes. Planes having similar indices are equivalent, e.g. faces of the cube (1
0 0), (0 1 0) and (0 0 1). This is termed as a family of planes and denoted as {1 0 0} which includes all the
(1 0 0) combinations including negative indices. Thus the planes of {1 0 0} are (1 0 0), (0 1 0), (0 0 1), (1̅ 0
0), (0 1̅ 0) and (0 0 1̅). So, multiplicity factor for {1 0 0} is 6. Multiplicity factor for {1 1 0} is 12 and the
planes are (1 1 0), (1 0 1), (0 1 1), (1̅ 1 0), (1̅ 0 1), (0 1̅ 1), (1 1̅ 0), (1 0 ̅1), (0 1 1̅), (1̅ 1̅ 0), (1̅ 0 1̅) and (0 ̅1 1̅).
Multiplicity factor for {1 1 1} is 8 and the planes are (1 1 1), (1̅ 1 1), (1 1̅ 1), (1 1 ̅1), (1̅ 1̅ 1), (1̅ 1 ̅1), (1 ̅1 1̅)
and (1̅ 1̅ 1̅).
Table 3: It’s a set of planes related by symmetry operations of the lattice.
Crystallography Note, Dr. Md. Muktadir Billah
Examples
For the following structures state (Fig. 14): a. the crystal system and b. the Miller indices of the directions
(remember to use right handed axes).
Figure 14: Crystallographic directions.
Answers:
a) Cubic , [1 0 1]
b) Cubic , [2 1 0]
c) Cubic , [1̅̅̅
1 1]
d) Tetragonal , [2 2̅ 1]
e) Tetragonal , [0 ̅1 0]
f) Tetragonal , [1 1 2]
g) Orthorhombic , [1̅ 1̅ 1̅]
h) Orthorhombic , [2̅ 0 1]
i) Orthorhombic , [1̅ ̅1 0]
j) Orthorhombic , [1̅ ̅̅̅
2 1]
Draw (as in Fig. 15) cells showing the [1 2 1] direction in each of the following lattices: i) cubic ii) tetragonal
where a = l, c = Z iii) orthorhombic where a = 1, b = 2, c = 3 (remember to use right handed axes).
Crystallography Note, Dr. Md. Muktadir Billah
Figure 15: Crystallographic directions.
Miller Indices are the reciprocals of the parameters of each crystal face. Thus in Fig. 16,
1
1
𝑥
𝑦
For pink face intercepts: = 1/1,
1
= 1/∞, = 1/∞
𝑧
So, 𝑥 = 1, 𝑦 = 0, 𝑧 = 0
Thus the plane is (1 0 0).
1
1
𝑥
𝑦
1
1
𝑥
𝑦
For green face intercepts: ( = 1/∞,
1
= 1/∞, = 1/1)
𝑧
So, 𝑥 = 0, 𝑦 = 0, 𝑧 = 1
Thus the plane is (0 0 1).
For yellow face intercepts: ( = 1/∞,
1
= 1/1, = 1/∞)
𝑧
So, 𝑥 = 0, 𝑦 = 1, 𝑧 = 0
Thus the plane is (0 1 0).
Figure 16: (1 0 0), (0 1 0) and (0 0 1) planes.
In Fig. 17, the plane of interest cuts two of the crystallographic axes.
1
1
1
𝑥
𝑦
𝑧
Intercepts: ( = 1/1, = 1/1, = 1/∞)
So, 𝑥 = 1, 𝑦 = 1, 𝑧 = 0
Thus the plane is (1 1 0).
Crystallography Note, Dr. Md. Muktadir Billah
Figure 17: (1 1 0) plane.
In Fig. 18, the plane cuts all three crystallographic axes.
1
1
1
𝑥
𝑦
𝑧
Intercepts: ( = 1/1, = 1/1, = 1/1)
So, 𝑥 = 1, 𝑦 = 1, 𝑧 = 1
Thus the plane is (1 1 1).
Figure 18: (1 1 1) plane.
In Fig 19, the plane cuts two of the reference axes, but not equidimensional.
1
1
1
𝑥
𝑦
𝑧
Intercepts: ( = 1/2, = 1/1, = 1/∞)
So, 𝑥 = 2, 𝑦 = 1, 𝑧 = 0, Thus the plane is (2 1 0).
Figure 19: (2 1 0) plane.
Crystallography Note, Dr. Md. Muktadir Billah
Fig. 20 lists all the (100) planes and Fig. 21 some of the (110) planes.
Figure 20: {1 0 0} plane.
Figure 21: {1 1 0} plane.
For the followings (Fig. 22), state a. the Bravais lattice b. the Miller Indices of the shaded plane
(remember to use right handed axes).
Crystallography Note, Dr. Md. Muktadir Billah
Figure 22: Crystallographic planes.
Answers:
a) Cubic , (1 1 1)
b) Cubic , (1 0 2)
c) Tetragonal , (1 1 0)
d) Triclinic , (1 0 0)
e) Orthorhombic , (0 0 1)
f) Orthorhombic , (0 1 2)
CLOSE PACKING OF SPHERES: FCC & HCP
The closest packing of spheres in two dimensions is realized by a hexagonal structure: each sphere is in
contact with six neighbored spheres (Fig. 38). In three dimensions one can now go ahead and add another
equivalent layer. However, for ideal packing it is necessary to shift this layer with respect to first one such
that it just fits into the first layer's gaps. Now the third layer can be either exactly above the first one or
shifted with respect to both the first and the second one.
Figure 37: CN = 12 for FCC.
So there are three relative positions of these layers possible (denoted by A, B and C). These could in
principle be combined arbitrarily but in nature the sequences A-B-A-B-A-B and A-B-C-A-B-C are most
common. The lattice that forms the latter one is just the FCC lattice which is one of the 14 Bravais lattices.
The other one is called HCP (hexagonal close packing) but not a Bravais lattice because the single lattice
sites (lattice points) are not completely equivalent! There are two different types of lattice sites which
have different environments. Therefore the hcp structure can only be represented as a Bravais lattice if a
two-atomic basis is added to each lattice site.
Crystallography Note, Dr. Md. Muktadir Billah
Figure 38: How close-packed structures of spheres can be constructed: In a first layer the spheres are
arranged in a hexagonal pattern, each sphere being surrounded by six others (A). Then a second layer with
the same structure is added. But this layer is slightly shifted and hence just filling the gaps of the first layer
(B). In a third step another equivalent layer is added filling the gaps just as before but now there are two
opportunities: Either this layer lies exactly above the first one (A) or it is shifted with respect to both A
and B and thus has its own position C.
The HCP is characterized by two nested hexagonal lattice that are shifted by three vectors (2/3, 1/3, ½)
(in the conventional unit cell basis) against each other (Fig. 39). The underlying lattice is not a Bravais
lattice since the individual lattice points are not equivalent with respect to their environments. But it can
be looked at as a hexagonal Bravais lattice with a two-atomic basis with the atoms sitting at the
positions (0, 0, 0,). Even though the term HCP is used for all those lattices, a close-packing of equal atoms
is only obtained for a certain ratio of the lattice constants,
𝑐
𝑎
= √8/3 = 1.633
Figure 39: (a) Position of the additional atom
(b) The hexagonal layers form a close-packed structure.
within
the
conventional unit
cell
and
Crystallography Note, Dr. Md. Muktadir Billah