PHYSICS OF CRYSTALS
1. Crystallography.
2. X-Ray Diffraction.
3. Defects in Crystals.
4. Diffusion.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 97
CRYSTAL STRUCTURE
10.1 Introduction
order. The crystalline structure, being orderly and
The matter, in general, can exist in three states
periodic arrangement of atoms, is characterized by
namely solid, liquid and gaseous. The matter can
sharp transition temperature (also called melting
be transformed from gaseous to liquid or liquid to
point) to liquid. However an amorphous solid,
solid by decreasing the motion of atoms through
when heated, softens over a range of temperature
cooling or any other process of retrieving energy
before
from the system. In a similar spirit, the transition
crystalline state is the lowest energy state of a
from solid to liquid or liquid to gaseous or solid to
solid.
getting
transformed
to
liquid.
The
gaseous state is possible by increasing the atomic
motion.
When liquid cools, two types of solids can be
formed. If the liquid is cooled slowly so that the
atoms get sufficient time to assume an orderly
arrangement, then a crystal structure results. On
the other hand if the temperature of liquid is
lowered abruptly thereby arresting the motion of
atoms before they reorganize themselves, then a
10.2 The Crystalline State
A solid is said to be in the crystalline state, if a
basic unit (which may be a single atom or a group
of atoms or an ion) is repeated periodically
throughout the volume of solid.
The idea of
periodicity can be had from the figure 10.1, shown
below.
mixed-up structure called glass or amorphous solid
may result. The glass is different from liquids in the
fact the relative motion of atoms is ceased, a
property which can be best described by viscosity.
In a crystalline structure, the orderly arrangement
or the periodicity of atoms is repeated over large
distance (i.e. very large number of inter-atomic
separations) which is also referred to as long range
order. In contrast to this, regularity or orderly
arrangement, if any, may exist in amorphous solid
but it does not persist beyond few inter-atomic
separations before it changes to another order of
arrangement. This is referred to a short range
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Figure 10.1: A crystalline solid with atoms
arranged periodically.
The distance between two nearest neighbours
along x-direction is a while that along y-direction is
b. The concept of exact periodicity is also
Page 98
expressed by saying that a crystal possesses the
the volume of the crystal. This is an inherent
translation symmetry, which means that if a crystal
limitation of a crystalline solid.
is translated by a vector joining two atoms as
(ii)
Another source of imperfection arises due to
shown in figure 10.1, its surroundings remains the
thermal vibration of atoms about their mean
exactly the same as it was before the translation
positions
operation. This periodicity, in crystalline solids,
vibration) at any finite temperature. These
prevails over a very large distance and is also
lattice vibrations distort the periodicity
referred to as long range order.
characterizing
The crystals can be grown from a gas by
continuously.
sublimation or from a liquid by crystallization.
(iii)
(also
referred
the
to
crystal
as
lattice
structure
The third kind of imperfection arises due to
When a crystal grows from a vapor or liquid, its
the impurity, which may be in traces but
continued growth depends upon the sufficient
remains in the crystal even when grown in
supply of growth units (i.e. the atoms required to
the best controlled conditions.
is
Hence the perfect crystal can be thought of an
overabundant, then crystal may grow extremely
approximation of very large crystal (whose ratio of
fast and in many directions at once thereby giving
surface area to volume is very small) maintained at
a branchlike appearance. Such a crystal, referred
very low temperature and having impurities in
to as poly-crystal, has many orderly arrangements
trace amounts. However these imperfections can,
of atoms, having different orientations, which
in certain cases, play a role in customizing the
interpenetrate into each other. In case the supply
material properties for technological applications.
make
up
its
structure).
If
the
supply
of growth units is insufficient then the crystal
starts dissolving back into its mother solution.
10.3 Basic Definitions
During the process of crystal growth, many errors
The
in periodicity occur. Strictly speaking one can not
grow a perfect crystal. The following can be
considered as limitations or imperfections in a
crystal:
(i)
study
of
crystal
structure
requires
acquaintance with certain terms which form the
language of crystallography. These terms are
usually referred to for one- and two-dimensional
cases for the simplicity of understanding but can
The surface of a crystal itself forms an
imperfection
as
the
experience
altogether
surface
a
atoms
different
environment than those lying deep within
be easily extended to the actual three dimensional
cases.
10.3.1 Crystal Lattice
In crystallography, only the geometrical properties
of the crystal are of interest, rather than those
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 99
arising from particular atoms (or group of atoms or
form one kind of lattice while A’, B’ and C’
molecule or ions) constituting the crystal. If each
constitutes another kind of lattice. These two non-
atom in the crystal is replaced by a geometrical
equivalent sets of points together constitute a
point located at its equilibrium position, then the
non-Bravais lattice.
resulting pattern is called a lattice.
Mathematically the lattice is periodic arrangement
10.3.2 Basis Vectors
of geometrical points in a line (linear or 1-
Consider the lattice shown in the figure 10.3
dimensional lattice) or plane (planar or surface or
below. Let A be the origin of coordinates for the
2-dimensional lattice) or space (space or 3-
lattice points. The position vector for any lattice
dimensional lattice).
point, in the plane, can be written as:
There are two different kinds of lattice:
Bravais lattice which has all lattice points
r
r
r
Rn = n1a + n2 b
(10.1)
equivalent and hence these points are occupied by
atoms of same kind. In the figure 10.1, the lattice
points A, B and C are equivalent and form the
Bravais lattice.
Non-Bravais lattice may have two or more sets of
equivalent lattice points, which implies that such a
lattice can be considered to be resulting due to
inter-penetration of two or more Bravais lattices.
In the figure10.2, the lattice points A, A’; B, B’; and
Figure 10.3: Bravais lattice demonstrating the
C, C’ form non-equivalent sets of points. Hence A,
basis vectors.
B and C
r
r
r
The a and b are called basis vectors. The Rn is
called
lattice
translation
vector which
can
accomplish translation from one lattice point to
another with the appropriate choice of integers set
(n 1 , n 2 ). Hence a Bravais lattice can be considered
to be invariant under group of all translations
Figure 10.2: The crystal lattice demonstrating the
non-Bravais lattice.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
expressed by equation (10.1). The choice of basis
vectors for a lattice is not unique but is guided by
Page 100
the convenience. For a three dimensional lattice
procedure
r r
r
a , b and c form the basis vectors and lattice
procedure is sequentially spelled out as follows:
case
of
two
Wigner-Seitz
cell.
The
through line segments to all its nearest neighbors.
(10.2)
Each line segment is bisected by a perpendicular
plane. These perpendicular bisector planes will
10.3.3 Unit Cell
In
called
Taking one lattice point as centre, it is connected
translation vector is expressed as:
r
r
r
r
Rn = n1a + n2 b + n3c
is
enclose a volume, which is referred to as Wignerdimensional
lattices,
the
parallelogram subtended by basis vectors as its
adjacent sides is called a unit cell. The unit cell is
usually the smallest area whose translation by
Seitz cell. This cell is characterized by occupying
minimum volume and also possessing one lattice
point per unit cell. This procedure is illustrated
diagrammatically in figure 10.4, shown below.
using equation (10.1) spans whole lattice once and
only once. The choice of the unit cell for one and
the same lattice is however not unique. However
all the unit cells representing same lattice must
obey the following characteristic features:
(i) They must have same area.
(ii) They should have one lattice point/unit cell.
The unit cell obeying the above mentioned
properties is called a primitive cell. However
certain choice of a unit cell can be such that (i)
area of unit cell is some multiple of that of
primitive cell and (ii) in addition to lattice points at
its vertices has one or more lattice points at body
or face centre. These are called non-primitive cell.
For a three dimensional lattice, the unit cell is a
parallelepiped in contrast to a parallelogram for
two dimensional case. Such a unit cell will be
characterized by three basis vectors.
The geometrical procedure of obtaining a primitive
cell from a given lattice was devised by Wigner and
Seitz. The primitive cell, so obtained from this
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Figure 10.4: The geometrical procedure to
construct Wigner Seitz cell.
10.5 Symmetries In Crystals
Crystals
have
a
regular
repetitive
internal
structure. The repetition of internal structure is
best described by concept of symmetry. The study
crystallography is concerned with exploring for
different symmetries exhibited by crystals and
cataloging them. This catalogue is further used to
infer crystal structure and also predict properties
on its basis. A symmetry operation is defined as
the action that leaves the system or object
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unchanged. A symmetry element can be a point or
symmetry is an essential requisite for a Bravais
line or plane through which the operation is
lattice.
performed
to
leave
the
crystal
structure
unchanged. The study of crystallography is based
10.5.2 Inversion Symmetry
upon the idea of translational, rotational and
The Bravais lattices usually have an inversion
inversion or mirror symmetry. Various kinds of
centre as there always exists a pair of lattice points
symmetries observed in crystals can be broadly
satisfying the mathematical relation:
classified in two catagories which are discussed in
r
r
R( n1 , n2 , n3 ) = − R( −n1 ,−n2 ,−n3 )
this section.
Point group consists of symmetry operations in
which at least one point remains fixed and
unchanged in space. The symmetry operations,
constituting point group, can be proper symmetry
if it doesn’t alter the handedness of the object
being operated otherwise it is referred to as
improper symmetry it is changes the handedness
of the object.
Space group comprises of both translational and
rotational symmetry operations of a crystal.
Each of the Bravais lattice is characterized with a
set of symmetry properties, which help in
understanding the underlying periodicity in lattice.
The various symmetries are discussed below:
10.5.1 Translation Periodicity
A lattice is said to possess translational symmetry
or periodicity if a translation represented by
r
r
r
r
Rn = n1a + n2 b + n3c
(10.3)
(10.4)
As can be understood from the translation
symmetry of bravais lattices, inversion centre will
always exist. The existence of inversion centre in a
non-bravais lattice is not an essential requisite.
10.5.3 Rotational Symmetry
As seen above, it is clearly evident that the
translational periodicity of lattice points is an
essential requirement for bravais lattice. Similarly
there exist one or more axes about which if the
crystal is rotated, it remains invariant (i.e the
environment, the basis atom or group of atoms
under rotation, observes before and after rotation
are identical). In case the character of the basis
does not change on rotation {i.e the right handed
(left handed) object gets repeated as right handed
(left handed)}, then the basis forms a congruent
set. If rotation of a crystal, leads to repetition of
congruent basis set, then it is called proper
rotation. However if rotation leads to change of
carries it from one lattice point in the crystal to
handedness of the basis (from right handed to left
another lattice point such that environment
handed character or vice versa), then it is referred
around two points is identical. The translational
to as an improper rotation.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 102
The axis of rotation is said to be n-fold if a rotation
Accordingly two opposite rotations by an amount
2π
keeps the unit cell
n
ϕ about two axes are shown in the figure 10.5.
Two new lattice points P and Q are produced
invariant. In bravais lattice 1-, 2- 3-, 4- and 6-fold
which must be equidistant from the original
rotation axes are allowed as per their symmetry.
lattice. The line joining P and Q must be parallel to
The
linear lattice translation t and also some integral
through an angle of
symbolic
representations
of
different
permitted rotational symmetries is given below:
P
Q
Figure 10.5: The diagram depicting the rotation of
lattice about a lattice position by an angle.
The 5 fold rotation and those higher than 6 folds
do not yield any symmetry. This is also
multiple of t. If it is not, then line joining P and Q is
qualitatively justifiable from the fact that if
not a translation of lattice and the linear array is
pentagonal unit cells or those with more than six
not periodic. Hence we can write, in general that
sides are packed then they result in either
PQ =mt which is related as:
overlapping or empty intermittent spaces. This
mt = t + 2t cos ϕ
m −1 N
=
cos ϕ =
2
2
clearly implies that such unit cells can’t repeat to
generate crystal structure.
10.5.3 Prohibited Rotational Symmetries
To discover the permissible throws that rotation
axes can have, let’s consider a linear lattice of
periodicity a and place an n-fold rotation axis at
each lattice point. Since n rotations by an amount
ϕ cause superposition, it does not matter
whether rotation proceeds in clockwise or
anticlockwise manner.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
m = 0,±1,±2,...
(10.5)
Since m is an integer so must be N. The possible
solutions of equation (10.5) will be exhausted by
range − 2 ≤ N ≤ 2 as cosine function’s value
can’t exceed unity. Hence we will have:
N
cosφ
Φ
N
-2
-1
180o
2
-1
-1/2
120o
3
0
0
90o
4
+1
+1/2
60o
+2
+1
o
6
o
360 or 0
1
Page 103
allowed rotational axes and are represented as
Hence 5 fold rotational symmetry and those higher
than 6 fold are not allowed.
(~ )
(~ )
one fold 1 , two-fold 2 , etc to distinguish from
proper rotation axes.
10.5.4 Reflection Symmetry
Roto-Inversion Symmetry: In the similar manner,
It is an imaginary plane passing through the unit
the improper rotation followed by inversion leads
cell, over which when a mirror reflection is
to another element of hybrid symmetry referred to
performed, it remains invariant. The triclinic lattice
as roto-inversion symmetry. Corresponding to five
has no reflection plane. The cubic lattice has in
rotational axes, there exist five roto-inversion axes
total nine reflection planes of which three are
parallel to faces while remaining six pass through
opposite edges.
10.5.5 Hybrid Symmetries
These are the symmetries arising out of complex
mathematical operations on the crystal lattice.
Such symmetries involve improper rotation as one
of the operations which results in change in
handedness of the object. The improper rotation
()
which are represented as one fold 1 , two-fold
(2 ) etc.
Screw Axis: A kind of hybrid symmetry is generated
by combining proper improper rotation with
translation parallel to the axis of rotation. The
combination of two operations is equivalent to a
screw motion which is shown for 3 1 screw axis in
the figure 10.6 below.
can be followed by another operation, which
restores the handedness of the object and hence
defines the symmetry. Some of these symmetries
are explained in the remaining portion of this
subsection:
Roto-Reflection Symmetry: The improper rotation
repeats a left handed object from right handed
one and vice versa. If the improper rotation
operation on a crystal is followed by reflection,
then crystal symmetry is observed. This is referred
to as roto-reflection symmetry. There exist rotoreflection axes corresponding to each of the five
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Figure 10.6: The diagram depicts the sequence of
operations underlying the 3 1 screw axis symmetry.
The overall operation involves the translation of
crystal lattice by 1/3 of basis vector magnitude
along the direction of screw axis followed by
rotation by 2π/3 of the lattice about the same axis.
Page 104
For an n-fold axis, each rotation through an angle
ϕ=
2π
n
must lead to translation T along
rotational axis. Hence one complete revolution
(involving n rotations) will result in integral
of the lattice in the direction of glide. Hence axial
glides
T=
will
correspond
r
r
r
r
translation vector Rn = n1 a + n 2 b + n3 c . Hence
we have:
correspond to
1
1
1
.
(a + c)
T = ( a + b) or (c + b) or
2
2
2
In a non-bravais lattice, each lattice point is
specified by the symmetry of basis, in addition to
(10.6)
As can be seen from equation (10.6), there are
screw
10.5.6 Point and Space Group Symmetries
basis or motif. The symmetry of such a lattice is
mRn
∴T =
n
possible
(
associated with the cluster of atoms called the
nT = mRn
eleven
translation
a
b
c
or or ) while diagonal glides will
2
2
2
number of the translations (nT), which further
must be equal to integral number (m) of lattice
to
axes
which
are
21 ,31 ,3 2 ,41 ,4 2 ,4 3 ,61 ,6 2 ,6 3 ,6 4 ,6 5 . In equation
(10.6), the subscript denotes the value of m and
number denotes the fold of axis of rotation. The
m=0 and m=n correspond to the cases of purely
proper rotations. The following symbol notations
are used to depict the screw axis symmetries.
the symmetry of bravais lattice. The symmetry of a
basis is called point-group symmetry, such as all
possible rotations, reflection and inversion centre,
which leave the basis invariant. There are possible
32 point-group symmetries which are consistent
with the requirement of translation symmetry for
the lattice as a whole. The limit on the number of
point-group symmetry is understood as the shape
and structure of basis can’t be arbitrarily complex
which would become incompatible with the
symmetry
of
inter-atomic
forces
operating
between the basis atoms (or atomic groups or
ions) at various lattice sites. The space-group
symmetries arise when rotation symmetries of
point-group
are
combined
with
translation
symmetry. This generates a large number of space
Glide Planes: Another kind of hybrid symmetry
group (72 exactly) symmetries. Further the
arises when reflection is combined with translation
consideration of hybrid symmetries leads to 230
parallel to the reflection plane. This forms the glide
space group symmetries.
plane. The translational component of T of a glide
plane is equal to one half of the normal translation
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 105
10.6 Characteristics of Unit Cell
10.6.1 Bravais Lattices in a Plane
A planar lattice which has all lattice positions
equivalent is referred to as a bravais lattice. Such a
lattice is characterized by two basis vectors (a and
b) and angle (θ) between them. The planar lattices
Centred Rectangle: It is characterized by basis
vectors a≠b which are orthogonal. It also forms a
non-primitive or conventional unit cell having
lattice points at centre of rectangle in addition to
those at four vertices. The figure 10.9 depicts the
centred rectangular lattice.
which obey the translation periodicity fall into five
broad classes:
Square Lattice: It is characterized by basis vectors
a=b which are orthogonal. It forms a primitive unit
cell and has lattice points only at vertices. Such a
lattice is shown in the figure 10.7 below:
Figure 10.9: The centred rectangular lattice with
its non-primitive unit cell.
Oblique: This unit cell is characterized by basis
Figure 10.7: The square planar lattice having a
primitive unit cell.
Rectangular lattice: It is characterized by basis
vectors a≠b which are orthogonal. It also forms a
vectors which may or may not be equal and angle
between them is other than 90o. This is a primitive
unit cell. The oblique lattice with its unit cell is
shown in the figure 10.10.
primitive unit cell and has lattice points at vertices.
Such a lattice is shown in the figure 10.8.
Figure 10.10: The oblique unit cell with its
Figure 10.8: The rectangular lattice with a
primitive unit cell.
primitive unit cell.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 106
Hexagonal: This is a planar lattice with a hexagonal
types of lattices. The unit cells of these lattices
unit cell characterized by a=b and θ=120o. This
obey the following:
shown in the figure 10.11.
•
They are simplest repeating unit in a crystal.
•
Its opposite faces are parallel.
•
Its edges are connected to equivalent points.
•
The lattices obey translation symmetry.
These lattices along with crystal systems are
tabulated below in table 10.1 along with their
Figure 10.11: The hexagonal planar lattice with
unit cell.
10.6.2 Bravais Lattices in Space
The unit cell in space is defined by its basis vectors
r r
r
a , b and c and inter-planar (inter-facial) angles
α , β and γ . These are shown for a three
dimensional unit cell in the figure 10.12 below:
Table 5.1: The tabulation of seven crystal systems,
falling into 14 Bravias lattices, along with their
shape characteristics and symmetries.
Figure 10.12: The characteristic parameters
characteristics and possible symmetries. The unit
describing the three dimensional unit cell.
cells of above mentioned fourteen lattices are of
three types which are: (a) Simple lattice having
10.6.3 Crystal Systems
lattice points only at vertices and is primitive in
In 1848, Auguste Bravais demonstrated that the
nature (b) Side centered having lattice points at
point symmetry operations in three dimensions
centre of a pair of opposite faces in addition to
lead to seven crystal systems having 14 distinct
those at vertices (c) Body centered having one
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 107
lattice point at body centre in addition to those at
a 2 = 4 RBCC
vertices. (d) Face centered having one lattice at
RBCC =
centre of each face in addition to those at vertices.
a 2
4
(10.8)
These 14 lattice types exhaust all the possibilities
In FCC, the length of face diagonal is exhausted by
for Bravais lattices. Any other type of lattice,
on radius each of the vertex atom and two radii of
reduces to one of them.
face centered atom. Hence we have:
a 3 = 4 RFCC
10.7 Cubic Unit Cell
RFCC =
The cubic unit cell is characterized by a=b=c and
a 3
4
(10.9)
α=β=γ=90o. Cubic unit cells exist in three forms as
10.7.2 Atoms per Unit Cell
(i) Simple Cubic Cell, which has one atom at each
In SCC, each atom at its vertex is shared by eight
vertex of the cube (ii) Body Centered Cubic Cell,
adjacent unit cells. Hence total number of atoms
which has eight atoms at vertices of the cube and
associated with each cell is 8 ×
one atom at the body centre of the cube (iii) Face
1
= 1.
8
centered Cubic Cell, which have eight atoms at
Similarly in BCC, vertex atoms together contribute
vertices of cube and one atom at centre of each
one atom per cell while the atom at the body
face of the cube.
centre is not shared by any other unit cell. Hence
number of atoms per unit cell are 8 ×
10.7.1 Atomic Radius
The atoms in a unit cell are closely packed and
hence touch each other. In a simple cubic cell, the
atoms forming the vertices of side of cube in same
plane touch each other. If a is the lattice constant
of the cell, then we can write:
1
+1 = 2.
8
In FCC unit cell, vertex atoms together contribute
one atom while atoms at center of each face are
shared by two adjacent unit cells. Hence atoms per
unit cell are 8 ×
1
1
+ 6 × = 4 . Hence FCC forms
2
8
most efficient way of arrangement of atoms.
a = 2 RSCC
RSCC =
a
2
(10.7)
The above expression gives the radius of atom in
SCC. Similarly for BCC, the length of the body
diagonal is exhausted by one radius each of vertex
atoms and two radii of body center atom. If lattice
10.7.3 Atomic Packing Factor (APF)
It is defined as the fraction of total volume of unit
cell which gets occupied by atoms. In general it can
be expressed mathematically as:
APF =
No. of atoms in unit cell × Vol. of atom
Volume of unit cell
(10.10)
constant of cell is a, then we have;
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 108
The APF for various types cubic cell are
triangular vertex pointing up are designated B,
( APF ) SCC = 0.52
( APF )
BCC = 0.68
( APF ) FCC = 0.74
while those with vertices pointing down are
(10.11)
labeled C as shown in the part (a) of the figure
10.13.
The above mentioned values show that atoms are
A second close-packed plane may be positioned
most efficiently packed in FCC lattice.
with the centers of its atoms over either B or C
type trigonal voids. Both of the arrangements are
10.8 Closest Packing Morphologies
Although their exists infinite number of possible
ways of closely packing the basis (which may be
atoms or ions or group of atoms or molecules) to
equivalent in all respects. Suppose that the B type
trigonal voids are chosen and the stacking
sequence is AB, which is shown part (b) of the
figure 10.13.
form a crystalline solid, only a very small number
of them get actually realized in nature. In fact most
of the closest packing morphologies observed in
nature are hexagonal and cubical close packing. All
the metallic crystals have their constituent atoms
arranged in one of these two structures, which are
the most efficient ways of packing of identical
spherical atoms. These two packing morphologies
can be conveniently depicted as close-packed
planes of atoms placed as layers one over the
other. The difference in two structures lies in the
Figure 10.13: The stacking sequence of closely
stacking sequence of stacking the closely packed
packed layer of atoms in hexagonal closed packing
atomic layers.
morphology.
10.8.1 Hexagonal Close Packing (HCP)
The real distinction lies in where the third close-
Let the centers of all the atoms lying in the one
packed plane is positioned. For HCP, the center of
close-packed plane be labeled A. Associated with
atoms of third layer is positioned directly over
this plane are two sets of equivalent triangular
original A positions. This leads to stacking
depressions or trigonal voids formed between
sequence ABABAB…… which repeats after every
three adjacent atoms, into which next close-
alternate layer of atoms.
packed plane of atoms may rest. Those having
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 109
10.8.2 Cubical Close Packing (CCP)
In other kind of close-packed structure, referred to
10.9.2 Number of Atoms per Unit Cell
as cubical (or Face Centered) close-packing, center
The top layer contains seven atoms. Each of the
of atoms of third layer are positioned over the C
atoms occupying the vertex position, in regular
sites of the first layer. This leads to a sequence
hexagon forming top or base plane, is shared by six
ABCABCABC….. wherein the stacking layers repeat
adjacent unit cells. The atom, lying at the centre of
after every third layer.
regular hexagon, is shared by two unit cells. The
three atoms with in the body solely contribute to
10.9 Hexagonally Close Packed Unit Cell
10.9.1 Structure of Hexagonal Unit Cell
In hexagonally closed packing, the top and bottom
face of the unit cell contains one atom at each of
the six vertices and an atom at centre of the
regular hexagonal face. Three more atoms are
positioned within the body which forms the
central plane of the cell. Each atom in top or
bottom plane (See figure 10.14) touches three
atoms in the layer just below or above its plane
and six atoms in its own plane. Further the atoms
touch each other along the edge of the regular
hexagonal unit cell are:
1
1
N = 12 × + 2 × + 3 × 1 = 6
6
2
(10.13)
10.9.3 Volume of Hexagonal Unit Cell
Let a and c be the edge and height of unit cell
respectively. Three atoms lie in a horizontal plane
at perpendicular distance of c/2 from the
orthocenter of alternate equilateral triangles
formed in the top or base plane of the hexagonal
cell. These three body atoms just rest in the
trigonal void formed by three atoms arranged at
hexagon. Thus we have
the vertex of the equilateral triangles formed in
a = 2R
⇒R=
the cell. Thus total number of atoms in a
a
2
base and top hexagonal planes of unit cell. The
(10.12)
figure 10.15 shown below, gives the bottom layer
of the HCP structure. The distance AZ between the
vertex atom A and body centered atom Z is
a = 2R .
In triangle ABY
cos 300 =
Figure 10.14: The atomic arrangement in a
hexagonal unit cell.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
AY
AB
AY = AB cos 300 =
a 3
2
(10.14)
Page 110
APF =
Va
π
=
= 0.74
V
3 2
(10.17)
It is surprising to observe that despite having
larger number of atoms per unit cell, the
hexagonal unit cell exhibits same APF as face
Figure 10.15: The geometry atomic arrangement in
centered cell involved in cubical close packing.
hexagonal plane and central plane.
10.10
In triangle AXZ, we have:
( AZ ) 2 = ( AX ) 2 + ( ZX ) 2
a
2
2a 3
⋅
=
AY =
3
3 2
3
AZ = a
ZX = c / 2
∴
in
Closest
Packed
Structures
AX =
a2 c2
+
a =
3
4
8
c
⇒ =
3
a
Voids
It has been observed that large number of known
inorganic compounds have structures which
slightly distorted closed packing or are related to
2
closest packing in another way. In such structures,
usually one kind of atoms are arranged as closed
(10.15)
packed structures while other kind of atoms
Volume of unit cell is given as the product of area
occupy voids created in closest packed structure of
of the hexagonal base and height of unit cell.
first ones. Common example of this kind of
Hence have
arrangement of atoms is metal oxides where
V = 6 × Area ( ∆AOB ) × c
oxygen atoms are arranged in closest packed
1
= 6 × AY × BO × c
2
structure while the metal atoms occupy voids in
this structure. Another good example is the steel
1
a 3 c
× ×a
= 6 × a ×
a
2
2
= 3 2a
3
alloy, where small sized foreign atoms (such as
carbon) get distributed in voids created in the
(10.16)
body centre cubic packing of iron. It should not be
borne in mind by the reader that the HCP or CCP
10.8.4 Calculation of Atomic packing Fraction
morphologies exhaust the efficient methods of
Volume of all the atoms in the hexagonal cell is
atomic packing in the solids. However it must be
given as
clearly borne in mind that small number of the
4
4 a
Va = 6 × πR 3 = 6 × π ( ) 3 = πa 3
3
3 2
available voids also get filled either through
The APF for the hexagonal cell is
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 111
natural arrangement of interstitial impurities or
imperfections.
10.10.2 Octahedral Void
Hence the voids play important role in the
If the trigonal void pointing up in one closest
formation of closely packed crystalline structure
packed layer is covered by another such void
and also influence the overall properties. Further
pointing down in an adjacent layer, then the void
the preference of alloys over pure metals can be
pointing upward gets surrounded by six atoms (see
owed to the better properties exhibited by alloys
figure 10.17). These six atoms are arranged at
due to occupation of voids by solute atoms,
corners of an octahedron and such a void is called
thereby freezing many degrees of freedom of
octahedral void.
neighbouring solvent atoms. This has play role in
customizing or tailoring the properties of crystals.
There are two kinds of voids that we come across
in closest packed structures. These voids and their
related aspects are discussed below:
10.10.1 Tetrahedral Voids
If the identical spheres are packed closely over a
plane then triangular voids are created between
the surfaces of three spheres in mutual contact
with each other. These are called trigonal voids. If
Figure 10.17: The arrangement of atoms around
another sphere is placed over this trigonal void
octahedral voids.
then it results in a void with four spheres around it
(see figure 8) and this is referred to as tetrahedral
10.10.3 Coordination Number of Voids
void. The centres of these four spheres occupy the
A sphere, in a closest packed layer, is surrounded
vertices of a tetrahedron.
by six trigonal voids of two kinds, which is clearly
revealed in the figure 10.17, of which three are
pointing in upward direction while remaining three
are pointing in downward direction. When the
second layer of atoms is added over this layer,
then atoms may occupy one of these two kinds of
Figure 10.16: The arrangement of atoms around a
tetrahedral void.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
voids in first layer. The voids occupied by second
layer
become
remaining
tetrahedral
three
in
unoccupied
nature
voids
while
assume
Page 112
octahedral shape. In similar manner, the layer
volume, hence one sphere in a closest packing
below the first layer will also result in three
belongs to the octahedral void. The radius of
tetrahedral and three octahedral voids. In addition
largest sphere that can be accommodated in the
to these voids, the considered sphere sits over a
octahedral void is 0.414R (R=radius of identical
tetrahedral void in both the layers below and
surrounding atoms).
above it. Hence an atom in the closest packing is
surrounded by eight tetrahedral voids and six
10.9.5 Voids in Body Centre Cubic Packing
octahedral voids. Since each tetrahedral void is
In the body centre cubic structure (see Figure
surrounded by 4 spheres so number of tetrahedral
10.18), the packing of atoms is not the closest but
voids per sphere is 2. In the similar sense, the
it becomes interesting to study the voids that
octahedral void is surrounded by six atoms leading
occur in such structures. Two kinds of voids occur
to one octahedral void per sphere in closest
in this type of close packing of atoms of which one
packing.
is octahedral in nature while the other is
tetrahedral.
10.10.4 Number of Voids around a Sphere
The number of spheres that are arranged around
each void is called its coordination number. If the
void is also represented as a sphere, then the
collection of spheres coordinating the sphere
(representing
void)
forms
the
coordination
polyhedron.
The tetrahedral void has a coordination number of
four spheres. Each of these spheres contributes its
one eighth of volume to the coordination
polyhedron which implies that one half of volume
of sphere belongs to a tetrahedral void. The radius
of largest sphere that can be accommodated in
tetrahedral void is 0.255R (R=radius of identical
surrounding atoms).
Figure 10.18: The BCC unit cell showing various
voids that exist.
The octahedral voids occur at the centre of face of
cubic cell (see figure 10.19). It is coordinated by
four atoms at vertices of the face and two atoms
located at the body centre of two adjacent cubes.
This void
The octahedral void is surrounded by six spheres
leading to coordination number of 6. Since each of
the six spheres contribute one-sixth of their
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 113
are four sphere surrounding each void and 24
voids surrounding each sphere, so the ratio of
spheres to voids is 1:6. These tetrahedral voids are
larger than octahedral voids.
Figure 10.19: Octahedral void in BCC unit cell.
forms irregular octahedron of atoms around it. It
can accommodate a largest sphere of radius
0.154R (R=radius of the coordinating sphere).
There are six spheres coordinating the void and 18
10.11
Indexing in Crystals
In describing physical phenomena, we require the
knowledge of directions and planes in crystals as
they are anisotropic in nature. The directions and
orientation of crystal planes are specified using
well defined prescription.
voids coordinating each sphere, so the ratio of
spheres to voids is 1:3.
10.11.1 Indexing of Crystal Directions
The second kind of void, tetrahedral in nature (See
Consider a straight line joining certain atoms in the
figure 10.20), exists in between two atoms forming
crystal. Consider one of the atoms as origin and
the edge of the cube and other two lying at the
translate from this atom to next using the
body
translation vector
r
r
r
r
Rn = n1 a + n2 b + n3 c . The
triplet of indices [n1 , n 2 , n3 ] in lowest possible
integer ratio gives the direction of line joining the
considered atoms. This direction does not mean a
particular straight line but a family of parallel lines.
Due to rotational symmetry of crystal, there may
exist several non-parallel directions which are
equivalent. For example in a cubic crystal the
Figure 10.20: The tetrahedral void in the BCC
unit cell.
centre positions of adjacent cubic cells. The
coordination of spheres around this void forms the
irregular tetrahedron. A sphere of largest radius
0.291R can be accommodated in this void. There
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
directions
[100], [010] and [001] are equivalent
and are represented together as 100 .
10.11.2 Indexing of Crystal Planes
The orientation of the plane in a lattice is specified
by its Miller indices.
Page 114
• Set up three coordinate axes along edges of the
inverted. This is useful as in the analysis of x-ray
diffraction spectrum of a crystal, one needs not
unit cell.
• Note the intercepts (x,y,z) of the plane (to be
indexed) on three coordinate axes.
the direct lattice but the reciprocal one, whose
lattice constants are related to these inverted
• Divide these intercept values with edge lengths
intercepts. The set of Miller’s indices are not
of unit cell along respective coordinate axis.
restricted to one plane but a family of parallel
x y z
, ,
a b c
planes. This is because all these planes scatter
incident x-rays in an identical manner as far as
• Invert the resultant triplet of ratios.
angle and phase of scattering are concerned and
lead to constructive interference of scattered x-
a b c
, ,
x y z
rays thereby yielding maxima of intensity in the
• Use a common multiplier (n) such that the
diffraction pattern.
triplet of inverted coordinates form the set of
lowest integers (h,k,l) given as:
10.11.3 Indexing in Hexagonal Cells
a
h = n x
k = n b
y
c
l = n
z
For hexagonal crystals, a problem arises in that
(10.18)
• Triplet (hkl) denotes the Miller’s indices for the
plane.
the some equivalent directions will not have same
set of indices. This is circumvented by using a set
of four indices instead of three and this system of
prescription is called Miller-Bravais indices. The
three axes a 1 , a 2 , a 3 are contained in the same
plane and are at 120o to each other (see figure
The set of Miller indices specify not one plane but
10.21). The z-direction is perpendicular to this
a set of parallel planes or family of planes which
basal plane formed by three axes.
are either parallel to each other or are equivalent
by virtue of symmetry. For example the planes
(100), (010), (001), ( 1 00), (0 1 0), (00 1 ) are equivalent
in a cubic cell and hence are represented in
condensed form as {100} .
The prescription of specifying Miller’s indices for
planes has good reasons. The intercepts of the
plane with three principal axes, expressed in units
of lattice constants along respective axes, are
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Figure 10.21: The axes for the Miller-Bravais
prescription of planes in hexagonal unit cell.
Page 115
The direction indices are obtained using same
procedure as for Miller’s indices. Out of the four
indices (h' , k ' , i ' , l ') ,
first
three
pertain
to
projections along three axes in basal plane and
fourth one is related to projection on the
perpendicular axis. These are related to three
index system of Miller as follows:
n
h ' = 3 ( 2h − k )
k ' = n ( 2k − h )
3
i ' = −( h + k )
l ' = nl
(10.19)
d hkl =
1
1
1
1
+ 2 + 2
2
z
x
y
(10.20)
The intercepts x,y and z are related to Miller’s
indices (h, k , l ) as
a
h = n x
k = n b
y
l = n c
z
(10.21)
Solving for x, y and z from above equation (10.21)
and inserting it in expression (10.20) for d hkl , we
The n denotes the integer which reduces the
get:
n
indices (h, k , l ) to smallest set of integers.
d hkl =
10.10.4 Inter Planar Spacing
where n denotes the order of the plane w.r.t the
In case of x-ray diffraction studies one need to
plane passing through origin having n=0.
h2 k 2 l 2
+
+
a 2 b2 c 2
(10.22)
know the spacing ( d hkl ) between the successive
planes labeled by same Miller indices (h, k , l ) . To
calculate this spacing, let’s consider another plane,
having same indices and oriented parallel to first
one, passing through the origin. The distance
between these two planes is d hkl , which is the
length of the normal between two planes. If this
normal makes angles α , β and γ with X, Y and Z
axes respectively, then we can write
d hkl = x cos α = y cos β = z cos γ
Since cos 2 α + cos 2 β + cos 2 γ = 1
2
d hkl d hkl d hkl
+
+
=1
x y z
2
2
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
10.11.5 Results from Miller’s Prescription
(a) The angle between orientation of two planes
(hkl) and (h’k’l’) is given as:
cos θ =
(h
2
1
h1h2 + k1k 2 + l1l2
+ k12 + l12 )(h22 + k 22 + l22 )
(b) The direction [uvw] is normal to the plane
(hkl) if u=h, v=k and l=w.
(c) The direction [uvw] is parallel to the plane
(hkl) if we have uh+vk+wl=0.
(d) Two planes (hkl) and (h’k’l’) are normal to
each other if they satisfy the condition that
hh’+kk’+ll’=0.
Page 116
(e) Two directions [uvw] and [u’v’w’] are normal
if
they
satisfy
the
condition
a = 2( RNa + RCl )
(10.23)
that
uu’+vv’+ww’=0.
10.12.2 Structure of CsCl
The crystal structure of CsCl, shown in figure
10.12 Simple Crystal Structures
10.23, has chlorine ions placed at corners of the
cube while the Cs+ ion is located at the body centre
10.12.1 Structure of NaCl
of unit cell. Each of the Cs+ and Cl- ions lie at vertex
The structure of NaCl is cubic and is such that
of its simple cubic cell but two unit cells are
along three principal directions, there is an
interpenetrated along lines directed along four
alternation of Na+ and Cl- ions as shown in the
body diagonals of the cube. There are, one Cs+ ion
figure 10.22 below. The cell is FCC in nature.
per unit cell located at point (0, 0, 0) and one Cl-
The position of four Na atoms are (0,0,0), (1/2,
ion per unit cell located at (1/2, ½, ½). This is a
½,0) , (1/2, 0, ½) and (0, ½, ½) while those of four
non-Bravais lattice composed of two simple cubic
chlorine atoms are (1/2, ½, ½), (0, 0, ½), (1/2, 0, 0)
lattices which are displaced relative to each other
and (0, ½, 0). The NaCl is a non-Bravais structure
by an amount equal to half the length of the body
composed of two interpenetrating lattices of Na+
diagonal.
and Cl-, which are displaced relative to each other
by a/2.
Figure 10.23: The crystal structure of CsCl.
10.11.3 Polymorphic Forms of Carbon
Diamond Crystal Structure
Figure 10.22: The Crystal Structure of NaCl.
In this crystal structure the unlike ions touch each
other and size of the unit cell can be obtained from
the relation given as:
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
The unit cell for the diamond crystal is FCC cell
(shown in figure 10.24) with a basis of two carbon
atoms associated with each lattice site. The
position of these two basis atoms are (0, 0, 0) and
(1/4, ¼, ¼). In this structure each atom has 4
Page 117
nearest neighbors, forming a regular tetrahedron
Structure of Graphite
having atom in question at its apex. Hence total
The graphite is a polymorph of carbon whose
number
are
crystal structure is different from that of diamond.
1 1
8 × + × 6 + 4 = 8 . The crystal structure is
8 2
The structure of graphite is more stable as
shown below in the figure 10.23.
temperature and pressure. The graphite is
To determine the APF of diamond crystal, the
composed of layers of hexagonally arranged
atomic radius is evaluated by considering that the
carbon atoms; within the layers each carbon atom
centre of two carbon atoms, located at vertex of
is bonded to three other coplanar neighboring
cube and at one fourth distance from this vertex
carbon atoms through strong covalent bonds. The
along body diagonal is equal to a. Hence we have:
fourth bonding electron participates in a weak van
3
a
4
3
∴R =
a
8
der Waals type of bond. It results in some of the
of
atoms
per
unit
cell
2R =
4
8 × πR 3
3
APF =
a3
π 3
=
= 0.34
16
compared to that of diamond at ambient
important
(10.24)
properties
of
graphite:
(i)
easy
interplanar cleavage (ii) excellent lubricant (iii) high
electrical conductivity along the direction parallel
to planes. The layered structure of graphite is
(10.25)
shown in the figure 10.25 below:
Figure 10.25: The arrangement of carbon atoms in
Figure 10.24: The arrangement of carbon atoms in
graphite structure.
the unit cell of diamond.
Structure of Fullerene
Although the unit cell of diamond crystal has 8
Another
carbon atoms per unit cell but still it is a loosely
discovered in 1985. It exists in discrete molecular
packed structure. This is owing to large unit cell of
form and consists of a hollow spherical cluster of
carbon.
60 carbons atoms which is represents a single
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
polymorphic
form
of
carbon
was
Page 118
molecule C 60 . Each molecule consists of group of
concentric cylinders of graphite sheets are also
carbon atoms that are bonded to one another to
found to exist.
form both hexagon (6 carbon atoms) and
pentagon
(5
carbon
atoms)
geometrical
configurations. One such molecule shown below in
the figure 10.26 consists of 20 hexagons and 12
pentagons, which are arrayed in such a manner
that no two pentagons share a common side. The
surface of this molecule appears to be like a soccer
ball and is
Figure 10.27: The structure of a carbon nanotube.
The nanotubes are extremely strong and stiff but
relatively ductile. These tubes have relatively low
densities due to which they are considered as
ultimate fiber having promising reinforcement
properties. The nanotubes have unique and
Figure 10.26: The arrangement of carbon atoms in
the fullerene molecule.
called
buckminsterfullerene
after
the
name
Buckminster Fuller who invented it.
Structure of Carbon Nanotube
Another molecular form of carbon has been
discovered and is fast turning into an important
material for future technological growth. It is
composed of a single graphite sheet rolled into a
cylinder and is capped on both its ends by C 60
fullerene hemispheres. This is called carbon
structure
sensitive
electrical
characteristics.
Depending upon the orientation of the hexagonal
units in the graphite plane with the tube axis, it
may behave as either metal or a semiconductor.
10.11.4 Structure of Zinc Blende (ZnS)
The crystal structure of ZnS is similar to diamond
except that now basis of two atoms is formed by
different atoms (Zn and S). Each cell contains four
ZnS molecules and each Zn or S finds itself at the
apex of tetrahedron formed by atoms of opposite
kind. The crystal structure of ZnS is shown in the
figure 10.28.
nanotube (See figure 10.27). Each nanotube is a
single molecule of millions of atoms and length of
this tube is much greater than its diameter (100nm
or less). Multiwalled nanotubes, consisting of
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 119
Silica
The simplest silicate material is silicon dioxide
(SiO 2 ). It is generated as a three dimensional
network of SiO44− tetrahedron when every oxygen
atom, lying at the vertex of a tetrahedron, is
shared by the adjacent tetrahedral structures. If
Figure 10.28: The arrangement of atoms of Zn and
S in unit cell of zinc blende.
these tetrahedrons are arranged in regular and
ordered manner, then it forms the crystalline
structure. There exist three crystalline polymorphs
10.11.5 Structure of Silica
of silica which are quartz, cristobalite and
Silicate Ceramic
tridymite. Their structures are complicated but
Silicates are materials primarily composed of
loosely packed leading to low relative density. The
silicon and oxygen and are found in soils, clays,
Si-O bond is highly strong which is reflected in high
rocks and sand. The basic unit of silicate ceramics
melting temperature of silica.
is SiO44− tetrahedron which has silicon sits at
centre of tetrahedron and is bonded to four
Silica Glasses
oxygen atoms lying at vertices of the tetrahedron
Silica can also exist as a non-crystalline solid or
(see figure 10.28). This arrangement has 4 negative
glass which has a degree of atomic randomness.
charges since each of the oxygen atoms requires
Such silica structures are also called fused or
one extra electron to achieve stable electronic
vitreous silica. These structures are also generated
configuration. This constitutes the basic unit which
by networking of SiO44− tetrahedron but results in
combines into one-, two- and three-dimensional
disorderly arrangement. Some other materials like
arrangements.
B 2 O 3 and GeO 2 also form glassy structures due to
their networking. Hence such materials (SiO 2 ,
GeO 2 , B 2 O 3 ) are also called network formers.
Certain oxides like CaO and Na 2 O when added in
glassy structures do not form polyhedral structures
but modify them. Hence they are also called
network modifiers. Still other oxides like Al 2 O 3 and
Figure 10.28: The structure of Silicate ion.
TiO 2 become part of polyhedral network and
stabilize
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
the
structure.
These
are
called
Page 120
intermediates. These are commonly used in
and 3Ǻ respectively, in a tetragonal crystal
amorphous silica to obtain different structures.
with c/a ratio of 1.5.
10. A plane includes points (0,0,0), (0.5,0.25,0) and
Numerical Problems Set I
1. Taking the edge of its unit cell of diamond to
be 3.57Ǻ long, evaluate its density.
2. Bunsenite (NiO) has FCC structure with O2- in
all face centre sites. The radii of Ni2+ and O2are 1.40Ǻ and 0.69Ǻ respectively. What is the
packing factor. What would be this factor if
radii ratio R(O)/R(Ni) = 0.41.
3. Titanium has HCP structure with a density of
4.5g/cm3. What is the volume of this cell.
4. Find the Miller indices of a set of parallel
planes which make intercept in the ratio 3a:4b
on the X and Y axes and are parallel to Z axis.
Given the lattice is simple cubic.
(0.5,0,0.5). What are its Miller indices.
11. In an orthorhombic unit cell a :b : c = 1:2:3.
The magnitude of a = 2Ǻ. What are the
intercepts of a plane having Miller indices
(230).
12. Calculate the density of MgO, which has
anions in FCC packing while cations are placed
in octahedral voids. The radii of Mg2+ and O2ions are 0.78Ǻ and 1.32Ǻ respectively.
13. A substance with FCC lattice has density
6250kg/m3
and
molecular
weight
60.2.
Calculate its lattice constant a.
14. Metallic iron changes from BCC to FCC at
910oC. The atomic radii in two structures are
5. A certain orthorhombic crystal has a ratio of a:
1.258Ǻ and 1.292Ǻ respectively. Calculate the
b: c of 0.428:1:0.376. Find Miller indices of the
volume change during this structural change.
faces with intercepts (i) 0.214:1:0.188 (ii)
Also calculate the change in density.
0.858:1:0.754 (iii) 0.429: ∞ : 0.126.
6. Draw (232) plane in the cubic crystal and (2 ī 1)
plane in orthorhombic crystal.
7. Lead has FCC structure with atomic radius
1.746Ǻ. Find the spacing between (200) and
(220) planes.
8. In a unit cell of SCC, find the angle between
the normal lines to the planes having Miller
indices (121) and (111).
9. Find the Miller indices of a plane that makes
intercepts on a, b and c axis equal to 3Ǻ, 4Ǻ
15. Zinc has HCP structure. The height of the unit
cell is 0.494nm. The nearest neighbor distance
is 0.27nm. The atomic weight of zinc is 65.37.
Calculate the volume of unit cell and density of
zinc.
16. Show that for
(a)
simple cubic system
d 100 : d 110 : d 111 = √6 : √3 : √2.
(b) face centred cubic system
d 100 : d 110 : d 111 = 1 : √2 : (√3)/2.
(c) body centred cubic system
d 100 : d 110 : d 111 = 1 : 1/√2 : √3.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 121
17. Show that for trigonal, tetrahedral and
accommodated in the octahedral void in the
octahedral voids can be just filled by spheres
BCC unit cell. Further prove that there is a
having radii not larger than 0.155r, 0.225r and
ratio of 1:3 spheres to octahedral voids in BCC
0.414r respectively, where r = radius of larger
unit cell.
sphere involved in closed packing of these
structures.
20. Show that a largest sphere of radius 0.291R
(R=radius of the coordinating sphere) can be
18. Show that the maximum radius of a sphere
accommodated in the tetrahedral void in the
which can just fit into the empty space at the
BCC unit cell. Further prove that there is a
centre of SCC structure is 0.732r, where r is the
ratio of 1:6 spheres to tetrahedral voids in BCC
radius of the atom.
unit cell.
19. Show that a largest sphere of radius 0.154R
(R=radius of the coordinating sphere) can be
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 122
X-RAY DIFFRACTION
intensity is mostly concentrated in central maxima
11.1 Introduction
When light passes through a fine aperture and falls
on a screen, it is observed that an interference
pattern is observed. This pattern is in the form of
central bright spot which represents the image of the
aperture and alternate dark and bright annular rings
of rapidly falling intensity. Such a pattern depicting
modification of light intensity distribution on passing
through an aperture is called diffraction and is
observed only if the size of aperture is comparable to
and falls off rapidly in secondary maxima.
The three dimensional crystal lattice has atoms
arranged in a regular and periodic manner which is
analogous to the system of multiple slits. The interplanar distance in crystals is of order comparable to
the wavelength of x-rays. When x-rays are incident
on the crystal, then they suffer diffraction from the
crystal lattice. The intensity distribution observed on
a screen is similar to that observed light illuminating
the wavelength of light. As interpreted by wave
a system of multiple fine slits.
theory of light, the diffraction phenomenon is
The diffraction pattern of x-rays, so observed, can be
described
as
modification
of
light
intensity
distribution resulting from interference between
light waves reaching the screen from different
portions of the same wave front. The typical
diffraction pattern observed on illumination of a
system of fine slits is shown in the figure 11.1
used to extract lot of information of crystal structure
such as shape and size of unit cell, arrangement of
contents in the unit cell, symmetries observed by the
unit cell and defects or imperfection or dislocations.
This has led to rapid growth of field of x-ray
diffraction in the pursuit to have insight into crystal
structures. Following x-rays as incident beam, the
electrons and neutrons have also been employed to
probe structure of crystals.
11.2 Bragg’s Law
Figure 11.1: The diffraction pattern observed due to
When monochromatic x-rays are incident on the
illumination of fine aperture.
surface of a crystal, reflection occurs only at certain
angles of incidence, which depend on lattice
It can be observed that the diffraction pattern differs
constant of crystal and wavelength of incident x-rays.
from that due to interference in the fact that light
This selective reflectivity can be understood by
considering the crystal as a set of parallel atomic
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 123
planes acting as partially reflecting mirrors. Each
have a common wave-front only at these angles.
atomic plane reflects the incident beam partially and
Consequently the amplitudes of scattered waves
reflected waves from different atomic planes
reinforce to yield maximum possible value. At other
interfere at the detector. The interference is
angles, the waves emanated from different atoms
constructive only if the path difference, between two
are not in phase with each other resulting in their
waves reflected from successive planes, is integral
destructive interference.
multiple of wavelength. The figure 11.2 shows the
The intensity of the beam decreases as order of
reflection from two successive atomic planes.
plane increases, which makes it more appropriate to
treat this process as diffraction rather than
interference. The wavelength (λ ) and glancing angle
(θ )
can be measured independently, hence the
Bragg’s law gives a method to evaluate lattice
constant (d ) of the crystal. The diffraction is possible
Figure 11.2: The diagram representing the reflection
only if λ < 2d , which implies that crystal diffraction
of x-rays from different atomic planes and
is not possible with optical waves.
subsequently interfering to produce the diffraction
However, the Bragg’s law is an oversimplified model
pattern.
to study the features of x-ray diffraction using crystal
The path difference
(∆ )
between rays 1 and 2
constituting the crystal, are considered to be
(shown in the figure 11.2) is given as:
∆ = AB + BC − AC ' = 2 AB − AC ' = 2d sin θ
Here d is the spacing between two successive planes
and θ is the glancing angle of incident ray with the
surface of crystal. The integer n = 1, 2, 3,.. refers to
the order of planes.
2d sin θ = nλ
as a three dimensional grating. The atomic planes,
(11.1)
The equation (11.1) is called the Bragg’s law. The
meaning of this equation is that the intensity of
scattered x-rays build up only at certain values of
angle θ , corresponding to a specific value of
continuous but the scattering of x-ray beam is
actually caused by discrete atoms. The diffraction
condition given in the Brigg’s law considers only the
path difference but phase component of these
waves are not included in the treatment. X-ray
diffraction from discrete atoms may not give
complete picture until and unless the structure of
atom, their arrangement in the unit cell and the
shape of the unit cell are incorporated appropriately
into the diffraction condition.
λ and d . This can be attributed to the fact that the
x-rays scattered from various atoms in the crystal
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 124
11.3 Theory of X-Ray Scattering
where f e is called scattering length of the electron,
11.3.1 Scattering from an Atom
D is the distance of observation point from the
In order to understand the phenomenon of
electron and k (=2π/λ) is the wave number of
diffraction of x-rays from a crystal structure, it is
scattered wave whose magnitude is same as k 0 due
instructive to study the scattering of x-rays by an
to elastic nature of scattering. The quantity D arises
atom. Further the knowledge of scattering from
in the denominator because of the fact that
individual atom can be utilized in order to assess the
scattered radiation will obey the inverse square law
situation arising from interference of radiation
of intensity.
scattered from different atoms constituting the
If we consider that incident wave acts on two
crystal.
electrons as shown in the figure 11.3, then both the
When x rays fall on an atom, the electrons suffer
electrons emit spherical waves and scattered
oscillations under the influence of alternating
intensity at any distant point is determined by sum
electric field of the radiation. These oscillating
of two partial fields scattered by each of them.
electrons behave like an oscillating dipole and emit
radiation in all possible directions. This is referred to
as scattering. As the electrons form a charge cloud
around the nucleus, so the scattering of x -rays from
an atom as a whole must account for phase of
radiation scattered from different portions of charge
cloud.
If we consider a single electron on which a
monochromatic plane wave is incident, then incident
Figure 11.3: The scattering of x-rays from the system
wave will have a form given as:
of two electrons.
u = Ae
r r
i ( k 0 ⋅ r − ωt )
(11.2)
This incident wave will be scattered uniformly in all
directions, thereby resulting in spherical wave font
emerging as a scattered wave which is expressed
The scattered wave observed at a distance D is given
as:
u' = f e
[
]
r r
A ikr ⋅ Dr
e
+ e i ( k ⋅ D + δ ) e − iωt
D
(11.4)
mathematically as:
u' = f e
A i ( kr ⋅ Dr −ωt )
e
D
(11.3)
where δ is the phase lag of electron 2, shown in the
figure 11.3, from that of electron 1 taken to be at
origin. The phase (δ ) is expressed as:
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 125
r
f = f e ∑ e irl ⋅ s
r
r r
δ = ( P1 M − P1 N )k = ( r ⋅ Sˆ − r ⋅ Sˆ0 )k
r
= r ⋅ ( Sˆ − Sˆ0 )k
(11.5)
The intensity I of the scattered wave is proportional
From the figure 11.3 shown above, we can write:
to the absolute square of scattering length and is
r r
r
s = ( Sˆ − Sˆ0 )k = k − k 0
expressed as:
s = k + k − 2kk 0 cos 2θ
2
2
0
(11.6)
r
r
Since scattering is elastic in nature, then k = k0
∴ s = 2k sin θ
r r
δ = r ⋅s
I = f
]
= f
2
e
∑e
r r
irl ⋅ s
2
(11.13)
The interference of waves, scattered from different
electrons, will be meaningful if and only if the
scattering electrons maintain a definite phase
The scattered wave amplitude is represented as
r r
A i ( kr ⋅ Dr − ωt )
u' = f e
e
1 + eir ⋅ s
D
2
l
(11.7)
(11.8)
[
(11.12)
l
relation with each other, otherwise time average will
(11.9)
smoothen the intensity variation of resultant
In deriving the equation (11.9), we have considered
scattered wave obtained by superposition. Further
that electron 1 is at origin of coordinates. If we chose
the electrons have been considered as discrete
an arbitrary position to be the origin of coordinates
points rather than spread of electronic charge cloud
r
r
such that electron 1 and 2 have positions r1 and r2
respectively,
then
the
above
equation
gets
generalized as:
u' = f e
A
e
D
r r
i ( k ⋅ D − ωt )
[e
r r
ir1 ⋅ s
+e
r r
ir2 ⋅ s
]
(11.10)
Now with the change of reference, the equation
over whole volume of the atom. If the continuous
distribution of electronic charge is included in this
treatment
r r
r rr
f = f e ∑ e irl ⋅s = f e ∫ ρ ( r )e is ⋅r d 3 r
The atomic scattering factor ( f a ) is defined as:
r
∫ ρ ( r )e
(11.10), representing scattered wave amplitude, can
fa =
be extended to any number of scattering electrons.
f = fe fa
Hence we can write for a system of electrons as:
u' = f e
r r
A i ( kr ⋅ Dr −ωt )
e
e irl ⋅ s
∑
D
l
r r
r r A
= f e ∑ e irl ⋅ s
e i ( k ⋅ D −ωt )
l
D
(11.14)
l
r r
is ⋅ r
d 3r
(11.15)
(11.16)
If the charge distribution around the nucleus of an
atom is spherically symmetric, then the integral
(11.15) representing the atomic scattering factor
(11.11)
Comparing equations (11.3) and (11.11), the
simplifies to the form:
fa =
∫
R
0
r Sinsr 3
4πr 2 ρ ( r )
d r
sr
(11.17)
scattering length of system of discrete electrons is
where R is the radius of atom. The atomic scattering
given as:
factor, given by equation (11.17), depends on the
following factors:
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 126
(i) Scattering angle through the factor s = 2kSin θ .
As 2 θ increases, so does s and hence atomic
(11.16) for a system of electrons, we can extend the
definition as:
f cr = f e ∑ eis ⋅ rl
r r
scattering factor decreases.
(ii) Wavelength of incident radiation: The term
Sinsr
is an oscillating in nature and its
sr
wavelength is inversely proportional to s. Hence
larger value of s yields shorter wavelength and
Sinsr
faster oscillations in
leading to a smaller
sr
(11.19)
l
In equation (11.19), the summation extends over all
the electrons in the crystal. However this summation
process is too complicated to be handled practically.
For this purpose, the equation (11.19) can be
rewritten
by
first
summing
the
scattering
contribution from all the electrons constituting single
atom and then summing over all the atoms in the
value of atomic scattering factor.
The above integral in equation (11.17) requires the
crystal lattice. The first summation can be introduced
knowledge of density function which can be
in (11.19) using the previously defined atomic
evaluated from considerations of forward scattering.
scattering factor and we get:
r r
f cr = f e ∑ f al e is ⋅Rl
In this limit we have:
(11.20)
l
θ = 0 , s = 0 and
fa =
∫ 4πr
2
Sinsr
=1
sr
ρ ( r )d 3 r = Z
The term f al denotes the atomic scattering factor
r
(11.18)
If density is expressed as the number of electrons
per unit volume, then atomic scattering factor for
forward scattering, given by equation (11.18), is
equal to number of electrons. This is because all the
radiations scattered from different electrons will
have the same phase and hence they will interfere
constructively.
11.3.3 Scattering from a Crystal
To obtain the amplitude of scattered waves from the
crystal, we have to consider the scattering
contribution from all the electrons constituting
various atoms of the crystal. In analogy with the
definition of scattering length given by equation
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
for the l th atom and Rl denotes the position of this
atom relative to some chosen reference or origin.
The equation (11.20) can more conveniently be
expressed as a product of two terms: Geometrical
structure factor (F) which is the sum of the
scattering contribution from atoms constituting a
unit cell and Lattice structure factor (S) denoting the
sum of scattering contribution from all the unit cells
comprising the crystal lattice. These factors are
expressed as:
F = f e ∑ f aj e
r r
is ⋅δ j
(11.21)
j
S =
∑e
r r
is ⋅Rl
(11.22)
l
In equation (11.21),
f aj and δ j represent the
atomic scattering factor and position for jth atom in
Page 127
cell. The factor S depends solely on the crystal
r r
Ns ⋅ a
Sin
2
S=
r r
s ⋅a
Sin
2
system while F depends upon geometrical shape as
The term S2 plays an important role in calculating the
well as contents of the unit cell. By segregation into
explicitly into structural properties of the lattice and
intensity of scattered x-rays and is expressed as:
r r
Ns ⋅ a
Sin 2
2
S2 =
(11.25)
r r
2s ⋅a
Sin
2
properties of constituent atoms.
It is a ratio of two oscillating functions with
unit cell relative to some chosen origin. In the
r
equation (11.22), Rl denotes the position of l th unit
geometrical and lattice structure factors, the
scattering
contributions
have
been
separated
(11.24)
numerator oscillating rapidly than denominator
11.3.3 Laue’s Equations
because of multiple N, which itself is a large number.
The lattice structure factor (S) plays an important
The quantity S2 depends on s ⋅ a as is shown in the
role in the discussion of x-ray scattering theory. Let’s
figure 11.5 below:
r r
evaluate the lattice structure factor for x-ray beam
scattered from a mono-atomic linear lattice with
basis vector a, as shown in the figure 11.4:
Figure 11.5: The diagram depicting the oscillations of
Figure 11.4: The scattering of x-rays from a
monoatomic linear lattice.
r r
This plot of S 2 vs s ⋅ a shows the following features
of interest:
The lattice structure factor becomes:
S=
N
∑e
r r
is ⋅la
S2 with s.a.
(i)
(11.23)
exist
principal
maxima
at
r r
r r
s ⋅ a = 0, 2π ,4π ,6π ,..... i.e. s ⋅ a = 2nπ
l =1
Here N is the total number of atoms constituting the
lattice. The series in equation (11.23) forms a
geometric progression with a common ratio e
There
r r
is ⋅ a
and
(ii)
In addition, there are present many secondary
maxima of much reduced intensity between
successive principal maxima.
its summation yields the resultant as:
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 128
(iii)
(iv)
If N is very large, as is expected for number of
will give rise to a diffracted beam on superposition.
atoms in a crystal, the secondary maxima have
The main problem at this juncture is to determine
negligible size as compared to the principal
the scattering vector
maxima.
conditions that:
The principal maxima get sharper in width as
r r
s ⋅ a = 2πh
r r
s ⋅ b = 2πk
sr ⋅ cr = 2πl
value of N increases.
As the principal maxima should exist with the
periodicity of 2π , we must have
r r
s ⋅ a = 2πh
h = any int eger
(11.26)
We can write:
(
)
r r 2π ˆ ˆ r
s ⋅a =
S − S 0 ⋅ a = 2πh
λ
(11.27)
for infinite number of directions forming a cone
whose axis lies along the lattice line (see in the figure
11.5 above). Hence (11.27) can be rewritten as:
(11.28)
where α 0 and α are the respective angles made
by incident and scattered x-rays with the lattice line.
Thus for a given value of h and α 0 , the beam
diffracts along all those directions for which α
satisfies the equation (11.28).
Similar arguments can be extended for three
dimensional lattice and obtain the conditions that:
a (cos α 0 − cos α ) = hλ
b(cos β 0 − cos β ) = kλ
c(cos γ − cos γ ) = lλ
0
(11.30)
11.4 Reciprocal Lattice
Let’s start with a lattice having basis vectors
The equation (11.27), for a given value of h, is valid
2πa
(cos α 0 − cos α ) = 2πh
λ
r
s , which satisfies the
(11.29)
The set of equations (11.29) gives Laue’s equations,
which define the various possible directions for the
constructive interference of partial waves scattered
r r
r
a , b and c and define another set of basis vectors
r r
r
A, B and C as following:
r
A =
r
B =
r
C =
r r
2π
r r r (b × c )
a ⋅ (b × c )
r r
2π
r r r (c × a )
a ⋅ (b × c )
r r
2π
r r r (a × b )
a ⋅ (b × c )
(11.31)
The vectors defined in equation (11.31) are called
reciprocal lattice vectors and the lattice so formed is
referred to reciprocal lattice. The translation vector
for such a basis of vectors is expressed as:
r
r
r
r
G = n1 A + n2 B + n3C
(11.32)
The reciprocal lattice vectors are related to direct
lattice vectors as:
r r r r
v
av ⋅ A = 2π
A⋅b = A⋅ c = 0
r r r r
r r
B ⋅ a = B ⋅ c = 0
b ⋅ B = 2π
r r r r
r r
C ⋅ b = C ⋅ a = 0
c ⋅ C = 2π
(11.33)
The following are the important properties of
reciprocal lattice:
from various constituent atoms of the crystal which
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 129
(i) Orthogonality of Basis Vectors: If the direct
lattice vectors form an orthogonal set of basis
vectors then reciprocal lattice vectors also form
crystal systems remains same although Bravais
structure may not be same.
(v) Interplaner Spacing: The reciprocal lattice vector
an orthogonal set with a parallel to A and so
r
r r r
Ghkl = hA + kB + lC is normal to the crystal plane
on.
defined by Miller’s indices
r
r
(ii) Generation of Reciprocal Lattice: The reciprocal
lattice corresponding to a direct lattice can be
formed by evaluating the reciprocal basis vectors
using equation (11.31). These basis vectors are
(d hkl )
(hkl ) .The
spacing
between two crystal successive planes
defined by Miller’s indices (hkl ) is related to
reciprocal lattice vector (G hkl ) as
r
used to obtain translation vector G , given by
d hkl =
equation (11.32), which further generates whole
2π
Ghkl
(11.35)
of the reciprocal lattice.
(iii) Brillouin
Zones:
The
primitive
cell
using
11.5 Diffraction Condition & Bragg’s Law
reciprocal basis vectors is formed by drawing
11.5.1: Bragg’s Law from Lattice Structure
basis vectors to connect the nearest lattice point
Factor
in their respective directions. Then planes are
The important applications of reciprocal lattice
drawn, which form perpendicular bisectors to
each basis vector. These planes enclose a
volume, which is the primitive cell for reciprocal
lattice, also called Brillouin zone. This primitive
cell is a parallelepiped of edges having length
2π 2π 2π
,
,
a b
c
and is also called the first
Brillouin zone (FBZ). In a similar manner higher
r
r
r
r
G hkl = hA + kB + lC
vector
lies in obtaining
summation involved in the evaluation of the lattice
factor S. For this purpose, let’s consider an arbitrary
r
r
r
r
lattice translation vector R = l1 a + l 2 b + l 3 c , then
we have:
N
S = ∑e
vr
iR ⋅Ghkj
j =1
= Nδ Rr ,Gr hkl
(11.36)
BZ can be constructed using multiple of
The quantity δ Rr ,Gr is called Kroneckor’s delta
reciprocal basis vectors in above prescription.
function which is non-vanishing and equal to unity
(iv) Reciprocal lattice for Various Crystal Systems:
The reciprocal lattice for SCC of lattice constant a
is SCC with lattice constant
2π
. The reciprocal
a
lattice for BCC and FCC are FCC and BCC
respectively. The reciprocal lattice for other
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
hkl
r
r
only when R = G hkl .
r r
Since S2=N2 when s ⋅ a = 0, 2π and lattice structure
factor (S) vanishes everywhere except when
r r
R = Ghkl . Hence we can write that scattering vector
(sr ) must satisfy the condition:
Page 130
r r
s = Ghkl
(11.37)
2π
s = 2k 0 sin θ = 2
λ
2π
= Ghkl =
d hkl
from certain family of planes is missing which is
sin θ
2d hkl sin θ = λ
However it is also observed that Bragg’s diffraction
because of vanishing value of geometrical structure
factor Fhkl .
(11.38)
Hence equation (11.38) gives the Bragg’s law.
Let’s
obtain
the
condition
using
geometrical structure factor:
Fhkl = f a ∑ e
r r
is ⋅δ j
(11.41)
j
r
r
r
r
r
s = Ghkl = hA + kB + lC
r
r
r
r
Let δ j = na + mb + zc
11.5.2 Bragg’s Diffraction Pattern
From the above analysis, the condition for the
diffraction is that when the magnitude of the
r
scattering vector s equals that of the reciprocal
( )
r
lattice vector G hkl then the lattice structure factor
assumes the value
(11.39)
given as:
2
(11.42)
(11.42) as:
The intensity of the scattered x-ray beam will be
2
Fhkl = f a ∑ e 2πi ( hn +km+lz )
position of atoms are (0,0,0) and (1/2, ½, ½), we get
Substituting this in equation given as:
I ≈ f cr ≈ N 2 Fhkl
cell. Hence we get general condition (11.41) as:
BCC Lattice: It has two atoms per unit cell, the
S hkl = N
f cr = Fhkl S hkl = NFhkl
r
where δ j denotes the position of atoms in the unit
(11.40)
Fhkl = f a ∑ (1 + eπi ( hn +km +lz ) )
(11.43)
The equation (11.43) gives non-zero values only
when (h+k+l) is even and zero if (h+k+l) is odd. Hence
for BCC lattice Bragg’s diffraction peaks are observed
The scattered x-ray intensity vanishes in all directions
for those planes for which the sum of Miller’s indices
except those in which lattice structure factor S is
(h+k+l) is even.
non-vanishing. These are therefore the directions
FCC Lattice: In a similar analysis for FCC lattice,
along which constructive interference between
Bragg’s diffraction peaks are observed for planes
scattered x-rays takes place. When the Bragg’s
having all the three Miller’s indices odd or all of them
condition is satisfied, then whole of the incident
are even.
beam is diffracted into a single beam producing a
spot on the film, which represents a set or family of
reflecting planes (hkl ) . If the crystal is rotated now,
a new set of planes will satisfy the Bragg’s condition
producing another spot representing this new set.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
11.5.3 Physical Meaning of Scattering Vector
We can give an important meaning to scattering
vector which is expressed mathematically as
r
r r r
s = k − k0 = G
Page 131
r r
r
k = k0 + G
r
r
r
hk = hk 0 + hG
certain limitations or demerits. These techniques are
discussed in subsections to follow:
(11.44)
According to equation (11.44), the term on LHS
denotes the momentum of scattered x-ray photons
while
the
terms
r
v
hk 0 and hG
denote
the
momentum of incident photon and momentum
associated with recoil of crystal due to collision of xray photon with it. Hence this equation represents
momentum conservation for interaction of x-ray
11.6.1 Rotating Crystal Method
This method is used for the analysis of structure of
single crystal. The experimental arrangement is
shown in figure 11.6 below. The crystal is usually
about 1mm in diameter and is mounted on a spindle
which can be rotated.
photon with the crystal lattice. Since the crystal is
massive, it carries negligible recoil energy and
collision is practically elastic in nature.
11.6 Experimental Techniques for X-Ray
Diffraction
The experimental techniques used to study crystal
structure have an objective of measuring the
following quantities in general:
(i) The scattering angle 2 θ between diffracted and
incident beams. By substituting sin θ in Bragg’s
law, one can determine inter-planar spacing and
the orientation of those crystal planes which play
an active role in diffraction.
(ii) The intensity of the diffracted beam, which
determines cell geometrical structure factor Fhkl
and hence yields information about arrangement
of atoms in the unit cell.
The different experimental techniques are employed
to quantitatively study the crystal structure. Each of
these techniques have their merits and also offer
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Figure 11.6: The experimental arrangement for
rotating crystal method of x-ray diffraction.
A photographic plate is placed on the inner side of
the cylinder concentric with the axis of rotation. A
monochromatic
beam
of
wavelength
(λ )
is
collimated and made to impinge on the crystal. The
specimen is then rotated, if necessary, until
diffraction
condition
is
λ and θ satisfy Bragg ' s law ).
satisfied
When
(i.e.
this
occurs, diffracted beam emerges from the crystal
and is recorded as a spot on the film. By recording
the diffraction pattern (both angle and intensities)
for various crystal orientations, one can determine
Page 132
the shape and size of unit cell and also the
part (b) of figure 11.7, the crystal exhibits 6-fold
arrangement of atoms inside the cell.
symmetry.
11.6.2 Laue’s Method
11.6.3 Powder Method
This method can be used for rapid determination of
This method is used to determine crystal structure
symmetry or orientation of a single crystal. The
even if the specimen is not a single crystal. The
experimental arrangement is shown in figure 11.7
specimen may be made up of fine grained powder
below. The white x-ray beam is made to fall on the
packed into a small cylindrical glass tube or it may be
crystal, which has a fixed orientation relative to the
polycrystalline having large number of micro-crystals
incident beam. Flat films are placed in front and
which are more or less randomly oriented. A
behind of the specimen. Since the incident x-ray
monochromatic x-ray beam impinges on the
have a continuous range of wavelengths, a family of
specimen and the diffracted beam spots are
crystal planes select a particular wavelength which
recorded on a cylindrical film surrounding it. Since
satisfies the Bragg’s law at the present orientation.
the crystallites are randomly oriented, there are
always enough of crystal planes which satisfy the
Bragg’s condition and hence diffracted beam emerge
at the corresponding angle. Since wavelength and
angle are measurable, the inter-planar spacing can
be determined. Other sets of planes lead to other
diffracted beams corresponding to different planar
spacing for the same wavelength. Thus one can
Figure 11.7: The experimental arrangement for
determine the lattice parameter accurately if the
Laue’s method.
crystal structure is known.
The diffracted beam emerges at a corresponding
Numerical Problem Set II
angle and gets recorded as a spot on the film. Since
1. A unit cell has the dimensions a=0.4nm,
the wavelength is not known, only ratios of interplanar spacing are measurable. These ratios enable
us to ascertain the shape of the unit cell but not its
size. If the direction of the beam is along the axis of
symmetry of the crystal, then the diffracted pattern
b=0.6nm, c=0.8nm,
α = β = 90 0 , γ = 120 0 .
Determine (i) lattice constants for the reciprocal
cell (ii) The volumes of reciprocal and real unit
cells (iii) The spacing between (210) planes. (iv)
Bragg angles for reflection from the (210) planes.
exhibits the symmetry of the crystal. As shown in
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 133
2. Show that the reciprocal lattice for FCC is BCC
monochromatic X-rays of λ = 0.58Ǻ. Bragg’s
and that for BCC is FCC.
3. Calculate the glancing angles of a cube (100) of a
rock salt (a=0.2814nm) corresponding to second
order
diffraction
maxima
6. An X-ray analysis of a crystal is made with
for
x-rays
of
wavelength 0.071nm.
reflections are observed at angles of 6.45o, 9.15o
and 13o. Calculate the inter-planer spacing.
7. From a powder camera of diameter 114.6mm,
using x-ray beam of wavelength 1.54Ǻ, the
4. Monochromatic X-rays of λ = 1.5Ǻ are incident
following s values (in mm) are obtained for a
on a crystal face having inter-planer spacing of
material : 86, 100, 148, 180, 188, 232 and 272.
1.61Ǻ. Find various orders at which Bragg’s
Determine the structure and lattice parameter of
reflection takes place.
the material.
5. A diffraction pattern of a cubic crystal structure
8. A diffraction pattern of a cubic crystal of lattice
of lattice parameter 3.16Ǻ is obtained with
parameter a = 3.16Ǻ is obtained with a
monochromatic X-ray beam of wavelength
monochromatic x-rays of wavelength 1.54Ǻ. The
1.54Ǻ. The first line on this pattern was observed
first four on this pattern were observed to have
o
at 20.3 . Determine the inter-planer spacing and
the respective values of 20.3o, 29.2o, 36.7o and
Miller’s indices of the reflecting plane.
43.6o. Determine the inter-planer spacing and
Miller indices of the reflecting planes.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 134
DEFECTS IN SOLIDS
defects are discussed in the remaining portion of this
12.1 Imperfections in Crystals
The perfect crystal is a mathematical idealization
which can be visualized as a regular and periodic
three dimensional array formed by repetition of a
section. Depending upon the type of abovementioned irregularity, the point defects are
classified as:
basis (which may an atom or ion or molecule). In real
12.2.1 Impurity or Interstitial Atoms
situations such crystals are difficult to grow and
This defect is caused by an irregularity in the crystal
employing
of crystal
structure localized to a lattice position or interstitial
growth can considerably reduce defects and
space, for example presence of a foreign or an
impurities but can never eliminate them completely.
impurity atom. Many impurity atoms of small size,
Certain defects such as binding surface of crystal,
present in atmosphere (such as hydrogen, oxygen
creeping of impurities or vacancies are associated
and nitrogen), get embedded in the crystal during
inherently with techniques employed in growing
the process of crystallization. Such atoms can occupy
crystals and metallurgical operations over them. In
regular lattice sites which are referred to as
many situations, the defects are induced in the
substitution impurities or they may lie in the
crystal to customize some of its properties.
interstitial spaces (called interstitial impurities). An
A defect is defined as a break or irregularity in the
appreciable number of substitution impurities may
long range order or periodicity observed in a crystal
be present if the size of impurity atoms is not far
structure. These defects can be classified as (i) point
different from that of regular atoms. In the similar
or zero order defects (ii) line or first order defects or
spirit, it can be argued that only small atoms can
dislocations (iii) surface or second order defects and
possibly be present in interstitial spaces.
sophisticated techniques
(iv) volume or third order defects.
12.2 Point Defects
These are defects which can be caused as a result of
(i) presence of foreign or impurity atoms in crystal,
(ii) presence of vacancies at lattice sites (iii)
displacement of regular lattice atoms to interstitial
Figure 12.1: The crystal structure showing different
spaces and (iv) formation of color centre. These
kinds point defects.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 135
12.2.2 Vacancies
crystal has N atoms, then number of vacancies, at
An empty lattice site, from where a regular atom has
any temperature T, are
been displaced or removed, leads to formation of
vacancy. These vacancies are created due to thermal
excitations at very high temperatures, where the
vibrating atoms acquire sufficient energy to leave
their regular lattice positions. Other possible sources
of vacancy production in the crystal are related to
the extensive plastic deformation and high energy
particle bombardment of crystals. The presence of
vacancy results in distortion in its immediate
neighborhood because of local lattice relaxation. It is
very common in the crystals to have large number of
vacancies in them at room temperatures.
The energy required to create a vacancy, called the
N v = Ne
−
Qv
kT
(12.1)
Here N o denotes the total number of lattice
positions in a unit volume of crystal, N v is the total
number of vacant positions in unit volume of the
crystal, Q v is the activation energy required for
formation of vacancy and k is the Boltzmann
constant. Hence as T increases, number of vacancies
increase
in
exponential
manner.
At
room
temperatures, the fraction of lattice position
remaining vacant may be as high as 0.1%.
12.2.3 Defects in Ionic Crystals
formation energy (Q v ), can be evaluated by the
In ionic crystals, the lattice positions are occupied by
knowledge of coordination number of atom when
ions. These ions can get dislocated from their regular
placed at regular site and average bond energy. The
lattice positions either due to thermal excitations or
average number of vacancies created in crystal
high energy particle bombardment. If the displaced
specimen, due to thermal excitations, can be
atom diffuses to the surface of crystal, then such a
estimated from the knowledge of the formation
defect is referred to as Schottky defect. In such a
energy of vacancy, temperature of crystal and total
defect the cation and anion vacancies are produced
number of atoms constituting the specimen. The
together so as to maintain charge neutrality.
vibrating atoms in the crystal, at any temperature T,
have an average energy 3kT which is usually
significantly smaller than energy Q v (formation
energy) for room temperatures. However the
vibrating atoms in the crystal can have all possible
energies and their distribution is guided by MaxwellBoltzmann distribution. Hence the probability that a
vibrating atom has energy Q v is e
−
Qv
kT
. Hence if a
Figure 12.2: The crystal lattice depicting ionic
defects.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 136
However in certain circumstances, the removed
due to anion (halogen ion) vacancy created by
atoms move through the interstitial spaces within
irradiation of the crystal. The valence electron of cat-
the deeper volume of the crystal and such a defect is
ion (alkali ion) suffers attraction and gets trapped in
called Frenkel defect. The Schottky defects are
vacant site. The orbital motion of this electron
characterized by presence of pair of ion vacancies
around vacancy (acting as positively charged) is
while the Frenkel defect is characterized by presence
quantized and has a series of discrete levels in
of ion vacancy which corresponds to missing ion and
optical frequency region. Hence such an orbiting
an interstitial ion displaced from the regular lattice
electron suffers excitation on absorption of light and
site. The Frenkel defects are energetically less
re-emits
favorable due to strain energy required in distorting
characteristic color to the crystal.
on
its
de-excitation
thereby
giving
the lattice at the site of vacancy as well as around
the atom displaced to the interstitial space. The
Frenkel defects are usually caused either thermally
at temperatures close to the melting point or due to
external radiation which may knock out an atom
from regular lattice site. The Schottky and Frenkel
defects are shown in the figure 12.2.
12.2.4 Color Centers
These set of defects arise because of charge
compensation. If a crystal is irradiated by neutrons,
electrons and gamma-rays, their color suffers a
12.3 Line Defects
The line defects, also called dislocations, arise due to
presence of additional linear array of atoms
extending over a considerable distance inside a
lattice. These defects are of two types (a) Edge
dislocation and (b) screw dislocation.
12.3.1 Edge Dislocation
The figure 12.3 shown below depicts the edge
dislocation
change. For example a diamond crystal turns blue on
electron bombardment and brown on neutron
irradiation. The color change takes place because of
crystal damage caused by irradiation, leading to
formation of various types of point defects. The color
Figure 12.3: The diagram depicts the edge
centre is a lattice defect that absorbs the visible
dislocation.
light.
The most extensively studied color centre is F-centre
The extra half plane of atoms AB has been
(F stands for Farbe meaning color in German) which
embedded into the upper half of the crystal as
is common in alkali halide crystals. They originate
shown in the cross-sectional view of crystal in the
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 137
figure 12.3. The half plane terminates at A, which in
of dislocation and it lies at end of the step BAB’. This
two dimensions, represents a linear array of atoms in
is referred to as screw dislocation because of the fact
direction normal to the plane of paper and this array
that if one moves in the atomic plane around the axis
form the edge referred to as the dislocation. The
of dislocation, the plane actually follows a spiral (see
region in the neighborhood of dislocation suffers
figure 12.5).
distortion which extends over many atomic lengths.
The upper region, having extra half atomic plane
introduced, suffers compression while lower region
is somewhat expanded. The strain caused by such a
dislocation fades far away from the site of
dislocation. The formation energy or elastic strain
energy for this dislocation is considerably high and
consequently such dislocations can’t be caused by
thermal means but they are formed as a result of
mechanical treatment such as hammering or
bending.
12.3.2 Screw Dislocation
The screw dislocation is observed in the figure 12.4
shown below. To visualize the nature of this
dislocation, let’s consider a planar cut ABCD made in
the crystal. The left side is then slipped up with
respect to the right side. The line AD is called the axis
Figure 12.5: The diagram depicts the spiral ramp
formed as a result of screw dislocation.
In such a dislocation, the crystal suffers shear strain
due to slippage of plane. The volume of the crystal
does not suffer any alteration unlike in the edge
dislocation. These dislocations involve large energy
of formation and are not possible by thermal means.
12.4 Surface Defects
In such defects, the crystalline irregularity extends in
plane and hence are referred to as surface defects.
The following figure 12.6 describes the classification
into different kinds of surface defects. The surface
defects may arise within pure crystal or even may
Figure 12.4: The diagram depicts the screw
arise because compositional changes observed
dislocation.
within in a crystal.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 138
The crystals can be grown from a gas by sublimation
or from a liquid by crystallization. When a crystal
grows from a vapor or liquid, its continued growth
depends upon the sufficient supply of growth units
(i.e. the atoms required to make up its structure). If
the supply is overabundant, then crystal may grow
extremely fast and in many directions at once
thereby giving a branchlike appearance. Such a
Figure 12.6: The diagram depicts different types of
crystal, referred to as poly-crystal, has many orderly
surface defects.
arrangements
of
atoms,
having
different
orientations, which interpenetrate into each other.
The surface defects are classified into the following
Each of the orderly arrangement is called a crystal
types:
grain and common interface of two adjacent grains is
12.4.1: External Surface Imperfections
referred to as a grain boundary. One such
The binding surface of the crystal itself, as shown in
polycrystalline structure composite of many grains in
the figure 12.7, is a surface defect as surface atoms
shown in the figure 12.8 below:
have much different environment than those lying in
deeper volume of the crystal.
Figure 12.7: The diagram depicts a different atomic
environment for surface atom which is a king of
imperfection.
12.4.2 Grain Boundaries
Most of the solids are not purely mono-crystalline
rather are polycrystalline in nature. A given crystal is
usually composed of large number of single crystal
grains which are joined together.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Figure 12.8: The polycrystalline structure composed
of main grains having different orientations.
The shape of the grain is usually influenced by
presence of its surrounding grains. The boundary
atoms in two randomly oriented grains can’t have
perfect complement of surrounding atoms and a
transition occurs in atomic packing which is the
cause of imperfection. At each grain boundary, the
crystal
suffers
change
in
orientation
and
Page 139
consequently this boundary behaves as a surface
12.4.3 Twin Boundaries
defect. The grain boundaries are the surfaces along
Another planar surface imperfection is a twin
which the crystal can be cleaved with little effort
boundary. The atomic arrangement on one side of
because of weak cohesivity. At such boundaries, the
twin boundary is the mirror reflection of the
atomic arrangement suffers a disorder and these
arrangement on the other side. Twin boundaries
regions are characterized by low atomic densities.
appear in pairs such that orientation change,
Due to disorder and low density, these regions have
introduced on one side, is restored on other side.
high mobility, diffusivity and chemical activity. These
The region between twin boundaries is called
defects have a strong influence on properties of a
twinned region. The twin boundary is shown in the
poly-crystal.
figure 12.10 below:
The boundaries where the crystal orientation suffers
a change of more than 10o are called high angle
grain boundaries. These boundaries have less
efficient packing of atoms and atoms forming the
boundary have higher energy than those within the
grains. In certain cases of surface defects, the
orientation difference between two adjacent grains
is less than 10o, which is referred to as low angle
grain or tilt boundaries. The tilted boundaries can be
described by an array of edge dislocations lying one
above the other as shown in the figure 12.9 below.
Figure 12.10: The diagram depicts the twin
boundary.
The phenomenon of twinning occurs when a crystal
planes suffer a slip because of applied shear stress
(mechanical twin which is quite common in BCC and
HCP
crystal
structures)
or
during
annealing
treatment (annealing twin observed in FCC crystals).
12.4.4 Stacking Faults
A stacking fault occurs when in the regular stacking
Figure 12.9: The low angle grain boundary can be
described as an array of close lying edge dislocations.
sequence of the crystal, one plane is out of
sequence, while the lattice on either side of the fault
is perfect. These faults occur during crystal growth
phase when close packed layers grow over one
another. It results when a layer starts incorrectly. For
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 140
example in FCC close packing, instead of regular
sequence of layers ABCABCABCA…. the sequence
12.5 Volume Defects
The volume or bulk defects in the crystals arise due
gets changed to ABCABABCA….. The plane of atoms
to the following reasons:
A after second B layer in latter sequence constitutes
• Occurrence of small electrostatic dissimilarity
a thin region of HCP stacking in FCC crystal which
lead to significantly weak cohesivity. Such type of
interruption of stacking is also referred to as
deformation fault.
between stacking sequence of close packed
planes in metals.
• Presence of a large void due to group or cluster of
atoms missing.
• Presence of group of atoms which are termed as
12.4.5 Slip in Crystals
precipitates.
The crystals, when subjected to shear stress, result in
restricted motion of certain atomic planes relative to
others. This is same as distortion produced in the
deck of cards when certain cards are forced to
protrude out. The phenomenon of slip occurs along a
plane in the crystal and this is referred to as slip
plane. If the atomic planes suffer shear stress
exceeding the critical value, then one or more atomic
plane get displaced relative to other. The slip in a
crystal is realized through motion of dislocations of
both type and this phenomenon plays an important
role in defining the plastic deformation in the
crystals.
It has been observed that the crystal planes having
low values of Miller’s indices are characterized by
greater atomic planar density and are more prone to
slip. The BCC and FCC systems have large number of
such planes and hence they can suffer extensive
plastic deformation resulting in greater malleability
and ductility. However the HCP crystal systems have
lesser number of such planes and hence are brittle in
12.6 Strength of Metals
The metals have many distinctive properties such as
elasticity, ductility and malleability, which make
them highly useful for industrial applications. The
metals in their pure state are soft and easily
deformable. We wish to understand the mechanical
strength of metals using the microscopic picture and
study the influence of dislocations and impurities on
their mechanical strength.
When a metal is subjected to small amount of stress,
they suffer strain but recover completely when the
imposed stress is withdrawn completely. This is
referred to as elasticity of metal. During application
of stress in the elastic region of metal, the atomic
planes are pulled apart and their separation is
proportional to the magnitude of stress applied. On
withdrawal of applied stress, these planes are
restored
to
their
equilibrium
positions
and
orientations.
nature.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 141
However when the metal is subjected to shear stress
crosses the atom of stationary plane, it has to
beyond the elastic limit, then it suffers shearing
overcome a potential barrier. Since the slippage of
strain and does not recover even when the stress is
one atomic plane over the other will lead to
plastic
overcoming very large number of such potential
deformation of metal. During plastic deformation,
barriers so the critical stress becomes much higher
various regions of crystal appear to slip against each
than observed experimentally. Consequently this
other. The crystalline regions undergoing slippage
picture is not capable to explain surprising softness
form slip bands which slide past each other.
and a general tendency of metals towards plastic
To understand the slip process, let’s consider that a
flow.
force F is applied over a surface having area of cross-
When a metal crystal is grown a large number of
section A such that it makes an angle θ with the
dislocations are formed in the process. If an edge
normal to surface. The shear force will be given by
dislocation is considered in the volume of crystal,
the component Fsin θ while the effective area over
then applied shear stress shifts the extra half plane
A
. Hence shear stress is given as:
cosθ
by one inter-atomic separation past the lower half
completely
removed.
which it acts is
τ=
τ=
This
is
called
F sin θ F
= sin θ cos θ
A
A
cos θ
σ
2
sin 2θ
plane. Consequently the half plane gets shifted to
surface of crystal (see figure 40).
(12.2)
Hence the shear stress is maximum when force is
inclined at an angle of 45o to the normal to the
surface. As a result, slippage must occur when
τ >
σ
2
along that plane. Since crystals are
anisotropic, there are plane which have lower critical
Figure 12.11: The slip of crystals described as motion
of edge dislocations under the influence of shear
stress.
value of shear stress to suffer slip and they usually
have higher atomic concentrations. They are called
easy slip planes.
If one plane of atoms is to slip over the other,
theoretically predicted values of critical stress are
much higher than those observed experimentally.
This is because each time atom of slipping plane
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Since the dislocation region is already strained so
they are not in stable equilibrium and lower stress is
required to shift the upper half plane. This requires
much lower shear stress than sliding of one whole
atomic plane over the other. Hence dislocations
make the crystal much softer and increase their
Page 142
tendency towards plastic flow. A metallic crystal, if
greater resistance to plastic deformation and is said
grown free of dislocations, will require very large
to be pinned down by impurities or defects. This
shearing stress for causing even very small plastic
phenomenon is called impurity hardening.
deformation. This explain as to why the perfect
Another interesting effect observed is that metals
crystals are highly resistant to plastic deformation.
become
The impure metals or alloys are usually stronger than
phenomenon is called work hardening. This happens
pure metals because of the fact the impurity atoms
because the strain produces very large number of
or defects have a strong tendency to accommodate
dislocations. These dislocations interfere with each
in the region of dislocations. The dislocation is now
other and prohibit the relative motion of each other.
hardened
after
being
strained. This
loaded with impurity atoms or defects and it
becomes difficult to move it readily as it has to carry
the impurities along with it. Such a crystal shows
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 143
DIFFUSION
13.1 Introduction
In a sample having non-uniform composition of
pharmaceutical applications is undisputed through
the drug therapy curing a specific portion of body.
certain kind of atoms, a concentration gradient is set
in, which leads to migration of atoms from region of
13.2 Basics of Diffusion
their higher concentration to the region of lower
The process of diffusion involves two components: (i)
concentration. This phenomenon is referred to as
solvent is the medium in which atoms of similar or
Diffusion and continues till atomic distribution
different kinds are to be diffused and (ii) solute are
becomes homogeneous throughout.
the atoms of one or more kinds which are to be
The diffusion process plays an important role in the
diffused into the environment of solvent. The solute
field of metallurgy and fabrication of extrinsic
atoms migrate and distribute themselves in solvent
semiconductors. The technology of controlled
in process of which they may become interstitial
diffusion plays a sensitive role in accomplishment of
impurity (occupying interstitial spaces in solvent
desired devices. The metals, in their pure form, are
lattice) or substitution impurity (substituting the
soft and do not possess large tensile strength. Their
solvent atom from its regular position in lattice).
properties can be modified by the process of alloying
The process of diffusion can be classified into two
in a controlled manner. The desired properties of
types which are (a) Self-diffusion in which solute and
alloys are achieved by proper choice of solute (atoms
solvent atoms are identical. For example diffusion of
to be diffused) and their amount to be diffused. This
copper in copper (b) Inter-diffusion in which the
is accomplished through sophisticated techniques
solute atoms are different from the solvent atoms as
and instruments. Further the dynamical processes of
in the case steel where solvent is iron whereas solute
bio-systems utilize the phenomenon of diffusion in
atoms are those of carbon and many other elements.
wide variety of ways which include the role of blood
The phenomenon of diffusion of solute atoms into
circulation leading to nourishment of various body
the solvent environment may occur through the
organs and also waste disposal from them. The
following mechanisms:
control of heart beat through hormonal diffusion and
(a) Vacancy mechanism where the solute atoms lie in
digestion of food through enzymatic action are some
the vacancy and moves to other available
of the examples. Further the role of diffusion in
vacancy. Atoms can move from one site to
another if there are vacancies and the atom has
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 144
sufficient energy to overcome a local activation
structure of solvent, imperfections in crystal and
energy barrier. The activation energy for diffusion
concentration of diffusing species.
is the sum of energy required to form a vacancy
The Fick’s law can be understood to a greater extent
and further cause its motion.
through a kinetic model. Let’s consider a one
(b) Interstitial Mechanism is applicable in case of
small atoms like hydrogen, helium, carbon,
dimensional flow of solute atoms in normal direction
through section S as shown in the figure 13.1 below:
nitrogen diffuse through the interstitial spaces in
a
a crystal. The activation energy for diffusion is the
energy required for these atoms to squeeze
through the small voids between the host lattice
S
atoms.
13.3 Macroscopic Model of Diffusion
c1
c2
13.3.1 Fick’s First Law
Figure 13.1: The diagram showing the one
The basic process of diffusion kinetics is described by
dimensional flow of atoms through the intermediate
Fick’s first law which states that the diffusion current
plane S through their jumping in either of two
(J) is directly proportional to the concentration
possible directions.
∂c
.
∂x
gradient
Hence
we
express
this
The concentration of solute atoms in adjacent planes
mathematically as:
J = −D
∂c
∂x
are indicated by c 1 and c 2 (c 1 > c 2 ) respectively. The
(13.1)
The negative sign indicates that solute atoms
migrate down the concentration gradient. The
diffusion current is defined as the mass of solute
atoms migrating in one second through a unit area of
cross-section placed normally to the direction of
flow. The constant D is called diffusion coefficient
atoms on both the planes jump randomly to right as
well as left. The atoms of plane 1 which jump to right
and those of plane 2, which jump to left will cross
the section S in opposite directions. Since the net
flow of solute atoms will be towards right so we can
write diffusion current as:
J =
1
1
n1v − n2 v
2
2
(13.2)
which is characteristic of solute atoms and solvent
In the equation (13.2), n 1 and n 2 represent planar
environment, depends upon temperature of solvent
densities of solute atoms at planes 1 and 2
medium, mechanism of diffusion, type of crystal
respectively while v represents the jump frequency.
The factor of half accounts for equal probability of
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 145
jumping in one of the two possible directions leading
13.3.2 Fick’s Second Law
to net rate of flow through section S. The
The diffusion of solute atoms in solvent environment
concentrations c 1 and c 2 are defined as number of
obeys the continuity equation given as:
solute atoms per unit volume at plane 1 and 2
respectively. Hence we can have relation between
atomic density and concentration of solute atoms at
a plane as n = ca. Using this relation in (13.2), we get:
J =
1
av ( c1 − c2 )
2
(13.3)
∂c ∂J
+
=0
∂t ∂x
(13.7)
Putting equation (13.1) in (13.7), we get:
∂c
∂
∂c
=−
− D
∂t
∂x
∂x
∂c
∂ 2c
=D
∂t
∂x 2
(13.8)
If the concentration does not vary rapidly between
The equation (13.8) is called the Fick’s Second law.
the planes 1 and 2, then we can write as:
The solution of equation (13.8) for the one
c1 − c2 = − a
∂c
∂x
dimensional case of diffusion being studied is
(13.4)
x2
A − 4 Dt
e
c( x, t ) =
t
Putting (13.4) in (13.3), we get:
∂c
1
J = − a 2v
∂x
2
(13.5)
(13.9)
The depth of diffusion as a function of time is
Comparing the equations (13.1) and (13.5), the
evaluated by root mean square value of x and is
diffusion coefficient D in one dimensional case can
given as:
∞
be expressed as:
D=
1 2
va
2
x=
(13.6)
∫x
0
2
∞
c( x, t )dx
(13.10)
∫ c( x, t )dx
If we consider the three dimensional case then the
atom can jump in any random direction and its
1
probability of crossing the section S will be
6
0
The above integral, with use of equation (13.9),
yields x as
x = 2Dt
(13.11)
instead of half and the diffusion coefficient will
Hence the diffusion front advances into the solvent
become:
environment with distance traversed proportional to
D=
1 2
va
6
square root of time. The diffusion coefficient being
(13.7)
-20
The values of D can range from 10
–10
-50
very small, the process of diffusion is an extremely
2
m /s
slow in nature.
which correspond to jumping frequencies ranging
from 10-20Hz to 1Hz. This indicates that the diffusion
is an extremely slow process.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 146
13.4 Microscopic Model of Diffusion
In the microscopic picture, the diffusion is migration
of atoms between adjacent lattice sites. The atomic
motion is random but leads to slow and organized
macroscopic
motion
down
the
R2
have x = y = z =
3
2
2
2
and equation (13.13)
takes the form as
x=
concentration
nd 2
3
(13.14)
gradient. To understand the atomic picture of the
Using equations (13.11) and (13.14), we get diffusion
diffusion process, let’s consider one dimensional
coefficient as:
lattice with an atom placed at position x=0. The atom
D=
hops to adjacent lattice site at frequency v but at
vd 3
6
(13.15)
every lattice site the direction of hopping is
The equation (13.15) gives same result as obtained
completely random. Hence it may jump to left or
using macroscopic model.
right with equal probability. If d is the distance
between adjacent sites, then position ( xn ) after n
13.5 Diffusion: Temperature Effects
jumps will be given as:
The diffusion in a solid may take place through any of
xn = d 1 + d 2 + .......... + d n
(13.12)
The d i may be +d or –d depending upon the
direction of hopping. Since the overall motion of
atoms leads to displacement so we can express it as
root mean square value of ( xn ) . Hence we can write
as:
the following mechanisms:
(a) The solute atom, if small in size that it can be
accommodated in interstitial spaces, jumps from
one interstitial space to another. This is common
case in the diffusion of carbon atoms in the
formation of steel alloy.
(b) The solute atoms, in case of substitution alloys or
x=
x n2 = nd 2 =
L
n
(13.13)
impurities, diffuse through the exchange of
positions with other neighbouring regular atoms.
where L=nd is the distance travelled in n jumps along
(c) The migration in case of vacancy is by succession
one direction only. For n very large, we have x << L
of many atoms filling the vacancy which migrates
which implies that diffusion is very slow due to
in this process in opposite direction.
random motion of atoms.
All these mechanisms may occur simultaneously too.
If the atom is allowed to move in all directions then
When the atom or vacancy moves from one position
x will be replaced by R since the atoms would now
to another, it pushes other atoms sideways resulting
move radially outwards. In symmetrical case we will
in local elastic strain. As a result, the process of atom
making a transition from one to other lattice site
requires at least energy equal to elastic strain
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 147
energy, which is also referred to atomic transition
energy (Em ) and is of the order of 1eV roughly.
Another energy required by atom to make transition
is that required to create vacancy at its initial site
(Ev )
and is called vacancy formation energy. Thus
total energy, also called activation energy, required
is sum of energy of formation of vacancy and atomic
transition energy i.e. Q = Ev + Em .
D = D0 e
−
Q
kT
ln D = ln D0 −
(13.16)
Q
kT
Hence one observes that the diffusion coefficient
varies rapidly with temperature. The plot of
1
gives D o as intercept and -Q/k as
T
log e D vs
slope.
The atom oscillates at frequency v0 around its
Numerical Problem Set III
equilibrium position and energy possessed by it is far
1. Calculate the equilibrium number of vacancies
less than that required to cause jumping to
per cubic meter for copper at 1000oC. The
neighbouring site. However for a fraction of time,
energy for vacancy formation is 0.9eV/atom; the
equal to Boltzmann factor e
−
Em
kT
, the atom gets
sufficient energy required for atomic transition.
Hence if the atom oscillates at frequency v0 , then
jumping frequency at any temperature is v = v0e
−
Em
kT
.
This atom, in three dimensional crystal, can jump to
any of its z neighbouring sites provided there is a
vacancy in them. The probability of a neighbouring
site having vacancy is e
−
Ev
kT
. Hence total probability
of jumping is
− Em
v = v0 e kT
= zv0 e
−
If
are 63.5 g/mole and 8.4 g/cm3 respectively. (2.2
x 1025 /m3)
2. Calculate the percentages of interstitials and
vacancies at the melting point in Cu (1356K). The
formation energy for these defects in Cu are
4.5eV and 1.5eV respectively.
3. A plate of iron is exposed to carburizing (carbon
rich)
atmosphere
on
one
side
and
a
decarburizing (carbon deficient) atmosphere on
the other side at 700oC. If the condition of steady
Eν
− kT
ze
(E +E )
− m ν
= zv0 e kT
state is achieved, calculate the diffusion flux of
carbon through the plate if the concentration of
carbon at position of 5mm and 10mm beneath
Q
kT
The diffusion coefficient D is given as:
D=
atomic weight and density (at 1000oC) for copper
−
Q
kT
1
zv0 a 2 e
6
1
D0 = zv0 a 2
6
the carburizing surface are 1.2 and 0.8kg/m3,
respectively. Assume diffusion coefficient= 3 x
10-11m2/s at this temperature. (J=2.4 x 10-9
kg/m2s)
then
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 148
4. The diffusion coefficients for diffusion of Cu in Al
at 500oC and 600oC are 4.8 x 10-14 m2/s and 5.3 x
10-13 m2/s that will produce the same diffusion
result (in terms of concentration of copper at
some specific point in Al) as a 10h heat
treatment at 600oC. (110.4h)
5. Nitrogen from a gaseous phase is to be diffused
into pure iron at 700oC. If the surface
concentration is maintained 0.1wt%N, what will
be concentration 1mm from the surface after 10
hours. The diffusion coefficient for nitrogen in
iron at 700oC is 2.5 x 10-11m2/s.
6. At approximately what temperature would a
specimen of γ − iron have to be carburized for 2
hours to produce same diffusion result as at
900oC for 15 hours.
7. The activation energy for the diffusion of Cu in
Ag is 193kJ/mole. Calculate the diffusion
coefficient at 1200K if diffusion coefficient at
1000K is 1.2x10-14m2/s.
8. The diffusion coefficients for carbon in nickel is
at 600oC and 700oC are 5.5 x 10-14 m2/s and 3.9 x
10-13 m2/s respectively. Find the value of Q and
D o . What is the diffusion coefficient at 850oC.
Dr. J.K.Goswamy’s Lecture Notes: Physics of Crystals.
Page 149
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