CRYSTALLOGRAPHY

1. BY PRASHANT KUMAR
ASST. PROFESSOR
MITS GWALIOR

3.  The group of atoms is called the basis; when repeated in space it forms the
crystal structure.
 The basis consists of a primitive cell, containing one single lattice point. Arranging
one cell at each lattice point will fill up the entire crystal.

5. A crystal is one in which atoms or molecules are in three
dimensional periodic arrangement.
 The periodicity may be same or different in different
directions.
The periodic positions of the atoms or molecules are
called space lattice or crystal lattice.
“The geometrical representation of a crystal structure in
terms of lattice points is called space lattice”

6. The smallest portion of the crystal which can generate the complete
crystal by repeating its own dimensions in various directions is
called UNIT CELL

7. Axial lengths: a, b, c
Interaxial angles : , ,  

8.  AUGUST BRAVAIS in 1848 mathematically proved that there are 14 distinct ways to
arrange points in space.These 14 lattices corresponding to 7 crystal system are
known as BRAVAIS LATTICE.

16.  CONTRIBUTION by atoms at CORNER = 1/8 atom
 Contribution by atoms at Face center =1/2 atom
 Contribution by atoms at Body center= 1 atom
 Contribution by atoms at Base Center/End Center=1/2 atom
 NO. of faces = 6
 NO. of corners=8
 NO. of Bases =2 (top/bottom)

17. Type of unit cell No. of atoms At
corners
No. of atoms In
faces
No. of atoms At the center Total
Simple Cubic 8 x 1/8 = 1 0 0 1
Body Centered
Cubic(B.C.C)
8 x 1/8 = 1 0 1 2
Face Centered
Cubic(F.C.C)
8 x 1/8 = 1 6 x 1/2 = 3 0 4
End Centred Cubic 8 x 1/8 = 1 2 x 1/2 = 1 0 2

20. Procedure for finding Miller Indices
Step 1: Determine the intercepts of the plane
along the axes X,Y and Z in terms of
the lattice constants a,b and c.
Step 2: Determine the reciprocals of these
numbers.

21. Plane ABC has intercepts of 2 units along X-axis,3
units alongY-axis and 2 units along Z-axis.
DETERMINATION OF ‘MILLER INDICES’
Step 1:The intercepts are 2,3 and 2 on the three axes.
Step 2:The reciprocals are 1/2, 1/3 and 1/2.
Step 3:The least common denominator is ‘6’.
Multiplying each reciprocal by lcd,
we get, 3,2 and 3.
Step 4:Hence Miller indices for the plane ABC is (3 2 3)

22. Plane parallel toY and Z axes
In this plane, the intercept along X axis is 1 unit.
The plane is parallel toY and Z axes. So, the intercepts
alongY and Z axes are .
Now the intercepts are 1, and .
The reciprocals of the intercepts are = 1/1, 1/ and
1/ .
Therefore the Miller indices for the above plane is (1 0
0).

 



25. In such cases (when plane is passing
through origin)the plane is translated
by a unit distance along the non zero
axis/axes
and the Miller indices are computed
1. Intercept of shifted plan =
( 1 ) 
2. Reciprocal 1 1 1
(
1
)
 
Miller Indices (0 1 0)

38.  BRAGG’S DIFFRACTION GRATING

40. • Why X-Ray is selected
for crystallography ?
Ans: because for proper
observation of
diffraction pattern Size of
Obstacle (here it is
Crystal) ~ should be
comparable to
wavelength of ray to be
diffracted
wavelength of X rays
~1Angstron
Atomic distances in
crystal~few angstron

41.  Diffraction is the slight bending of light as it passes around the edge of an object.
 The amount of bending depends on the relative size of the wavelength of light to
the size of the opening.
 If the opening is much larger than the light's wavelength, the bending will be
almost unnoticeable.

46.  DEFNITION Point defects,as the name implies, are imperfect point-like regions in the
crystal.Typical size of a point defect is about 1-2 atomic diameters
 As the name suggests the size of the defect is very small.
 It is denoted by a point having no dimension.
 Dimension of point defects = 1-2 atomic diameters

47.  A vacancy is a vacant lattice position from where the atom is missing.
 It is usually created when the solid is formed by cooling the liquid.
 There are other ways of making a vacancy, but they also occur naturally as a result
of thermal excitation, and these are thermodynamically stable at temperatures
greater than zero.
 At equilibrium, the fraction of lattice sites that are vacant at a given temperature (T)
are
So no of vacancy is dependent on temperature T
exp
o of vacancy
of lattice point
Q=
constant
T=temp in kelvin
V L
V
L
the energy required to move an atom from the interior of a crystal to its surface
Q
N N
kT
N n
N
k boltzmann
no
 
  
 




48.  An interstitial atom or interstitialcy is an atom that occupies a place outside the normal
lattice position.It may be of the same type of atom as the rest surrounding it (self interstitial)
or a foreign impurity atom.
 Interstitialcy is most probable if the atomic packing factor is low.
 Probablity of interstitialcy is highest in S.C > BCC > FCC (Bcz atomic packing factor of
SC<BCC<FCC)
 Diameter of foreign atom < diameter of lattice atom/host atom
 Valency of foreign atom > valency of host atom
 Eg. Steel carbon occupies interstitial sites in Fe (iron) lattice increases strength of resulting
material as number of bonds increases

49.  Atom exchange with impurities
 Rate of substitution depends on
 i) Activation energy to exchange
Condition for substitution
 Diameter of impurity/dopant ~ diameter of host atom
 Valency of impurity/dopant >/< vacancy of host atom
Eg Doping in Semiconductor
1. Acceptor impurity doping in silicon by 13group
elements (Al, Ga, In)
2. Donar impurity doping in Silicon by 15th group
element (Phosphorus)
• Eg.2 if we add copper to nickel, copper
atoms will occupy crystallographic sites
where nickel atoms would normally be
present.
• The substitutional atoms will often
increase the strength of the metallic
material.

50.  In ionic crystals, existence of point defects is subjected to the condition of charge
neutrality.
 There are two possibilities for point defects in ionic solids.
 - when an ion displaced from a regular position to an interstitial position creating a
vacancy, the pair of vacancy-interstitial is called Frenkel defect. Cations are
usually smaller and thus displaced easily than anions. Closed packed structures
have fewer interstitials and displaced ions than vacancies because additional
energy is required to force the atoms into the interstitial positions.
 - a pair of one cation and one anion can be missing from an ionic crystal, without
violating the condition of charge neutrality when the valency of ions is equal.The
pair of vacant sites, thus formed, is called Schottky defect.This type of point
defect is dominant in alkali halides.These ion-pair vacancies, like single vacancies,
facilitate atomic diffusion.

51. Frenkel Defects: some ions
,generally cations due to
smaller size, move from
regular lattice point to
interstitial sites
Schottky defects:
When equal no of cations
and anions are missing from
regular lattice point.
Both these defects maintain
charge neutrality.

52. PARAMETERS SchottKy defects Frenkel defects
REASON OF OCCURENCE Missing of equal no of cations
& anions from lattice sites
(Vacancy)
due to missing of
ions(generally cations) from
lattice sites & these occupy
Interstitial sites
(Interstial)
EFEECT ON DENSITY OF
CRYSTAL
DECREASES NO EFFECT
GENERAL OCCURENCE FOUND IN STONG IONIC
COMPOUNDS
Eg. NaCl
CaCl
Found in crystal with Low
coordination No.
Eg. ZnS AgI
Differences:

53. 1.Usually dislocations have a mixed character and Edge
and Screw dislocations are the ideal extremes
2.Some types of dislocations can be visualized as
being caused by the termination of a plane of atoms in
the middle of a crystal.
CAUSE OF
DISLOCATIONS:
1. SHEAR STRESS

54. STEPS OF MODELLING
SCREW DISLOCATIONS
1.TAKE A PERFECT
CRYSTAL
2. CUT THE CRYSTAL
HALF WAY
3.SKEW/SHEAR THE
CRYSTAL BY ONE ATOM

55.  If we follow a crystallographic plane one revolution around the axis
on which the crystal was skewed, starting at point x and traveling
equal atom spacings in each direction, we finish at point y one atom
spacing below our starting point.
 If a screw dislocation were not present, the loop would close. The
vector required to complete the loop is the Burgers vector b.
 If we continued our rotation, we would trace out a spiral path. The
axis, or line around which we trace out this path, is the screw
dislocation.
 The Burgers vector is parallel to the screw dislocation.

58.  Dislocation is the boundary between slipped and unslipped part of the crystal
lying over slip plane

60. POSTIVE EDGE
DISL.
( EXTRA HALF
PLANE INSERTED
IN UPPER HALF)
NEGATIVE EDGE DISLOCATIONS
(EXTRA HALF PLANE INSERTED IN
LOWER HALF PLANE)
STEPS TO
MODEL EDGE
DISLOCATION
1.TAKE PERFECT
CRYSTAL
2.CUT CRYSTAL
INTO TWO
EQUAL HALVES
3.INSERT AN
EXTRA HALF
PLANE OF
ATOMS

61.  The bottom edge of this inserted plane represents the edge dislocation line.
 If we describe a clockwise loop around the edge dislocation,starting at point x and traveling
an equal number of atom spacings in each direction,we finish at point y one atom spacing
from the starting point.
 If an edge dislocation were not present, the loop would close.The vector required to
complete the loop is, again,the Burgers vector (b). In this case, the Burgers vector is
perpendicular to the dislocation.
 Top half plane of
atoms=compression
 Bottom half plane of
atoms=tension
 Burger vector is
perpendicular to
dislocation

64.  The dislocation moves due to the change in the position of the top half of the atoms
relative to the bottom half. Because the motion of the atoms results in
thermodynamically stable states as the dislocation moves, the structural changes
do not reverse when the shear stress is removed.To put it in continuum mechanics
terms: we've plastically deformed our material because we moved the dislocation.
This mechanism of plastic deformation by motion of dislocations is called slip

65.  Non Stoichiometric defect: In non-Stoichiometric defect the ratio of number of
cations and anions is disturbed by addition or removal of the ions.
 They are of two types:
 Metal deficiency defect: In this the solids have less number of metals relative to the
described Stoichiometric proportion.
 Metal excess defect: There are two types of metal excess defect:
 Metal excess defect due to anionic vacancies:This kind of defect occurs due to
absence of anions from its original lattice site in crystals.This position is occupied
by electrons.
 Metal excess defect due to presence of extra cations at interstitial sites: In this type
of defect extra cations are released by the compound on heating .This extra cation
occupies the interstitial sites in crystals and the equal number of electrons goes to
neighbouring interstitial sites.

66.  1 Dimensional Defect or Line defect
 Atoms surrounding the defects are Misaligned.
 EDGE DISLOCATION: 1.An edge dislocation is a defect where an extra half-plane
of atoms is introduced midway through the crystal, distorting nearby planes of
atoms.
 2.MOVEMENT OF EDGE DISLOACTION:When enough force is applied from one
side of the crystal structure, this extra Half plane passes through planes of atoms
breaking and joining bonds with them until it reaches the grain boundary.

68. 1.External Surface:
 Surface atoms have unsatisfied atomic bonds, and higher surface energies, than the bulk atoms.
 To reduce surface free energy, material tends to minimize its surface areas against the surface tension
(e.g. liquid drop).
2.Grain boundaries:
 regions between two adjacent grains
 slightly disordered
 low density in grain boundaries
 high mobility
 high diffusivity
 high chemical reactivity

69. 3.Tilt Boundary:
 A Tilt Boundary, between two slightly
mis-aligned grains appears as an array of edge
dislocations.
 Rotation axis is parallel to the boundary plane
4.Twist Boundary:
 Rotation axis is perpendicular to the boundary plane
 Low angle grain boundaries, that appear as an array of Screw
dislocations.

70. 5.Twin boundaries:
 These are the boundaries in the grains at which the atomic arrangement on one
side of the boundary is the mirror image of the atoms on the other side .
 May be produced by shear deformation.
 The region between the pair of boundaries is called the twinned region.

71. 6.Stacking Faults:
 Formed by fault in the stacking sequence of atomic planes in crystals.
 Considering stacking arrangement in FCC: (ABC sequence repeats)
ABCABCABCABCABC
ABCABC BCABCABC
 This thin region is a surface imperfection and is called Stacking faults.

72.  Volume defects in crystals are three-dimensional aggregates of atoms or vacancies.
 It is common to divide them into four classes in an imprecise classification that is
based on a combination of the size and effect of the particle.
 The four categories are:
 precipitates: which are a fraction of a micron in size and decorate the crystal;
 dispersant: which vary in size from a fraction of a micron to the normal grain size
(10-100µm), but are intentionally introduced into the microstructure;
 Inclusions: vary in size from a few microns to macroscopic dimensions, and are
relatively large, undesirable particles that entered the system as dirt or formed by
precipitation;
 voids, which are holes in the solid formed by trapped gases or by the
accumulation of vacancies.

73.  Laue method
 Rotating crystal method
 Powder method
METHOD
LAUE
METHOD
Variable Fixed
ROTATING
CRYSTAL
METHOD
Fixed Variable
Powder
method
Fixed Variable


2 sinn d 
What are the ways in which we can
use Bragg’s law ?
Thus, there are two handles that we
can vary - λ and θ. Based on these
handles there are essentially 3
methods of doing diffraction,as
tabulated in 1 The difference
between the Rotating crystal and
Powder method is in how the angle
is varied.

74. X RAY DIFFRACTION
TECHNIQUES
LAUE METHOD
Used to find crystal
ORIENTATION
SINGLE CRYSTAL
FIXED ANGLE
POLYCHROMATIC BEAM
(wavelength varied)
ROTATING CRYSTAL
TYPE
Used to find LATICE
CONSTANT
SINGLE CRYSTAL
VARIABLE ANGLE (varied
using rotation)
MONOCHROMATIC BEAM
POWDER METHOD
Used to find Lattice
parameters
Polycrystal
Monochromatic beam
Variable angle (varied
using orientation)

75.  Laue method was the first diffraction method used.
 A single crystal(fixed orientation) is required for this method.
 The technique use white radiation (range of wavelengths), typically unfiltered
radiation from a X-ray tube.This falls on a single crystal that is fixed. Thus, the
diffraction angle,θ, is fixed for every diffraction plane and for that particular
θ value a certain wavelength will get diffracted, using Bragg’s equation .Thus,
each diffracted beam will correspond to a certain wavelength which then falls on
the film.
 Based on how the film is positioned there are two types in Laue method

76. • Used to find LATICE
CONSTANT (d)
• SINGLE CRYSTAL
• VARIABLE ANGLE (varied
using rotation)
• MONOCHROMATIC BEAM
In the rotating crystal
method, a single crystal
is mounted with an axis
normal to a
monochromatic x-ray
beam. A cylindrical film
is placed around it & the
crystal is rotated about
the chosen axis.
• As the crystal rotates,Sets of
lattice planes will at some
point make the correct Bragg
angle
for the monochromatic incident
beam, & at that point a diffracted
beam will be formed.
2 sind n 

77. If a powdered specimen is used, instead of a single crystal, then there is no need to
rotate the specimen, because there will always be some crystals at an orientation
for which diffraction is permitted. Here a monochromatic X-ray beam is incident on
a powdered or polycrystalline sample.
This method is useful for samples that are difficult to obtain in single crystal
form

78. The powder method is used to determine the value of the lattice parameters
accurately. Lattice parameters are the magnitudes of the unit vectors a, b and c
which define the unit cell for the crystal.
For every set of crystal planes, by chance, one or more crystals will be in the
correct orientation to give the correct Bragg angle to satisfy Bragg's equation.
Every crystal plane is thus capable of diffraction. Each diffraction line is made up of
a large number of small spots, each from a separate crystal. Each spot is so small as
to give the appearance of a continuous line.

79.  If a monochromatic X-ray beam is
directed at a single crystal, then only
one or two diffracted beams may result.
See figure
 For a sample of several randomly
orientated single crystals, the diffracted
beams will lie on the surface of several
cones.The cones may emerge in all
directions, forwards and backwards.
See figure
 For a sample of hundreds of crystals
(powdered sample), the diffracted
beams form continuous cones. A circle
of film is used to record the diffraction
pattern as shown. Each cone intersects
the film giving diffraction lines.The
lines are seen as arcs on the film.
See figure
The Powder Method
79

80. ADVANTAGES & DISADVANTAGES OF XRD
COMPARED TO OTHER METHODS
Advantages
 X-Rays are the least expensive, the most convenient & the most widely used
method to determine crystal structures.
 X-Rays are not absorbed very much by air, so the sample need not be in an
evacuated chamber.
Disadvantages
 X-Rays do not interact very strongly with lighter elements.

83.  [1] http://nptel.ac.in/courses/113106032/6%20-%20Lattice%20defects.pdf
 [2] http://nptel.ac.in/courses/113105023/Lecture10.pdf
 [3] http://nptel.ac.in/courses/Webcourse-contents/IISc-
BANG/Material%20Science/pdf/Lecture_Notes/MLN_03.pdf
 [4] https://sparkyswordscience.blogspot.in/2013/12/introduction-to-crystal-
structure.html

84.  characterization of crystalline materials
 identification of fine-grained minerals such as clays and mixed layer clays that are
difficult to determine optically
 determination of unit cell dimensions
 measurement of sample purity