Materials Science and

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ME 112 – MSE
Module 3
The Structure of Crystalline Solids

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Materials Science and
Engineering Modules/chapters
Introduction
Atomic Structure and Interatomic Bonding
The Structure of Crystalline Solids
Imperfections in Solids
Diffusion
Mechanical Properties of Metals
Failure
Phase Diagrams
Phase Transformations: Development of Microstructure
Applications and Processing of Metals and Alloys
Ceramics
Polymers
Composites
Biomaterials/Electronic Materials
Properties of Materials (Electric, Thermal, Magnetic and Optical) NITK, Surathkal
Materials Science and
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Crystal Structure
Fundamental concepts
Unit cells

Metallic crystal structure

The face-centered cubic crystal structure
The body-centered cubic crystal structure
The hexagonal close-packed crystal structure
Density computation
Polymorphism and allotropy
Crystal systems
Crystallographic points, directions and planes
Atomic arrangement
Linear and Planar density
Closed-pack crystal structure
Polycrystalline Materials
Anisotropy
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Fundamental concept
• Atoms are situated in a repeating or
periodic array over atomic distance.
• Upon solidification the atoms will
position themselves in repetitive 3D
pattern.
• Each atom is bonded to its nearest
neighbor atom
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• Properties of crystalline solids depend on the

crystal structure of the material, the manner in
which atoms, ions, or molecules are spatially
arranged.

• In crystalline structure atoms or ions represents

as solid spheres having well defined diameters.

• This is termed the atomic hard-sphere model in

which spheres representing nearest-neighbor
atoms touch one another. NITK, Surathkal
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Unit cell • Small repeat entities in a
crystal structure.
• Unit cells for most crystal
structures are
parallelepipeds or prisms
having three sets of
parallel faces.
• Represent the symmetry
of the crystal structure
Defines the crystal structure by
virtue of its geometry and the atom
positions within. NITK, Surathkal
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Metallic crystal structure
Metallic bonding occurs when group
of metal atoms shares a cloud of
valance electron.

Non directional in nature

The hard-sphere model for the crystal

structure, each sphere represents an
ion core.
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https://www.chemistrylearner.com/quantum-numbers.html
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Face-Centered cubic crystal structure

The crystal structure found for many metals has a unit cell of cubic
geometry, with atoms located at each of the corners and the centers
of all the cube faces. Examples copper, aluminum, silver, and gold
etc.

(a) A hard sphere unit cell representation (b) A reduced sphere unit cell (c) an aggregate of many atoms
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These spheres or ion cores touch one another across a face diagonal;
the cube edge length a and the atomic radius R are related through
a = 2R√2

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For FCC crystal structure, there are eight corner atoms Nc =8, six
face atoms Nf = 6 and no interior atom Ni =0
𝑁𝑓 𝑁𝑐
𝑁 = 𝑁𝑖 + +
2 8

N= 0+6/2+8/8
=4
A total of four whole atoms may be assigned to a given unit cell

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Number of atoms associated with each unit cell computed using

following formula
𝑁𝑓 𝑁𝑐
𝑁 = 𝑁𝑖 + +
2 8

where Ni = the number of interior atoms

Nf = the number of face atoms
Nc = the number of corner atoms

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Coordination number: the number of nearest neighbor atoms or

ions surrounding an atom or ion

https://www.toppr.com/ask/question/the-coordination-number-of-an-atom-in-a-fcc-lattice-is/
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Atomic packing factor: the sum of sphere volumes of all atoms with
in the unit cell (assuming the atomic hard-sphere model) divided by
the unit cell volume

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑖𝑛 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙

𝐴𝑃𝐹 =
𝑡𝑜𝑡𝑎𝑙 𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒

For the FCC structure, the atomic packing factor is 0.74, which is the
maximum packing possible for spheres all having the same diameter
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Body Centered cubic crystal structure

A cubic unit cell with atoms located at all eight corners and a single
atom at the cube center

(a) A hard sphere unit cell representation (b) A reduced sphere unit cell (c) an aggregate of many atoms
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Center and corner atoms touch one another along cube diagonals,
and unit cell length a and atomic radius R are related through
4𝑅
𝑎=
3

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Each BCC unit cell has eight corner atoms and a single center atom,
which is wholly contained within its cell;
𝑁𝑓 𝑁𝑐
𝑁 = 𝑁𝑖 + +
2 8

N= 1+0+8/8
=2

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Coordination number: 8

Atomic packing factor: 0.68

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Hexagonal Closed-packed crystal structure

• The top and bottom faces of the unit cell consist of six atoms that form regular hexagons
and surround a single atom in the center.
• Another plane that provides three additional atoms to the unit cell is situated between the
top and bottom planes
• The atoms in this midplane have as nearest neighbors atoms in both of the adjacent two
planes

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Each HCP unit cell has eight corner atoms and a single center
atom, which is wholly contained within its cell
𝑁𝑓 𝑁𝑐
𝑁 = 𝑁𝑖 + +
2 6

N= 3+2/2+12/6
=6
a and c represent, respectively, the short and long unit cell dimensions, the
c/a ratio should be 1.633; however, for some HCP metals, this ratio
deviates from the ideal value.
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The coordination number and the atomic packing factor for the HCP
crystal structure are the same as for FCC: 12 and 0.74, respectively.
The HCP metals include cadmium, magnesium, titanium, and zinc

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https://old.amu.ac.in/emp/studym/100007300.pdf
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Problem 1

Determination of FCC Unit Cell

Volume Calculate the volume of an
FCC unit cell in terms of the atomic
radius R

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Computation of the Atomic Packing

Factor for FCC Show that the
atomic packing factor for the FCC
crystal structure is 0.74

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Computation of the Atomic Packing

Factor for simple cubic

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Density computation
Theoretical density for metals.
𝑛𝐴
𝜌=
𝑉𝑐 𝑁𝐴
Where as n = number of atoms associated with each unit cell
A = atomic weight
Vc = volume of the unit cell
NA= Avogadro’s number (6.022 x 1023 atoms/mol)

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Problem 4
Theoretical Density Computation for Copper Copper has an atomic
radius of 0.128 nm, an FCC crystal structure, and an atomic weight of
63.5 g/mol. Compute its theoretical density, and compare the answer
with its measured density

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Solution 4
Number of atoms per unit cell =4
Atomic weight of copper = 63.5 g/mol
Unit cell volume = 16R3 √2 where R is atomic radius

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Polymorphism and allotropy
When some metals as well as may have more than one crystal structure
then the phenomenon known as polymorphism.

When it is present in elemental solid, the condition is known as

allotropy.

Crystal structure depends on the temperature and external pressure.

Example : carbon: graphite is the stable polymorph at ambient conditions, whereas

diamond is formed at extremely high pressures
pure iron has a BCC crystal structure at room temperature, which changes to FCC
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iron at 912°C (1674°F)
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Tin (its allotropic transformation)

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Materials Science and
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• Based on the unit cell geometry i.e. the shape of
appropriate unit cell parallelepiped without
regard to the atomic positions in the cell.

• The unit cell geometry is completely defined in

terms of six parameters: the three edge lengths
a, b, and c, and the three interaxial angles 𝛼, 𝛽,
and 𝛾.

• These parameters are known as lattice A unit cell with x, y and z coordinate axes, showing axial
lengths (a, b, and c) and interaxial angles (α, β, and γ)
parameters NITK, Surathkal
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Materials Science and
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directions and planes
• Point coordinates: Lattice position is defined in terms of three
lattice position coordinates, which are associated with the x, y, and z
axes
position coordinates in unit cell P (Px, Py, and Pz). These
indices are the fractional multiples of a, b, and c unit length

𝑃𝑥 = 𝑞𝑎
𝑃𝑦 = 𝑟𝑏
𝑃𝑧 = 𝑠𝑐

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Materials Science and
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Location of Point Having Specified Coordinates for the unit cell
shown below locate the point having indices( ¼ 1 ½).

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Problem 6
Specify indices for all numbered points of the unit cell in the
following figure

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Crystallographic direction
A crystallographic direction is defined as a line directed
between two points or a vector. There are the following
steps to determine the three-dimensional indices

1. A right-handed x-y-z coordinate system is first constructed. As a

matter of convenience, its origin may be located at a unit cell
corner.

2. The coordinates of two points that lie on the direction vector

(referenced to the coordinate system) are determined—for
example, for the vector tail, point 1: x1, y1, and z1; whereas for the
vector head, point 2: x2, y2, and z2. NITK, Surathkal
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3. Tail point coordinates are subtracted from head point components—that

is, x2 − x1, y2 − y1, and z2 − z1

4. These coordinate differences are then normalized in terms of (i.e., divided

by) their respective a, b, and c lattice parameters—that is,
(x2 − x1)/a, ( y2 − y1)/b, and (z2 − z1)/c
which yields a set of three numbers.

5. If necessary, these three numbers are multiplied or divided by a common

factor to reduce them to the smallest integer values

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6. The three resulting indices, not separated by commas, are enclosed in square
brackets, thus: [uvw]. The u, v, and w integers correspond to the normalized
coordinate differences referenced to the x, y, and z axes, respectively

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For each of the three axes, there are both positive and negative
coordinates.

Thus, negative indices are also possible, which are represented by a

bar over the appropriate index.

For example, the [11ത 1] direction has a component in the −y

direction. Also, changing the signs of all indices produces an
ത 1]
antiparallel direction; that is, [11 ത is directly opposite to [1 11].
ത

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Problem 8
Determine the indices for the direction shown in the
following figure

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Solution 8

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Problem 9
ത direction with its
Within the following unit cell draw a [110]
tail located at the origin of the coordinate system, point O.

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Solution

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Direction in hexagonal
Four axis or Miller-Bravais
crystal
coordinate system

The three a1, a2, and a3 axes are

all contained within a single plane
(called the basal plane) and are at
120° angles to one another.

The z axis is perpendicular to this

basal plane NITK, Surathkal
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Conversion from the three-index system to the four-index system

as
𝑈𝑉𝑊 = 𝑢𝑣𝑡𝑤

by using following formula

Whereas uppercase U,V and W indices are associated with

three index scheme, where as the lower case u,v,t and w
correlate with Millar-Bravai’s four index system.

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Calculation of U, V, and W indices from the three-axis a1, a2 and z coordinate

system.

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 Materials Science and
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For the direction shown in the
following figure, do the following:
(a) Determine the directional
indices referenced to the three-
axis coordinate system of
(b)Convert these indices into an
index set referenced to the four-
axis scheme

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(a) Vector passes through the origin, then

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(b) Convert these indices into an index set referenced to the four-axis scheme

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Materials Science and
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Crystallographic planes are specified by three Miller indices as (hkl). Any
two planes parallel to each other are equivalent and have identical
indices. The procedure used to determine the h, k, and l index numbers
is as follows:

1. If the plane passes through the selected origin, either another parallel plane must
be constructed within the unit cell by an appropriate translation, or a new origin
must be established at the corner of another unit cell.
2. At this point, the crystallographic plane either intersects or parallels each of the
three axes. The coordinate for the intersection of the crystallographic plane with
each of the axes is determined. These intercepts for the x, y, and z axes will be
designed by A, B, and C, respectively.
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1. The reciprocals of these numbers are taken. A plane that parallels an axis is
considered to have an infinite intercept and therefore a zero index.

2. The reciprocals of the intercepts are then normalized in terms of (i.e., multiplied
by) their respective a, b, and c lattice parameters.

3. If necessary, these three numbers are changed to the set of smallest integers by
multiplication or by division by a common factor.

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Finally, the integer indices, not separated by commas, are enclosed within parentheses,
thus: (hkl). The h, k, and l integers correspond to the normalized intercept reciprocals
referenced to the x, y, and z axes, respectively.
In summary, the h, k, and l indices may be determined using the following equations.

An intercept on the negative side of the origin is indicated by a bar or minus sign
positioned over the appropriate index. Furthermore, reversing the directions of all
indices specifies another plane parallel to, on the opposite side of, and equidistant
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from the origin.
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Determine the Miller indices for the plane shown in the following
figure

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Solution 11

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Construct a (101) plane with in the following unit cell

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For direction (101) miller indices will be

n = 1 because all the indices are integers

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Determine the Miller–Bravais indices for the

plane shown in the hexagonal unit cell

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Materials Science and A, B, and C to represent intercepts on the respective a1, a2, and z
Engineering
axes, normalized intercept reciprocals may be written as

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 Materials Science and
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The atomic arrangement for a crystallographic plane, which is often of
interest, depends on the crystal structure

(a) Reduced-sphere BCC unit cell with the (110) (a) Reduced-sphere FCC unit cell with the (110) plane. (b)
plane. (b) Atomic packing of a BCC (110) plane Atomic packing of an FCC (110) plane.

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A “family” of planes contains all planes that are crystallographically

equivalent that is, having the same atomic packing; a family is
designated by indices enclosed in braces—such as {100}. For example,
in cubic crystals, the (111), (1 1 1), (111), (1 1 1), (111), (1 11), (111), and
(111) planes all belong to the {111} family.

In the cubic system only, planes having the same indices, irrespective of
order and sign, are equivalent. For example, both (123) and (312)
belong to the {123} family.
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Linear density (LD) is defined as the number of atoms per unit length
whose centers lie on the direction vector for a specific crystallographic
direction
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑐𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑜𝑛 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
𝐿𝐷 =
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟

The units of linear density are reciprocal length m-2, nm-2

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(a) Reduced-sphere FCC unit cell

with the [110] direction indicated. (b) The bottom
face-plane of the FCC unit cell in (a) on which is
shown the atomic spacing in the [110] direction,
through atoms labeled X, Y, and Z.
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Planer density is the number of atoms per unit area that are
centered on a particular crystallographic plane
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑐𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑜𝑛 𝑝𝑙𝑎𝑛𝑒
𝑃𝐷 =
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒

Unit of planer density are m-2, nm-2

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(a) Reduced-sphere FCC unit cell with the (110) plane. (b)
Atomic packing of an FCC (110) plane.

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Close-Packed crystal structure
In crystals, close packing refers to the efficient arrangement of
constituent particles in the lattice

(a) A portion of a close packed plane of atoms; A, B, and C positions are indicated. (b) The AB stacking sequence for
close-packed atomic planes

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Close-pack plane stacking sequence for the hexagonal close-packed

structure

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(a) Close-packed stacking sequence for the face-centered cubic structure. (b) A corner has
been removed to show the relation between the stacking of close-packed planes of atoms
and the FCC crystal structure; the heavy triangle outlines a (111) plane.

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Most crystalline solids are composed of a collection of many small
crystals or grains; such materials are termed polycrystalline

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Schematic diagrams of the various

stages in the solidification of a
polycrystalline material; the square
grids depict unit cells.
(a) Small crystallite nuclei.
(b) Growth of the crystallites; the
obstruction of some grains that
are adjacent to one another is
also shown.
(c) Upon completion of
solidification, grains having
irregular shapes have formed.
(d) The grain structure as it would
appear under the microscope;
dark lines are the grain
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The physical properties of a single crystal depends on the
crystallographic direction in which it is measured.

This directionality of properties is termed anisotropy, and it is

associated with the variance of atomic or ionic spacing with
crystallographic direction.

Substances in which measured properties are independent of the

direction of measurement are isotropic.
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The extent and magnitude of anisotropic effects in crystalline

materials are functions of the symmetry of the crystal structure; the
degree of anisotropy increases with decreasing structural
symmetry—triclinic structures normally are highly anisotropic

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Materials Science and
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Noncrystalline solids lack a systematic and regular arrangement of
atoms over relatively large atomic distances

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Crystalline or amorphous solid forms depend on the ease with

which a random atomic structure in the liquid can transform to an
ordered state during solidification.

Rapidly cooling through the freezing temperature favors the

formation of a noncrystalline solid because little time is allowed for
the ordering process.

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Thank You

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