Chapter 3 - Atomic and Ionic Arrangements.pdf

1
The Science and Engineering
of Materials, 4th ed
Donald R. Askeland – Pradeep P. Phulé
Chapter 3 – Atomic and Ionic Arrangements

2
Chapter 3 Outline
 3.1 Short-Range Order versus Long-Range Order
 3.2 Amorphous Materials: Principles and Technological Applications
 3.3 Lattice, Unit Cells, Basis, and Crystal Structures
 3.4 Points, Directions, and Planes in the Unit Cell
 3.5 Interstitial Sites

3
 No Order - In monoatomic gases, such as argon (Ar) or plasma
created in a fluorescent tubelight, atoms or ions have no orderly
arrangement.
Section 3.1
Short-Range Order versus Long-Range Order

4
 Short-range order (SRO) - The regular and predictable arrangement
of the atoms over a short distance - usually one or two atom spacings
(water vapor, nitrogen gas, amorphous silicon, silicate glass, and
many polymers).
 Long-range order (LRO) - A regular repetitive arrangement of atoms
in a solid which extends over a very large distance (metals, alloys,
many ceramics, and some polymers).

5
Classification of materials based on the type of atomic order.

6
 Amorphous materials - Materials, including glasses, that have no long-range
order, or crystal structure.
 In general, most materials want to form periodic arrangements since this
configuration maximizes the thermodynamic stability of the material.
Amorphous materials tend to form when, for one reason or other, the
kinetics of the process by which the material was made did not allow for the
formation of periodic arrangements.
Section 3.2
Amorphous Materials: Principles and Technological
Applications

7
Atomic arrangements in crystalline silicon and amorphous silicon. (a) Amorphous
silicon. (b) Crystalline silicon.
Note the variation in the inter-atomic distance for amorphous silicon.

8
 Lattice - A collection of points that divide space into smaller equally sized
segments.
 Lattice points - Points that make up the lattice. The surroundings of each
lattice point are identical.
 Lattice parameters - The lengths of the sides of the unit cell and the angles
between those sides. The lattice parameters describe the size and shape of
the unit cell.
Section 3.3
Lattice, Unit Cells, and Crystal Structures
A one-dimensional lattice.

9
 Unit cell - A subdivision of the lattice that still retains the overall
characteristics of the entire lattice.
A two-dimensional crystal
The crystal with an overlay
of a grid that reflects the
symmetry of the crystal.
The repeat unit of the grid
known as the unit cell.
Each unit cell has its own
origin.

10
 Atomic radius - The apparent radius of an atom, typically calculated from
the dimensions of the unit cell, using close-packed directions (depends upon
coordination number).
 Close-packed directions - Directions in a crystal along which atoms are in
contact.
 Coordination number - The number of nearest neighbors to an atom in its
atomic arrangement.
 Packing factor - The fraction of space in a unit cell occupied by atoms.

11
Definition of the lattice parameters and their use in cubic, orthorhombic,
and hexagonal crystal systems.

15
The models for simple cubic (SC), body centered cubic (BCC), and face-centered cubic
(FCC) unit cells, assuming only one atom per lattice point.
Illustration showing sharing of face and corner atoms.

16
Illustration of the coordination number in (a) SC and (b) BCC unit cells. Six atoms touch
each atom in SC, while the eight atoms touch each atom in the BCC unit cell.

17
Determine the number of lattice points per cell in the cubic crystal systems. If
there is only one atom located at each lattice point, calculate the number of
atoms per unit cell.
Example 3.1 SOLUTION
In the SC unit cell: lattice point / unit cell = (8 corners)1/8 = 1
In BCC unit cells: lattice point / unit cell = (8 corners)1/8 + (1 center)(1) = 2
In FCC unit cells: lattice point / unit cell = (8 corners)1/8 + (6 faces)(1/2) = 4
The number of atoms per unit cell would be 1, 2, and 4, for the simple cubic,
body-centered cubic, and face-centered cubic, unit cells, respectively.
Example 3.1
Determining the Number of Lattice Points in Cubic
Crystal Systems

18
Determine the relationship between the atomic radius and the lattice
parameter in SC, BCC, and FCC structures when one atom is located at
each lattice point.
Example 3.2
Determining the Relationship between Atomic Radius
and Lattice Parameters
r
a 2
0 
3
4
0
r
a 
2
4
0
r
a 

19
Example 3.3
Calculating the Packing Factor
74
.
0
18
)
2
/
4
(
)
3
4
(4)(
Factor
Packing
2
4r/
cells,
unit
FCC
for
Since,
)
3
4
)(
atoms/cell
(4
Factor
Packing
3
3
0
3
0
3








r
r
r
a
a
Calculate the packing factor for the FCC cell.
In a FCC cell, there are four lattice points per cell; if there is one atom per lattice
point, there are also four atoms per cell. The volume of one atom is 4πr3/3 and
the volume of the unit cell is .
3
0
a

20
The hexagonal close-packed (HCP) structure (left) and its unit cell (right).

22
 Miller indices - A shorthand notation to describe certain crystallographic
directions and planes in a material. Denoted by [ ] brackets. A negative
number is represented by a bar over the number.
 Directions of a form (directions of a family) - Crystallographic
directions that all have the same characteristics, although their ‘‘sense’’
is different. Denoted by〈〉brackets.
Section 3.4
Points, Directions, and Planes in the Unit Cell
Coordinates of selected points in
the unit cell. The number refers to
the distance from the origin in
terms of lattice parameters.

23
Determine the Miller indices of directions A, B, and C.
Example 3.4
Determining Miller Indices of Directions
Crystallographic directions and coordinates.

24
Direction A
1. Two points are 1, 0, 0, and 0, 0, 0
2. 1, 0, 0, -0, 0, 0 = 1, 0, 0
3. No fractions to clear or integers to reduce
4. [100]
Direction B
1. Two points are 1, 1, 1 and 0, 0, 0
2. 1, 1, 1, -0, 0, 0 = 1, 1, 1
3. No fractions to clear or integers to reduce
4. [111]
Direction C
1. Two points are 0, 0, 1 and 1/2, 1, 0
2. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1
3. 2(-1/2, -1, 1) = -1, -2, 2
2]
2
1
[
.
4

25
 Because directions are vectors, a direction and its negative are not
identical; [100] is not equal to [100]. They represent the same line, but
opposite directions.
 A direction and its multiple are identical; [100] is the same direction as
[200].
 We expect a material to have the same properties in each of these twelve
directions of the form 〈110 〉.

26
Equivalency of crystallographic directions of a form in cubic systems.

28
 Repeat distance - The distance from one lattice point to the adjacent
lattice point along a direction.
 Linear density - The number of lattice points per unit length along a
direction.
 Packing fraction - The fraction of a direction (linear-packing fraction) or
a plane (planar-packing factor) that is actually covered by atoms or ions.

29
Determine the Miller indices of planes A, B, and C.
Example 3.5
Determining Miller Indices of Planes
Crystallographic planes and intercepts.

30
Plane A
1. x = 1, y = 1, z = 1
2. 1/x = 1, 1/y = 1 , 1/z = 1
3. No fractions to clear
4. (111)
Plane B
1. The plane never intercepts the z axis, so x = 1, y = 2, and z =
2. 1/x = 1, 1/y =1/2, 1/z = 0
3. Clear fractions:
1/x = 2, 1/y = 1, 1/z = 0
4. (210)
Plane C
1. We must move the origin, since the plane passes through 0, 0, 0. Let’s move
the origin one lattice parameter in the y-direction. Then, x = , y = -1, and z =
2. 1/x = 0, 1/y = -1, 1/z = 0
3. No fractions to clear.
)
0
1
0
(
.
4




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 Planes and their negatives are identical (this was not the case for
directions) because they are parallel. Therefore, (020)=(020).
 Planes and their multiples are not identical (again, this is the opposite of
what we found for directions).
 In cubic systems, a direction that has the same indices as a plane is
perpendicular to that plane.
 In each unit cell, planes of a form or family represent groups of
equivalent planes that have their particular indices because of the
orientation of the coordinates. We represent these groups of similar
planes with the notation {}.
 The planar density is the number of atoms per unit area with centers that
lie on the plane.
 The planar packing fraction is the fraction of the area of that plane
actually covered by these atoms.

33
Calculate the planar density and planar packing fraction for the (010) and (020)
planes in simple cubic polonium, which has a lattice parameter of 0.334 nm.
Example 3.6
Calculating the Planar Density and Packing Fraction
The planer densities of the (010) and (020) planes in SC unit cells are not identical.

34
The total atoms on each face is one. The planar density is:
2
14
2
2
atoms/cm
10
96
.
8
atoms/nm
96
.
8
)
334
.
0
(
face
per
atom
1
face
of
area
face
per
atom
(010)
density
Planar





The planar packing fraction is given by:
79
.
0
)
2
(
)
(
)
(
atom)
1
(
face
of
area
face
per
atoms
of
area
(010)
fraction
Packing
2
2
2
0
2




r
r
r
a


However, no atoms are centered on the (020) planes. Therefore, the
planar density and the planar packing fraction are both zero. The (010)
and (020) planes are not equivalent!

35
Example 3.7
Drawing Some Planes

36
Miller-Bravais indices are obtained for crystallographic planes in HCP unit cells by
using a four-axis coordinate system.
Typical directions in the HCP unit cell, using both three-and-four-axis systems. The
dashed lines show that the [1210] direction is equivalent to a [010] direction.

37
Determine the Miller-Bravais indices for planes A and B and directions C and D.
Example 3.8
Determining the Miller-Bravais Indices for Planes and
Directions

38
Plane A
1. a1 = a2 = a3 = , c = 1
2. 1/a1 = 1/a2 = 1/a3 = 0, 1/c = 1
4. (0001)
Plane B
1. a1 = 1, a2 = 1, a3 = -1/2, c = 1
2. 1/a1 = 1, 1/a2 = 1, 1/a3 = -2, 1/c = 1
4.
Direction C
1. Two points are 0, 0, 1 and 1, 0, 0.
2. 0, 0, 1, -1, 0, 0 = 1, 0, 1
3. No fractions to clear or integers to reduce.
4.

)
1
2
11
(
113]
2
[
or
]
01
1
[

39
Example 3.8 SOLUTION (Continued)
Direction D
1. Two points are 0, 1, 0 and 1, 0, 0.
2. 0, 1, 0, -1, 0, 0 = -1, 1, 0
3. No fractions to clear or integers to reduce.
4. 100]
1
[
or
]
10
1
[

40
 Close-packed (CP) structure - Structures showing a packing fraction
of 0.74 (FCC and HCP).

41
The ABABAB stacking sequence of close-packed planes produces the HCP
structure.
The coordination number is twelve.

42
The ABCABCABC stacking sequence of close-packed planes produces the FCC
structure.

43
 Interstitial sites - Locations between the ‘‘normal’’ atoms or ions in a crystal
into which another - usually different - atom or ion is placed. Typically, the
size of this interstitial location is smaller than the atom or ion that is to be
introduced.
 Cubic site - An interstitial position that has a coordination number of eight.
An atom or ion in the cubic site touches eight other atoms or ions.
Section 3.5
Interstitial Sites

44
 Octahedral site - An interstitial position that has a coordination number of
six. An atom or ion in the octahedral site touches six other atoms or ions.
 Tetrahedral site - An interstitial position that has a coordination number of
four. An atom or ion in the tetrahedral site touches four other atoms or ions.

45
Example 3.9
Calculating Octahedral Sites
Calculate the number of octahedral sites that uniquely belong to one FCC unit cell.
The octahedral sites include the 12 edges of the unit cell and the center position.

47
 Atoms with radii smaller than the radius of the hole are not allowed to fit
into the interstitial site because the ion would “rattle” around in the site.
 If the interstitial atom becomes too large, it prefers to enter a site having a
larger coordination number.
 Interstitial atoms or ions whose radii are slightly larger than the radius of the
interstitial site may enter that site, pushing the surrounding atoms slightly
apart.

48
We wish to produce a radiation-absorbing wall composed of 10,000 lead balls,
each 3 cm in diameter, in a face-centered cubic arrangement. We decide that
improved absorption will occur if we fill interstitial sites between the 3-cm balls
with smaller balls. Design the size of the smaller lead balls and determine how
many are needed.
Example 3.10
Design of a Radiation-Absorbing Wall

49
First, we can calculate the diameter of the octahedral sites located between the
3-cm diameter balls.
Length AB = 2R + 2r = 2 R
r = R – R = ( - 1)R
r/R = 0.414
Since r/R = 0.414, the radius of the small lead balls is
r = 0.414 * R = (0.414)(3 cm/2) = 0.621 cm.
From Example 3-11, we find that there are four octahedral sites in the FCC
arrangement, which also has four lattice points. Therefore, we need the same
number of small lead balls as large lead balls, or 10,000 small balls.
2
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