Download Packing of atoms in solids

MME131: Lecture 6
Packing of atoms in solids
A. K. M. B. Rashid
Professor, Department of MME
BUET, Dhaka
Today’s topics
 Atomic arrangements in solids
 Points, directions and planes in unit cell
References:
1. Callister. Materials Science and Engineering: An Introduction.
2. Askeland. The Science and Engineering of Materials.
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Atomic arrangement in solids
No order
 No regular relationship between atoms
 Atoms randomly fill up the space to which the gas is confined
 Example: monatomic gases
Ar
no order in argon gas
Short-range order
 special arrangement of atoms extends only to the atom’s nearest neighbours
 Each water molecule has a short-range order due to covalent bonding between
hydrogen and oxygen atoms at an angle of 104.5°, but each water molecule has
no special arrangement but instead randomly fill up the space available to them.
 Similar situation exists in ceramic glasses
and non-crystalline polymers
H
O
short range order in steam
short range order in glass
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Long-range order
 Atoms rearrange themselves in an identical, regular, repetitive three dimensional
grid-like pattern, called a lattice.
 Lattice is a collection of points, arranged in a periodic pattern so that the surroundings
of each lattice point in the lattice are identical.
 Each atom has both short-range order,
since the surroundings of each lattice points
are identical, and long-range order,
since the lattice extends periodically
throughout the entire material.
 A lattice differs from material to material
in both shape and size, depending on
the size of atoms and the type of bonding
between the atoms.
regular long range order in metal
Concept of ordering
Crystalline
 both short- and long-range ordered,
repetitive three-dimensional, geometric
arrangement
 common in metallic, ceramic and some
polymeric materials
Non-crystalline, or Amorphous
 random, short-range ordered, non-repetitive
 common in glassy and some metallic
materials due to complex chemistry and
rapid cooling
ordered and disordered form of silica
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silicon single crystal
Liquid crystal display
amorphous but can undergo localized crystallization
in response to an external electric field
polycrystalline stainless steel
showing grains and grain boundaries
Classification of materials
based on the type of atomic order
Monatomic Gases
Liquid Crystals
Crystalline Materials
No order
Long-range order and
short-range order in
small volumes
Short and long-range order
Example: Argon gas
Examples: LCD Polymers
Amorphous Materials
No long-range order,
only short-range order
Examples: Amorphous Si,
Glasses, Plastics
Single Crystals
Examples: Si, GaAs
Poly Crystals
Examples: Metals & alloys,
and most Ceramics
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Lattice and the unit cell

Lattice - A collection of points that divide space into smaller equally sized
segments.

Unit cell - A sub-division of the lattice that still retains the overall characteristics
of the entire lattice, and stacked together endlessly to form the lattice.

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).

Packing factor - The fraction of space in a unit cell occupied by atoms
unit cell
Unit cell parameters
• dimensions (a, b, c)
• angles (a, b, g)
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The fourteen types of Bravais lattices grouped in seven crystal systems.
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Unit cells commonly found in metallic materials
The cubic unit cells
Simple (SC)
Body Centered (BCC)
Face Centered (FCC)
Example:
Po
Example:
a-Fe, Cr, Mo, V
Example:
g-Fe, Al, Cu, Ni
The hexagonal unit cell
Example: Mg, a-Ti, Zn, Zr
c
c/2
a
Hexagonal Close-Packed (HCP)
True HCP unit cell
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Illustration showing
sharing of face and
corner atoms
Illustration of co-ordinations 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.
Example
Atomic packing factor (APF) for Body-centred cubic (BCC)
closed-packing direction
D
C
r
A
a
CD2
CD2
(4r)2
a0 =
B
= AD2 + AC2
= AD2 + (AB2 + BC2)
= a02 + (a02 + a02) = 3a02
(4/3) r
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atom volume
Atomic Packing Factor (APF) =
=
cell volume
( # atom / cell ) x ( volume / atom )
cell volume
APF for BCC Unit Cell
=
(2) x (4/3 p r3)
a3
=
(2) x (4/3 p r3)
(4r/3)3
= 0.68
The relationships between the atomic radius and the lattice parameter in cubic systems
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Unit cell properties: A summary
Lattice
Parameter
Co-ordination
Number
Number of Atoms
per Unit Cell
Atomic Packing
Factor
SC
a0 = 2 r
6
1
0.52
BCC
a0 = (4/3) r
8
2
0.68
FCC
a0 = (22) r
12
4
0.74
HCP
a0 = 2 r
c = 1.633 a0
12
6
0.74
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Points, directions, and planes
in the unit cell
 Miller indices - A shorthand notation to describe certain crystallographic directions
and planes in a material. Distance is measured in terms of lattice parameters along
x, y and z coordinates using right-hand coordinate system.
 Planes – Crystallographic planes are denoted by first ( ) brackets.
A negative number is represented by a bar over the number.
 Directions - Crystallographic directions are denoted by square [ ] brackets.
A negative number is represented by a bar over the number.
 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.
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Determining Miller indices of directions
1. Determine the coordinates of the two points (head and tail)
2. Subtract coordinates of tail from the coordinates of head
3. Clear fractions and/or reduce results to the lowest integer
4. Enclose the number in square brackets.
If a negative sign is resulted, show the negative sign with a
bar over the number.
Example
Determining Miller indices of directions
Determine the Miller indices of directions A, B, and C in the following figure.
1.
2.
3.
4.
Determine the coordinates of head and tail
Subtract coordinates of tail from that of head
Clear fractions and/or reduce to the lowest integer
Enclose the number in square brackets.
Direction A
1.
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]
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Direction B
1.
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.
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
4.
[122]
family of directions
Indicated by < > brackets
Equivalency of crystallographic directions of a form in cubic systems.
Some other points about direction:
• A direction and its negative are not identical: [100] is not equal to [100].
They represents the same but opposite directions.
• A direction and its multiple are identical: [100] is the same direction as [200].
We just forgot to reduce to lowest integers.
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Example
Determining repeat distance, linear density and packing
fraction of [110] direction in FCC copper
Distance of two corner atoms in a FCC structure
is a02.
So the repeat distance (between two adjacent
atoms) is = ½(a02) = ½(3.6151x10–8 2)
= 2.5563x10–8 cm
The linear density is reciprocal of the repeat
distance, i.e. linear density
= 1 / 2.5563x10–8 = 3.912x107 lattice points/cm
Packing fraction of a particular direction is the fraction actually covered by atoms.
So here, packing fraction is the linear density times 2r, or
Packing fraction = (3.912x107 lattice point/cm ) x 2 (1.278x10–8 cm) = 1.0
i.e., atoms lie continuously along [110] direction (since this is the close-packed direction).
Determining Miller indices of planes
 Identify the coordinate points at which the plane intercept
the x, y, and z axes in terms of the number lattice parameters
If the plane passes through the origin, the origin must be moved !!!
 Take reciprocals of these intercepts
 Clear fractions, but do not reduce them to lowest integers
 Enclose the resulting numbers in parentheses ( ).
Negative numbers should be written with a bar over the number.
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Example
Determining Miller indices of planes
Determine the Miller indices of planes A, B, and C in the following figure.
•
•
•
•
Identify the coordinate points
Take reciprocals of these intercepts
Clear fractions
Enclose the numbers in parentheses ( ).
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.
(1 1 1)
Plane B
1.
The plane never intercepts
the z axis, so x = 1, y = 2, z = ∞
2.
1/x = 1, 1/y =1/2, 1/z = 0
3.
Clear fractions (by x2):
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 right in the y-direction.
Then, x = ∞ , y = -1, z = ∞
2.
1/x = 0, 1/y = -1, 1/z = 0
3.
No fractions to clear.
4.
(0 1 0)
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z’
z
[020]
y
y’
family of planes
indicated by { } brackets
x
x’
Some other points about planes:
•
Planes and their negatives are identical: [020] is identical to [020].
•
Planes and their multiples are not identical: [100] is not same as [200].
Example
Calculating the planar density and packing fraction
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.
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SOLUTION
The total atoms on each face is one. The planar density is:
Planar density (010) 
atom per face
1 atom per face

2
area of face
(0.334)
 8.96 atoms/nm  8.96  10 atoms/cm
2
14
2
The planar packing fraction is given by:
Packing fraction (010) 

area of atoms per face
(1 atom) (pr 2 )

2
area of face
(a 0)
pr 2
( 2r )
2
 0.79
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!
Example
Drawing direction and plane
Draw (a) [121] direction and (b) (210) plane
in a cubic unit cell.
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SOLUTION
(a) [ 1 2 1 ] direction
Because we know that we will need to move in the negative y-direction,
let’s locate the origin at 0, +1, 0.
The “tail” of the direction will be located at this new origin.
A second point on the direction can be determined by moving +1 in the
x-direction, 2 in the y-direction, and +1 in the z-direction.
(a) ( 2 1 0 ) plane
To draw the plane, first take reciprocals of the indices to obtain the intercepts,
that is:
x = 1/-2 = -1/2
y = 1/1 = 1
z = 1/0 = ∞
Since the x-intercept is in a negative direction, and we wish to draw the plane
within the unit cell, let’s move the origin +1 in the x-direction to 1, 0, 0.
Then we can locate the x-intercept at 1/2 and the y-intercept at +1. The plane will
be parallel to the z-axis.
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Next Class
MME131: Lecture 7
Packing sequences in
crystals
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