Transcript Tutorial 1

CENG151 Introduction to Materials
Science and Selection
Tutorial 1
14th September, 2007
Teaching Assistants

Candy Lin
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Rm 7250
Email: [email protected]
Bryan Wei
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
Rm 7111
Email: [email protected]
Crystal
In chemistry and mineralogy, a
crystal is a solid in which the
constituent atoms, molecules, or ions
are packed in a regularly ordered,
repeating pattern extending in all
three spatial dimensions.
Insulin crystals
Synthetic bismuth crystal
Gallium, a metal that easily
forms large single crystals
Quartz crystal
7 crystal system &14 Bravais lattice
triclinic
monoclinic
orthorhombic
cubic
hexagonal
rhombohedral
tetragonal
In all, there are 14
possible Bravais
lattices that fill 3D
space, and they
classified by 7 kind
of crystal system.
Comparing Different Atomic Arrangement
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Different atom arrangement can alter
chemical and physical properties.
Carbon is a very good example of this!
Carbon - Graphite
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Large parallel sheets of hexagonal rings.
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The sheets are held together by weak Van der Waals forces.
Can be broken easily.
Because the layers of carbon rings can rub over each other,
graphite is a good lubricant.
Carbon - Diamond
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Each carbon atom covalently bonded to four other carbons in a
tetrahedral arrangement to form a tight tetrahedron lattice.
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Very strong bonds  strongest natural material on earth.
(a) Tetrahedron
(b) The diamond
cubic (DC) unit
cell.
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Diamond can be cleaved along its planes,
but it cannot flake apart into layers
because of this tetrahedral arrangement
of carbons.
Carbon Buckminsterfullerene
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The 1996 Nobel Prize in
Chemistry has been
awarded to three chemists
(R. Smalley, R. Curl, H.
Kroto) for discovery of
fullerenes.
Shaped like a soccer ball,
with 20 hexagons and 12
pentagons.
A total of 60 carbons in this molecules.
Molecule is extremely stable and can withstand
very high temperatures and pressures.
Can react with other atoms and molecules.
Packing Factor (Packing Density)
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Packing – geometrical arrangement of these
(atomic) spheres.
Packing Factor/Density – proportion of space
occupied by these solid spheres in a unit cell.
Determining the Relationship between
Atomic Radius and Lattice Parameters
SC, BCC, and FCC structures when one atom is located
at each lattice point.
(c) 2003 Brooks/Cole Publishing / Thomson
Learning™
The relationships between the atomic radius and the Lattice
parameter in cubic systems.
Atoms touch along the edge of the cube in SC
structures.
a0
 2r
In BCC structures, atoms touch along the body diagonal.
There are two atomic radii from the center atom and one
atomic radius from each of the corner atoms on the body
diagonal, so
a0
4r

3
In FCC structures, atoms touch along the face diagonal
of the cube. There are four atomic radii along this
length—two radii from the face-centered atom and one
radius from each corner, so:
a0
4r

2
Packing Factor
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Packing Factor – the fraction of space occupied
by atoms, assuming that atoms are hard spheres
sized so that they touch their closest neighbor.


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 number of atoms / cell  volume of each atom 


Packing factor  
volume of unit cell
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

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
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

number
of
atoms
/
cell
atomic
mass
Density (of a material) – can be
 calculated
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Packing density 
properties of the crystal structure:
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 volume of unit cell  Avogadro' s number 




using
 number of atoms / cell  atomic mass 


Density    
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
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 volume of unit cell  Avogadro' s number 

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Coordination Number =
no. of touching neighbors
HCP and FCC
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Greatest packing density obtainable (closest
packed) with spheres of equal size is 74%.
1 atom surrounded by 12 others of equal
size, all touching the reference sphere along
the diameter.
2 related crystal structures obtain this
packing density:
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Hexagonal close packed (HCP)
Face centered cubic (FCC)
Lattice Positions
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Coordinates of
selected points in
the unit cell.
The number
refers to the
distance from the
origin in terms of
lattice parameters.
In Class Exercise
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Try to find A,
B, C and D.
Answers
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A = 1, 0, 0
B = 0, 0, 1
C = ½, 1, 0
D = 1, 1, 0
Miller Indices for directions
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Miller Indices are shorthand notations used
to describe directions. Follow these rules:
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Using right-handed coordinate system,
determine the coordinates of the two points that
lie on the direction.
Substract the coordinates of the “tail” point from
the coordinates of the “head” point.
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“head” – “tail”
Clear fractions and/or reduce the results
obtained from the substraction to lowest
integers.
Enclose the numbers in square brackets []. If
negative sign is produced, put a bar over the
number.
Lattice Directions
Example: Determine the direction of A in the figure.
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1.
2.
3.
4.
(c) 2003 Brooks/Cole Publishing /
Thomson Learning™
Direction A
Two points are
1, 0, 0, and 0, 0, 0
1, 0, 0, - 0, 0, 0
= 1, 0, 0
No fractions to
clear or integers to
reduce
[100]
Lattice Directions
Determine the direction of B in the figure.
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1.
2.
3.
4.
(c) 2003 Brooks/Cole Publishing /
Thomson Learning™
Direction B
Two points are
1, 1, 1, and 0, 0, 0
1, 1, 1, - 0, 0, 0
= 1, 1, 1
No fractions to clear
or integers to
reduce
[111]
Lattice Directions
Determine the direction of C in the figure.
Direction C
1. Two points are
0, 0, 1, and ½, 1, 0
2. 0, 0, 1, - ½, 1, 0
= -½, -1, 1
3. Fractions for clearing:
2(-½, -1, 1) = -1, -2, 2

(c) 2003 Brooks/Cole Publishing /
Thomson Learning™
4. [ 1 22]
Exercise: Try to determine the directions for
A, B, C, and D.
(c) 2003 Brooks/Cole Publishing / Thomson Learning
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A: 0,0,1 – 1,0,0
= -1,0,1 = [101]
B: 1,0,1 – ½,1,0
= ½,-1,1 = [122]
C: 1,0,0 – 0,¾,1
= 1,-¾,-1 = [434]
D: 0,1,½ – 1,0,0
= -1,1,½ = [221]
Miller Indices for planes
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Miller Indices are used to describe planes
as well. Follow these rules:
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Identify the points at which the plane intercepts
the x, y, z coordinates. If the plane passes
through the origin, the origin must be moved!
Take reciprocals of these intercepts.
Clear fractions and do NOT reduce to lowest
integers.
Enclose the numbers in parentheses (). If
negative sign is produced, put a bar over the
number, like before.
Lattice Planes
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Intercepting points at:
x= ∞, y=1, z= ∞
Take reciprocals: x=0,
y=1, z=0
No fractions to clear,
Enclose numbers in
parentheses: (010)
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Intercepting points at:
x=-1, y= ∞, z= ∞
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Take reciprocals: x=1, y=0, z=0
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No fractions to clear,
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Enclose numbers in
parentheses: (100)
Lattice positions  no
brackets
Lattice directions 
square brackets
Lattice planes 
parentheses
Plane
1. x = 1, y = 1, z = ∞
2.1/x = 1, 1/y = 1, 1 /z = 0
3. No fractions to clear
4. (110)
Plane
1. X = ⅔, y = 1, z = ∞
2. 1/x = 3/2, 1/y=1, 1/z=0
3. Clear fractions:
1/x = 3,
1/y = 2,
1/z = 0
4. (320)
Plane (origin in blue)
1. x = 1, y = 1, z = ∞
2.1/x = 1, 1/y = 1, 1/z = 0
3. No fractions to clear
4. (110)
Plane (origin in red)
1. x = -1, y = -1, z = ∞
2.1/x = -1, 1/y = -1, 1/z = 0
3. No fractions to clear
4. (110)
y
Plane (origin in green)
1. x = -1, y = 1, z = ∞
2.1/x = -1, 1/y = 1, 1/z = 0
3. No fractions to clear
4. (110)
Plane (origin in orange)
1. x = 1, y = -1, z = ∞
2.1/x = 1, 1/y = -1, 1/z = 0
3. No fractions to clear
4. (110)
Exercise: Determining Miller Indices of Planes
Determine the Miller indices of planes A, B, and C in Figure.
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)
Exercise: Determining Miller Indices of Planes
Determine the Miller indices of planes A, B, and C in Figure.
Plane C (origin in red)
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.
4. (010)
(c) 2003 Brooks/Cole Publishing /
Thomson Learning™
More Exercise! Try these!
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Plane A (origin in red)
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 (origin in orange)
1. x = ∞, y = ⅓, and z =∞
2.1/x = 0, 1/y =3, 1/z = 0
3. No fractions to clear
4. (030)
Plane C (origin in green)
1.
2.
3.
Note: When origin in orange  4.
x=1, y=∞, and z=-½
1/x = 1, 1/y = 0, 1/z = -2
No fractions to clear.
(102), (102)
End of Tutorial 1
Thank you!