ENGR 151: Materials of Engineering LECTURES #5 & 6: HCP, POSITIONS, DIRECTIONS, AND PLANES ·...
Transcript of ENGR 151: Materials of Engineering LECTURES #5 & 6: HCP, POSITIONS, DIRECTIONS, AND PLANES ·...
LECTURES #5 & 6: HCP, POSITIONS,
DIRECTIONS, AND PLANES
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ENGR 151: Materials of Engineering
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Hexagonal Close-Packed Structure (HCP
– another view)
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Hexagonal Close-Packed Structure (HCP
– top view)
• Coordination # = 12
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HEXAGONAL CLOSE-PACKED STRUCTURE
(HCP)
• Coordination # = 12
• ABAB... Stacking Sequence
• APF = 0.74
• 3D Projection • 2D Projection
6 atoms/unit cell
ex: Cd, Mg, Ti, Zn
• c/a = 1.633
c
a
A sites
B sites
A sites Bottom layer
Middle layer
Top layer
ABAB... Stacking Sequence
THEORETICAL DENSITY, R
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where n = number of atoms/unit cell
A = atomic weight
VC = Volume of unit cell = a3 for cubic
NA = Avogadro’s number
= 6.022 x 1023 atoms/mol
Density = =
VC NA
n A =
Cell Unit of Volume Total
Cell Unit in Atoms of Mass
THEORETICAL DENSITY, R
Ex: Cr (BCC)
A (atomic weight) = 52.00 g/mol
n = 2 atoms/unit cell
R = 0.125 nm
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theoretical
a = 4R/ 3 = 0.2887 nm
actual
a R
= a 3
52.00 2
atoms
unit cell mol
g
unit cell
volume atoms
mol
6.022 x 1023
= 7.18 g/cm3
= 7.19 g/cm3
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SUMMARY
• Atoms may assemble into crystalline or amorphous structures.
• We can predict the density of a material, provided we know the
atomic weight, atomic radius, and crystal geometry (e.g., FCC,
BCC, HCP).
• Common metallic crystal structures are FCC, BCC and HCP.
Coordination number and atomic packing factor are the same
for both FCC and HCP crystal structures.
• Interatomic bonding in ceramics is ionic and/or covalent.
SINGLE CRYSTALS
For a crystalline solid, when the periodic and
repeated arrangement of atoms extends
throughout without interruption, the result is
a single crystal.
The crystal lattice of the entire sample is
continuous and unbroken with no grain
boundaries.
For a variety of reasons, including the
distorting effects of impurities,
crystallographic defects and dislocations,
single crystals of meaningful size are
exceedingly rare in nature, and difficult to
produce in the laboratory under controlled
conditions. 10
Huge KDP (monopotassium phosphate)
crystal grown from a seed crystal in a
supersaturated aqueous solution at LLNL.
Below, silicon boule.
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CRYSTALS AS BUILDING BLOCKS
• Some engineering applications require single crystals:
• Properties of crystalline materials
often related to crystal structure.
-- Ex: Quartz fractures more easily
along some crystal planes than
others.
-- diamond
Single crystals for abrasives
-- turbine blades
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GLASS STRUCTURE
• Quartz is crystalline SiO2:
• Basic Unit: Glass is noncrystalline (amorphous)
• Fused silica is SiO2 - no impurities
have been added.
• Other common glasses contain
impurity ions such as Na+, Ca2+,
Al3+, and B3+
(soda glass)
Si 4+
Na +
O 2 -
Silicate glass tetrahedron SiO4
ANISOTROPY
The physical properties of single crystals of some substances depend on the crystallographic direction in which measurements are taken.
For example, the elastic modulus, electrical conductivity, and the index of refraction may have different values in the [100] and [111] directions.
The directionality of the properties is termed anisotropy and is associated with the atomic spacing.
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ISOTROPIC
If measured properties are independent of the direction of measurement then they are isotropic.
For many polycrystalline materials, the crystallographic orientations of the individual grains are totally random.
So, though, a specific grain may be anisotropic, when the specimen is composed of many grains, the aggregate behavior may be isotropic.
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POLYCRYSTALS
Most crystalline solids are composed of many small crystals
(also called grains).
Initially, small crystals (nuclei) form at various positions.
These have random orientations.
The small grains grow and begin to impinge on one another
forming grain boundaries.
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Micrograph of a polycrystalline
stainless steel showing grains
and grain boundaries
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POLYCRYSTALS • Most engineering materials are polycrystals.
• Nb-Hf-W plate with an electron beam weld.
• Each "grain" is a single crystal.
• If grains are randomly oriented, overall component properties are not directional.
• Grain sizes typical range from 1 nm to 2 cm
(from a few to millions of atomic layers).
1 mm
Isotropic
Anisotropic
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POLYMORPHISM
Two or more distinct crystal structures for the same
material (allotropy/polymorphism)
titanium
, -Ti
carbon
diamond, graphite
BCC
FCC
BCC
1538ºC
1394ºC
912ºC
-Fe
-Fe
-Fe
liquid
iron system
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SINGLE VS POLYCRYSTALS
• Single Crystals
-Properties vary with
direction: anisotropic.
-Example: the modulus
of elasticity (E) in BCC iron:
• Polycrystals
-Properties may/may not
vary with direction.
-If grains are randomly
oriented: isotropic.
(Epoly iron = 210 GPa)
200 mm
E (diagonal) = 273 GPa
E (edge) = 125 GPa
CRYSTAL SYSTEMS
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7 crystal systems
14 crystal lattices
Unit cell: smallest repetitive volume that
contains the complete lattice pattern of a crystal.
a, b and c: lattice constants
The fourteen (14) types of Bravais lattices grouped in seven (7) systems.
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
Unit Cells Types
Primitive Face-Centered
Body-Centered End-Centered
A unit cell is the smallest component of the crystal that reproduces the whole
crystal when stacked together.
• Primitive (P) unit cells contain only a single lattice point.
• Internal (I) unit cell contains an atom in the body center.
• Face (F) unit cell contains atoms in the all faces of the planes composing the cell.
• Centered (C) unit cell contains atoms centered on the sides of the unit cell.
Combining 7 Crystal Classes (cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic,
trigonal) with 4 unit cell types (P, I, F, C) symmetry allows for only 14 types of 3-D lattice.
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
Lattice parameters in cubic, orthorhombic and hexagonal crystal systems.
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Cubic
Hexagonal
Tetragonal
Rhombohedral
Orthorhombic
Monoclinic
Triclinic
THE 7 TYPES OF CRYSTAL SYSTEMS
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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.
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.
Point Coordinates
• Each unit cell is a reference or basis.
• The length of an edge is normalized as a unit of
measurement.
• E.g. if the length of the edge of the unit cell along the
X-axis is a, then ALL measurements in the X-direction
are referenced to a (e.g. a/2).
Point Coordinates – Contd.
Point coordinates for unit cell center are
a/2, b/2, c/2 ½ ½ ½
Point coordinates for unit cell corner are 111
Translation: integer multiple of lattice constants identical position in another unit cell
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z
x
y a b
c
000
111
y
z
2c
b
b
Point Coordinates
EXAMPLE PROBLEMS
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EXAMPLE PROBLEMS CONTD.
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EXAMPLE PROBLEMS CONTD.
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Determine the Miller indices of directions A, B, and C.
Miller Indices, Directions
(c) 2003 Brooks/Cole Publishing /
Thomson Learning™
GENERAL APPROACH FOR MILLER INDICES
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SOLUTION
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]21[ .4
(c) 2003 Brooks/Cole Publishing
/ Thomson Learning™
CRYSTALLOGRAPHIC DIRECTIONS
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1. Vector repositioned (if necessary) to pass
through origin.
2. Read off projections in terms of
unit cell dimensions a, b, and c
3. Adjust to smallest integer values
4. Enclose in square brackets, no commas
[uvw]
ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]
-1, 1, 1
z
x
Algorithm
where overbar represents a
negative index
[ 111 ] =>
y
Problem 3.6, pg. 58
EXAMPLE PROBLEMS
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EXAMPLE PROBLEMS, CONTD.
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EXAMPLE PROBLEMS, CONTD.
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Exam problems – make
sure you translate and
scale vector to fit WITHIN
unit cube if asked!
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HCP CRYSTALLOGRAPHIC DIRECTIONS
Hexagonal Crystals
4 parameter Miller-Bravais lattice coordinates are
related to the direction indices (i.e., u'v'w') as follows.
=
=
=
' w w
t
v
u
) v u ( + -
) ' u ' v 2 ( 3
1 -
) ' v ' u 2 ( 3
1 - =
] uvtw [ ] ' w ' v ' u [
Fig. 3.8(a), Callister & Rethwisch 8e.
- a3
a1
a2
z
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HCP CRYSTALLOGRAPHIC DIRECTIONS
Fig. 3.8(a), Callister & Rethwisch 8e.
- a3
a1
a2
z
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HCP CRYSTALLOGRAPHIC DIRECTIONS
1. Vector repositioned (if necessary) to pass
through origin.
2. Read off projections in terms of unit
cell dimensions a1, a2, a3, or c
3. Adjust to smallest integer values
4. Enclose in square brackets, no commas
[uvtw]
[ 1120 ] ex: ½, ½, -1, 0 =>
Adapted from Fig. 3.8(a),
Callister & Rethwisch 8e.
dashed red lines indicate
projections onto a1 and a2 axes a1
a2
a3
-a3
2
a 2
2
a 1
- a3
a1
a2
z
Algorithm
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REDUCED-SCALE COORDINATE AXIS
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PROBLEM 3.8
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PROBLEM 3.8, CONTD.
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PROBLEM 3.8, CONTD.
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PROBLEM 3.8, CONTD.
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PROBLEM 3.8, CONTD.
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PROBLEM 3.8, CONTD.
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PROBLEM 3.9
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PROBLEM 3.9, CONTD.
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DRAWING HCP CRYSTALLOGRAPHIC
DIRECTIONS (I)
1. Remove brackets
2. Divide by largest integer so all values
are ≤ 1
3. Multiply terms by appropriate unit cell
dimension a (for a1, a2, and a3 axes)
or c (for z-axis) to produce
projections
4. Construct vector by placing tail at
origin and stepping off these
projections to locate the head
Algorithm (Miller-Bravais coordinates)
Adapted from Figure 3.10,
Callister & Rethwisch 9e.
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DRAWING HCP CRYSTALLOGRAPHIC
DIRECTIONS (II)
Draw the direction in a hexagonal unit cell.
[1213]
4. Construct Vector
1. Remove brackets -1 -2 1 3
Algorithm a1 a2 a3 z
2. Divide by 3 1 3
1
3
2
3
1
3. Projections
proceed –a/3 units along a1 axis to point p
–2a/3 units parallel to a2 axis to point q
a/3 units parallel to a3 axis to point r
c units parallel to z axis to point s
[1 2 13]
p
q r
s
start at point o
Adapted from p. 72,
Callister &
Rethwisch 9e.
[1213] direction represented by vector from point o to point s
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1. Determine coordinates of vector tail, pt. 1:
x1, y1, & z1; and vector head, pt. 2: x2, y2, & z2.
in terms of three axis (a1, a2, and z)
2. Tail point coordinates subtracted from head
point coordinates and normalized by unit cell
dimensions a and c
3. Adjust to smallest integer values
4. Enclose in square brackets, no commas,
for three-axis coordinates
5. Convert to four-axis Miller-Bravais lattice
coordinates using equations below:
6. Adjust to smallest integer values and
enclose in brackets [uvtw]
Adapted from p. 72, Callister &
Rethwisch 9e.
Algorithm
DETERMINATION OF HCP CRYSTALLOGRAPHIC
DIRECTIONS (II)
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4. Brackets [110]
1. Tail location 0 0 0
Head location a a 0c
1 1 0 3. Reduction 1 1 0
Example a1 a2 z
1/3, 1/3, -2/3, 0 => 1, 1, -2, 0 => [ 1120 ]
6. Reduction & Brackets
Adapted
from p. 72,
Callister &
Rethwisch
9e.
DETERMINATION OF HCP CRYSTALLOGRAPHIC
DIRECTIONS (II)
Determine indices for green vector
2. Normalized
FAMILIES OF DIRECTIONS <UVW>
For some crystal structures, several
nonparallel directions with different
indices are crystallographically equivalent;
this means that atom spacing along each
direction is the same.
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CRYSTALLOGRAPHIC PLANES
Adapted from Fig. 3.10,
Callister & Rethwisch 8e.
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CRYSTALLOGRAPHIC PLANES
Miller Indices: Reciprocals of the (three) axial intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices.
Algorithm 1. Read off intercepts of plane with axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce to smallest integer values 4. Enclose in parentheses, no commas i.e., (hkl)
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CRYSTALLOGRAPHIC PLANES z
x
y a b
c
4. Miller Indices (110)
example a b c z
x
y a b
c
4. Miller Indices (100)
1. Intercepts 1 1
2. Reciprocals 1/1 1/1 1/
1 1 0 3. Reduction 1 1 0
1. Intercepts 1/2
2. Reciprocals 1/½ 1/ 1/
2 0 0 3. Reduction 1 0 0
example a b c
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CRYSTALLOGRAPHIC PLANES z
x
y a b
c
4. Miller Indices (634)
example 1. Intercepts 1/2 1 3/4
a b c
2. Reciprocals 1/½ 1/1 1/¾
2 1 4/3
3. Reduction 6 3 4
(001) (010),
Family of Planes {hkl}
(100), (010), (001), Ex: {100} = (100),
FAMILY OF PLANES
Planes that are crystallographically equivalent
have the same atomic packing.
Also, in cubic systems only, planes having the
same indices, regardless of order and sign,
are equivalent.
Ex: {111}
= (111), (111), (111), (111), (111), (111), (111), (111)
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(001) (010), (100), (010), (001), Ex: {100} = (100),
_ _ _ _ _ _ _ _ _ _ _ _
FCC UNIT CELL WITH (110) PLANE
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BCC UNIT CELL WITH (110) PLANE
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CRYSTALLOGRAPHIC PLANES (HCP)
In hexagonal unit cells the same idea is used
example a1 a2 a3 c
4. Miller-Bravais Indices (1011) (hkil)
1. Intercepts 1 -1 1 2. Reciprocals 1 1/
1 0
-1
-1
1
1
3. Reduction 1 0 -1 1
a2
a3
a1
z
Adapted from Fig. 3.8(b),
Callister & Rethwisch 8e.
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DISTINCTION BETWEEN CRYSTALLOGRAPHIC
DIRECTIONS AND MILLER INDICES
Crystallographic Direction Crystallographic Planes
Associated with point
coordinates in 3D space
Associated with planar
structures in 3D space
Denoted by enclosing in
square brackets as [abc]
Denoted by enclosing in
regular parentheses (hkl)
EXAMPLE PROBLEMS
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EXAMPLE PROBLEMS, CONTD.
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Example of
family of
planes
EXAMPLE PROBLEMS, CONTD.
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Don’t
forget to
slide plane
back within
unit cell!
EXAMPLE PROBLEMS, CONTD.
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Strategy
Find answer in
3 coordinates
and then
convert to 4-
coordinate
representation
EXAMPLE PROBLEMS, CONTD.
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ex: linear density of Al in [110]
direction
a = 0.405 nm
LINEAR DENSITY
Linear Density of Atoms LD =
a
[110]
Unit length of direction vector
Number of atoms
# atoms
length
3.5 atoms/nm a 2
2 LD = =
Adapted from Fig. 3.1(a),
Callister & Rethwisch 8e.
2 atoms per line, sharing computed
along vector length
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PLANAR DENSITY OF (100) IRON Solution: At T < 912ºC iron has the BCC structure.
(100)
Radius of iron R = 0.1241 nm
R 3
3 4 a =
Adapted from Fig. 3.2(c), Callister & Rethwisch 8e.
2D repeat unit
= Planar Density = a 2
1
atoms
2D repeat unit
= nm2
atoms 12.1
m2
atoms = 1.2 x 1019
1
2
R 3
3 4 area
2D repeat unit
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PLANAR DENSITY OF (111) IRON
Solution (cont): (111) plane 1 atom in plane/ unit surface cell
3 3 3
2
2
R 3
16 R
3
4
2 a 3 ah 2 area =
= = =
atoms in plane
atoms above plane
atoms below plane
a h 2
3 =
a 2
1
= = nm2
atoms 7.0
m2
atoms 0.70 x 1019
3 2 R 3
16 Planar Density =
atoms
2D repeat unit
area
2D repeat unit
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X-RAY DIFFRACTION
Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation.
Can’t resolve spacings
Spacing is the distance between parallel planes of atoms.
X-RAY DIFFRACTION
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Xray-Tube
Detector
Sample
Bragg´s Equation
d = λ/2 sinθ
d – distance between the same atomic planes λ – monochromatic wavelength θ – angle of diffracto- meter
Xray-Tube
Detector
Metal Target
(Cu or Co)
X-Ray Diffractometer
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X-RAYS TO DETERMINE CRYSTAL STRUCTURE
X-ray intensity (from detector)
q
q c
d = n
2 sin q c
Measurement of
critical angle, qc,
allows computation of
planar spacing, d.
• Incoming X-rays diffract from crystal planes.
Adapted from Fig. 3.20,
Callister & Rethwisch 8e.
reflections must be in phase for a detectable signal
spacing between planes
d
q
q
extra distance travelled by wave “2”
CONSTRUCTIVE INTERFERENCE
Occurs at angle θ to the planes, if path length
difference is equal to a whole number, n, of
wavelengths (n equals order of reflection):
Bragg’s Law
BRAGG’S LAW
If Bragg’s Law is not satisfied then we get non-
constructive interference, which will yield a very
low-intensity diffracted beam.
For cubic unit cells:
BRAGG’S LAW
Specifies when diffraction will occur for unit
cells having atoms positioned only at the cell
corners
Atoms at different locations act as extra
scattering centers and results in the absence of
some diffracted beams.
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X-RAY DIFFRACTION PATTERN
Adapted from Fig. 3.22, Callister 8e.
(110)
(200)
(211)
z
x
y a b
c
Diffraction angle 2q
Diffraction pattern for polycrystalline -iron (BCC)
Inte
nsity (
rela
tive
)
z
x
y a b
c
z
x
y a b
c
EXAMPLE PROBLEM
84
SUMMARY
• Atoms may assemble into crystalline or
amorphous structures.
• We can predict the density of a material, provided we
know the atomic weight, atomic radius, and crystal
geometry (e.g., FCC, BCC, HCP).
• Common metallic crystal structures are FCC, BCC, and
HCP. Coordination number and atomic packing factor
are the same for both FCC and HCP crystal structures.
• Crystallographic points, directions and planes are
specified in terms of indexing schemes.
Crystallographic directions and planes are related
to atomic linear densities and planar densities.
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• Some materials can have more than one crystal
structure. This is referred to as polymorphism (or
allotropy).
SUMMARY
• Materials can be single crystals or polycrystalline.
Material properties generally vary with single crystal
orientation (i.e., they are anisotropic), but are generally
non-directional (i.e., they are isotropic) in polycrystals
with randomly oriented grains.
• X-ray diffraction is used for crystal structure and
interplanar spacing determinations.
ANNOUNCEMENTS
Homework 3 (Due Wed, 4/8/17)
3.20, 3.24, 3.29, 3.30, 3.32, 3.35
Quiz on Wednesday, March 1
Topic: Theoretical Density