Organizacion Atomica
description
Transcript of Organizacion Atomica
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Organizacin Atmica
Materiales Avanzados
Ing. Humberto Gmez. Ph.D
Departamento de Ingeniera Mecnica
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Non dense, random packing
Dense, ordered packing
Dense, ordered packed structures tend to have lower energies.
Energy and Packing Energy
r
typical neighbor bond length
typical neighbor bond energy
Energy
r
typical neighbor bond length
typical neighbor bond energy
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atoms pack in periodic, 3D arrays
Crystalline materials...
-metals -many ceramics -some polymers
atoms have no periodic packing
Noncrystalline materials...
-complex structures -rapid cooling
crystalline SiO2
noncrystalline SiO2 "Amorphous" = Noncrystalline Adapted from Fig. 3.22(b), Callister 7e.
Adapted from Fig. 3.22(a), Callister 7e.
Materials and Packing
Si Oxygen
typical of:
occurs for:
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Single Crystals vs. Polycrystalline Materials
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Czochralski process
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Melt spinning and rapid solidification
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Definitions Lattice
Collection of points that divide space into smaller equally sized segments
Basis
Group of atoms associated with a lattice point
Unit cell
Subdivision of the lattice that still retains the overall characteristics of the entire lattice
Atomic radius
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.
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Crystal Structure
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Unit Cell
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The fourteen types of Bravais lattices grouped in seven crystal systems.
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Metallic Crystal Structures
Metals are usually (poly)crystalline; although formation of amorphous metals are possible by rapid cooling As we learned in Chapter 2, the atomic bonding in metals is non-directional no restriction on numbers or positions of
nearest-neighbor atoms large number of nearest neighbors and dense atomic packing
Atom (hard sphere) radius, R, defined by ion core radius -
typically 0.1 - 0.2 nm
The most common types of unit cells are the faced-centered cubic (FCC), the body-centered cubic (BCC) and the hexagonal
close-packed (HCP).
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Tend to be densely packed.
Reasons for dense packing:
- Typically, only one element is present, so all atomic radii are the same. - Metallic bonding is not directional. - Nearest neighbor distances tend to be small in order to lower bond energy. - Electron cloud shields cores from each other
Have the simplest crystal structures.
We will examine three such structures...
Metallic Crystal Structures
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Rare due to poor packing (only Po has this structure) Close-packed directions are cube edges.
Coordination # = 6 (# nearest neighbors)
(Courtesy P.M. Anderson)
Simple Cubic Cell (SCC)
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APF for a simple cubic structure = 0.52
APF =
a 3
4
3 p (0.5a) 3 1
atoms
unit cell
atom
volume
unit cell
volume
Atomic Packing Factor (APF)
APF = Volume of atoms in unit cell*
Volume of unit cell
*assume hard spheres
Adapted from Fig. 3.23, Callister 7e.
close-packed directions
a
R=0.5a
contains 8 x 1/8 = 1 atom/unit cell
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Coordination # = 8
Adapted from Fig. 3.2, Callister 7e.
Atoms touch each other along cube diagonals.
--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.
Body Centered Cubic Structure (BCC)
ex: Cr, W, Fe (), Tantalum, Molybdenum
2 atoms/unit cell: 1 center + 8 corners x 1/8
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Atomic Packing Factor: BCC
a
APF =
4
3 p ( 3 a/4 ) 3 2
atoms
unit cell atom
volume
a 3
unit cell
volume
length = 4R =
Close-packed directions:
3 a
APF for a body-centered cubic structure = 0.68
a R
Adapted from Fig. 3.2(a), Callister 7e.
a 2
a 3
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Coordination # = 12
Adapted from Fig. 3.1, Callister 7e.
Atoms touch each other along face diagonals.
--Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.
Face Centered Cubic Structure (FCC)
ex: Al, Cu, Au, Pb, Ni, Pt, Ag
4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8
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APF for a face-centered cubic structure = 0.74
Atomic Packing Factor: FCC
maximum achievable APF
APF =
4
3 p ( 2 a/4 ) 3 4
atoms
unit cell atom
volume
a 3
unit cell
volume
Close-packed directions:
length = 4R = 2 a
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell a
2 a
Adapted from Fig. 3.1(a), Callister 7e.
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A sites
B B
B
B B
B B
C sites
C C
C A
B
B sites
ABCABC... Stacking Sequence 2D Projection
FCC Unit Cell
FCC Stacking Sequence
B B
B
B B
B B
B sites
C C
C A
C C
C A
A
B C
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Coordination # = 12
ABAB... Stacking Sequence
APF = 0.74
3D Projection 2D Projection
Adapted from Fig. 3.3(a), Callister 7e.
Hexagonal Close-Packed Structure (HCP)
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
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Compounds: Often have similar close-packed structures.
Close-packed directions --along cube edges.
Structure of NaCl
(Courtesy P.M. Anderson) (Courtesy P.M. Anderson)
Structure of Compounds: NaCl
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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.
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.
Determining the Number of Lattice Points in Cubic Crystal Systems
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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.
Determining the Relationship between Atomic Radius and Lattice Parameters
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
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We find that atoms touch along the edge of the cube in SC structures.
3
40
ra
In FCC structures, atoms touch along the face diagonal of the cube. There are four atomic radii along this lengthtwo radii from the face-centered atom and one radius from each corner, so:
2
40
ra
ra 20
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
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Example: Copper Data from Table inside front cover of Callister (see next slide):
crystal structure = FCC: 4 atoms/unit cell atomic weight = 63.55 g/mol (1 amu = 1 g/mol) atomic radius R = 0.128 nm (1 nm = 10 cm) -7
Compare to actual: rCu = 8.94 g/cm3
Result: theoretical rCu = 8.89 g/cm3
Theoretical Density
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Density Calculation
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Determining the Density of BCC Iron
Determine the density of BCC iron, which has a lattice parameter of 0.2866 nm.
SOLUTION
Atoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cm
Atomic mass = 55.847 g/mol
Volume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24 cm3/cell
Avogadros number NA = 6.02 1023 atoms/mol
3
0a
3
2324/882.7
)1002.6)(1054.23(
)847.55)(2(
number) sadro'cell)(Avogunit of (volume
iron) of mass )(atomicatoms/cell of(number Density
cmg
r
r
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Element Aluminum Argon Barium Beryllium Boron Bromine Cadmium Calcium Carbon Cesium Chlorine Chromium Cobalt Copper Flourine Gallium Germanium Gold Helium Hydrogen
Symbol Al Ar Ba Be B Br Cd Ca C Cs Cl Cr Co Cu F Ga Ge Au He H
At. Weight (amu) 26.98 39.95 137.33 9.012 10.81 79.90 112.41 40.08 12.011 132.91 35.45 52.00 58.93 63.55 19.00 69.72 72.59 196.97 4.003 1.008
Atomic radius (nm) 0.143 ------ 0.217 0.114 ------ ------ 0.149 0.197 0.071 0.265 ------ 0.125 0.125 0.128 ------ 0.122 0.122 0.144 ------ ------
Density
(g/cm3) 2.71 ------ 3.5 1.85 2.34 ------ 8.65 1.55 2.25 1.87 ------ 7.19 8.9 8.94 ------ 5.90 5.32 19.32 ------ ------
Adapted from
Table, "Charac-
teristics of
Selected
Elements",
inside front
cover,
Callister 6e.
Characteristics of Selected Elements at 20C
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rmetals rceramics rpolymers
Why? Metals have... close-packing (metallic bonding)
large atomic mass
Ceramics have... less dense packing (covalent bonding)
often lighter elements
Polymers have... poor packing (often amorphous)
lighter elements (C,H,O)
Composites have... intermediate values Data from Table B1, Callister 6e.
Densities of Material Classes
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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 typ. range from 1 nm to 2 cm
(i.e., from a few to millions of atomic layers).
Adapted from Fig. K, color inset pages of Callister 5e. (Fig. K is courtesy of Paul E. Danielson, Teledyne Wah Chang Albany)
1 mm
Polycrystals
Isotropic
Anisotropic
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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) -If grains are textured, anisotropic.
200 mm
Data from Table 3.3, Callister 7e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.)
Adapted from Fig. 4.14(b), Callister 7e. (Fig. 4.14(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC [now the National Institute of Standards and Technology, Gaithersburg, MD].)
Single vs Polycrystals E (diagonal) = 273 GPa
E (edge) = 125 GPa
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Allotropy
The characteristic of an element being able to exist in more than one crystal structure, depending on temperature and pressure.
Polymorphism
Compounds exhibiting more than one type of crystal structure.
Allotropic or Polymorphic Transformations
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Polymorphism and Allotropy
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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
1538C
1394C
912C
-Fe
-Fe
-Fe
liquid
iron system
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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.
Directions of a form - Crystallographic directions that all have the same characteristics, although their sense is different. Denoted by h i brackets.
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.
Points, Directions, and Planes in the Unit Cell
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Positions and Directions
x, y, z u, v, w X
Y
Z
Lattice Positions and Directions: 1) Always establish an origin 2) Determine the coordinates of the lattice points of interest 3) Translate the vector to the origin if required by drawing a
parallel line or move the origin. 4) Subtract the first point from the second: u2-u1,v2-v1,w2-w1 5) Clear fractions and reduce to lowest terms 6) Write direction with square brackets [uvw] 7) Negative directions get a bar over them.
Crystallography Concepts
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Figure: Coordinates of selected points in the unit cell. The number refers to the distance from the origin in terms of lattice parameters.
Miller Point Indices
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Miller indices- h,k,l- for naming points in the crystal lattice. The origin has been arbitrarily selected as the bottom left-back corner of the unit cell.
1 2
3
4
5
1)
2)
3)
4)
5)
Miller Direction Indices
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Miller Indices
Miller indices are a notation for describing directions and labelling planes in lattices and crystals. The basis for determining the index is the unit cell. It is important to be clear about the unit cell being used. A direction is expressed in terms of its ratio of unit vectors in the form [uvw] where u, v, w are integers. A family of crystallographically equivalent directions is expressed as . A Miller index for a plane is expressed as (hkl), where h, k and l are integers. A family of crystallographically equivalent planes is expressed as {hkl}.
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Determine the Miller indices of directions A, B, and C in below Figure.
Determining Miller Indices of Directions
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
Figure: Crystallographic directions and coordinates
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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
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(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure: Equivalency of crystallographic directions of a form in cubic systems.
Direction Equivalence
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Drawing Plane in the Unit Cell
1) Identify the locations where the plane intercepts the x, y, z axes as the fractions of the unit cell edge lengths a, b, c.
2) Infinity if the plane is parallel. 3) Take the reciprocal of the intercepts. 4) Clear any fraction but do not reduce to lowest terms. 5) Example: 1/3,1/3,1/3 is (333) not (111)!!! 6) Use parentheses to indicate planes (hkl) again with a line over the negative indices. 7) Families are indicated by {hkl}
Remember Terminology: Defined coordinate system: x, y, z Respective unit cell edge lengths: a, b, c Direction: Denoted by [uvw] Family of direction(s): Denoted by: Plane: Denoted by: (hkl) Family of Plane(s): Denoted by: {hkl}
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1) Identify the locations where the plane intercepts the x, y, z axes as the fractions of the unit cell edge lengths a, b, c.
2) Infinity if the plane is parallel. 3) Take the reciprocal of the intercepts. 4) Clear any fraction but do not reduce to lowest terms. 5) Example: 1/3,1/3,1/3 is (333) not (111)!!! 6) Use parentheses to indicate planes (hkl) again with a line over the negative indices. 7) Families are indicated by {hkl}
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Determine the Miller indices of planes A, B, and C in below Figure.
Determining Miller Indices of Planes
Figure: Crystallographic planes and intercepts
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
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Example 3.8 SOLUTION
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. Lets 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.
)010( .4
(c) 2003 Brooks/Cole Publishing / Thomson Learning
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Important Concepts..
Directions are always perpendicular to their respective planes, i.e. [111]direction is
perpendicular to the (111) plane (for cubic systems, not true in all systems) . Families of equivalent planes are equal with respect to symmetrical structures, they do not have to be parallel. Equivalent planes must be translated to the correct atomic positions in order to maintain the proper crystal symmetry. Families of directions are equivalent in absolute magnitude.
(222) planes are parallel to the (111) planes but not equal. Intercepts for the (222) planes are 1/2,1/2,1/2
Intercepts for the (333) planes are 1/3,1/3,1/3, remember this is in what we call reciprocal space. If you draw out the (333) plane it is parallel to the (111) plane but not equivalent.
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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.
Calculating the Planar Density and Packing Fraction
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
Figure: The planer densities of the (010) and (020) planes in SC unit cells are not identical
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SOLUTION
The total atoms on each face is one. The planar density is:
2142
2
atoms/cm 1096.8atoms/nm 96.8
)334.0(
faceper atom 1
face of area
faceper atom (010)density Planar
The planar packing fraction is given by:
79.0)2(
)(
)( atom) 1(
face of area
faceper atoms of area (010)fraction Packing
2
2
20
2
r
r
r
a
p
p
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!
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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: , , -1, 0 =>
Adapted from Fig. 3.8(a), Callister 7e.
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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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 7e.
- a3
a1
a2
z
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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)
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(a), Callister 7e.
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Planar Density of (100) Iron Solution: At T < 912C 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 7e.
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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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Chapter 3: Atomic and Ionic Arrangements
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Diffraction
The constructive interference, or reinforcement, of a beam of x-rays or electrons interacting with a material. The diffracted beam provides useful information concerning the structure of the material.
Braggs law
The relationship describing the angle at which a beam of x-rays of a particular wavelength diffracts from crystallographic planes of a given interplanar spacing.
In a diffractometer
a moving x-ray detector records the 2 angles at which the beam is diffracted, giving a characteristic diffraction pattern
Diffraction Techniques for Crystal Structure Analysis
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What is radiation?
Radiation is energy in the form of waves or particles. Radiation high enough in
energy to cause ionization is called ionizing radiation. It includes particles and
rays given off by radioactive material, stars, and high-voltage equipment. Ionizing
radiation includes x-rays, gamma-rays, beta particles, alpha particles, and
neutrons.
Without the use of monitoring equipment, humans are not able to "find" ionizing radiation. In contrast to heat, light, food, and noise, humans are not able to see, feel, taste, smell, or hear ionizing radiation.
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What are X-rays?
X rays are a form of electromagnetic radiation which arises as electrons are deflected from their
original paths or inner orbital electrons change
their orbital levels around the atomic nucleus. X
rays, like gamma rays, are capable of traveling
long distances through air and most other
materials. Like gamma rays, x rays require more
shielding to reduce their intensity than do beta or
alpha particles. X and gamma rays differ primarily
in their origin: x rays originate in the electronic shell, gamma rays originate in the nucleus.
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X-rays
X-rays were discovered in 1895 when Wilhelm
Conrad Roentgen observed that a screen coated
with a barium salt fluoresced when placed near a
cathode ray tube. Roentgen concluded that a form
of penetrating radiation was being emitted by the
cathode ray tube and called the unknown rays, X-
rays.
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X-ray tube
An x-ray tube requires a source of electrons, a means to accelerate
the electrons, and a target to stop the high-speed electrons.
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X-ray interactions
In passing through matter, energy is transferred from the incident x-ray
photon to electrons and nuclei in the target material. An electron can
be ejected from the atom with the subsequent creation of an ion. The
amount of energy lost to the electron is dependent on the energy of the
incident photon and the type of material through which it travels. There
are three basic methods in which x-rays interact with matter:
photoelectric effect, Compton scattering, and pair production.
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Characteristic x-rays are emitted from heavy elements when their electrons make transitions between the lower atomic energy levels. The characteristic x-rays emission which shown as two sharp peaks in the illustration at left occur when vacancies are produced in the n=1 or K-shell of the atom and electrons drop down from above to fill the gap. The x-rays produced by transitions from the n=2 to n=1 levels are called K-alpha x-rays, and those for the n=3->1 transiton are called K-beta x-rays. Transitions to the n=2 or L-shell are designated as L x-rays (n=3->2 is L-alpha, n=4->2 is L-beta, etc. ). The continuous distribution of x-rays which forms the base for the two sharp peaks at left is called "bremsstrahlung" radiation.
Characteristic X-Rays
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X-ray production typically involves bombarding a metal target in an x-ray tube with high speed electrons which have been accelerated by tens to hundreds of kilovolts of potential. The bombarding electrons can eject electrons from the inner shells of the atoms of the metal target. Those vacancies will be quickly filled by electrons dropping down from higher levels, emitting x-rays with sharply defined frequencies associated with the difference between the atomic energy levels of the target atoms. The frequencies of the characteristic x-rays can be predicted from the Bohr model . Moseley measured the frequencies of the characteristic x-rays from a large fraction of the elements of the periodic table and produces a plot of them which is now called a "Moseley plot". Characteristic x-rays are used for the investigation of crystal structure by x-ray diffraction. Crystal lattice dimensions may be determined with the use of Bragg's law in a Bragg spectrometer.
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Electron transitions to lower atomic levels in heavy atoms have quantum energies which place them in the x-ray region of the electromagnetic spectrum. The x-ray emissions associated with these transitions are called characteristic x-rays. The labels on the illustration show the historical labeling of characteristic x-ray transitions.
X-ray Transitions
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The diffraction Phenomenon
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X-Ray Diffraction and Braggs Law
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= +
= sin + sin
= 2 sin
=
2 + 2 + 2
-
Incoming X-rays diffract from crystal planes.
Measurement of: Critical angles, qc, for X-rays provide
atomic spacing, d.
Adapted from Fig.
3.2W, Callister 6e.
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X-Ray Diffraction Pattern
(110)
(200)
(211)
z
x
y a b
c
Diffraction angle 2q
Diffraction pattern for polycrystalline -iron (BCC)
Inte
nsi
ty (
rela
tive
)
z
x
y a b
c
z
x
y a b
c
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The results of a x-ray diffraction experiment using x-rays with = 0.7107 (a radiation obtained from molybdenum (Mo) target) show that diffracted peaks occur at the following 2 angles:
Examining X-ray Diffraction
Determine the crystal structure, the indices of the plane producing each peak, and the lattice parameter of the material.
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Department of Mechanical Engineering -95-
Solution
We can first determine the sin2 value for each peak, then divide through by the lowest denominator, 0.0308.
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Department of Mechanical Engineering -96-
SOLUTION (Continued)
We could then use 2 values for any of the peaks to calculate the interplanar spacing and thus the lattice parameter. Picking peak 8:
2 = 59.42 or = 29.71 868.2)4)(71699.0(
71699.0)71.29sin(2
7107.0
sin2
2224000
400
lkhda
dq
This is the lattice parameter for body-centered cubic iron.
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Department of Mechanical Engineering -97-
Figure Photograph of a transmission electron microscope (TEM) used for analysis of the microstructure of materials. (Courtesy of JEOL USA, Inc.)
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Department of Mechanical Engineering -98-
Figure: A TEM micrograph of an aluminum alloy (Al-7055) sample. The diffraction pattern at the right shows large bright spots that represent diffraction from the main aluminum matrix grains. The smaller spots originate from the nano-scale crystals of another compound that is present in the aluminum alloy. (Courtesy of Dr. Jrg M.K. Wiezorek, University of Pittsburgh.)