crystal structure
TRANSCRIPT
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Crystal StructureTypes of Solids
� Amorphous: Solids with considerable disorder in their structures
(e.g., glass). Amorphous: lacks a systematic atomic arrangement.
� Crystalline: Solids with rigid a highly regular arrangement of
their atoms. (That is, its atoms or ions, self-organized in a 3D
periodic array). These can be monocrystals and polycrystals.
Amorphous crystalline polycrystalline
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To discuss crystalline structures it is useful to consider atoms as being
hard spheres with well-defined radii. In this hard-sphere model, the
shortest distance between two like atoms is one diameter.
Lattice: A 3-dimensional system of points that designate the positions
of the components (atoms, ions, or molecules) that make up the
substance.
Unit Cell: The smallest repeating unit of the lattice.
The lattice is generated by repeating the unit cell in all three
dimensions
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Crystal SystemsCrystallographers have shown that only
seven different types of unit cells are
necessary to create all point lattice
Cubic a= b = c ; α = β = γ = 90
Tetragonal a= b ≠ c ; α = β = γ = 90
Rhombohedral a= b = c ; α = β = γ ≠ 90
Hexagonal a= b ≠ c ; α = β = 90, γ =120
Orthorhombica≠ b ≠ c ; α = β = γ = 90
Monoclinic a≠ b ≠ c ; α = γ = 90 ≠ β
Triclinic a≠ b ≠ c ; α ≠ γ ≠ β ≠ 90
Bravais Lattices
Many of the seven crystal systems have variations of the basic unit
cell. August Bravais (1811-1863) showed that 14 standards unit
cells could describe all possible lattice networks.
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Principal Metallic Structures
Most elemental metals (about
90%) crystallize upon
solidification into three densely
packed crystal structures: body-
centered cubic (BCC), face-
centered cubic (FCC) and
hexagonal close-packed (HCP)
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Structures of Metallic Elements
Ru
H
Li
Na
K
Rb
Cs
Fr
Be
Mg
Ca
Sr
Ba
Ra
Sc
Y
La
Ac
Ti
Zr
H f
V
Nb
Ta
Cr
Mo
W
Mn
Tc
Re
Fe
Os
Co
Ir
Rh
Ni
Pd
Pt
Cu
Ag
Au
Zn
Cd
Hg
B
Al
Ga
In
Tl
C
Si
Ge
Sn
Pb
N
P
As
Sb
Bi
O
S
Se
Te
Po
F
Cl
Br
I
At
Ne
Ar
Kr
Xe
Rn
He
Primitive Cubic
Body Centered Cubic
Cubic close packing(Face centered cubic)
Hexagonal close packing
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0.1470.35840.2950Titanium
0.1250.40690.2507Cobalt
0.1600.52090.3209Magnesium
0.1330.56180.2665Zinc
HCP
0.1440.409Silver
0.1250.352Nickel
0.1440.408Gold
0.1280.361Copper
FCC
BCC
Structure
0.1370.316Tungsten
0.1860.429Sodium
0.2310.533Potassium
0.1360.315Molybdenum
0.1240.287Iron
0.1250.289Chromium
0.1430.405Aluminum
c, nma, nm
Atomic
Radius, nm
Lattice ConstantMetal
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• Rare due to poor packing (only Po has this structure)
• Close-packed directions are cube edges.
• Coordination # = 6 (# nearest neighbors)
SIMPLE CUBIC STRUCTURE (SC)
• Number of atoms per unit cell= 1 atom
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6
APF = Volume of atoms in unit cell*
Volume of unit cell
*assume hard spheres
• APF for a simple cubic structure = 0.52
APF =
a3
4
3π (0.5a)31
atoms
unit cellatom
volume
unit cell
volumeclose-packed directions
a
R=0.5a
contains 8 x 1/8 = 1 atom/unit cell
Atomic Packing Factor (APF)
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• Coordination # = 8
• Close packed directions are cube diagonals.--Note: All atoms are identical; the center atom is shaded differently only for ease of
viewing.
Body Centered Cubic (BCC)
aR
Close-packed directions: length = 4R
= 3 a
Unit cell contains: 1 + 8 x 1/8 = 2 atoms/unit cell
APF =
a3
4
3π ( 3a/4)32
atoms
unit cell atom
volume
unit cell
volume• APF for a BCC = 0.68
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• Coordination # = 12
• Close packed directions are face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded differently only for
ease of viewing.
Face Centered Cubic (FCC)
a
APF =
a3
4
3π ( 2a/4)34
atoms
unit cell atom
volume
unit cell
volume
Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell
Close-packed directions: length = 4R
= 2 a
• APF for a FCC = 0.74
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Hexagonal Close-Packed (HCP)The APF and coordination number of the HCP structure is the same
as the FCC structure, that is, 0.74 and 12 respectively.
An isolated HCP unit cell has a total of 6 atoms per unit cell.
12 atoms shared by six cells = 2 atoms per cell2 atoms shared by two cells = 1
atom per cell
3 atoms
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Close-Packed StructuresBoth the HCP and FCC crystal structures are close-packed structure.
Consider the atoms as spheres:
�Place one layer of atoms (2 Dimensional solid). Layer A
�Place the next layer on top of the first. Layer B. Note that there are
two possible positions for a proper stacking of layer B.
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A B A : hexagonal close packed A B C : cubic close packed
The third layer (Layer C) can be placed in also teo different
positions to obtain a proper stack.
(1)exactly above of atoms of Layer A (HCP) or
(2)displaced
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A B C : cubic close pack
A
B C
A
120°
90°A
A
B
A B A : hexagonal close pack
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Interstitial sitesLocations between the ‘‘normal’’ atoms or ions in a crystal into
which another - usually different - atom or ion is placed.
o 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.
o 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.
o 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.
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Crystals having filled Interstitial Sites
FCC Lattice has:
3 [=12(¼)] Oh sites at edge centers
+ 1 Oh site at body center
Octahedral, Oh, Sites
FCC Lattice has:
8 Th sites at ¼, ¼, ¼ positions
Tetrahedral, Th, Sites
Interstitial sites are important because we can derive more structures from these basic
FCC, BCC, HCP structures by partially or completely different sets of these sites
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Density CalculationsSince the entire crystal can be generated by the repetition of the
unit cell, the density of a crystalline material, ρ = the density of the
unit cell = (atoms in the unit cell, n ) × (mass of an atom, M) / (the
volume of the cell, Vc)
Atoms in the unit cell, n = 2 (BCC); 4 (FCC); 6 (HCP)
Mass of an atom, M = Atomic weight, A, in amu (or g/mol) is
given in the periodic table. To translate mass from amu to grams
we have to divide the atomic weight in amu by the Avogadro
number NA = 6.023 × 1023 atoms/molThe volume of the cell, Vc = a3 (FCC and BCC)
a = 2R√2 (FCC); a = 4R/√3 (BCC)where R is the atomic radius.
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Density Calculation
ACNV
nA=ρ
n: number of atoms/unit cell
A: atomic weight
VC: volume of the unit cell
NA: Avogadro’s number
(6.023x1023 atoms/mole)
Example Calculate the density of copper.
RCu =0.128nm, Crystal structure: FCC, ACu= 63.5 g/mole
n = 4 atoms/cell, 333 216)22( RRaVC ===
3
2338/89.8
]10023.6)1028.1(216[
)5.63)(4(cmg=
×××=ρ
8.94 g/cm3 in the literature
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Example
Rhodium has an atomic radius of 0.1345nm (1.345A) and a density of 12.41g.cm-3.
Determine whether it has a BCC or FCC crystal structure. Rh (A = 102.91g/mol)
Solution
ACNV
nA=ρ
n: number of atoms/unit cell A: atomic weight
VC: volume of the unit cell NA: Avogadro’s number
(6.023x1023 atoms/mole)
structure FCC a has Rhodium
01376.04
)1345.0(627.22
627.22)2
4( and 4 n then FCC is rhodium If
0149.02
)1345.0(316.12
316.12)3
4( and 2 n then BCC is rhodium If
01376.0103768.1.10023.6.41.12
.91.102
333
333
333
333
3323
1233
13
nmnmx
n
a
rra
nmnmx
n
a
rra
nmcmxmoleatomsxcmg
molg
N
A
n
a
n
V
A
c
==
===
==
===
===== −
−−
−
ρ
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Linear And Planar Atomic Densities
Linear atomic density:
= 2R/Ll
Planar atomic density:
Ll
Crystallographic direction
= 2π R2/(Area A’D’E’B’)
3
4RaLl == = 0.866
A’ E’