Crystal Structure Types 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, selforganized in a 3D periodic array). These can be monocrystals and polycrystals.
Amorphous
crystalline
polycrystalline
To discuss crystalline structures it is useful to consider atoms as being hard spheres with welldefined radii. In this hardsphere model, the shortest distance between two like atoms is one diameter. Lattice: A 3dimensional 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
Crystal Systems Crystallographers 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 Orthorhombic a≠ 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 (18111863) showed that 14 standards unit cells could describe all possible lattice networks.
Principal Metallic Structures Most elemental metals (about 90%) crystallize upon solidification into three densely packed crystal structures: bodycentered cubic (BCC), facecentered cubic (FCC) and hexagonal closepacked (HCP)
Structures of Metallic Elements H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
Fr
Ra
Ac
Prim itive Cubic
Cubic close packing (Face centered cubic)
Body Centered Cubic
Hexagonal close packing
Structure
Metal
Lattice Constant a, nm
BCC
FCC
HCP
c, nm
Atomic Radius, nm
Chromium
0.289
0.125
Iron
0.287
0.124
Molybdenum
0.315
0.136
Potassium
0.533
0.231
Sodium
0.429
0.186
Tungsten
0.316
0.137
Aluminum
0.405
0.143
Copper
0.361
0.128
Gold
0.408
0.144
Nickel
0.352
0.125
Silver
0.409
0.144
Zinc
0.2665
0.5618
0.133
Magnesium
0.3209
0.5209
0.160
Cobalt
0.2507
0.4069
0.125
Titanium
0.2950
0.3584
0.147
SIMPLE CUBIC STRUCTURE (SC) • Rare due to poor packing (only Po has this structure) • Closepacked directions are cube edges. • Coordination # = 6 (# nearest neighbors)
• Number of atoms per unit cell= 1 atom
Atomic Packing Factor (APF) Volume of atoms in unit cell* APF = Volume of unit cell *assume hard spheres
• APF for a simple cubic structure = 0.52
atoms unit cell
a R=0.5a closepacked directions contains 8 x 1/8 = 1 atom/unit cell
APF =
volume atom 4 π (0.5a)3 1 3 a3
volume unit cell 6
Body Centered Cubic (BCC)
• Close packed directions are cube diagonals.
Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.
• Coordination # = 8
Unit cell ontains: c 1 + 8 x 1/8 = 2 atoms/unit ce Closepacked directions: length = 4R = 3a
R
a
atoms volume 4 3 π ( 3a/4) 2 unit cell atom 3 APF = • APF for a BCC = 0.68 volume a3 unit cell
Face Centered Cubic (FCC) • Close packed directions are face diagonals.
Note: All atoms are identical; the facecentered atoms are shaded differently only for ease of viewing.
• Coordination # = 12 Closepacked directions: length = 4R = 2a Unit cell ontains: c 6 x 1/2 + 8 x 1/8 a = 4 atoms/unit ce
atoms volume 4 3 π ( 2a/4) 4 unit cell atom 3 APF = • APF for a FCC = 0.74 volume 3 a unit cell
Hexagonal ClosePacked (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. 2 atoms shared by two cells = 1 atom per cell
3 atoms
12 atoms shared by six cells = 2 atoms per cell
ClosePacked Structures Both the HCP and FCC crystal structures are closepacked 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.
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
A B A : hexagonal close packed
A B C : cubic close packed
A B
C A
A B C : cubic close pack A B A : hexagonal close pack 90°
A
B A
120°
Interstitial sites Locations 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.
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
Density Calculations Since 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/mol The volume of the cell, Vc = a3 (FCC and BCC) a = 2R√2 (FCC); a = 4R/√3 (BCC) where R is the atomic radius.
Density Calculation n: number of atoms/unit cell
nA ρ= VC N A
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,
VC = a 3 = (2 R 2 ) 3 = 16 2 R 3
(4)(63.5) 3 ρ= = 8 . 89 g / cm [16 2 (1.28 × 108 ) 3 × 6.023 × 10 23 ] 8.94 g/cm3 in the literature
Example Rhodium has an atomic radius of 0.1345nm (1.345A) and a density of 12.41g.cm3. Determine whether it has a BCC or FCC crystal structure. Rh (A = 102.91g/mol) Solution
nA ρ= VC N A
n: number of atoms/unit cell VC: volume of the unit cell
A: atomic weight NA: Avogadro’s number (6.023x1023 atoms/mole)
Vc a 3 A 102.91g .mol −1 = = = = 1.3768 x10 − 23 cm 3 = 0.01376nm 3 −3 −1 23 n n ρN A 12.41g .cm 6.023x10 atoms.mole If rhodium is BCC then n = 2 and a 3 = (4r ) 3 = 12.316r 3 3 a 3 12.316 x(0.1345nm) 3 = = 0.0149nm 3 2 n If rhodium is FCC then n = 4 and a 3 = (4r ) 3 = 22.627 r 3 2 a 3 22.627 x(0.1345nm) 3 = = 0.01376nm 3 4 n Rhodium has a FCC structure
Linear And Planar Atomic Densities
Crystallographic direction
Linear atomic density: 4R Ll = a = 3
= 2R/Ll = 0.866
Planar atomic density:
A’
E’
Ll = 2π R2/(Area A’D’E’B’)
Polymorphism or Allotropy Many elements or compounds exist in more than one crystalline form under different conditions of temperature and pressure. This phenomenon is termed polymorphism and if the material is an elemental solid is called allotropy. Example:
Iron (Fe – Z = 26) liquid above 1539 C. δiron (BCC) between 1394 and 1539 C. γiron (FCC) between 912 and 1394 C. αiron (BCC) between 273 and 912 C.
α iron BCC
912oC
γ iron FCC
1400oC
δ iron BCC
1539oC
liquid iron
Another example of allotropy is carbon. Pure, solid carbon occurs in three crystalline forms – diamond, graphite; and large, hollow fullerenes. Two kinds of fullerenes are shown here: buckminsterfullerene (buckyball) and carbon nanotube.
Crystallographic Planes and Directions Atom Positions in Cubic Unit Cells A cube of lattice parameter a is considered to have a side equal to unity. Only the atoms with coordinates x, y and z greater than or equal to zero and less than unity belong to that specific cell.
z 0,0,1 1,0,1
0,1,1 1,1,1
½, ½, ½
1,0,0
0,1,0
0,0,0 1,1,0
x
y
0 ≤ x, y , z < 1
Directions in The Unit Cell For cubic crystals the crystallographic directions indices are the vector components of the direction resolved along each of the coordinate axes and reduced to the smallest integer.
z
Example direction A 0,0,1
1,0,1
0,1,1 1,1,1
½, ½, ½
1,0,0
0,1,0
0,0,0 A 1,1,0
x
y
a) Two points origin coordinates 0,0,0 and final position coordinates 1,1,0 b) 1,1,0  0,0,0 = 1,1,0 c) No fractions to clear d) Direction [110]
z 0,0,1
C B
0,0,0
x
Example direction B a) Two points origin coordinates 1,1,1 and final position coordinates 0,0,0 b) 0,0,0  1,1,1 = 1,1,1 c) No fractions _to_ clear _ d) Direction [111] Example direction C 1,1,1 a) Two points origin coordinates ½,1,0 and final position coordinates 0,0,1 y b) 0,0,1  ½,1,0 = ½,1,1 ½, 1, 0 c) There are fractions to clear. Multiply times 2. 2( ½,1,1) = _ _ 1,2,2 d) Direction [1 2 2]
Notes About the Use of Miller Indices for Directions A direction and its negative are not identical; [100] is not equal to [bar100]. They represent the same line but opposite directions. . direction and its multiple are identical: [100] is the same direction as [200]. We just forgot to reduce to lowest integers. Certain groups of directions are equivalent; they have their particular indices primarily because of the way we construct the coordinates. For example, a [100] direction is equivalent to the [010] direction if we redefine the coordinates system. We may refer to groups of equivalent directions as directions of the same family. The special brackets < > are used to indicate this collection of directions. Example: The family of directions consists of six equivalent directions
< 100 > ≡ [100], [010], [001], [010], [001 ], [ 100]
Miller Indices for Crystallographic planes in Cubic Cells ¾ Planes in unit cells are also defined by three integer numbers, called the Miller indices and written (hkl). ¾ Miller’s indices can be used as a shorthand notation to identify crystallographic directions (earlier) AND planes. Procedure for determining Miller Indices locate the origin identify the points at which the plane intercepts the x, y and z coordinates as fractions of unit cell length. If the plane passes through the origin, the origin of the coordinate system must be moved!
take reciprocals of these intercepts clear fractions but do not reduce to lowest integers enclose the resulting numbers in parentheses (h k l).
Again, the negative numbers should be written with a bar over the number.
z
A
x
Example: Miller indices for plane A a) Locate the origin of coordinate. b) Find the intercepts x = 1, y = 1, z = 1 c) Find the inverse 1/x=1, 1/y=1, 1/z=1 d) No fractions to clear y e) (1 1 1)
More Miller Indices  Examples c
c
c 1/5
0.5
a
2/3
b
b
b
a
0.5
a
c
c
c b
b a
a
b
a
Notes About the Use of Miller Indices for Planes A plane and its negative are parallel and identical. Planes and its multiple are parallel planes: (100) is parallel to the plane (200) and the distance between (200) planes is half of the distance between (100) planes.
Certain groups of planes are equivalent (same atom distribution); they have their particular indices primarily because of the way we construct the coordinates. For example, a (100) planes is equivalent to the (010) planes. We may refer to groups of equivalent planes as planes of the same family. The special brackets { } are used to indicate this collection of planes. In cubic systems the direction of miller indices [h k l] is normal o perpendicular to the (h k l) plane. in cubic systems, the distance d between planes (h k l ) is given by the formula where a is the lattice a d= constant. h2 + k 2 + l 2 Example: The family of planes {100} consists of three equivalent planes (100) , (010) and (001)
A “family” of crystal planes contains all those planes are crystallographically equivalent. • Planes have the same atomic packing density • a family is designated by indices that are enclosed by braces.  {111}: (111 ), (111 ), (111), (111), (111), (111), (111), (111)
• Single
Crystal • Polycrystalline materials • Anisotropy and isotropy
Two Types of Indices in the Hexagonal System a1 ,a2 ,and c are independent, a3 is not!
c
a3 =
a3
Miller: a2
(hkl)
(a1 + a2)

(same as before)
MillerBravais: (hkil) → i =  (h+k)
a1
(001) = (0001) c


c
(110) = (1100)
a3
a3 a2
a2 a1
a1

(110) = (1100) (100) = (1010)
XRay Diffraction XRay Spectroscopy
Visible light: 0.40.7m~ 6000A
Mo: 35KeV ~ 0.020.14nm
E = hν = hc / λ
XRay Diffraction from a Crystal • Electromagnetic radiation is wavelike: Electric field
+
+ 
+ 
+ 
+ 
+ 
•Waves can add constructively or destructively: Electric field
Sum
=
+
+ 
+ 
+ 
+ 
+ 
Direction of motion of xray photon
Bragg’s Law For constructive interference
For cubic system:
d hkl =
nl = DE + EF nl = dhkl sinθ + dhkl sinθ nl = 2dhkl sinθ
a 2 2 + + h k l 2
But no all planes diffract !!!
For the BCC structure the first two sets of principal diffracting planes are {110} and {200}. For the FCC structure the first two sets of principal diffracting planes are {111} and {200}.
sin 2 θ A hA2 + k A2 + l A2 = 2 2 sin θ B hB + k B2 + lB2 sin 2 θ A = 0.5 ( BCC ); = 0.75 ( FCC ) 2 θ sin B
Ex: An element, BCC or FCC, shows diffraction peaks at 2θ: 40, 58, 73, 100.4 and 114.7. Determine: (a) Crystal structure? (b) Lattice constant? (c) What is the element?
From W.F.. Smith, Fundamentals of materials Science And Engineering, McGrawHill, 1992, Ex. 3.16.