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, self-organized 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 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
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 (1811-1863) 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: body-
centered cubic (BCC), face-
centered cubic (FCC) and
hexagonal close-packed (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 Atomic
a, nm c, nm Radius, nm
Chromium 0.289 0.125
Iron 0.287 0.124
Molybdenum 0.315 0.136
BCC
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
FCC 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
HCP
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)
• Close-packed 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
volume
atoms atom
a 4
unit cell 1 π (0.5a)3
3
R=0.5a APF =
a3 volume
close-packed directions
unit cell
contains 8 x 1/8 =
1 atom/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
Close-packed directions:
length = 4R R
= 3a a
atoms volume
4 3
unit cell 2 π ( 3a/4)
3 atom
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 face-centered atoms are shaded differently only for
ease of viewing.
• Coordination # = 12
Close-packed directions:
length = 4R
= 2a
Unit cell ontains:
c
6 x 1/2 + 8 x 1/8
= 4 atoms/unit ce a
atoms volume
4 3
unit cell 4 π ( 2a/4)
3 atom
APF = • APF for a FCC = 0.74
3 volume
a
unit cell
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.
2 atoms shared by two cells = 1
atom per cell 12 atoms shared by six cells = 2 atoms per cell
3 atoms
Close-Packed Structures
Both 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.
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 A: atomic weight
ρ= VC: volume of the unit cell
VC N A
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)
ρ= = 8 . 89 g / cm 3
[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.cm-3.
Determine whether it has a BCC or FCC crystal structure. Rh (A = 102.91g/mol)
Solution
n: number of atoms/unit cell A: atomic weight
nA
ρ=
VC N A VC: volume of the unit cell 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
n n ρN A 12.41g .cm 6.023x10 atoms.mole
−3 23 −1
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
n 2
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
n 4
Rhodium has a FCC structure
Linear And Planar Atomic Densities
Crystallographic direction
Linear atomic density: A’ E’
4R = 2R/Ll
Ll = a =
3 = 0.866
Ll
Planar atomic density: = 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.
δ-iron (BCC) between 1394 and 1539 C.
γ-iron (FCC) between 912 and 1394 C.
α-iron (BCC) between -273 and 912 C.
912oC 1400oC 1539oC
α iron γ iron δ iron liquid iron
BCC FCC BCC
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 0,1,1
0 ≤ x, y , z < 1
1,0,1 1,1,1
½, ½, ½
y
0,1,0
0,0,0
1,0,0
1,1,0
x
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 0,1,1 a) Two points origin coordinates
0,0,0 and final position
1,0,1 1,1,1
coordinates 1,1,0
½, ½, ½
b) 1,1,0 - 0,0,0 = 1,1,0
y
0,0,0 A
0,1,0 c) No fractions to clear
1,0,0
1,1,0 d) Direction [110]
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
z c) No fractions _to_ clear
_
d) Direction [111]
0,0,1
C Example direction C
1,1,1
a) Two points origin coordinates
B
½,1,0 and final position
y coordinates 0,0,1
0,0,0 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 _ _
x
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 co-
ordinates. For example, a [100] direction is equivalent to the [010]
direction if we re-define the co-ordinates 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 <100> 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 co-ordinate 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 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
A d) No fractions to clear
y e) (1 1 1)
x
More Miller Indices - Examples
c c c
1/5
2/3
b b b
0.5
a a a
0.5
c c
c
b b b
a a 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 co-ordinates. 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 a where a is the lattice
constant. d=
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 crystallo-
graphically 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)
• SingleCrystal
• Polycrystalline materials
• Anisotropy and isotropy
Two Types of Indices in the Hexagonal System
a1 ,a2 ,and c are independent, a3 is not!
c
a3 = - (a1 + a2)
a3 Miller: (hkl) (same as before)
a2
a1 Miller-Bravais: (hkil) → i = - (h+k)
(001) = (0001)
c - -
(110) = (1100)
c
a3 a3
a2 a2
a1 a1
- -
(110) = (1100)
-
(100) = (1010)
X-Ray Diffraction
X-Ray Spectroscopy Visible light: 0.4-0.7m~ 6000A
Mo: 35KeV ~ 0.02-0.14nm E = hν = hc / λ
X-Ray Diffraction from a Crystal
• Electromagnetic radiation is wave-like:
Electric
Direction
field + + + + + + of motion
- - - - - of x-ray
photon
•Waves can add constructively or destructively:
Electric
field + + + + + +
- - - - -
Sum
=
Bragg’s Law
For constructive interference nl = DE + EF
nl = dhkl sinθ + dhkl sinθ
For cubic system:
nl = 2dhkl sinθ
a
d hkl =
+22
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
sin θ B hB + k B2 + lB2
2
sin 2 θ A
= 0.5 ( BCC ); = 0.75 ( FCC )
2
sin B θ
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, McGraw-Hill, 1992, Ex.
3.16.
