Crystal Structure - UPRMacademic.uprm.edu/pcaceres/Courses/MatEng/MSE3-3.pdf · ... M = Atomic...
Transcript of Crystal Structure - UPRMacademic.uprm.edu/pcaceres/Courses/MatEng/MSE3-3.pdf · ... M = Atomic...
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
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 SystemsCrystallographers have shown that only seven different types of unit cells are necessary to create all point latticeCubic a= b = c ; α = β = γ = 90Tetragonal a= b ≠ c ; α = β = γ = 90Rhombohedral a= b = c ; α = β = γ ≠ 90Hexagonal a= b ≠ c ; α = β = 90, γ =120Orthorhombica≠ b ≠ c ; α = β = γ = 90Monoclinic a≠ b ≠ c ; α = γ = 90 ≠ β Triclinic a≠ b ≠ c ; α ≠ γ ≠ β ≠ 90
Bravais LatticesMany 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 StructuresMost 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
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
Lattice ConstantStructurea, nm c, nm
Chromium 0.289 0.125Iron 0.287 0.124Molybdenum 0.315 0.136Potassium 0.533 0.231Sodium 0.429 0.186
BCC
Copper 0.361 0.128Gold 0.408 0.144Nickel 0.352 0.125Silver 0.409 0.144Zinc 0.2665 0.5618 0.133Magnesium 0.3209 0.5209 0.160Cobalt 0.2507 0.4069 0.125Titanium 0.2950 0.3584 0.147
HCP
FCC
Tungsten 0.316 0.137Aluminum 0.405 0.143
Atomic Radius, nm
Metal
• 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
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 cellvolume
close-packed directions
a
R=0.5a
contains 8 x 1/8 = 1 atom/unit cell
Atomic Packing Factor (APF)
• 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 ce
APF = a3
4
3π ( 3a/4)32
atoms
unit cell atomvolume
unit cell
volume • APF for a BCC = 0.68
• 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 atomvolume
unit cell
volume
Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit ce
Close-packed directions: length = 4R
= 2 a
• APF for a FCC = 0.74
Hexagonal Close-Packed (HCP)The APF and coordination number of the HCP structure is the sameas 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
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 APlace the next layer on top of the first. Layer B. Note that there are
two possible positions for a proper stacking of layer B.
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
A B C : cubic close pack
A
B CA
120°
90°A
A
B
A B A : hexagonal close pack
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 sixother 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 sitesare importantbecause we canderive morestructures fromthese basic FCC,BCC, HCPstructures bypartially orcompletelydifferent sets ofthese sites
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.
Density Calculation
AC NVnA
=ρ
n: number of atoms/unit cellA: 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 ===
32338
/89.8]10023.6)1028.1(216[
)5.63)(4( cmg=×××
=ρ
8.94 g/cm3 in the literature
ExampleRhodium 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
AC NVnA
=ρn: 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
33231233
13
nmnmxna
rra
nmnmxna
rra
nmcmxmoleatomsxcmg
molgNA
na
nV
A
c
==
===
==
===
===== −−−
−
ρ
Linear And Planar Atomic Densities
Linear atomic density:
= 2R/Ll
Planar atomic density:
Ll
Crystallographic direction
= 2π R2/(Area A’D’E’B’)
34RaLl == = 0.866
A’ E’
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.
α iron γ iron δ iron912oC 1400oC 1539oC
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 DirectionsAtom Positions in Cubic Unit CellsA 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.
y
z
x
1,0,1
1,0,0
1,1,1
½, ½, ½
0,0,0
0,0,1 0,1,1
1,1,0
0,1,0
1,,0 <≤ zyx
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.
y
z
x
1,0,1
1,0,0
1,1,1
½, ½, ½
0,0,0
0,0,1 0,1,1
1,1,0
0,1,0A
Example direction A
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]
y
z
x
1,1,1
0,0,0
0,0,1
B
C
½, 1, 0
Example direction Ba) Two points origin coordinates
1,1,1 and final position coordinates 0,0,0
b) 0,0,0 - 1,1,1 = -1,-1,-1c) No fractions to cleard) Direction ]111[
___
Example direction Ca) Two points origin coordinates
½,1,0 and final position coordinates 0,0,1
b) 0,0,1 - ½,1,0 = -½,-1,1c) There are fractions to clear.
Multiply times 2. 2( -½,-1,1) = -1,-2,2
d) Direction ]221[__
Notes About the Use of Miller Indices for DirectionsA 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
00]1[],1[000],1[0[001],[010],[100], >100< ≡
Miller Indices for Crystallographic planes in Cubic CellsPlanes in unit cells are also defined by three integer numbers,
called the Miller indices and written (hkl).Miller’s indices can be used as a shorthand notation to identify
crystallographic directions (earlier) AND planes.Procedure for determining Miller Indices
locate the originidentify 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 interceptsclear fractions but do not reduce to lowest integersenclose the resulting numbers in parentheses (h k l). Again, the
negative numbers should be written with a bar over the number.
y
z
x
A
Example: Miller indices for plane Aa) Locate the origin of coordinate.b) Find the intercepts x = 1, y = 1, z = 1c) Find the inverse 1/x=1, 1/y=1, 1/z=1d) No fractions to cleare) (1 1 1)
More Miller Indices - Examples
ab
c
0.50.5
ab
c
2/31/5
ab
c
ab
c
ab
c
ab
c
Notes About the Use of Miller Indices for PlanesA 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 where a is the lattice constant.Example: The family of planes {100} consists of three equivalent planes (100), (010) and (001)
222 lkhad
++=
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(
• Single Crystal• Polycrystalline materials• Anisotropy and isotropy
c
(110) = (1100)- -
(100) = (1010)-
a1
a2
a3
(001) = (0001)
(110) = (1100)- -
Two Types of Indices in the Hexagonal System
a1
a2
a3
c
a1
a2
a3
c
Miller: (hkl) (same as before)
Miller-Bravais: (hkil) → i = - (h+k)
a3 = - (a1 + a2)a1 ,a2 ,and c are independent, a3 is not!
X-Ray Spectroscopy
λν /hchE ==35KeV ~ 0.02-0.14nmMo:
X-Ray Diffraction
Visible light: 0.4-0.7m~ 6000A
X-Ray Diffraction from a Crystal• Electromagnetic radiation is wave-like:
Electric field + + + + +
- - - - -+
Direction of motion of x-ray photon
•Waves can add constructively or destructively:Electric
field + + + + +- - - - -
+
Sum
=
Bragg’s Law
For cubic system:
But no all planes diffract !!!
For constructive interference
θlθθl
l
sin2sinsin
hkl
hklhkl
dnddn
EFDEn
=+=
+=
222 lkhadhkl ++
=
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}.
222
222
2
2
sinsin
BBB
AAA
B
A
lkhlkh
++++
=θθ
)(75.0);(5.0sinsin
2
2
FCCBCCB
A ==θθ
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.