3. Types of Solids
• Solids can be divided into two groups
based on the arrangement of the atoms.
These are
–Crystalline
–Amorphous
4. Types of Solids
• In a Crystalline solid, atoms are arranged
in an orderly manner. The atoms are
having long range order.
– Example : Iron, Copper and other metals,
NaCl etc.
• In an Amorphous solid, atoms are not
present in an orderly manner. They are
haphazardly arranged.
– Example : Glass
5. ENERGY AND PACKING
• Non dense, random packing
Energy
typical neighbor
bond length
typical neighbor
bond energy
• Dense, regular packing
r
Energy
typical neighbor
bond length
typical neighbor
bond energy
r
Dense, regular-packed structures tend to have lower energy.
5
6. Some Definitions
Crystalline material is a material comprised of one or many crystals. In each
crystal, atoms or ions show a long-range periodic arrangement.
Single crystal is a crystalline material that is made of only one crystal (there
are no grain boundaries).
Grains are the crystals in a polycrystalline material.
Polycrystalline material is a material comprised of many crystals (as
opposed to a single-crystal material that has only one crystal).
Grain boundaries are regions between grains of a polycrystalline material.
6
7. Atomic PACKING
Crystalline materials...
• atoms pack in periodic, 3D arrays
• typical of:
-metals
-many ceramics
-some polymers
crystalline SiO2
Adapted from Fig. 3.18(a),
Callister 6e.
Noncrystalline materials...
Si
Oxygen
• atoms have no periodic packing
• occurs for:
-complex structures
-rapid cooling
"Amorphous" = Noncrystalline
noncrystalline SiO2
Adapted from Fig. 3.18(b),
Callister 6e.
From Callister 6e resource CD.
7
8. Space Lattice
A Space LATTICE is an infinite, periodic array of mathematical points, in
which each point has identical surroundings to all others.
LATTICE
A CRYSTAL STRUCTURE is a periodic arrangement of atoms in the crystal that can
be described by a LATTICE
8
9. Space Lattice
Important Note:
• Lattice points are a purely mathematical concept,
whereas atoms are physical objects.
• So, don't mix up atoms with lattice points.
• Lattice Points do not necessarily lie at the center of atoms.
In Figure (a) is the 3-D periodic arrangement of atoms, and
Figure (b) is the corresponding space lattice.
In this case, atoms lie at the same point as the space lattice.
9
10. Crystalline Solids: Unit Cells
Unit Cell: It is the basic structural unit of a crystal
structure. Its geometry and atomic positions
define the crystal structure.
Fig. 3.1 Atomic configuration in
Face-Centered-Cubic
Arrangement
R
R R
A unit cell is the smallest part of the unit cell,
which when repeated in all three directions,
reproduces the lattice.
R
a
10
11. Unit Cells and Unit Cell Vectors
Lattice parameters
axial lengths: a, b, c
interaxial angles: α, β, γ
unit vectors:
a b c
c
b
a
All unit cells may be described via
these vectors and angles.
11
15. Unit Cells Types
A unit cell is the smallest component of the crystal that reproduces the whole
crystal when stacked together with purely translational repetition.
• Primitive (P) unit cells contain only a single lattice point.
• Internal (I) unit cell contains an atom in the body center.
• Face (F) unit cell contains atoms in the all faces of the planes composing the cell.
• Centered (C) unit cell contains atoms centered on the sides of the unit cell.
Primitive
Body-Centered
Face-Centered
End-Centered
KNOW THIS!
15
17. Orderly arrangement of atoms in Crystals
Atom
Translation Vectors
a1, a2 ,a3
a3
a2
a1
17
18. CLASSIFICATION OF SOLIDS BASED ON ATOMIC
ARRANGEMENT
AMORPHOUS
CRYSTALLINE
There exists at least one crystalline state of lower energy (G) than
the amorphous state (glass)
The crystal exhibits a sharp melting point
“Crystal has a higher density”!!
19. Factors affecting the formation of the
amorphous state
When the free energy difference between the crystal and the glass is
small ⇒ Tendency to crystallize would be small
Cooling rate → fast cooling promotes amorphization
• “fast” depends on the material in consideration
• Certain alloys have to be cooled at 106 K/s for amorphization
• Silicates amorphizes during air cooling
21. Types of Lattices
• There are 14 lattice types
• Most common types:
– Cubic:
Li, Na, Al, K, Cr, Fe, Ir, Pt, Au etc.
– Hexagonal Closed Pack (HCP):
Mg, Co, Zn, Y, Zr, etc.
– Diamond:
C, Si, Ge, Sn (only four)
22. Unit Cell
• Unit Cell is the smallest part of the lattice
which when repeated in three directions
produces the lattice.
• Unit Cell is the smallest part of the lattice
which represents the lattice.
23. Metals
• Most of the metals are crystalline in solid
form. They normally have the following
crystal structures.
– Body Centered Cubic (BCC)
– Face Centered Cubic (FCC)
– Hexagonal Close Packed (HCP) or Close
Packed Hexagonal (CPH)
24. Unit Cell of BCC Lattice
•Fe (Up to 9100 C and from 14010C to
Melting Point), W, Cr, V,
25. A less close-packed structure is Body-Centered-Cubic (BCC).
Besides FCC and HCP, BCC structures are widely adopted by metals.
• Unit cell showing the full cubic symmetry of the BCC arrangement.
• BCC: a = b = c = a and angles α = β =γ= 90°.
• 2 atoms in the cubic cell: (0, 0, 0) and (1/2, 1/2, 1/2).
25
26.
27. Unit Cell of FCC lattice
• Al, Cu, Ni, Fe (9100 C-14010 C)
28.
29. ABCABC.... repeat along <111> direction gives Cubic Close-Packing (CCP)
• Face-Centered-Cubic (FCC) is the most efficient packing of hard-spheres of any lattice.
• Unit cell showing the full symmetry of the FCC arrangement : a = b =c, angles all 90°
• 4 atoms in the unit cell: (0, 0, 0) (0, 1/2, 1/2) (1/2, 0, 1/2) (1/2, 1/2, 0)
Self-Assessment: Write FCC crystal as BCT unit cell.
29
33. Unit Cell of Hexagonal closed packed (HCP)
• Zn, Mg
34. ABABAB.... repeat along <111> direction gives Hexagonal Close-Packing (HCP)
• Unit cell showing the full symmetry of the HCP arrangement is hexagonal
• Hexagonal: a = b, c = 1.633a and angles α = β = 90°, γ = 120°
• 2 atoms in the smallest cell: (0, 0, 0) and (2/3, 1/3, 1/2).
34
36. Comparing the FCC and HCP Planes Stacking
Looking down (0001) plane
FCC
HCP
Looking down (111) plane!
36
37. Important Properties of unit cell
• Number of atoms required to represent a
unit cell
• Effective number of atoms per unit cell
• Coordination number
• Atomic packing factor
38. Number of atoms required to
represent a unit cell
• It is the number of atoms required to show a unit
cell
39. Number of atoms required to represent
a unit cell of BCC = 9
40. Number of atoms required to represent
a unit cell of FCC = 14
41. Number of atoms required to represent
a unit cell of HCP = 17
42. Effective Number of Atoms
• It is the total number of atoms belonging
to a unit cell
43. Counting Number of Atoms Per Unit Cell
Counting Atoms in 3D Cells
Atoms in different positions are shared by differing numbers of unit cells.
• Corner atom shared by 8 cells => 1/8 atom per cell.
• Edge atom shared by 4 cells => 1/4 atom per cell.
• Face atom shared by 2 cells => 1/2 atom per cell.
• Body unique to 1 cell => 1 atom per cell.
Simple Cubic
8 atoms but shared by 8 unit cells. So,
8 atoms/8 cells = 1 atom/unit cell
How many atoms/cell for
Body-Centered Cubic?
And, Face-Centered Cubic?
43
45. Effective Number of atoms per unit cell for FCC
It will be 8*1/8 + 6*1/2 = 4
William D. Callister, Jr., Materials Science and Engineering, An Introduction, John Wiley & Sons, Inc. (2003)
46. Effective Number of atoms per unit cell for
Hexagonal closed packed (HCP)
•It will be 6*1/6 + 2*1/2 + 3*1 = 6
William D. Callister, Jr., Materials Science and Engineering, An Introduction, John Wiley & Sons, Inc. (2003)
47. Coordination Number
• It is the number of nearest equidistant
neighbours of an atom in the lattice
48. Coordination Number of a Given Atom
Consider a simple cubic structure.
Six yellow atoms are are touching the blue atom.
Hence, the blue atom is having six nearest neighbors.
Therefore, for a Simple cubic lattice: coordination number, CN = 6
48
51. Atomic Packing Factor
• Atomic Packing Factor is the fraction of
volume occupied by atoms in a unit cell.
52. Atomic Packing Factor (APF)
APF = vol. of atomic spheres in unit cell
total unit cell vol.
Face-Centered-Cubic
Arrangement
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell
Depends on:
• Crystal structure.
• How “close” packed the atoms are.
• In simple close-packed structures with hard sphere atoms,
independent of atomic radius
52
53. Basic Geometry for FCC
Geometry along close-packed direction give relation between a and R.
2 2R
2R
2 2R
2=R
a 4
a
a
Vunit _ cell = a = (2 2 R ) = 16 2 R
3
3
4
Vatoms = 4 πR 3
3
3
Geometry:
a = 2 2R
4 atoms/unit cell
Coordination number = 12
53
54. Atomic Packing Factor for FCC
Face-Centered-Cubic
Arrangement
APF = vol. of atomic spheres in unit cell
total unit cell vol.
How many spheres (i.e. atoms)?
4/cell
What is volume/atom?
4π R3/3
What is cube volume/cell?
a3
How is “R” related to “a”?
2=R
a 4
atoms
volume
4
3
π ( 2a/4)
4
unit cell
atom
3
APF =
volume
a3
unit cell
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell
= 0.74
Independent of R!
54
55. Atomic Packing Factor for BCC
Again, geometry along close-packed direction give relation between a and R.
2a
Geometry:
4R ≡ 3
a
2 atoms/unit cell
a
4R ≡ 3
a
V
APF = atoms =
Vcell
a 3 3
4
2 π
3 4
a3
=
3π
=0.68
8
55
56. Important Properties of unit cell
SC
BCC
FCC
HCP
Relation between atomic radius
(r) and lattice parameter (a)
a = 2r
Effective Number of atoms per
unit cell
1
2
4
6
Coordination Number (No. of
nearest equidistant neighbours)
6
8
12
12
0.52
0.68
0.74
0.74
Packing Factor
3a = 4r 2a = 4r
a = 2r
57. Miller Indices
• Miller Indices are used to represent the directions and
the planes in a crystal
• Miller Indices is a group of smallest integers which represent a
direction or a plane
z
z
c
[111]
[100]
[110]
x
b
[021]
a
[1 1 1]
y
y
[01 2 ]
x
58. Miller Indices of a Direction
• Select any point on the direction line other than
the origin
• Find out the coordinates of the point in terms of
the unit vectors along different axes
• Specify negative coordinate with a bar on top
• Divide them by the unit vector along the
respective axis
• Convert the result into the smallest integers by
suitable multiplication or division and express as
<uvw>
59. Miller Indices of directions
z
z
c
[111]
[100]
[110]
x
b
[021]
a
[1 1 1]
y
y
[01 2 ]
x
60. Miller Indices of planes
- Find the intercepts of the plane with the three axes:
(pa, qb, rc)
- Take the reciprocals of the numbers (p, q, r)
- Reduce to three smallest integers (h, k, l) by suitable
multiplication or division.
- Miller indices of the plane: (hkl)
- Negative indices are indicated by a bar on top
- Same indices for parallel planes
- A family of crystallographically equivalent planes (not
necessarily parallel) is denoted by {hkl}
62. Example: determine Miller indices
of a plane z
• Intercepts: (a/2, 2b/3, ∞)
• p= ½, q= 2/3, r= ∞
• Reciprocals: h = 2, k = 3/2, l = 0
• Miller indices: (430)
c
a
x
b
y
65. Hexagonal indices
• A special case for hexagonal crystals
• Miller-Bravais indices: (hkil)
h= Reciprocal of the intercept with a1-axis
k= Reciprocal of the intercept with a2-axis
i = -(h + k)
l= Reciprocal of the intercept with c-axis
c
c
(1 1 00)
(11 2 0)
a2
a1
(0001)
a2
a1
66. Packing Densities in Crystals: Lines and Planes
Concepts
FCC
Linear and Planar Packing Density
which are independent of atomic radius!
Also, Theoretical Density
66
67. Linear Density in FCC
LD =
Number of atoms centered on a direction vector
Length of the direction vector
Example: Calculate the linear density of an FCC crystal along [1 1 0].
ASK
a. How many spheres along blue line?
b. What is length of blue line?
ANSWER
a. 2 atoms along [1 1 0]
in the cube.
b. Length = 4R
2atoms
1
LD110 =
=
4R
2R
XZ = 1i + 1j + 0k = [110]
Self-assessment: Show that LD100 = √2/4R.
67
68. Planar Density in FCC
PD =
Number of atoms centered on a given plane
Area of the plane
Example: Calculate the PD on (1 1 0) plane of an FCC crystal.
• Count atoms within the plane: 2 atoms
• Find Area of Plane: 8√2 R2
a = 2 2R
4R
Hence,
2
1
PD =
=
8 2R 2 4 2R 2
68
69. Planar Packing Density in FCC
PPD =
Area of atoms centered on a given plane
Area of the plane
Example: Calculate the PPD on (1 1 0) plane of an FCC crystal.
• Find area filled by atoms in plane: 2π R2
• Find Area of Plane: 8√2 R2
a = 2 2R
4R
Hence,
2π 2
R
π
PPD =
=
=0.555
8 2R 2 4 2
Always independent of R!
Self-assessment: Show that PPD100 = π /4 = 0.785.
69
70. Density of solids
• Mass per atom:
mA =
atomic weight from periodic table (g/mol)
6.02 ×1023 (atoms/mol)
4 3
• Volume per atom: VA = πRA
3
• Number of atoms per unit cell (N)
2 for bcc, 4 for fcc, 6 for hcp
Volume occupied by atoms in a unit cell = NVA
• Volume of unit cell (VC)
- Depends on crystal structure
• Mass density:
NmA
ρ=
VC
Packing Factor =
A
NV
V
C
71. Theoretical Density
# atoms/unit cell
ρ= nA
VcNA
Volume/unit cell
(cm3/unit cell)
Atomic weight (g/mol)
Avogadro's number
(6.023 x 10 23 atoms/mol)
Example: Copper
• 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
-
Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10-23cm3
Result: theoretical ρCu = 8.89 g/cm 3
Compare to actual: ρCu = 8.94 g/cm3
71
72. Characteristics of Selected Elements at 20 C
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
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
-----------
Crystal
Structure
FCC
-----BCC
HCP
Rhomb
-----HCP
FCC
Hex
BCC
-----BCC
HCP
FCC
-----Ortho.
Dia. cubic
FCC
-----------
Atomic radius
(nm)
0.143
-----0.217
0.114
-----Adapted from
-----Table, "Charac0.149 teristics of
0.197 Selected
0.071 Elements",
inside front
0.265 cover,
Callister 6e.
-----0.125
0.125
0.128
-----0.122
0.122
0.144
----------72
73. DENSITIES OF MATERIAL CLASSES
ρmetals > ρceramics > ρpolymers
Metals have...
• close-packing
Metals/
Alloys
30
20
10
(metallic bonds)
Platinum
Gold, W
Tantalum
Silver, Mo
Cu,Ni
Steels
Tin, Zinc
Ceramics have...
• less dense packing
(covalent bonds)
• often lighter elements
Polymers have...
• poor packing
ρ (g/cm3)
• large atomic mass
(often amorphous)
5
4
3
2
1
• lighter elements (C,H,O)
Composites have...
• intermediate values
Titanium
Aluminum
Magnesium
Graphite/
Ceramics/ Polymers
Semicond
Composites/
fibers
Based on data in Table B1, Callister
*GFRE, CFRE, & AFRE are Glass,
Carbon, & Aramid Fiber-Reinforced
Epoxy composites (values based on
60% volume fraction of aligned fibers
in an epoxy matrix).
Zirconia
Al oxide
Diamond
Si nitride
Glass-soda
Concrete
Silicon
Graphite
Glass fibers
PTFE
Silicone
PVC
PET
PC
HDPE, PS
PP, LDPE
0.5
0.4
0.3
GFRE*
Carbon fibers
CFRE*
Aramid fibers
AFRE*
Wood
Data from Table B1, Callister 6e.
73
74. Slip Systems
• Slip system is a combination of a slip plane and
a slip direction along which the slip occurs at
minimum stress
• Slip plane is the weakest plane in the crystal
and the slip direction is the weakest direction in
that plane.
• The most densely packed plane is the weakest
plane and the most densely direction is the
weakest direction in the crystal.
77. An FCC unit cell and its slip system
{111} - type planes (4 planes) have
all atoms closely packed.
Slip occurs along <110> - type directions
(A-B,A-C,D-E) within the {111} planes.
FCC has 12 slip systems - 4 {111} planes and each has 3 <110> directions
78. Shown displaced for clarity
HCP
+
A
+
B
=
A
A plane
B plane
A plane
Note: Atoms are coloured differently but are the same
HCP
79. * A crystal will be anisotropic (exhibit directional properties) if the slip systems are less
than 5. It will be isotropic (have same properties in all the directions) if the slip
systems are more than 5.