Chapter 3-Crystal Structure ceramic .pdf

Crystalline Structure Perfection
Chapter 3
MENG310 LIU-BIU
M.B

Contents
3.1- Types of solids
3.2- Crystal structures
3.3- Seven systems and fourteen lattices
3.4- Metal structures
3.5- Ceramic structures
3.6- Polymeric Structures
3.7- Semiconductor Structures
3.8- Lattice positions, directions, and planes
2

3
3.1- Types of solids
 Solids can be classified depending on the arrangement of their atoms or molecules.
 Crystalline: has highly defined and repeatable arrangements of molecular chains.
 Polycrystalline: consists of many small crystals (grains) that are separated by grain
boundaries and normally have random crystallographic orientations.
 Amorphous: one kind of non-equilibrium material; its characteristic of atomic
arrangement is more like liquid and has no long-range periodicity.
 The crystal structure of materials are presented in the following table:
Material Crystalline Polycrystalline Non-crystalline
Metals
Glasses
Ceramics
Polymers
Semiconductors

4
3.2- Crystal structure
 Crystal structure is a unique arrangement of atoms or molecules in a crystalline liquid
or solid
 The crystal structure of a material can be described in terms of its unit cell.
 Unit cell is a small box containing one or more atoms arranged in 3D.
Hard sphere unit cell representation
Aggregate (group) of many atoms
Reduced-sphere unit cell

5
3.2- Crystal Structure
 The length of unit-cell edges and the angles between crystallographic axes are referred
to as lattice constants, or lattice parameters.
 The key feature of the unit cell is that it contains a full description of the structure as a
whole because the complete structure can be generated by the repeated stacking of
adjacent unit cells face to face throughout three-dimensional space.
 Geometry of the unit cell
 a, b, and c are the unit length along x, y and z
axes, respectively. α, β, and γ are the angles
between the crystallographic axes.

6
 There are only seven unique unit-cell shape that can be stacked together to fill three-
dimensional space. These are called Seven Crystal Systems.
 The points where the atoms are stacked in a given unit-cell are called lattice points.
There are 14 possible ways of arrangement of the points, called 14 Bravais’ Lattices.

7
 The crystalline structures of most metals belong to one of three relatively simple types.
 Ceramics which have variety of chemical compositions, exhibit variety of crystalline
structures.
 Glass is non-crystalline, and polymers may have as much as 50 to 100% of their
volume non-crystalline (polycrystalline).

8
3.4- Metal structure
 Most Metals have one of three crystal structures:
(HCP)
Hexagonal Compact Packet
Examples:
Be, Ti, Zn, Zr, Mg, Cd
(BCC)
Body Centered Cubic
Examples:
Fe, V, Cr, Mo, W
(FCC)
Face Centered Cubic
Examples:
Al, Ni, Ag, Cu, Au
Metal Crystal lattice examples

9
3.4- Metal Structure
Atomic Packing Factor (APF) is the fraction of volume in a crystal structure that is
occupied by atoms.
 APF is determined by assuming that atoms are rigid spheres.
 APF is represented mathematically as:
 The most dense arrangement of atoms has an APF of about 0.74
 In general, metals with a high atomic packing factor will have a higher malleability
or ductility.
 Atomic packing factor
 Coordination number
Coordination Number is the number of adjacent atoms surrounding a reference atom.
cell
unit
atoms
atoms
V
V
N
APF



10
 Body Centered Cubic crystal structure
Body Centered Cubic (BCC) unit cell has atoms at each of the eight corners plus one
atom in the center of the cube. .
 Atoms are in contact along body diagonal of the unit cell.
 The side of the cube is a and the radius of an atom is R.
1/8 atom 1 atom
2
a
3
a
a
a
a
3
4
4
3
R
a
R
a 


cell
unit
inside
atoms
Natoms 2
1
8
1
8 



68
.
0
3
4
3
8
3
4
2
3
3
3
3













R
R
a
R
V
V
N
volume
Total
volume
Atomic
APF
cell
unit
atom
atoms



11
 Face Centered Cubic crystal structure
Face Centered Cubic (FCC) unit cell has atoms at each of the eight corners plus one
atom at the center of each face of the cube. .
 Atoms are in contact along face diagonal of the unit cell.
 The side of the cube is a and the radius of an atom is R.
1/8 atom 1/2 atom
2
a
3
a
a
a
a
2
4
4
2
R
a
R
a 


cell
unit
inside
atoms
Natoms 4
2
1
6
8
1
8 




74
.
0
2
4
3
8
3
4
4
3
3
3
3













R
R
a
R
V
V
N
volume
Total
volume
Atomic
APF
cell
unit
atom
atoms



12
Example: 3.1
Copper Cu has the FCC crystal structure with an atomic radius of 0.128 nm and
of molar mass equal to 63.55 g/mole.
Determine its atomic packing factor APF and the density of copper
a
Solution
Atoms inside unit cell = 6×1/2 + 8×1/8 = 4 atoms
362
.
0
2
4
4
2 



R
a
R
a
 
74
.
0
2
2
3
16
3
4
4
3
3
3
3





R
R
a
R
volume
Total
volume
atomic
APF


 
3
23
3
7
3
/
89
.
8
10
023
.
6
10
362
.
0
55
.
63
4
cm
g
N
a
M
N
V
m
A












The density of FCC unit cell is:
A
A N
M
N
m
N
N
M
m
n






13
 Hexagonal Close-Packed crystal structure
Hexagonal Close-Packet (HCP) unit cell has atoms at each of the twelve corners, three
atoms inside the unit cell and one on each base.
 Atoms are in contact along the side of the unit cell.
 The side of the unit cell is a and the radius of an atom is R.
1/2 atom 1/6 atom
1 atom
74
.
0
2 
 APF
and
R
a
cell
unit
inside
atoms
Natoms 6
6
1
12
2
1
2
3 






14
3.4- Metal Structure
 Coordination number
Coordination Number (CN) is the number of adjacent atoms surrounding a reference
atom.
 The coordination number of the central metal atom/ion may be used to deduce the
molecular geometry of the coordination compound.
(BCC)
CN = 8
(FCC)
CN = 12 (HCP)
CN = 12

15
3.5- Ceramic structure
 Ceramics usually have a combination of ionic bonds.
 Ionic bond occurs between a metal and nonmetal with transfer of electrons.
 Since the bonding is ionic, the packing factor will be Ionic Packing Factor (IPF).
cell
unit
ions
ions
V
V
N
IPF


 Ionic packing factor
IPF represents the fraction of the volume of ions occupied in a unit cell over the total unit
cell volume.
 Types of crystal structures
1. MX
2. MX2
3. M2X3
M: Metallic element
X: Non-metallic element

16
 MX Ceramic Structures: CsCl
 CsCl can be thought of as two interpenetrating simple cubic arrays where the corner of
one cell sits at the body center of the other.
 Simple cubic Bravais lattice (8 lattice points at the corner of the cube).
 The structure of CsCl is not BCC. It is a simple cubic Bravais lattice.
 The number of complete ions inside unit cell is: 1CS+ and 1Cl-
 Materials similar to CsCl structure are: CsBr, CsI, CsCN, TiCl, TiBr, and TiCN.

17
 MX Ceramic Structures: NaCl
 NaCl has a cubic unit cell. It is best thought of as a face-centered cubic array of anions
with an interpenetrating fcc cation lattice (or vice-versa).
 Whether one starts with anions or cations on the corners, the cell has the same
structure.
 The number of complete ions inside unit cell is: 4Na+ and 4Cl-
 Materials similar to NaCl structure are: MgO, CaO, FeO, and NiO.
1/8 ion 1/4 ion
1/2 ion

Example: 3.2
MgO has the NaCl crystal structure.
a- Calculate the Ionic packing factor IPF of MgO.
b- Determine the density of MgO
Given: rMg
2+ =0.078 nm, rO
2- =0.132 nm, 16O and 24.31Mg
Solution
nm
r
r
a O
Mg
42
.
0
2
2 2
2 

 

465
.
0
3
4
4
3
4
4
3
3
3
2
2








a
r
r
volume
Total
volume
Ionic
IPF
Mg
O


 
 
3
23
3
7
3
/
61
.
3
10
023
.
6
10
42
.
0
16
4
31
.
24
4
cm
g
N
a
M
N
V
m
A














18

19
 MX2 Ceramic Structures: CaF2
 There are 12 ions (4 Ca2+ and 8 F-) per unit cell.
 The cations Ca2+occupy the 8 corners of the unit cell and the centers of the faces while
the anions F- occupy the corners of a cube inside the unit cell.
 Materials similar to CaF2 structure are: UO2 ThO2 and TeO2

20
 MX2 Ceramic Structures: CaF2
 Ions F- (black spheres) are located on the body diagonal of the cubic unit cell.
 Their centers are placed on the quarter of body diagonals and in contact with Ca2+
𝑑
𝑑
4
 



 




 F
Ca
F
Ca
r
r
a
r
r
a
d
2
2
3
4
4
3
4
𝑎
𝑎
𝑎

21
3.6- Polymeric structure
Polyethylene unit cell
 Polymers have long chain polymeric molecules
 The arrangement of these long molecules into a regular pattern is difficult.
 Plastics tends complex crystalline structure
 Polyethylene (C2H4)n and plastics are examples of polymers.

22
3.6- Polymeric structure
 Crystallinity of Polymeric Structures
 Percentage of crystallinity can be computed by dividing
the amount of the crystalline phase by the total amount
of the material and multiplying by 100.
 The fraction of the ordered molecules in polymer is
characterized by the degree of crystallinity, which
typically ranges between 10% and 80%.
 Strength Polymeric Structures
 Tensile strength often increases with molecular weight and percentage of crystallinity

Example: 3.3
For a density of 0.9979 g/cm3 Calculate the number of C and H
atoms in the polyethylene unit cell.
Given the volume of unit cell is 0.0933 nm3, 1.008H and 12.01C.
Solution
 
g
N
N
m H
C
H
C 23
23
10
66
.
4
10
023
.
6
)
008
.
1
(
4
)
01
.
12
(
2
4
2
4
2 






3
3
21
23
/
997
.
0
10
0933
.
0
10
66
.
4
4
2
cm
g
cm
g
N
V
m H
C





 


atoms
N
and
atoms
N
N
H
C
H
C
8
4
2
4
2





23

24
3.7- Semiconductor structure
 Elemental semiconductors: Si, Ge and gray Sn share diamond its cubic structure.
1/8 atom
1/2 atom
1 atom
 The number of complete atoms inside unit cell is: 4 + 6 ×
1
2
+ 8 ×
1
8
= 8 𝑎𝑡𝑜𝑚𝑠
 Gray atoms are atoms located at the center of faces and corners. Colored atoms are
located at the body diagonals of the cube.
 The complete atoms inside the unit cell are placed on the quarter of the body diagonal.
 Pure semiconductors

25
 The semiconducting compounds are composed of pairs of elements from columns III
and V (e.g., GaAs) or from columns II and VI (e.g., ZnS).
3.7- Semiconductor structure
 Compound semiconductors
 Materials similar to ZnS structure are: ZnSe, CdS, and HgTe
 Materials similar to GaS structure are: AIP, and InSb
 The number of complete atoms inside unit cell is:
4𝑆 + 6 ×
1
2
+ 8 ×
1
8
𝑍𝑛 = 4𝑆 + 4𝑍𝑛

Example: 3.4
Silicon Si has a diamond cubic structure.
a- Calculate the Atomic packing factor APF.
b- Determine the density of Si.
Given: rSi =0.117 nm, and 28.09Si.
Solution
Si
Si r
a
a
r
3
8
4
3
2 


34
.
0
3
8
3
4
8
3
4
8
3
3
3
3
3












Si
Si
Si
r
r
a
r
volume
Total
volume
Atomic
APF


 
3
23
3
7
3
/
36
.
2
10
023
.
6
10
54
.
0
09
.
28
8
cm
g
N
a
M
N
V
m
A












26

27
3.8- Lattice positions, directions and planes
 In a crystal lattice, each atom, molecule or ions (constituent particle) is represented by a
single point. These points are called lattice site or lattice point.
 Lattice positions: same as the position in the coordinate system xyz. They are expressed
as fraction or multiples of the unit cell dimensions.
 Lattice positions
a, b and c are side lengths of the unit cell
along x, y and z axes respectively. They are
the unit-cell dimensions.
How to determine the Lattice Position of P:
- Coordinates of P in the coordinate system
xyz is: P(1a, 2b,1c)
- Divide the coordinates by a, b and c
- Lattice Position of P is: 121
P

28
 Crystal directions are all straight lines passing through two nodes in the lattice usually
one of them is the origin and the other one is the identified point.
 A direction is expressed as set of integers, which are obtained by identifying the
smallest integer positions intercepted by a parallel line from the origin of the
crystallographic axes.
 Lattice directions
Notation for lattice directions.
Note that parallel [uvw]
directions (e.g., [111]) share the
same notation because only the
origin is shifted.
Note that the line from the origin
of the crystallographic axes
through the
1
2
1
2
1
2
body centered
position can be extended to
intercept the 111 unit cell corner
position.

29
 A family of directions includes any directions that are equivalent in length and types of
atoms encountered.
 Lattice directions – Family of directions
Family of directions, <111>, representing all body diagonals for adjacent unit cells in
the cubic system.
               
1
11
,
1
1
1
,
11
1
,
111
,
1
11
,
1
1
1
,
11
1
,
111
111 

30
 It is important to know the angle between two lattice directions and planes since that
affect the mechanical properties of material.
 Lattice directions – Angle between two lattice directions

cos
D
D
D
D 



Let 𝐷 = 𝑢𝑎 + 𝑣𝑏 + 𝑤𝑐 and ሖ
𝐷 = ƴ
𝑢𝑎 + ƴ
𝑣𝑏 + ƴ
𝑤𝑐 to be two lattice directions. The angle
θ between them is defined as:
      

























 
2
2
2
2
2
2
1
cos
cos
w
v
u
w
v
u
w
w
v
v
u
u
D
D
D
D


Example: 3.5
What is the angle between the [110] and [111] directions in the cubic system.
Solution
     
0
1
2
2
2
2
2
2
1
5
.
35
3
2
0
1
1
cos
cos 





 























 

w
v
u
w
v
u
w
w
v
v
u
u


31
 A lattice plane is a plane whose intersections with the lattice (or any crystalline
structure of that lattice) are periodic (i.e. are described by 2d Bravais lattices).
 Miller Indices is a method of describing the orientation of a plane or set of planes
within a lattice in relation to the unit cell. They are represented by three integers.
 Lattice planes-Miller Indices (hkl)
Steps for computing Miller Indices
1- Take the intersections of plane with the
Cartesian axis x,y,z
2- Take the reciprocals and make sure that
the three numbers are of the smallest
integers (no fractions or multiples)
3- The obtained set is a miller indices
plane representation (hkl)
 Knowledge of slip or cleavage planes can facilitate the fundamental understanding of
mechanical properties and engineering operations, such as tableting and milling.

32
 Below are some examples on miller indices:
 Miller plane examples

33
Example: 3.6
Find the miller indices of the following planes

34
 A family of planes contains all the planes that are crystallo-graphically equivalent.
 Every family of lattice planes can be described by a set of integer Miller indices that
have no common divisors
 Miller indices: Family of lattice planes
Family of planes, {100}, representing all faces of unit cells in the cubic system.
  )
1
00
(
),
0
1
0
(
),
00
1
(
),
001
(
),
010
(
),
100
(
100 

35
 The linear density is the number of atoms occupied in a unit length.
 The intersection of lattice direction with sphere can be either a radius or a diameter
which represents half and one atom along the line respectively.
 If the intersection is not a radius or diameter then no atoms count along the line.
 Linear density
line
of
length
line
along
atoms
of
nb
density
Linear 
atom
1
atom
2
1

36
 The planar density is the number of atoms occupied in a unit area
 The intersection of lattice plane with sphere can be either a disk or part of a disk.
 The number of atoms in plane can be counted as follows:
 Planar density
plane
of
Area
plane
in
atoms
of
nb
density
Planer 
atom
4
1
atom
2
1
atom
1

atom
360


Example: 3.7
Consider FCC Aluminum crystal structure
rAl = 0.143 nm.
a- Find the linear densities along [110], and [111]
b- Find the planar densities in planes (100), (110), and (111)
Solution
37
Length
atoms
of
nb
density
Linear 
[110]
nm
atoms
atoms
a
LD /
49
.
3
2
404
.
0
2
2
1
1
2
1
2
]
110
[ 





nm
atoms
atom
a
LD /
443
.
1
3
404
.
0
1
3
2
1
2
]
111
[ 



[111]
nm
a
r
a 404
.
0
4
2 


2
a
3
a

38
 
2
2
2
)
100
( /
25
.
12
404
.
0
2
1
4
1
4
nm
atoms
atoms
a
PD 




Area
atoms
of
nb
density
Planar 
(100)
2
2
)
110
( /
66
.
8
2
404
.
0
2
2
2
1
2
4
1
4
nm
atoms
atoms
a
a
PD 






a
a
a
2
a
(110)
2
)
111
( /
14
.
14
2
60
sin
2
2
2
1
3
6
1
3
nm
atoms
a
a
PD 





(111)
2
a 2
a
2
a

Problem 1
Calculate the IPF for UO2 which has the CaF2 structure. Given rU
4+ = 0.105 nm and
rO
2- = 0.132 nm. Hint the oxygen atoms are located at one-quarter of the body
diagonal.
Solution
588
.
0
3
4
8
3
4
4
3
3
3
2
4








a
r
r
volume
cell
Unit
volume
Ionic
IPF
O
U


nm
a
a
r
r O
U
548
.
0
4
3
2
4 


 

40

Problem 2
Calculate the density of UO2. Given 238.03U and 16O.
Solution
The volume of unit cell is computed in problem 2.
There exist 4 atoms of Uranium and 8 atoms of Oxygen in unit cell.
 
 
3
23
3
7
3
/
9
.
10
10
023
.
6
10
548
.
0
)
16
(
8
03
.
238
4
cm
g
N
a
M
N
V
m
A












41

Problem 3
Calculate the APF for polyethylene. Given the volume of unit cell is 0.09333 nm3, rC
=0.077 nm and rH =0.046nm. Hint the unit cell contains 4 carbon atoms.
Solution
There exist 4 carbon atoms → 8 hydrogen atoms.
12
.
0
0933
.
0
3
4
8
3
4
4 3
3






H
C r
r
volume
Total
volume
Atomic
APF


42

Problem 4
Calculate the IPF for the zinc blende (ZnS) structure. Given rZn
2+ = 0.083 nm and rS
2-
= 0.174 nm. Hint the sulfur atoms are located at one-quarter of the body diagonal.
Solution
There exist 4 zinc atoms and 4 sulfur atoms.
 
468
.
0
594
.
0
3
4
4
3
4
4
3
3
3
3
2
2








S
Zn
r
r
a
volume
Ionic
IPF


nm
a
a
r
r S
Zn
594
.
0
4
3
2
2 


 

43

Problem 5
Benefiting from problem 5, calculate the density of zinc blende (ZnS) structure.
Given 65.38Zn and 32.06S.
Solution
There exist 4 zinc atoms and 4 sulfur atoms in unit cell.
 
 
3
23
3
7
3
/
09
.
3
10
023
.
6
10
594
.
0
)
06
.
32
(
4
38
.
65
4
cm
g
N
a
M
N
V
m
A
cell
unit
ZnS












44

x
y
z
Problem 6
(a) Sketch, in a cubic unit cell, a [111] and a [112] lattice direction.
(b) Use trigonometric calculation to find the angle between these two directions.
(c) Use equation 3.3 to determine the angle between these two directions.

[111]
1
l
[112]
2
l
45
Solution
(b) 6
,
3 2
1 
 l
l
Using cosine law
  0
2
1
2
2
2
1
2
5
.
19
cos
2
1 



 

l
l
l
l
(c)
     2
2
2
2
2
2
2
1
2
1
cos
w
v
u
w
v
u
w
w
v
v
u
u
l
l
l
l

















18
4
2
1
1
1
1
1
2
1
1
cos
2
2
2
2
2
2









 
0
5
.
19
18
4
arccos 









Problem 7
(a) Calculate the linear densities along [010], [101], and [111] direction in UO2.
(b) Calculate the planar densities in the following planes (010), (101), (002), (111),
and (220). Given rU
4+ = 0.105 nm and rO
2- = 0.132 nm.
46
Solution
Length
ions
of
nb
density
Linear 
nm
U
U
a
U
LD /
828
.
1
547
.
0
1
2
1
2 4
4
4
]
010
[







nm
a
r
r
a
O
U 547
.
0
4
3 2
4



 

[010]
a

47
  nm
U
U
a
U
LD /
585
.
2
2
547
.
0
2
2
1
2
1
2 4
4
4
]
101
[








[101]
[111]
nm
O
nm
U
a
O
U
LD /
11
.
2
/
055
.
1
3
2
2
1
2 2
4
2
4
]
111
[









2
a
3
a
 
 
2
4
2
4
2
4
)
010
( /
684
.
6
547
.
0
2
1
4
1
4
nm
U
U
a
U
PD 







Area
ions
of
nb
density
Planar 
(010)
a
a

2
2
2
4
2
4
)
101
( /
45
.
9
/
72
.
4
2
4
2
nm
O
nm
U
a
a
O
U
PD 








(101)
 
 
2
4
2
4
4
)
002
( /
684
.
6
547
.
0
2
2
1
4
nm
U
U
a
a
U
PD 







(002)
a
a
  2
4
4
)
111
( /
718
.
7
2
60
sin
2
2
2
1
3
6
1
3
nm
U
a
a
U
PD 







(111)
2
a 2
a
2
a
 
 
2
4
2
4
)
220
( /
726
.
4
547
.
0
2
2
1
2
nm
U
l
a
U
PD 






(220)
a
2
a
l 
a
2
a

Problem 8
(a) Calculate the linear densities of ions along [101], and [111] in zinc blende (ZnS).
(b) Calculate the planar densities of ions along the (110) and (020) plane. Given rZn
2+
= 0.083 nm and rS
2- = 0.174 nm.
49
Solution
nm
a
r
r
a
S
Zn 593
.
0
4
3 2
2



 

Length
ions
of
nb
density
Linear 
  nm
Zn
Zn
a
Zn
LD /
38
.
2
2
593
.
0
2
2
2
1
2
1 2
2
2
]
101
[








[101]
2
a

50
[111]
nm
S
nm
Zn
a
S
Zn
LD /
97
.
0
/
97
.
0
3
1
2
1
2 2
2
2
2
]
111
[









3
a
Area
ions
of
nb
density
Planar 
2
2
2
2
2
2
)
110
( /
02
.
4
/
02
.
4
2
2
2
nm
S
nm
Zn
a
a
S
Zn
PD 








(110)
 
 
2
2
2
2
2
)
020
( /
687
.
5
593
.
0
2
2
1
4
nm
Zn
Zn
a
a
Zn
PD 







(020)
a
a
a
2
a

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