Phy351 ch 3
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Transcript of Phy351 ch 3
CRYSTAL STRUCTURE
Chapter 3
PHY351
Long-Range-Order (LRO)
Most metals and alloys, semiconductors, ceramics and some polymers have a crystalline structure (where the atoms or ions display in a long-range-order (LRO).
The atoms or ions in these material form a regular repetitive, grid like pattern in three dimension.
These materials are referred as a crystalline structure.
2
Question 1:
a. Define the following terms:
i. SRO atom/ion arrangement
ii. no order atom/ion arrangement
iii. crystalline solid
iv. crystal structure
v. amorphous
b. Give 2 example material of
i. SRO atom/ion arrangement
ii. no order atom/ion arrangement
3
Properties of SOLIDS depends upon:
bonding force
crystal structure
The space lattice and unit cells
4
Space lattice
An IMAGINARY NETWORK of lines with atoms at lines intersection that representing the arrangement of atoms is called space lattice.
Unit cells
Unit cell is that block of atoms which REPEATS itself to form space lattice.
Unit Cell
Space Lattice
5
Question 2:
a. Define the following terms:
i. Lattice point
ii. Motif
iii. Lattice constant
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Crystal Systems and Bravais Lattice
Only SEVEN different types of unit cells are necessary to create all point lattices which:
Cubic
Tetragonal
Orthorhombic
Rhombohedral
Hexagonal
Monoclinic
Triclinic
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According to Bravais (1811-1863), fourteen standard unit cells can describe all possible lattice networks based on FOUR basic types of unit cells which are:
Simple
Body Centered
Face Centered
Base Centered
8
i. Cubic Unit Cell a = b = c
α = β = γ = 900
Simple Body Centered Face centered
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ii. Tetragonal a =b ≠ c
α = β = γ = 900
Simple Body Centered
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iii. Orthorhombic a ≠ b ≠ c
α = β = γ = 900
Simple Base CenteredFace CenteredBody Centered
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iv. Rhombohedral a =b = c
α = β = γ ≠ 900
Simple
v. Hexagonal a = b ≠ c α = β = 900
γ = 1200
Simple
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vi. Monoclinic a ≠ b ≠ c
α = γ = 900 ≠ β
vii. Triclinic a ≠ b ≠ c
α ≠ β ≠ γ ≠ 900
SimpleSimple
Base
Centered
13
Principal Metallic Crystal Structures
90% of the metals have either Body Centered Cubic (BCC), Face Centered Cubic (FCC) or Hexagonal Close Packed (HCP) crystal structure.
HCP is denser version of simple hexagonal crystal structure.
BCC Structure FCC Structure HCP Structure14
Body Centered Cubic (BCC) Crystal Structure
Represented as:
one atom at each corner of cube
one at the center of cube.
Examples :-
Chromium (a=0.289 nm)
Iron (a=0.287 nm)
Sodium (a=0.429 nm)
(b) Hard-sphere unit cell
Figure 3.4: BCC crystal structure
(a) Atomic-site unit cell (c) Isolated unit cell15
Each unit cell has:
eight 1/8 atom at corners
1 full atom at the center.
Therefore each unit cell has (8 x 1/8) + 1 = 2 atoms
Atoms contact each other at cube diagonal. Therefore;
3
4Ra
16
Face Centered Cubic (FCC) Crystal Structure
FCC structure is represented as one atom each
at the corner of cube
at the center of each cube face.
Examples :- Aluminum (a = 0.405) Gold (a = 0.408)
(b) Hard-sphere unit cell
Figure 3.6: FCC crystal structure
(a) Atomic-site unit cell (c) Isolated unit cell17
Each unit cell has:
eight 1/8 atom at corners
six ½ atoms at the center of six faces
Therefore each unit cell has (8 x 1/8) + (6 x ½) = 4 atoms
Atoms contact each other across cubic face diagonal. Therefore;
2
4Ra
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Hexagonal Close-Packed Structure (HCP)
The HCP structure is represented as
1 atom at each of 12 corners of a hexagonal prism
2 atoms at top and bottom face
3 atoms in between top and bottom face.
Examples:- Zinc (a = 0.2665 nm, c/a = 1.85) Cobalt (a = 0.2507 nm, c/a = 1.62)
Figure 3.8: HCP crystal structure:(a) Schematic of the crystal structure (larger cell)
(b) Hard-sphere model19
Each atom has:
six 1/6 atoms at each of top and bottom layer
two ½ atoms at top and bottom layer
3 full atoms at the middle layer
Therefore each HCP unit cell has:
Ideal HCP c/a ratio is 1.633.
(2 x 6 x 1/6) + (2 x ½) + 3 = 6 atoms
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Isolated HCP unit cell also called the primitive cell.
The atoms at locations marked:
‘1’ contribute 1/6 of an atom
‘2’ contribute 1/12 of an atom
‘3’ contribute 1 atom
Therefore each HCP unit cell (primitive cell) has
(4 x1/6) + (4 x 1/12) + 1 = 2 atoms
Figure 3.8: HCP crystal structure:
(c) Isolated unit cell schematic
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APF = Volume of atoms in unit cell
Volume of unit cell
Atomic Packing Factor (APF)
Formula:
22
APF = 8.723 R3
12.32 R3 = 0.68
Vatoms =
= 8.373R3
3
3
4
R
= 12.32 R3
V unit cell = a3 =
3
4.2
3R
APF for BCC structure:
But;
Therefore;
APF = Volume of atoms in unit cell
Volume of unit cell
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Question 3
3.1 Determine the atomic packing factor (APF) for:a. Face centered cubic structure
b. Hexagonal Close-Packed Structure (HCP)
3.2 Calculate the volume of the zinc crystal structure unit cell by using the following data: pure zinc has HCP crystal structure with lattice constant a = 0.2665 nm and c = 0.4947nm.
3.3 By using data in question 2.2, find the volume of the larger cell.
3.4 What are the three most common metal crystal structures? List five metals that have each of these crystal structures.
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Volume density
Formula volume density of metal:
vMass per Unit cell
Volume per Unit cell=
25
Example:-
Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm. Calculate the volume density.
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v Mass per Unit cell Volume per Unit cell
=
Solution:
Known FCC unit cell has 4 atoms.
Therefore;
Mass of unit cell = m
= 4 (63.54) 6.02 x 1023
= 4.22 x 10-22g
2
4Ra = =
2
1278.04 nm= 0.361 nm
Volume of unit cell = V= a3
But;
Therefore;Volume of unit cell = (0.361nm)3
= 4.7 x 10-29 m3
v V
m 4.22 x 10-22 =8.95 x 106 g/m3
4.7 x 10-29
4 x 0.1278 x 10-9
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Atom Positions in Cubic Unit Cells
Cartesian coordinate system is use to locate atoms.
In a cubic unit cell
y axis is the direction to the right.
x axis is the direction coming out of the paper.
z axis is the direction towards top.
Negative directions are to the opposite of positive directions.
Atom positions are located using unit distances along the axes.
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Figure3.9 Atomic position in a BCC unit cell.
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Direction
In cubic crystals, direction indices are vector components of directions resolved along each axes, resolved to smallest integers.
Direction indices are position coordinates of unit cell where the direction vector emerges from cell surface, after converted to integers.
Fig 3.1: Some directions in cubic unit cells30
(0,0,0)
(1,1/2,1)
zProduce the direction vector till it emerges from surface of cubic cell
Determine the coordinates of pointof emergence and origin
Subtract coordinates of point of Emergence by that of origin
(1,1/2,1) - (0,0,0)= (1,1/2,1)
Are all areintegers?
Convert them to
smallest possible
integer by multiplying
by an integer.
2 x (1,1/2,1)= (2,1,2)
Are any of the directionvectors negative?
Represent the indices in a square bracket without comas with a over negative index (Eg: [121])
Represent the indices in a square bracket without comas (Eg: [212] )
The direction indices are [212]
x
y
YES
NO
YESNO
PROCEDURE TO FIND DIRECTION INDICES
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Example:
Determine the direction indices of the cubic direction shown below:
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Tips:The vector representing EF can be found by subtracting the coordinate of the tips of the vector F from the coordinate of the tail E.
Exercise 3.4
i. Draw the following direction vectors in cubic unit cells:
a. [100] and [110]
b. [112]
c. [1 1 0]
d. [3 2 1]
ii. Determine the direction indices of the cubic direction between the position coordinates (3/4 , 0 , ¼) and (1/4 , ½ , ½)
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Lattice plane
Miller Indices are are used to refer to specific lattice planes of atoms.
They are reciprocals of the fractional intercepts (with fractions cleared) that the plane makes with the crystallographic x,y and z axes of three nonparallel edges of the cubic unit cell.
z
x
y
Miller Indices = (111)
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Choose a plane that does not pass through origin
Determine the x,y and z intercepts
of the plane
Find the reciprocals of the intercepts
Fractions? Clear fractions bymultiplying by an integer
to determine smallest set of whole numbers
Enclose in parenthesis (hkl)where h,k,l are miller indicesof cubic crystal plane
forx,y and z axes. Eg: (111)
Place a ‘bar’ over theNegative indices
Yes
MILLER INDICES - PROCEDURE
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z
x
y
Miller Indices = (111)
Miller Indices - Examples
Intercepts of the plane at x,y & z axes are 1, ∞ and ∞
Taking reciprocals we get (1,0,0).
Miller indices are (100).
Intercepts are 1/3, 2/3 & 1.
taking reciprocals we get (3, 3/2, 1).
Multiplying by 2 to clear fractions, we get (6,3,2).
Miller indices are (632).
xx
y
z
(100)
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Plot the plane (101)
Taking reciprocals of the indices we get (1 ∞ 1).
The intercepts of the plane are x=1, y= ∞ (parallel to y) and z=1.
Plot the plane (2 2 1)
Taking reciprocals of the indices we get (1/2 1/2 1).
The intercepts of the plane are x=1/2, y= 1/2 and z=1.
Figure EP3.7 a
Figure EP3.7 c37
Plot the plane (110)
• The reciprocals are (1,-1, ∞)
• The intercepts are x=1, y=-1 and z= ∞ (parallel to z axis)
Note:
To show this plane in a single unit cell, the origin is moved along the positive
direction of y axis by 1 unit.
Figure EP3.7 b
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Exercise 3.7
1. Draw the following crystallographic planes in cubic unit cells:
a. (101)
b. (110)
c. (221)
d. (110)
2. Determine the miller index for the plane given cubic unit cell.
39 (a) (b)
Planar Atomic Density
Formula planar atomic density:
p =
Equivalent number of atoms whose
centers are intersected by selected area
Selected area
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Example:-
In Iron (BCC, a=0.287 nm), the (110) plane intersects center of 5 atoms (four ¼ and 1 full atom). Calculate the planar atomic density.
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Solution:
- Equivalent number of atoms = (4 x ¼ ) + 1 = 2 atoms
- Area of 110 plane =
Therefore;
222 aaa
p 2287.02
2=
2
13
2
1072.12.17
mm
atoms
nm
atoms
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1.72 x 1019 atoms/m2
Linear Atomic Density
Formula linear atomic density:
=
Number of atomic diameters intersected by
selected length of line in direction of interest
Selected length of line
43
l
Example:-
For a FCC copper crystal (a=0.361 nm), the [110] direction intersects 2 half diameters and 1 full diameter.
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Solution:
- The number of atomic diameters intersected by this length of line are:
½ + ½ + 1 = 2 atomic diameters
- The length of the line is:
Therefore;
mm
atoms
nm
atoms
nm
atoms 61092.392.3
361.02
2
l
nm361.02
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3.92 x 109 atoms/m
Interplanar Spacing
The distance between two adjacent parallel planes of atoms with the same Miller indices is called the interplanar spacing, dhkl .
The interplanar spacing in cubic materials is given by the general equation:
46
Crystal Structure Analysis –X-ray Diffraction
Information about crystal structure are obtained using X-Rays diffraction technique.
The X-rays source:
- The wavelength (0.05-0.25 nm) which same as distance between crystal lattice planes.
47
Figure 3.22 Schematic diagram of the cross section of a sealed-off filament X-
ray tube.
48
Crystal planes of target metal act as mirrors reflecting X-ray beam.
If rays leaving a set of planes are out of phase (as in case of arbitrary angle of incidence) no reinforced beam is produced.If rays leaving are in phase, reinforced beams are produced.
Figure 3.25 The reflection of an X-ray beam by the (hkl) planes of a crystal
a) No reflected beam is produced at an arbitrary angle of incidence.
b) At the Bragg angle , the reflected rays are in phase and reinforce one another.
49
For rays reflected from different planes to be in phase, the extra distance traveled by a ray should be a integral multiple of wave length λ .
c) Similar to b) except that the wave representation has been omitted.
50
Let us consider incident X-rays 1 and 2 as figure 3.25c.
For these rays to be in phase, the extra distance of travel of ray 2 is equal to MP + PN which must be an integral number of wavelength.
Thus;
nλ = MP + PN
where;
n = order of the diffraction
= 1,2,3,…..
51
Since both MP and PN equal to dhklsin, the condition for constructive interference (the production of a diffraction peak of intense radiation) must be:
nλ = 2 dhkl sinθ
This equation known as Bragg’s law.
Bragg’s law gives the relationship among the angular positions of the reinforced diffracted beams in terms of the wavelength of the incoming X-ray radiation and of the interplanar spacing dhkl of the crystal planes.
52
Example 3.15:
A sample of BCC iron was placed in an X-ray diffractometer using incoming X-rays with a wavelength of 0.1541 nm. Diffraction from the planes was obtained at 2 = 44.70. Calculate a value for the lattice constant, a of BCC iron. (Assume first-order diffraction with n = 1).
Answer: 0.287 nm
110
References
A.G. Guy (1972) Introduction to Material Science, McGraw Hill.
J.F. Shackelford (2000). Introduction to Material Science for Engineers, (5th Edition), Prentice Hall.
W.F. Smith (1996). Priciple to Material Science and Engineering, (3rd Edition), McGraw Hill.
W.D. Callister Jr. (1997) Material Science and Engineering: An Introduction, (4th Edition) John Wiley.
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