This document discusses the atomic arrangement and properties of crystalline solids such as metals. It begins by describing the long-range order in crystalline solids compared to the short-range order in amorphous solids. It then discusses various crystal structures including cubic, hexagonal, and body-centered cubic. It provides examples of calculating properties like atomic packing factor and theoretical density based on crystal structure. Finally, it discusses using X-ray diffraction to determine crystal structure by measuring spacing between crystal planes.

1. Structure of Metals and their
Physical Properties
1

2. 2
Atomic arrangement of solids
Source: The Principles of Engineering Materials [Craig R. Barrett, Alan S. Tetelman, William D. Nix
Solids
Crystalline solids
Amorphous solids
 Long range order
 Repetitive pattern
 Periodic arrangement
 of atoms
Bricks on a wall
 Random arrangement
 Short range order
-metals
-ceramics
-some polymers

3. 3
The basic principles of many materials characterization
techniques such as X-ray diffraction (XRD), are based
on crystallography.
Crystalline cristoballite
Silica glass

4. Energy of Crystalline and Amorphous
Structures
• Non-dense, random packing
• Dense, ordered packing
Dense, ordered packed structures tend to have lower energies.
Energy
r
typical neighbor
bond length
typical neighbor
bond energy
Energy
r
typical neighbor
bond length
typical neighbor
bond energy
4

5. Energy
r
typical neighbor
bond length
typical neighbor
bond energy
5
• Atomic radii of metallic atoms do not differ
dramatically
• Metallic bonding is non-directional
• Nearest neighbor distances tend to be small
• Electron cloud shields cores from each other
Smallest
repeating
structure as
unit cell
Dense ordered packing in metallic structures.

6. 6
Space Lattice
It is a three-dimensional array of points.
Also defined as periodic configuration of points in space, defines 3D shapes that fill
space
Bravais Lattice
14 different ways of arranging lattice points so that each lattice point has
identical surrounding.
14 different lattices are called ‘Bravais lattices’

7. 7
14 different unit cells which are grouped to form 7 crystal systems.
7 systems of Crystal symmetry--- Crystal Systems
Crystal Systems Bravais Lattices
Cubic 3 Bravais Lattice
Tetragonal 2 Bravais Lattice
Orthorhombic 4 Bravais Lattice
Monoclinic 2 Bravais Lattice
Triclinic 1 Bravais Lattice
Hexagonal 1 Bravais Lattice
Rhomdohedral 1 Bravais Lattice

8. 7 systems of crystal symmetry
Source: R.E. Smallman, Ch1, F1.7
8

9. Seven Crystal Systems
Source: R.E. Smallman, Ch1, F1.7
9

10. 10
Cubic Crystal Systems – 3 Bravais Lattice arrangement
Source: Solid-State Physics: Introduction to the Theory ; By James Patterson, Bernard Baile
 Simple Cubic Lattice
 Body Centered Cubic Lattice
 Face Centered Cubic Lattice

11. Enforcing Terminology
Unit Cell of Crystal Structure: Describes the smallest repetitive pattern and
explains the complete lattice pattern of the crystal.
• a, b, and c are the lattice parameters
• α, β, γ are the angles between axes
11

12. Using Crystal Structure to Predict
Properties
For each of the cubic based crystal structures, we will now identify the:
 Closed packed directions (where the hard spheres can touch)
 Identify the number of nearest neighbors to each atom
 Calculate the atomic packing factor
 Atomic coordination number
 Calculate the theoretical density
APF =
Volume of atoms in unit cell*
Volume of unit cell
12

13. Unit Cells and Percentage of Atoms
13

14. • Rare due to low packing density (only Po has this structure)
• Close-packed directions are cube edges.
(Courtesy P.M. Anderson)
Simple Cubic Structure (SC)
14
Atomic coordination number = ?

15. Atomic Packing Factor (APF) for SC
close-packed directions
a
R=0.5a
15
Step 1: Volume of Atoms in the unit cell :
Given Atomic radii is ‘R’
Information required ->
Number of atoms in the unit cell
Each corner contains 1/8th of the atom
So, total number of atoms in the unit cell =
8 x 1/8 = 1 atom per unit cell
Volume of atoms =1 atom x ( 4/3 𝝅𝑹 𝟑)
Step 2: Identify the close-packed direction in SC unit cell i.e, ‘a’ = 2R
Volume of unit cell = a3=(2R)3 = 8R3
Atomic Packing Factor =
4
3
𝜋𝑅3
8𝑅3 = 0.52

16. Body Centered Cubic
Source: R.E. Smallman, Ch1, F1.15
16
• Atoms touch each other along cube diagonals.
--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.
ex: Cr, W, Tantalum, Molybdenum
Atomic coordination number = 8

17. Atomic Packing Factor (APF) for BCC
a
aR
Adapted from
Fig. 3.2(a), Callister 7e.
a2
a3
17
Step 1: Number of atoms per unit cell =
1 atom at center + 1/8 th of 8 corner atoms = 2 atoms
Step 2: Identify close packed direction
3 𝑎 = 4𝑅
Step 3: Volume of unit cell =𝑎3=(
4𝑅
3
)3
Atomic Packing Factor =
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒖𝒏𝒊𝒕 𝒄𝒆𝒍𝒍
=
𝟐𝒙
𝟒
𝟑
𝝅𝑹 𝟑
𝒂 𝟑 = 0.68

18. • Coordination # = 12
Adapted from Fig. 3.1, Callister 7e.
(Courtesy P.M. Anderson)
• Atoms touch each other along face diagonals.
--Note: All atoms are identical; the face-centered atoms are shaded
differently only for ease of viewing.
Face Centered Cubic Structure (FCC)
ex: Al, Cu, Au, Pb, Ni, Pt, Ag
4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8
18

19. Atomic Packing Factor (APF) for FCC
19
Close packed direction length, 4R= a. 2 => a=
4𝑅
√2
Atomic Packing Factor=
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒖𝒏𝒊𝒕 𝒄𝒆𝒍𝒍
=
𝟒𝒙
𝟒
𝟑
𝝅𝑹 𝟑
𝒂 𝟑 = 0.74
Step 1: Number of atoms per unit cell = 6 x ½ + 8 x 1
8
= 4 atoms per unit cell
Step 2: Close packed direction in FCC are along the
face diagonals.

20. Hexagonal Close Packed
For ideally packed hcp, c/a=1.633 c/a(Zn)=1.856, c/a(Ti)=1.587
Source: R.E. Smallman, Ch1, F1.15
20
Coordination number =12
APF=0.74 with 6 atoms/unit cell
 ABAB… stacking
sequence

21. 21
• The face-centered atom and the three
mid-layer atoms form a tetrahedron
MNOP which has sides equal to a (as
atoms at vertices touch each other) and
height of c/2.
• Using this tetrahedron it can be shown
that for an ideal hexagonal crystal c/a
ratio = 1.633
Hexagonal Lattice

22. Stacking: Difference between HCP
(0.74) and FCC (0.74)
FCC: ABCABCABC
HCP: ABABAB
Source: R.E. Smallman, Ch1, F1.16
22

23. FCC vs. HCP
23

24. Two or more distinct crystal structures for the same material
(allotropy/polymorphism)
 Each structure is stable under
different conditions such
as temperature or pressure
 the crystal structure with the
lowest free energy will be
stable
BCC
FCC
BCC
1538ºC
1394ºC
912ºC
Fe
Gamma -Fe
alpha-Fe
liquid
iron system
24
Carbon
Diamond
(high pressure)
Graphite
(stable at
normal
conditions)

25. 25
Where,
n= number of atoms/unit cell
A=atomic weight (g/mol)
Vc= Volume of unit cell = a3 for cubic
NA= Avogadro’s number= 6.023 x 1023atoms/mol
Density =  =
VC NA
n A
 =
CellUnitofVolumeTotal
CellUnitinAtomsofMass
Theoretical Density
Example: Chromium - with BCC crystal structure.
A=52 g/mol and atomic radius R=0.125 nm
a=4R/ 3 𝜌 =
2×52.0
6.023×1023
×𝑎3

26. Examples of Metallic Crystal Structure Unit Cells:
SC, BCC, FCC, HCP
z
x
y
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27. Point Coordinates: Determining where atoms sit.
Point coordinates for unit cell center are
(½ , ½, ½)
Point coordinates for unit cell corner are
(1, 1, 1)
Translation: integer multiple of lattice
constants  identical position in
another unit cell
Notation for point coordinates -> (a,b,c)
z
x
y
a b
c
000
111
y
z
·
2c
·
·
·
b
b
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28. Crystallographic Directions: Moving Between Atomic
Positions
1. Vector repositioned (if necessary) to pass
through origin.
2. Read off projections in terms of
unit cell dimensions a, b, and c
3. Adjust to smallest integer values
z
x
Algorithm
y
1/19/2017 28
Example
1, 0, ½ => 2,0,1 => [201]
-1,1,1 => [111] , the overbar represents negative index.
Notation for crystallographic direction
Enclose in square brackets, no commas
[uvw]

29. 29
Example 2: The vector is obtained by subtracting the coordinates
of the tip of the vector from the coordinates of tail
-4, 1, 2
families of directions <uvw>
z
x
where the overbar represents a
negative index
[412]=>
y
Step 1:
pt. 1 x1 = a, y1 = b/2, z1 = 0
pt. 2 x2 = -a, y2 = b, z2 = c
=> -2, 1/2, 1
pt. 2
head
pt. 1:
tail
Step 2: Multiplying by 2 to eliminate the fraction
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30. 1/19/2017 30
Example 3: In a cubic unit cell find the vector with indices [123]
Step 1: Divide all indices with the largest integer.
X component = 1/3
Y component = 2/3
Z component = 3/3
Step 2: Note the coordinates (1/3, 2/3, 1)
Step 3: Draw the vector

31. Crystallographic Planes
Adapted from Fig. 3.9, Callister 7e.
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32. 32
Crystallographic Planes
Miller Indices:
(hkl) is a specific crystallographic plane.
{hkl} is family of planes with each having same atomic
arrangement.
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33. 33
Example 1 z
x
y
a b
c
Read off
intercepts of
plane with X, Y
and Z axes in
terms of a, b, c
Intercepts a b c
1 1 1
Take reciprocals
of intercepts
Reciprocals 1/1 1/1 1/ꝏ
Reduce to
smallest integer
values
Reduction 1 1 0
Enclose in
parentheses, no
commas (hkl)
Miller indices (110)
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34. 34
Identify the Miller indices of given plane
z
x
y
a b
c
·
·
·
4. Miller Indices (634)
example
1. Intercepts 1/2 1 3/4
a b c
2. Reciprocals 1/½ 1/1 1/¾
2 1 4/3
3. Reduction 6 3 4
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35. a b c
z
x
y
a b
c
4. Miller Indices (100)
1. Intercepts 1/2  
2. Reciprocals 1/½ 1/ 1/
2 0 0
3. Reduction 2 0 0
Example 2
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36. Linear Density
Linear Density of Atoms: LD =
a
[110]
Unit length of direction vector
Number of atoms
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37.  Properties vary with direction:
anisotropic.
-Example: the modulus
of elasticity (E) in BCC iron:
Importance of Linear/ Planar Densities
E (diagonal) = 273 GPa
E (edge) = 125 GPa
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38. Examples of Linear Density
a
[110]
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Linear density of Al in [110] direction
a = 0.405 nm
# atoms
length
1
3.5 nm
a2
2
LD -
==

39. Planar Density of (100) Iron
At T < 912 C iron has the BCC structure.
(100)
R
3
34
a =
Adapted from Fig. 3.2(c), Callister 7e.
2D repeat unit
1/19/2017 39
=Planar Density =
a2
1
atoms
2D repeat unit
=
nm2
atoms
12.1
m2
atoms
= 1.2 x 1019
1
2
R
3
34
area
2D repeat unit

40. 40
Planar Density of (111) Iron-BCC
Solution (cont): (111) plane 1 atom in plane/ unit surface cell
atoms in plane
atoms above plane
atoms below plane
ah
2
3
=
a2
1/2
= =
nm2
atoms
7.0
m2
atoms
0.70 x 1019
3
2
R
8
Planar Density =
atoms
2D repeat unit
area
2D repeat unit
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41. Spacing between equivalent planes
dhkl =
a
h2
+ k2
+ l2
What is the spacing between (111) planes in
FCC Nickel? The atomic radius is 0.124 nm.
a) 0.20 nm
b) 0.35 nm
c) 0.30 nm
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42. 42
X-Rays to Determine Crystal Structure
X-ray
intensity
(from
detector)
θ
θc
d =
nλ
2 sinθc
Measurement of
critical angle,  c,
allows computation of
planar spacing, d.
• Incoming X-rays diffract from crystal planes.
Adapted from Fig. 3.22,
Callister & Rethwisch 9e.
reflections must
be in phase for
a detectable signal
spacing
between
planes
d
θ
λ
θ
extra
distance
travelled
by wave “2”
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43. 43
X-Ray Diffraction Pattern
Adapted from Fig. 3.22, Callister 8e.
(110)
(200)
(211)
z
x
y
a b
c
Diffraction angle 2θ
Diffraction pattern for polycrystalline α-iron (BCC)
Intensity(relative)
z
x
y
a b
c
z
x
y
a b
c
dhkl-spacing increases
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