Remember Miller Indices?
• For directions:
– Determine coordinates
for “head” and “tail” of
the direction
– “head”-”tail”
– Clear fraction/reduce
results to lowest
integers.
– Enclose numbers in []
and a bar over negative
integers.
• For planes:
– Identify points at which
the plane intercepts the
x, y, z axis.
– Take reciprocals of
these intercepts.
– Clear fractions and do
NOT reduce to the
lowest integers.
– Enclose the numbers in
parentheses () and a bar
over negative integers.
Chapter 3 -
Special note for directions…
• For Miller Indices of directions:
– Since directions are vectors, a direction and its
negative are not identical!
• [100] ≠ [100] Same line, opposite directions!
– A direction and its multiple are identical!
• [100] is the same direction as [200] (need to reduce!)
• [111] is the same direction as [222], [333]!
– Certain groups of directions are equivalent; they have
their particular indices because of the way we
construct the coordinates.
• Family of directions: <111>=[111], [111],[111],[111],…
Chapter 3 -
Special note for planes…
• For Miller Indices of planes:
– Planes and their negatives are identical (not the case
for directions!)
• E.g. (020) = (020)
– Planes and their multiples are not identical (Again,
different from directions!) We can show this by
defining planar densities and planar packing fractions.
• E.g. (010) ≠ (020) See example!
– Each unit cell, equivalent planes have their particular
indices because of the orientation of the coordinates.
• Family of planes: {110} = (110),(110),(110),(101), (101),…
– In cubic systems, a direction that has the same
indices as a plane is perpendicular to that plane.
Chapter 3 -
Example: Calculating the Planar Density
Calculate the planar density for the (010) and (020) planes in
simple cubic polonium, which has a lattice parameter of 0.334 nm.
a0
a0
(c) 2003 Brooks/Cole Publishing /
Thomson Learning™
Chapter 3 -
SOLUTION
The total atoms on each face is one. The planar density is:
atom per face
1 atom per face
Planar density (010) 

area of face
(0.334) 2
(a0)2  8.96 atoms/nm 2  8.96  1014 atoms/cm 2
However, no atoms are centered on the (020) planes.
Therefore, the planar density is zero. The (010) and (020)
planes are not equivalent!
Chapter 3 -
Planar Density of (100) Iron
Solution: At T < 912C iron has the BCC structure.
2D repeat unit
(100)
a
4 3
R
3
Radius of iron R = 0.1241 nm
atoms
2D repeat unit
Planar Density =
area
2D repeat unit
1
a2
=
1
4 3
R
3
atoms
atoms
19
= 1.2 x 10
2 = 12.1
2
nm
m2
Chapter 3 - 6
Planar Density of (111) Iron
There are only (3)(1/6)=1/2 atoms in the plane.
2a
h
2a
h
3
a
2
2
atoms
2D repeat unit
4 3 
16 3 2
2
2area 2 ah  3 a  3 
R  
R
3
 3

0.5
atoms =
= 7.0
2
Planar Density =
area
2D repeat unit
16 3
6
R
2
nm
0.70 x 1019
atoms
m2
Chapter 3 - 7
In-Class Exercise 1: Determine planar density
Determine the planar density for BCC lithium in the (100), (110),
and the (111) planes.
atomic radius for Li = 0.152 nm
Chapter 3 -
Solution for plane (100)
rLi  0.152nm

4r 40.152nm 


a0 BCC  

 0.3510nm 3.510  
3
3


For (100):
planar _ density 
1atom
3.510 10
8
cm

2
 8.115 1014 atoms / cm 2
Chapter 3 -
Solution for plane (110)
For (110):
It is important to visualize how the plane is cutting
across the unit cell – as shown in the diagram!
planar _ density 

2atoms
2 3.510 10 8 cm

2
 1.148 1015 atoms / cm 2
Chapter 3 -
Solution for plane (111)
For (111):
Note: Since the (111) does NOT pass through the
center of the atom in the middle of the BCC unit cell,
we do not count it!
1
1
plane _ area  bh 
2
2
planar _ density 
2a0



 3 
3a0  a0   0.866a02
 2 
1 / 2atom
0.866 3.510 10 8 cm

2
 4.686 1014 atoms / cm 2
2a0
Chapter 3 -
In-Class Exercise 2: Determine planar density
Determine the planar density for FCC nickel in the (100),
(110), and (111) planes.
atomic radius for Nickel= 0.125 nm
Remember when
visualizing the plane, only
count the atoms that the
plane passes through the
center of the atom. If the
plane does NOT pass
through the center of that
atom, we do not count it!
Chapter 3 -
Solution for plane (100)
rNi  0.125nm

4r 40.125nm 


a0 FCC  

 0.3536nm 3.536  
2
2


a0
For (100):
planar _ density 
2atoms
3.536 10
8
cm

2
 1.600 1015 atoms / cm 2
Chapter 3 -
Solution for plane (110)
For (110):
It is important to visualize how the plane is cutting
across the unit cell – as shown in the diagram!
planar _ density 
a

2atoms
2 3.536 10 8 cm

2
 1.1311015 atoms / cm 2
0
2a0
Chapter 3 -
Solution for plane (111)
For (111):
Again try to visualize the plane, count the number of
atoms in the plane:
2a0
1
1
plane _ area  bh 
2
2
planar _ density 


2atoms
8

 3 
2a0  a0   0.866a02
 2 
0.866 3.536 10 cm

2
 1.847 1015 atoms / cm 2
2a0
2a0
Chapter 3 -
Home Exercise: Determine planar density
Determine the planar density for (0001) plane
for an HCP unit cell Titanium
atomic radius for titanium is 0.145 nm
16
Chapter 3 -
Crystals as Building Blocks
• Some engineering applications require single crystals:
--diamond single
crystals for abrasives
--turbine blades(Co and Ni superalloys)
(Courtesy Martin Deakins,
GE Superabrasives,
Worthington, OH. Used with
permission.)
Fig. 8.33(c), Callister 7e.
(Fig. 8.33(c) courtesy
of Pratt and Whitney).
• Properties of crystalline materials
often related to crystal structure.
--Ex: Quartz fractures more easily
along some crystal planes than
others.
Chapter 3 - 17
Poly crystal Material
Single crystal
Grains
Chapter 3 -
Polycrystals
• Most engineering materials are polycrystals.
Anisotropic
Adapted from Fig. K,
color inset pages of
Callister 5e.
(Fig. K is courtesy of
Paul E. Danielson,
Teledyne Wah Chang
Albany)
1 mm
• Nb-Hf-W plate with an electron beam weld.
• Each "grain" is a single crystal.
• If grains are randomly oriented,
Isotropic
overall component properties are not directional.
• Grain sizes typ. range from 1 nm to 2 cm
(i.e., from a few to millions of atomic layers).
Chapter 3 - 19
Single vs Polycrystals
• Single Crystals
E (diagonal) = 273 GPa
Data from Table 3.3,
Callister 7e.
(Source of data is R.W.
Hertzberg, Deformation
and Fracture Mechanics
of Engineering
Materials, 3rd ed., John
Wiley and Sons, 1989.)
-Properties vary with
direction: anisotropic.
-Example: the modulus
of elasticity (E) in BCC iron:
• Polycrystals
-Properties may/may not
vary with direction.
-If grains are randomly
oriented: isotropic.
(Epoly iron = 210 GPa)
-If grains are textured,
anisotropic.
E (edge) = 125 GPa
200 mm
Adapted from Fig.
4.14(b), Callister 7e.
(Fig. 4.14(b) is courtesy
of L.C. Smith and C.
Brady, the National
Bureau of Standards,
Washington, DC [now
the National Institute of
Standards and
Technology,
Gaithersburg, MD].)
Chapter 3 - 20
Section 3.6 – Polymorphism
• Two or more distinct crystal structures for the same
material (allotropy/polymorphism)
iron system
titanium
liquid
, -Ti
1538ºC
-Fe
BCC
carbon
1394ºC
diamond, graphite
-Fe
FCC
912ºC
BCC
-Fe
Chapter 3 - 21
Section 3.16 - X-Ray Diffraction
• Diffraction gratings must have spacings comparable to
the wavelength of diffracted radiation.
• Can’t resolve spacings  
• Spacing is the distance between parallel planes of
atoms.
Chapter 3 - 22
(c) 2003 Brooks/Cole Publishing / Thomson Learning
(a) Destructive (out of phase) xray beam gives a weak signal.
(b)Reinforcing (in phase)
interactions between x-rays
and the crystalline material.
Reinforcement occurs at
angles that satisfy Bragg’s
law.
Chapter 3 -
X-Rays to Determine Crystal Structure
• Incoming X-rays diffract from crystal planes.
extra
distance
travelled
by wave “2”
q
q

d
Measurement of
critical angle, qc,
allows computation of
planar spacing, d.
reflections must
be in phase for
a detectable signal
Adapted from Fig. 3.19,
Callister 7e.
spacing
between
planes
X-ray
intensity
(from
detector)
n
d
2 sin qc
q
qc
Chapter 3 - 24
(c) 2003 Brooks/Cole Publishing / Thomson Learning
(a) Diagram of a
diffractometer, showing
powder sample, incident
and diffracted beams.
(b) (b) The diffraction
pattern obtained from a
sample of gold powder.
Chapter 3 -
X-Ray Diffraction Pattern
z
z
Intensity (relative)
c
a
x
z
c
b
y (110)
a
x
c
b
y
a
x (211)
b
(200)
Diffraction angle 2q
Diffraction pattern for polycrystalline -iron (BCC)
Adapted from Fig. 3.20, Callister 5e.
Chapter 3 - 26
y
Bragg’s Law:
Bragg’s Law:
n  2d hkl sin q
Where q is half the angle
between the diffracted
beam and the original
beam direction
is the wavelength of X-ray
d is the interplanar spacing
Interplanar
spacing:
d
d hkl 
a0
h2  k 2  l 2
Miller
Chapter 3 -
SUMMARY
• Atoms may assemble into crystalline or
amorphous structures.
• Common metallic crystal structures are FCC, BCC, and
HCP. Coordination number and atomic packing factor
are the same for both FCC and HCP crystal structures.
• We can predict the density of a material, provided we
know the atomic weight, atomic radius, and crystal
geometry (e.g., FCC, BCC, HCP).
• Crystallographic points, directions and planes are
specified in terms of indexing schemes.
Crystallographic directions and planes are related
to atomic linear densities and planar densities.
Chapter 3 - 28
SUMMARY
• Materials can be single crystals or polycrystalline.
Material properties generally vary with single crystal
orientation (i.e., they are anisotropic), but are generally
non-directional (i.e., they are isotropic) in polycrystals
with randomly oriented grains.
• Some materials can have more than one crystal
structure. This is referred to as polymorphism (or
allotropy).
• X-ray diffraction is used for crystal structure and
interplanar spacing determinations.
Chapter 3 - 29