MME 2001
MATERIALS SCIENCE 1
27.10.2015
outline
Planar density
Interplanar spacing
Structure-property relations
X-ray diffraction
Imperfections in solids
point defects
vacancies
impurities
solid solutions
line defects
dislocations
Quiz
at 14:50
planar densities
The parameter corresponding to the linear density
for crystallographic planes is planar density, and
planes having the same planar density values are
also equivalent.
(001) / {001}
planar density
Planar Density of Atoms 
number of atoms inside the plane
PD =
area of the plane
1/4
A
B
1/2
(110)
[110]
C
a
D
x
Although six atoms have
centers that lie on this plane,
only one-quarter of each of
atoms A, C, D, and F, and onehalf of atoms B and E, for a
total equivalence of just 2
atoms, are on that plane.
planar density
Planar density of (110) planes of FCC crystal
(110)
[110]
a
area of (110) plane is
equal to the product of
its length and width.
length = 4R
width (vertical
dimension)= 2R2
area of this plane =
(4R)( 2R2) = 8R2 2
planar density
Planar Density of Atoms  PD = Number of atoms
Area
of
the
plane
(110)
[110]
area of (110) plane = 8R2 2
PD110 = 2 atoms / 8R2 2
= 1 / 4R22
a
Corner atoms (A, C, D and F) only quarter
of each is in the plane: 4 x ¼ = 1
Face atoms (B and E): only half of each is
in the plane: 2x1/2: 1
Total number of atoms inside the plane = 2
planar density
Calculate the planar atomic density for the (110)
plane in FCC copper (a = 0.3615 nm) in
atoms/cm2.
(110)
[110]
a
A, C, D, F: corner atoms;
shared by 4 unit cells
B and E: edge atoms; shared
by 2 unit cells
planar density
Number of atoms on the (110) plane in FCC:
4 atoms at the corners, each shared by 4 unit cells,
2 atoms at the edges, each shared by 2 unit cells
4x1/4 + 2x1/2 = 2 atoms
(110)
[110]
area of (110) plane = a x a2
a = 0.3615 nm
a
PD110 = 2 atoms /a22
= 2/a2
= 1.082 x 1015 atoms/cm2
Planar Density of (100) Iron
Solution: At T < 912C iron has the BCC structure
(100) plane
(100)
atoms
PD=
area
4
R
3
a=
Radius of iron R = 0.1241 nm
1
a2
=
1
4
R
3
2
= 12.1
atoms
nm2
= 1.2 x
atoms
1019
m2
Planar Density of (111) Iron
(111) plane
1 atom in plane/unit surface cell
2a
atoms in plane
atoms above plane
atoms below plane
h=
3
a
2
2
 4 3  16 3 2
area = 2 ah = 3 a = 3 
R  =
R
3
 3

2
atoms
1
Planar Density =
area
16
3
3
=
R2
7.0
atoms =
nm2
0.70 x 1019
Radius of iron R = 0.1241 nm
atoms
m2
planar density
(111) Plane in a FCC crystal:
(3 corner atoms x 1/6) + (3 side atoms x ½) = 2 atoms
Planar density of (111) plane in the FCC:
2
4
=
PD(111) =
1 2.a x 2.a 3 3.a2
2
2
PD110 = 2/a2
PD(111) > PD (110)
(111) planes are more
densely packed
Linear and planar densities
Linear and planar densities are important
considerations relative to the process of
slip—that is, the mechanism by which metals
plastically deform.
Slip occurs on the most densely packed
crystallographic planes and, in those planes,
along directions having the greatest atomic
packing.
Spacing between crystal planes
Relation between planes and
directions
In the cubic system planes and directions having same
indices are perpendicular to each other,i.e. İf [uvw] direction
is perpendicular to (hkl) plane, then h=u, k=v and l=w
Ex: {100} planes and <100> directions are perpendicular to
each other.
If [uvw] direction is parallel to (hkl), that is if [uvw] lies in
the plane (hkl) then hu + kv + lw = 0.
For example,
_
[110] lies in the plane (111)
Since
1x(-1) + 1x1 + 1x0 = 0
Structure property correlation
Al: ductile
FCC
Fe: not ductile Mg: not ductile
BCC
HCP
Plastic deformation via SLIP
Sliding of crystal planes over one another!
SLIP occurs on most densely packed planes in the
most closely packed planes!
slip plane + slip direction: SLIP SYSTEM
Structure property correlation
In FCC,
{111} planes are closely packed and there are
4 unique {111} planes.
Each of these planes contains
3 closely packed <111> directions.
4x3=12
slip
systems
In HCP,
The basal plane (0001) is the close packed.
It contains 3 close packed <1120> directions.
1x3=3
slip
systems
Higher number of slip systems allows greater plastic
deformation before fracture imparting ductility to FCC
metals!
Structure property correlation
4x3=12 slip systems
1x3=3 slip systems
Structure property correlation
Close packed planes are planes with greatest
interplanar spacing. This allows slip to takeplace
easily on these planes.
BCC structure has 48 possible slip systems.
However, there is no close packed plane. Hence
plastic deformation before fracture is not
substantial.
Slip might occur in {110}, {112} and {123} planes in
the <111> directions.
Learning check
—
Sketch within a cubic unit cell the plane (312):
learning check
Which one
shows the
plane (221)
correctly
drawn?
learning check
Which one
shows the
(212) plane
correctly
drawn?
learning check
Which one
shows
the
_
(111) plane
correctly
Drawn?
learning check
122
131

213

331
 
123

121
learning check
Sketch within a cubic unit cell the following
planes:
Determine the Miller indices for the planes in the
following unit cell:
A: (403)

B: (112)
learning check
Determine the Miller indices for the planes in the
following unit cell:


A: (322) B: (202)

A: (324) B: (221)
X-ray diffraction
waves 1 and 2 have the same wavelength and
remain in phase after a scattering event.
The amplitudes of the scattered waves add together
in the resultant wave.
X-ray diffraction
Waves 3 and 4 have the same wavelength and
become out of phase after a scattering event.
The amplitudes of the two scattered waves
cancel one another.
+
_
X-ray diffraction
n = SQ + QT
n  = dhkl sin  + dhkl sin 
n  = 2dhkl sin 
X-Rays to Determine Crystal Structure
• Incoming X-rays diffract from crystal planes.
extra
distance
travelled
by wave “2”



d
reflections must
be in phase for
a detectable signal
spacing
between
planes
X-ray diffraction
The magnitude of the distance between two
adjacent and parallel planes of atoms (i.e., the
interplanar spacing d) is a function of the Miller
indices (h, k, and l) as well as the lattice
parameter(s).
For cubic symmetry,
in which a is the lattice parameter (unit cell edge
length).
X-ray diffraction
Schematic diagram of an
x-ray diffractometer;
T x-ray source,
S specimen,
C detector, and
O the axis around which
the specimen and
detector rotate.
interplanar spacing computations
diffraction angle computations
For BCC iron, compute the diffraction angle for the (220) set
of planes. The lattice parameter for Fe is 0.2866 nm. Also,
assume that monochromatic radiation having wavelength of
0.1790 nm is used and the order of reflection is 1.
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.
X-Ray Diffraction Pattern
z
z
Intensity (relative)
c
a
x
z
c
b
y (110)
c
y
a
y
a
b
x
b
x
(211)
(200)
Diffraction angle 2
Diffraction pattern for polycrystalline a-iron (BCC)
imperfections in solids
imperfections in solids
imperfections in solids
homogenization/annealing
rolling/extrusion
bonding
X’tal structure
Thermomechanical
processing
Atomic
composition
Microstructure:
material
properties
Addition and manipulaton
of defects
imperfections in solids
● perfect order is assumed to exist throughout
crystalline materials on an atomic scale.
● However, such an idealized solid does not
exist; all contain large numbers of various
defects.
● Crystalline defect refers to a lattice
irregularity having one or more of its
dimensions on the order of an atomic
diameter.
imperfections in solids
● Many of the properties of materials are
profoundly sensitive to deviations from
crystalline perfection!
● the impact is not always adverse!
● often specific characteristics are
deliberately fashioned by the introduction
of controlled quantity of defects.
Types of Imperfections
• vacancy atoms
• interstitial atoms
• substitutional atoms
Point defects
0-dimensional
• dislocations
Line defects
1-dimensional
• grain boundaries
Area defects
2-dimensional
• cavities/porosity
volume defects
3-dimensional
Point Defects / vacancies
vacant atomic site/missing atom
Vacancy
distortion
of planes
All crystalline solids contain vacancies.
vacancies increase the entropy (the randomness) of the
crystal: thermodynamically favorable!
the number of vacancies increases exponentially with T!
For most metals, Nv/N at  Tm is on the order of 104; one
lattice site out of 10,000 will be empty.
Point Defects / self-interstitial
● own atom crowded into an interstitial site!
distortion
of planes
selfinterstitial
● introduces relatively large distortions in the
surrounding lattice as the atom is much larger than
the interstitial position.
● is not highly probable!
● exists in very small concentrations
● significantly lower than for vacancies.
Point Defects
Most common defects in crystalline solids are point
defects.
At high temperatures,
atoms frequently and
randomly change their
positions leaving behind
empty lattice sites.
In general, diffusion
(mass transport by atomic
motion) can only occur
because of vacancies.
energy
atom
Em
vacancy
distance
Point Defects in ionic crystals
Charge neutrality must be
maintained!
Schottky defect
Schottky imperfection
Formation of equivalent
(not necessarily equal)
numbers of cationic and
anionic vacancies
Frenkel imperfection
Formation of an ion
vacancy and an ion
interstitial
Frenkel defect
Equilibrium Concentration:
Point Defects
• Equilibrium concentration varies with temperature!
No. of defects
No. of potential
defect sites
Activation energy
Q v
Nv
= exp 

 k T
N




Boltzmann's constant
Temperature
(1.38 x 10 -23 J/atom-K)
(8.62 x 10 -5 eV/atom-K)
Each lattice site is a potential vacancy site
Measuring Activation Energy
• We can get Qv from
an experiment.
• Measure this...
Nv
Q v 
Nv



= exp 
 k T 
N
 Qv 1
Nv   Q v
ln
=
=
T
N
kT  k
• Replot it...
N
exponential
dependence!
ln
Nv
N
slope
-Q
v /k
T
defect concentration
1/ T
Estimating Vacancy Concentration
Equilibrium # of vacancies in 1 m3 of Cu at 1000C?
 = 8.4 g / cm 3
A Cu = 63.5 g/mol
Q v = 0.9 eV/atom
N A = 6.02 x 1023 atoms/mol
Q v
Nv =

exp 
 k T
N
For 1
m3 ,
NA
0.9 eV/atom


= 2.7 x 10-4

1273K
8.62 x 10-5 eV/atom-K
6 cm3
10
x
N=  x
A Cu
1 m3
= 8.0 x 1028 sites/m3
N v = (2.7 x 10-4)(8.0 x 1028) sites = 2.2 x 1025 vacancies
Estimating Vacancy Concentration
Equilibrium # of vacancies in 1 m3 of Fe at 850C?
 = 7.65 g/ cm 3
A Fe = 55.85 g/mol
Q v = 1.08 eV/atom
N A = 6.02 x 1023 atoms/mol
Q v

Nv = N exp 
 k T
1.08 eV/atom




N
Q
v
A


 = 
exp 
 k T 
A Fe

1123K
8.62 x 10-5 eV/atom-K
N=
(6.022x1023 atoms/mol) (7.65g/cm3) exp 1.08 eV/atom
55.85 g/mol
(8.62x10-5 eV/atom-K)(1123K)
v
N v = 1.18 x 1024 vacancies
Estimating Vacancy Concentration
fraction of atom sites that are vacant for lead at its
melting temperature of 327°C?
Q v = 0.55 eV/atom
Q v
Nv =

exp 
 k T
N
N A = 6.02 x 1023 atoms/mol
0.55 eV/atom


= 2.41 x 10-5

600K
8.62 x 10-5 eV/atom-K
see you next week!