Packing density

1. Introduction to Materials
Science and Selection

2. © 2003 Brooks/Cole Publishing / Thomson Learning™
© 2003 Brooks/Cole Publishing / Thomson Learning™
Metals
Ceramic
combination
with a metal
Associated
with polymers
Semiconductor
materials

3. Representative strengths of various
categories of materials
© 2003 Brooks/Cole Publishing / Thomson Learning™

4. Atomic Bonding
Primary bonding:Metallic bond, Covalent
bond, Ionic bond.
Secondary bonding: Van der Waals
interactions (London forces, Debye
interaction, Keesom interaction)
Intermetallic compoundis a compound such
as Al
3V formed by two or more metallic
atoms

5. Metallic Bonding
The metallic bond forms
when atoms give up their
valence electron, then
forms an electron sea.
The positively charged
atom cores are bonded
by mutual attraction to
the negatively charged
electrons.
When voltage is applied,
the sea of electrons can
move freely to produce a
current.
© 2003 Brooks/Cole Publishing / Thomson Learning™

6. Covalent Bonding
Covalent bonding
requires that
electrons be shared
between atoms in
such a way that
each atom has its
outer sp orbital filled.
In silicon, with a
valence of four, four
covalent bonds must
be formed.

7. Ionic Bonding
An ionic bond is the result of electron transferfrom one
atom to another.
When sodium donates its valence electron to chlorine, each
becomes an ion; attraction occurs, and the ionic bond is
formed.
It is important to knot that ionic bonds are nondirectional. A
positively charged Na+ will attract any adjacent Cl-equally
in all directions!
© 2003 Brooks/Cole Publishing / Thomson Learning™

8. Ionic Bonding
© 2003 Brooks/Cole Publishing /
Thomson Learning™
When voltage is
applied to an ionic
material, entire ions
must move to cause
a current to flow.
Ion movement is
slow and the
electrical
conductivity is poor.

9. Atomic Arrangement
Levels of atomic arrangements
in materials:
(a) Inert monoatomic gases
have no regular ordering of
atoms
(b, c) Some materials,
including water vapor,
nitrogen gas, amorphous
silicon and silicate glass
have short-range order.
(d) Metals, alloys, many
ceramics and some
polymers have regular
ordering of atoms/ions that
extends through the
material.
(c) 2003 Brooks/Cole Publishing / Thomson
Learning™

10. Example: LCD
Liquid crystal display.
These materials are
amorphous in one
state and undergo
localized crystallization
in response to an
external electrical field
Widely used in liquid
crystal displays.
(Courtesy of Nick
Koudis/PhotoDisc/Gett
yImages.)

11. Lattice, Unit Cells, and Crystal
Structures
Lattice -A collection of points that divide space into
smaller equally sized segments.
Unit cell–The simplest repeating unit of any structure
that can be stacked to fill space. All atoms must be
the same in every unit cell.
Atomic radius-The apparent radius of an atom,
typically calculated from the dimensions of the unit
cell, using close-packed directions (depends upon
coordination number).
Packing factor-The fraction of space in a unit cell
occupied by atoms.

12. Definition of the lattice parameters
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
Their use in
cubic,
orthorhombic,
and hexagonal
crystal systems

13. Hexagonal close packed
structure (HCP)
The hexagonal close-packed (HCP)
structure (left) and its unit cell.
(c) 2003 Brooks/Cole Publishing / Thomson
A
B
A

14. 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.

15. 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],…

16. 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.

17. Calculate the planar density and planar packing fraction
for the (010) and (020) planes in simple cubic polonium,
which has a lattice parameter of 0.334 nm.
Example: Calculating the Planar Density and
Packing Fraction
(c) 2003 Brooks/Cole Publishing /
Thomson Learning™ a
0a
0

18. 2142
2
atoms/cm 1096.8atoms/nm 96.8
)334.0(
faceper atom 1
face of area
faceper atom
(010)density Planar

 SOLUTION
The total atoms on each face is one. The planar density
is:
The planar packing fraction is given by:79.0
)2(
)(
)( atom) 1(
face of area
faceper atoms of area
(010)fraction Packing
2
2
2
0
2


r
r
r
a


However, no atoms are centered on the (020) planes.
Therefore, the planar density and the planar packing
fraction are both zero. The (010) and (020) planes are
not equivalent!
(a
0)
2

19. In-Class Exercise 1:Determine planar
density and packing fraction.
Determine the planar density and packing fraction for FCC
nickel in the (100), (110), and (111) planes. Which, if
any, of these planes is close-packed?
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!

20. First, find atomic radius for Nickel from Appendix 2, page 797 of
Textbook (6
th
Ed. Shackleford) to calculate the lattice parameter:
Solution for plane (100)
 









536.33536.0
2
125.04
2
4
125.0
0
nm
nmr
FCCa
nmr
Ni
a
0 For (100): 

 
7854.0
2/4
2
_
/10600.1
10536.3
2
_
2
2
215
2
8





r
r
fractionpacking
cmatoms
cm
atoms
densityplanar


21. Solution for plane (110) 

 
5554.0
2/42
2
_
/10131.1
10536.32
2
_
2
2
215
2
8





r
r
fractionpacking
cmatoms
cm
atoms
densityplanar

For (110):
a
002a
It is important to visualize how the plane is cutting
across the unit cell –as shown in the diagram!

22. Solution for plane (111)02a 02a
For (111):
Again try to visualize the plane, count the number of
atoms in the plane:02a 
2
000
866.0
2
3
2
2
1
2
1
_ aaabhareaplane 








  
215
2
8
/10847.1
10536.3866.0
2
_ cmatoms
cm
atoms
densityplanar 


  
9069.0
866.0
4/22
_
2
0
2
0

a
a
fractionpacking

Therefore, plane (111) is close-packed!

23. In-Class Exercise 2:Determine planar
density and packing fraction.
Determine the planar density and packing fraction for
BCC lithium in the (100), (110), and the (111) planes.
Which, if any, of these planes is close packed?

24. Solution for plane (100)
 









510.33510.0
3
152.04
3
4
152.0
0 nm
nmr
BCCa
nmr
Li
First, find atomic radius for Nickel from Appendix 2, page 797 of
Textbook (6
th
Ed. Shackleford) to calculate the lattice parameter:
For (100): 
 
5890.0
4/3
_
/10115.8
10510.3
1
_
2
0
2
0
214
2
8





a
a
fractionpacking
cmatoms
cm
atom
densityplanar


25. 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! 
 
8330.0
2
4/32
_
/10148.1
10510.32
2
_
2
0
2
0
215
2
8





a
a
fractionpacking
cmatoms
cm
atoms
densityplanar


26. Solution for plane (111)02a
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!
2
000
866.0
2
3
3
2
1
2
1
_ aaabhareaplane 








  
214
2
8
/10686.4
10510.3866.0
2/1
_ cmatoms
cm
atom
densityplanar 


  
3401.0
866.0
4/32/1
_
2
0
2
0

a
a
fractionpacking

Therefore, there is no close-pack plane in BCC!
There are only (3)(1/6)=1/2 atoms in the plane.02a

27. X-Ray Diffraction (XRD)
Max von Laue (1879-1960) won the Nobel Prize in
1912 for his discovery related to the diffraction of
x-rays by a crystal.
William Henry Bragg (1862-1942) and his son
William Lawrence Bragg (1890-1971) won the
1915 Nobel Prize for their contributions to XRD.
Diffraction –The constructive interference, or
reinforcement, of a beam of x-rays or electrons
interacting with a material. The diffracted beam
provides useful information concerning the
structure of the material.
Used widely as an equipment to determine crystal
structures of various materials.

28. (c) 2003 Brooks/Cole Publishing / Thomson Learning
(a)Destructive (out of
phase) x-ray 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.

29. Bragg’s Law: sin2
hkldn
Bragg’s Law:222
0
lkh
a
d
hkl


Interplanar
spacing:
d
Miller Indices
Where is half the angle
between the diffracted
beam and the original
beam direction
is the wavelength of X-ray
d is the interplanar spacing

30. Interplanar spacing (d-spacing)
The distance between two adjacent
parallel planes of atoms with the same
Miller Indices (d
hkl).
The interplanar spacing in cubic materials
is given by the general equation:
where a
0is the lattice parameter and h,kand l
represent the Miller Indices of the adjacent
planes being considered.222
0
lkh
a
d
hkl



31. (c) 2003 Brooks/Cole Publishing / Thomson Learning
(a) Diagram of a
diffractometer, showing
powder sample, incident
and diffracted beams. (b)
The diffraction pattern
obtained from a sample of
gold powder.

32. The results of a x-ray diffraction experiment using x-
rays with λ= 0.7107 Å(a radiation obtained from
molybdenum (Mo) target) show that diffracted peaks
occur at the following 2θangles:
Example: Examining X-ray Diffraction
Determine the crystal structure, the indices of the
plane producing each peak, and the lattice parameter
of the material.

33. EXAMPLE SOLUTION
We can first determine the sin2 θvalue for each peak,
then divide through by the lowest denominator, 0.0308.

34. EXAMPLE SOLUTION (Continued)
We could then use 2θvalues for any of the peaks to
calculate the interplanar spacing and thus the lattice
parameter. Picking peak 8:
2θ =59.42 or θ= 29.71
Å
Å868.2)4)(71699.0(
71699.0
)71.29sin(2
7107.0
sin2
222
4000
400


lkhda
d


This is the lattice parameter for body-centered cubic iron.

35. In-Class Exercise 3: Repeat Distance
In a FCC unit cell, how many d111 are present
between the 0,0,0 point and 1,1,1 point?03a 0
2 2 2
hkl
a
d
h k l


The distance between the 0,0,0 and 1,1,1 point is:
Answer:
The interplanar spacing is:3111
0
222
0
111
aa
d 


Now, the number of interplanar spacings (d
111) between the
specified points are:

36. End of Tutorial 2