Dobson Chapter 3 crystals

Dobson Chapter 3 crystals - CRYSTALLINE

SOLIDS • How...

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Unformatted text preview: CRYSTALLINE

SOLIDS • How do atoms assemble into solid structures? • How does the density of a material depend on its structure? • Why do material properties sometimes vary with the material’s orientation? 3‐1 Materials
and
Packing/Arrangement Crystalline
materials... •

atoms
pack
in
periodic,
3D
arrays •

typical
of: ‐metals ‐many
ceramics ‐some
polymers crystalline
SiO2 Adapted
from
Fig.
3.23(a), 
Callister
&
Rethwisch
8e.
 Noncrystalline
materials... Si Oxygen •

atoms
have
no
periodic
packing •

occurs
for: ‐complex
structures ‐rapid
cooling "Amorphous"
=
Noncrystalline noncrystalline
SiO2 Adapted
from
Fig.
3.23(b), 
Callister
&
Rethwisch
8e.
 • A crystalline solid that exhibits only one orientation of arrangement is referred to as a single crystal • Solids that exhibit more than one orientation are referred to as polycrystalline. • For crystalline materials, we will describe structure in terms of a crystal structure. • Crystal structures can be systematically differentiated in terms of their: • Lattice- a periodic and geometric arrangement of atoms or ions in space. • The specific atom or ion type (identity and number) is known as the motif or basis of the crystal structure. • Crystal= lattice+basis. hTp://vimeo.com/15751112 The
Unit
Cell The
basic
building
block
of
the
crystal
structure Atomic
hard
sphere
model Metallic
Crystal
Structures •

Tend
to
be
densely
packed. •

Reasons
for
dense
packing: Typically,
only
one
element
is
present,
so
all
atomic
radii
are
the same. Metallic
bonding
is
not
direcXonal. Nearest
neighbor
distances
tend
to
be
small
in
order
to
lower bond
energy •

Have
the
simplest
crystal
structures. Metallic Crystal Structures Simple cubic Structures (SC) Face Centered Cubic Structures (FCC) Body Centered Cubic Structures (BCC) Hexagonal ClosePacked Structures (HCP) QuanXtaXve
CharacterizaXon Cube edge length (a) a=2R












(SC) a = 2 R 2 (FCC) Atomic packing factor Coordination number Density a = 4R 3 (BCC) QuanXtaXve
CharacterizaXon Cube edge length (a) Atomic packing factor Coordination number Density 3‐9 APF = volume of atoms in unit cell volume of unit cell volume of atoms in unit cell APF = volume of unit cell • The number of atoms per unit cell refers to the contribution/presence of an atom or ion to/within the unit cell. • Essential to calculations of packing factors and density. 3‐10 The coordination number refers to the number of nearest neighbors (“touching” atoms) for a specific lattice point. SCC
–
CoordinaXon
#=6 BCC
–
CoordinaXon
#=8 3‐11 QuanXtaXve
CharacterizaXon Cube edge length (a) Atomic packing factor Coordination number Density (g/cm3) nA != Vc N A n = number of atoms associated with each unit cell A = atomic weight (g/mol) Vc = volume of unit cell N A = Avogadro's number 3‐12 Simple
Cubic
Structure •

Rare
due
to
low
packing
density •

Close‐packed
direcXons
are
cube
edges. •

Coordina0on
#
=
6 


(#
nearest
neighbors) 13 Atomic
Packing
Factor
(SCC) APF
=
 Volume
of
atoms
in
unit
cell* Volume
of
unit
cell *assume
hard
spheres volume atoms a unit
cell R=0.5a APF
=
 1 4 3 a3 close‐packed
direcXons contains
8
x
1/8
=
 1
 atom/unit
cell Adapted
from
Fig.
3.24, 
Callister
&
Rethwisch
8e.
 π (0.5a) 3 atom volume unit
cell •

APF
for
a
simple
cubic
structure
=
0.52 Body
Centered
Cubic
Structure •

Atoms
touch
each
other
along
cube
diagonals. ‐‐Note:

All
atoms
are
idenXcal;
the
center
atom
is
shaded 


differently
only
for
ease
of
viewing. ex:
Cr,
W,Fe
(α),
Tantalum,
Molybdenum •

Coordina0on
#
=
8 Click
once
on
image
to
start
animaXon (Courtesy
P.M.
Anderson) Adapted
from
Fig.
3.2, 
Callister
&
Rethwisch
8e.
 2
atoms/unit
cell:

1
center
+
8
corners
x
1/8 Atomic
Packing
Factor
(BCC) •

APF
for
a
body‐centered
cubic
structure
=
0.68 3a a 2a Adapted
from Fig.
3.2(a),
Callister
& Rethwisch
8e. Close‐packed
direcXons: R atoms unit
cell APF
=
 length
=
4R
= a 2 4 3 π
 (
 3 a/4 ) 3 a3 volume atom volume unit
cell 3
a Face
Centered
Cubic
Structure •
Atoms
touch
each
other
along
face
diagonals. ‐‐Note:

All
atoms
are
idenXcal;
the
face‐centered
atoms
are
shaded 


differently
only
for
ease
of
viewing. ex:
Al,
Cu,
Au,
Pb,
Ni,
Pt,
Ag •

Coordina0on
#
=
12 Adapted
from
Fig.
3.1,
Callister
&
Rethwisch
8e. 4
atoms/unit
cell:
6
face
x
1/2
+
8
corners
x
1/8 17 Atomic
Packing
Factor
(FCC) maximum achievable APF Close‐packed
direcXons:
 length
=
4R
= 2
a
 2a Unit
cell
contains: 




6
x
1/2
+
8
x
1/8

 

=
 4
atoms/unit
cell a Adapted
from Fig.
3.1(a), Callister
&
Rethwisch 8e. atoms unit
cell APF
=
 4 4 3 π (
 2 a/4 ) 3 a3 •

APF
for
a
face‐centered
cubic
structure
=
0.74 volume atom volume unit
cell Hexagonal
Close‐Packed
Structure
(HCP) • APF
=
0.74 • Coordina0on
#
=
12 • 6
atoms/unit
cell ex: Cd, Mg, Ti, Zn 19 Polymorphism • Two
or
more
disXnct
crystal
structures
for
the
same material
(allotropy/polymorphism) iron system 








Xtanium liquid 










α,
β‐Ti 1538ºC δ-Fe BCC 








carbon 1394ºC diamond,
graphite γ-Fe FCC 912ºC BCC α-Fe Close‐Packed
Crystal
Structures The FCC and HCP crystal structures both possess a packing fraction of 0.74, consistent with similar local atomic arrangements. However, these structures differ in structure; this difference can be illustrated through a “stacking” representation. 3‐21 HCP involves ABABA stacking FCC involves ABCABC stacking Example: Calculate the volume of an FCC unit cell in terms of the atomic radius R. Example: Calculate the volume of an FCC unit cell in terms of the atomic radius R. a2+a2
=
(4R)2 a
=
2R
√2 V
=
a3
=
(2R
√2)
3
=
16R3
√2 Example: Calculate the unit cell volume of FCC aluminum (r=0.1431 nm). Example: Calculate the unit cell volume of FCC aluminum (r=0.1431 nm). a
=
2R
√2 a
=
0.4047
nm Volume
=
a3 Volume
=
0.662
nm3 E L 1 TheoreXcal
Density
(ρ) Mass 
of 
Atoms 
in
Unit 
Cell Density
=

ρ

= Total 
Volume 
of 
Unit 
Cell ρ

= nA VC NA Where






n
=
number
of
atoms/unit
cell 






A
=
atomic
weight 






VC
=
Volume
of
unit
cell
=
a3
for
cubic 






NA
=
Avogadro’s
number 











=
6.022
x
1023
atoms/mol EXAMPLE: Calculate the theoretical density of copper, which exists in an FCC structure with a radius of 0.128 nm and an atomic weight of 63.5 g/mol. EXAMPLE: Calculate the theoretical density of copper, which exists in an FCC structure with a radius of 0.128 nm and an atomic weight of 63.5 g/mol. Density
=

ρ

= Mass 
of 
Atoms 
in
Unit 
Cell Total 
Volume 
of 
Unit 
Cell a a = 2R 2 atoms 4 63.5 unit
cell Unit
cell
volume nA VC NA • Ex:
Cu
Copper
(FCC) A
=
63.5
g/mol R
=
0.128
nm n
=
4
atoms/unit
cell 2a ρ
=
 ρ

= a3 6.022
x
1023 g Vc = 16R3 2 mol ρtheoretical atoms mol =
8.89
g/cm3 ρactual =
8.94
g/cm3 28 DensiXes
of
Materials
Classes In general ρ Metals/
 Alloys ρ ρ metals >
 ceramics >
 polymers 30 Why? 20 10 

 




•
less
dense
packing 




•
osen
lighter
elements 
Polymers
have... 




•
low
packing
density 








(osen
amorphous) 




•
lighter
elements
(C,H,O) 
Composites
have... 




•
intermediate
values ρ 
Ceramics
have... 3 (g/cm

) 
Metals
have... 




•
close‐packing 







(metallic
bonding) 




•
osen
large
atomic
masses PlaXnum 

 Gold,
W 

 Tantalum 

 Silver,
Mo 

 Cu,Ni 

 Steels 

 Tin,
Zinc 

 5 

 4 

 3 

 2 1 0.5 

 0.4 

 0.3 

 Graphite/
 Ceramics/
 Semicond Polymers Composites/
 fibers B ased
on
data
in
Table
B1,
Callister
 *GFRE,
CFRE,
&
AFRE
are
Glass, Carbon,
&
Aramid
Fiber‐Reinforced Epoxy
composites
(values
based
on 60%
volume
fracXon
of
aligned
fibers in
an
epoxy
matrix). 

 
 
 
 
 Zirconia 

 Titanium 

 Al
oxide 

 Diamond 

 Si
nitride 

 Aluminum 

 Glass ‐ soda 

 Concrete 

 PTFE 

 Silicon 

 Magnesium 

 G raphite 

 Silicone 

 PVC 

 PET 

 PC 

 H DPE,
PS 

 PP,
LDPE 

 Glass
fibers 

 GFRE* 

 Carbon
 fibers 

 CFRE *

 A ramid
fibers 

 AFRE *

 Wood 

 Data
from
Table
B.1,
Callister
&
Rethwisch,
8e. Crystal
Systems Unit
cell:

smallest
repeXXve
volume
which
contains
the complete
latce
paTern
of
a
crystal. Unit
cell
geometry
defined
by: •3
Edge
lengths
a,b
and
c
(latce
constants) •3
Interaxial
angles
α,
β,
and
γ 7 crystal systems 14 crystal lattices Fig.
3.4,
Callister
&
Rethwisch
8e. Seven
Crystal
Systems SC BCC FCC HCP 3‐31 Least
Symmetry 3‐32 Crystallographic
Points,
DirecXons
&
Planes Coordinates are used to identify points, directions, and planes with respect to a crystal structure. Points are identified by their location within a right-handed Cartesian coordinate system. Crystallographic
Points,
DirecXons
&
Planes q,r,s q
–
distance
qa
(fracXonal
length
of
a)
along
x
axis r
–
distance
rb
(fracXonal
length
of
b)
along
y
axis s
–
distance
sc
(fracXonal
length
c)
along
z
axis a,b
and
c
are
unit
cell
length c b a Example: Specify point coordinates for all atom positions for a BCC unit cell. 3‐35 Example: For the unit cell shown below, locate the point having the coordinates ¼, 1, ½ Example: For the unit cell shown below, locate the point having the coordinates ¼, 1, ½ q=
¼
•
0.48nm
=
0.12nm r=
1•
0.46nm
=
0.46nm s=
½
•
0.40nm
=
0.20nm Crystallographic
Points,
DirecXons
&
Planes Coordinates are also used to identify directions within a crystal structure. The following guidelines are used to identify the direction: 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 4. Enclose in square brackets, no commas c a [uvw] b ex: 1, 0, ½ => 2, 0, 1 => [ 201 ] -1, 1, 1 => [ 111 ] where overbar represents a negative index hTp://www.youtube.com/watch?v=mQy2CdKYqX0&feature=related Example: Determine the indicies for the vector shown in green below Steps: 1.ReposiXon
through
origin 2.ProjecXons
onto
x,y
and
z 1. ½(a) 2. 1(b) 3. 0(c) 3.ReducXon
to
smallest
integer 1 2


0 4.Enclosure
[120] Identify the directions of the vector shown in blue. Steps: 1.ReposiXon
through
origin 2.ProjecXons
onto
x,y
and
z 3.ReducXon
to
smallest
integer 4.Enclosure Identify the directions of the vector shown in blue. Steps: 1.ReposiXon
through
origin 2.ProjecXons
onto
x,y
and
z 1. 0(a) 2. ‐1(b) 3. 1(c) 3.ReducXon
to
smallest
integer 4.Enclosure
[011] Other notation systems exist for hexagonal structures – will not be covered Crystallographic
Points,
DirecXons
&
Planes In addition to directions, specific planes within a crystal structure hold particular significance. For example: •Metals deform easily along close-packed directions •Crystal terminations are defined in terms of surface planes Procedure for writing plane notation: 1. 2. 3. 4. 5. Shift coordinate system if plane passes through origin Identify intersections along x,y,z coordinate axes. A plane parallel to axis has ∞ intercept, 0 index Take reciprocal of intercepts Clear fractions but don’t reduce integers Enclose numbers in parentheses, using overscores for negative #’s (hkl) Crystallographic
Planes Adapted
from
Fig.
3.10,
Callister
& Rethwisch
8e. Crystallographic
Planes z 3. Reduction a








b







c 1








1







∞ 1/1





1/1




1/∞ 1








1







0 1








1







0 4. Miller Indices (110) example 1. Intercepts 2. Reciprocals 3. Reduction Miller Indices a x a








b







c 1/2






∞






∞ 1/½




1/∞



1/∞ 2







0








0 2







0








0 4. c (200) example 1. Intercepts 2. Reciprocals b y z c b a x hTp://www.youtube.com/watch?v=pBSKjvoFfac&feature=mfu_in_order&list=UL y Crystallographic
Planes Example z a








b







c 1. Intercepts 1/2







1






3/4 2. Reciprocals 1/½




1/1




1/¾ 3. Reduction 4. Miller Indices • c • a 2








1





 4/3 
 6








3







4 (634) b x • y Example: Determine the Miller indices for the plane shown below. Example: Determine the Miller indices for the plane shown below. 1. 2. 3. 4. 5. Shift coordinate system if plane passes through origin Identify intersections along x,y,z coordinate axes. A plane parallel to axis has ∞ intercept, 0 index Take reciprocal of intercepts Clear fractions but don’t reduce integers Enclose numbers in parentheses, using overscores for negative #’s (hkl) 1. 2. Intercepts Reciprocals 3. Reduction a








b







c ∞







‐1 





1/2 1/∞





1/‐1




2/1 0










‐1







2 0










‐1







2 4. Miller Indices (012) Linear
Density
&
Planar
Density Linear density (LD) and planar density (PD) fraction calculations are useful in describing atomic densities with respect to crystallographic direction or crystallographic plane. LD = number of atoms centered on direction vector length of direction vector Units
are
reciprocal
length PD = number of atoms centered on plane area of plane Units
are
reciprocal
area 3‐50 Linear
Density • Linear
Density
of
Atoms
≡
LD
= [110] Number
of
atoms
 Unit
length
of
direcXon
vector ex: linear density of Al in [110] direction a = 0.405 nm #
atoms a Adapted
from Fig.
3.1(a), Callister
&
Rethwisch 8e. length = 2 = 3.5
nm −1 LD 2a Planar
Density
of
(100)
Iron SoluXon:


At
T
<
912ºC
iron
has
the
BCC
structure. 2D repeat unit 4R a= 3 (100) Radius of iron R = 0.1241 nm Adapted
from
Fig.
3.2(c),
Callister
&
Rethwisch
8e. atoms 2D
repeat
unit Planar
Density
=
 area 2D
repeat
unit 1 a2 1 =
 4R 3 2 =
 12.1 atoms atoms =
1.2
x
1019
 nm2 m2 E L 2 Seven
Crystal
Systems SC BCC FCC HCP 3‐53 Least
Symmetry 3‐54 Carbon
Crystals Diamond: Diamond Cubic Structure Graphite: Hexagonal Structure Bucky Ball (Buckminsterfullerene): C60 Carbon Nanotube Strongest known material / low density TheoreXcal
Density
(ρ) Mass 
of 
Atoms 
in
Unit 
Cell Density
=

ρ

= Total 
Volume 
of 
Unit 
Cell ρ

= nA VC NA Where






n
=
number
of
atoms/unit
cell 






A
=
atomic
weight 






VC
=
Volume
of
unit
cell
=
a3
for
cubic 






NA
=
Avogadro’s
number 











=
6.022
x
1023
atoms/mol TheoreXcal
Density
for
mixed
atomic
crystals Example: NaCl ρ

= nA VC NA ∑ANa = 22.99 g/mol ∑ACl = 35.45 g/mol NaCl is FCC, so n = 4 TheoreXcal
Density
for
mix
atomic
crystals Example: NaCl ρ

= nA VC NA ∑ANa = 22.99 g/mol ∑ACl = 35.45 g/mol a = 2rNa+ + 2rClV = a3 = (2rNa+ + 2rCl-)3 TheoreXcal
Density
for
mix
atomic
crystals Example: NaCl ρ

= nA VC NA ∑ANa = 22.99 g/mol ∑ACl = 35.45 g/mol a = 2rNa+ + 2rClV = a3 = (2rNa+ + 2rCl-)3 ρ=



























4
(22.99
+
35.45) 






[2(0.102
x
10‐7)
+
2(0.181
x
10‐7)]3
(6.02
x
1023) TheoreXcal
Density
for
mix
atomic
crystals Example: NaCl ρ

= nA VC NA ∑ANa = 22.99 g/mol ∑ACl = 35.45 g/mol a = 2rNa+ + 2rClV = a3 = (2rNa+ + 2rCl-)3 ρ
=



























4
(22.99
+
35.45) 






[2(0.102
x
10‐7)
+
2(0.181
x
10‐7)]3
(6.02
x
1023) 


=
2.14
g/cm3 Experimental
value
=
2.16
g/cm3 Crystalline
vs.
Non‐crystalline • A crystalline solid that exhibits only one orientation of arrangement is referred to as a single crystal • Solids that exhibit more than one orientation are referred to as polycrystalline. • Regions of similar orientation are known as grains. • These regions are typically separated by amorphous structures known as grain boundaries. SolidificaXon
of
Polycrystalline
Materials Grain boundaries 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 Isotropic •

Niobium‐Hafnium‐Tungsten
(Nb‐Hf‐W)
plate
with
an
electron
beam
weld •

Each
"grain"
is
a
single
crystal. •

If
grains
are
randomly
oriented, 




overall
component
properXes
are
not
direcXonal. •

Grain
sizes
typically
range
from
1
nm
to
2
cm 





(i.e.,
from
a
few
to
millions
of
atomic
layers). Single
vs.
Polycrystals •

Single
Crystals E
(diagonal)
=
273
GPa ‐ProperAes
vary
with 

direcXon:

anisotropic. ‐Example:

the
modulus 

of
elasXcity
(E)
in
BCC
iron: •

Polycrystals ‐Bulk
properXes
may/may
not 

vary
with
direcXon. ‐If
grains
are
randomly 

oriented:
isotropic. 

(Epoly
iron
=
210
GPa) ‐If
grains
are
textured, 

anisotropic. E
(edge)
=
125
GPa 200
µm Anisotropy The property of anisotropy describes a variation in atomic or materials property with respect to crystallographic direction. X‐ay

DiffracXon X‐ray
DiffracXon X-ray diffraction (XRD) is commonly used to characterize crystal structure. The technique is based on Bragg’s law, which relates angles corresponding to constructive interference to the inter-plane spacing (d). • DiffracXon
graXngs
must
have
spacings
comparable
to
the wavelength
of
diffracted
radiaXon. • Can’t
resolve
spacings
<
λ • Spacing
is
the
distance
between
parallel
planes
of
atoms. 66 X‐ray
DiffracXon:
Interference
/
Phase X‐ray
DiffracXon ” θ o θ ” “2 “1 ” “2 i go ut or “1 g in m co s in ray X‐ extra
 distance
 travelled
 by
wave
“2” s ay ‐r X g

 n ct te de ” •

Incoming
X‐rays
diffract
from
crystal
planes. λ d Measurement
of
criXcal angle,
θc,
allows computaXon
of
planar spacing,
d. Bragg’s Law: nλ = 2d sin θ reflecXons
must
 be
in
phase
for
 a
detectable
signal Adapted
from
Fig.
3.20, Callister
&
Rethwisch
8e. spacing
 between
 planes X‐ray
 intensity
 (from
 detector) d
 = nλ 2
sin θc 
 θ θc X‐ray
DiffracXon The
magnitude
of
the
distance
(d)
between
two
parallel
planes
in
a
cubic
crystal
is
a funcXon
of
the
Miller
indices
(h,k,l)
as
well
as
the
latce
parameters. Interplanar separation for a plane having indices h, k, and l dhkl = a h2 + k2 + l2 X‐ray
DiffracXon
PaTern z z Intensity (relative) c a x z c b y (110) a x c b y a x (211) b (200) Diffraction angle 2θ Diffraction pattern for polycrystalline α-iron (BCC) Adapted
from
Fig.
3.22,
Callister
8e. hTp://www.youtube.com/watch?v=lwV5WCBh9a0 y X‐ray
DiffracXon
PaTern Summary •

Atoms
may
assemble
into
crystalline
or
amorphous
structures. •

Common
metallic
crystal
structures
are
FCC,
BCC,
and
HCP. 



CoordinaXon
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,
direcXons
and
planes
are
specified
in 




terms
of
indexing
schemes.
Crystallographic
direcXons
and 




planes
are
related
to
atomic
linear
densiXes
and
planar 




densiXes. Summary •

Materials
can
be
single
crystals
or
polycrystalline. 




Material
properAes
generally
vary
with
single
crystal 




orientaXon
(i.e.,
they
are
anisotropic),
but
are
generally 




non‐direcXonal
(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
diffracXon
is
used
for
crystal
structure
and 




interplanar
spacing
determinaXons. ...
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