3.Metal and Ceramic Structures

8/25/2008

MSE 280: Introduction to Engineering Materials

Metal and Ceramic Structures Reading: Chapter 3 –

– – –

Defining ordered atoms atoms in crystalline solids. Unit cells, unit vectors, Coordinates, directions, planes, close packing… Densities: Crystal Crystal density, density, Line density, density, Planar density density Crystal symmetry and families Crystallography

Part Part 2 – – – –

Ionic crystals. Lattice energy. energy. Silica & silicates. silicates. Carbon 1

© 2007, 2008 Moonsub Shim, University of Illinois

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ENERGY AND PACKING • Non dense, random packing

Energy typical neighbor

typical neighbor bond energy

• Dense, regular regular packing packing

r

Energy typical neighbor bond length r

typical neighbor bond energy

Dense, regular-packed structures tend to have lower energy. 2


From Callister 6e resource CD.

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Packing atoms together Crystalline materials... • atoms pack in periodic, 3D arrays • yp ca o : -metals -many ceramics -some polymers

crystalline SiO2 Adapted from Fig. 3.18(a), Callister 6e.

Noncrystalline materials... • atoms have no periodic packing • occurs for: -complex structures -rapid cooling " Amorphous Amorphous" = Noncrystalline © 2007, 2008 Moonsub Shim, University of Illinois

Si

Oxygen

noncrystalline SiO2 Adapted from Fig. 3.18(b), Callister 6e.

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Crystalline materials: Unit Cell unit of a crystal structure. Its geometry and atomic positions define the crystal structure. Note: more than one unit cell can be chosen for a given crystal but by convention/convenience the one with the highest symmetry is chosen.

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Packing atoms together Crystalline materials... • atoms pack in periodic, 3D arrays • yp ca o : -metals -many ceramics -some polymers

crystalline SiO2 Adapted from Fig. 3.18(a), Callister 6e.

Noncrystalline materials... • atoms have no periodic packing • occurs for: -complex structures -rapid cooling " Amorphous Amorphous" = Noncrystalline © 2007, 2008 Moonsub Shim, University of Illinois

Si

Oxygen

noncrystalline SiO2 Adapted from Fig. 3.18(b), Callister 6e.

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Crystalline materials: Unit Cell unit of a crystal structure. Its geometry and atomic positions define the crystal structure. Note: more than one unit cell can be chosen for a given crystal but by convention/convenience the one with the highest symmetry is chosen.

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Unit cells and unit vectors Lattice parameters axial lengths: a, b, c interaxial angles: α, v β,vγ v unit vectors: a b c

c

v

v b

In general:

a≠b≠c α ≠ β ≠ γ

a

v

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Unit Cells: Bravais Lattices

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(trigonal)

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Unit cell types Primitive

Body-centered

Face-centered

End-centered

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Number of atoms in a unit cell

Simple cubic 8 atoms bu t each atom is shared by 8 unit cells. →

1 atom per unit cell

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Coordination Number Number of nearest neighbor atoms

Simple cubic: coordination number = 6 10


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Atomic Packing Factor (APF) APF =

Vol. of atoms in unit cell Vol. of unit cell

Depends on: • Crystal structure. • How “close” acked the atoms are. • In simple close-packed structures with hard sphere atoms, independent of atomic radius

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Close packing of atoms Consider atoms as hard spheres.

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Face-Centered Cubi c (FCC)

Atoms at the corners of the cube + Atoms at the center of each face

a

a = ||unit vector||

4 R

R = atomic radius

a 2 + a 2 = ( 4 R )2 a

a = 2 2 R 13


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Face-Centered Cubi c (FCC)

Adapted from Fig. 3.1(a), Callister 6e.

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Number of atom s in an FCC uni t cell

• Each corner atom . There are 8 corner atoms in an FCC unit cell. • Each face atom contributes as 1/2. There are 6 face atoms.

1

1

× 8 ( corner _ atoms ) + × 6 ( face _ atoms ) = 4 atoms / unit _ cell

8

2

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Coordination numb er for FCC

2 2 R R 2

6

9

10

5 2

2 2 R

3

1

4 7

12

11

8

Total 12 nearest neighbor atoms Coordination number = 12 16


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At om ic pac ki ng fac to r (A PF) fo r FCC 2 2 R R 2

2 2 R a

4 R

V atoms V unit _ cell = a 3

a

APF =

V atoms V unit _ cell

⎛ 16 π R 3 ⎞ ⎜ ⎟ π 3 ⎝ ⎠ = = = 0.74 3 16 2 R

4 ⎞ = 4⎛ ⎜ π R3 ⎟

= (2 2 R) 3 = 16 2 R 3

Independent o f R and a!

3 2

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Summary for FCC

||unit vector|| = a

=2

2 R

4 atoms/unit cell a

4 R

Coordination number = 12 APF = 0.74

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Bod y-Centered Cubi c (BCC)

2 a Atoms at the corners of the cube + Atom at the center of the cube a

4R

a 2 + a=

(

2 a

)

2

= (4 R )2

4 R 3 19


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Bod y-Centered Cubi c (BCC)

From Callister 6e resource CD.

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Number of atoms in a BCC unit c ell

• Each corner atom . There are 8 corner atoms in an FCC unit cell. • The center atom contributes as 1. There is only 1 center .

1

× 8 ( corner _ atoms ) + 1 ×1( center _ atom ) = 2 atoms / unit _ cell

8

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Coordination number for B CC

1

4

6

5

Total 8 nearest neighbor atoms Coordination numb er = 8

7 8

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At om ic pac ki ng fac to r (A PF) fo r B CC V atoms

2 a

4 ⎞ = 2⎛ ⎜ π R 3 ⎟ ⎝ 3 ⎠

2

a= a

4R

2

=

2

4 R 3

V unit _ cell = a

3

=(

4 R 3

)

=

3

64 R 3 3 3

8

APF =

V atoms V unit _ cell

⎜ π R 3 ⎟ 3 ⎠ = = ⎝ 64 R

3

3π 8

= 0.68

3 3 23


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Summary for BCC a

||unit vector|| = a

a

4 R

=

4 R 3

2 atoms/unit cell Coordination number = 8 APF = 0.68

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Hexagonal Close-Packing (HCP)

Find a, c, number of atoms/unit cell, coordination number, and APF for HW. 25


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Point Coordinates to define a point within a unit cell…. Essentially same as Cartesian coordinates except values of x, y, and z are expressed as fractions of the magnitude of unit vector(s) (and x, y, . pt. coord. z x ( a) y ( b) z ( c) b a

c

A

0

0

0

B

1

0

0

C

1

1

1

D

1/2

0

1/2

= point “A” = origin = point “B” x

= point “C” = point “D”


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Crystallographic Directions To define a vector: 1. Start at the origin. . (or fractions) of a, b, and c. 3. Multiply or divide by a common factor to reduce the lengths to the smallest integer values. 4. Enclose in square brackets: [u v w] where u, v, and w are integers. Along unit vectors: a

b

c

Note: in any of the 3 directions there are both positive and negative directions. Negative directions are denoted with a “bar” ove r the number. e.g. [1 1 1] 27


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Example 1: Assign the coordinates to the following crystallographic direction Along x: 1 a

z

Along y: 1 b

b

Along z: 1 c

a

c

y

111 x

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Example 2: draw [1 1 0] direction. 1 unit vector length along x -1 unit vector length along y z

0 unit vector length along z b

a

[1 1 0]

x

origin 29


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Note: for some crystal structures, different directions can be equivalent.

e.g. For cubic crystals: [1 0 0], [ 10 0], [0 1 0], [0 1 0], [0 0 1], [0 0 1 ] are all equivalent Families of crystallographic di rections e.g. <1 0 0>

Angled brackets denote a family of crystallographic directions.

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Families and Symmetry • • •

Which directions belong in the same family is determined by the crystal symmetry. What is symmetry? c one s more symmetr c c a a a a

•

b

The more symmetry operations there are the more symmetric it is. 1

3

1

1

etc…

and 2

3

•

2

1 equivalent

2

3

3

2

equivalent

Which is more symmetric, cubic or tetragonal? 31


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Families and Symmetry z

Cubic symmetry

z

[010]

Rotate 90o about z-axis

y

y [100] x

x z

Rotate 90o about -axis Symmetry operation can generate all the directions within in a family. x © 2007, 2008 Moonsub Shim, University of Illinois

y

Similarly for other equivalent directions 32

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Crystallographic directions for hexagonal crystals Consider vectors q and r…. If we keep the 3-coordinate system: q = [1 0 0] v r = [1 1 0]

v

z

Different set of indices. However, these two vectors are equivalent by symmetry (i.e. via 60 o rotation).

y

r

v

q

v x

Choose a 4-coordinate system! 33


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Crystallographic directions for hexagonal crystals In the Miller-Bravais (4 axes) coordinate system, We have:

q

v

= [2 1 1 0]

z

a2

q

a3

v

2 along a1 -1 along a2

a1

Although the length is different the direction is the same.

-1 along a3


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Crystallographic directions for hexagonal crystals Similarly for r , in the Miller-Bravais (4 axes) coordinate system, we now have: v r = [1 1 2 0]

v

z

a2

-

r

3

v

a3

1 along a1

a1

1 along a2

Again, consistent direction.

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Crystallographic directions for hexagonal crystals Miller-Bravais (4 axes) coordinate system We now have: z

q = [2 1 1 0] v r = [1 1 2 0]

v

a2

r

v

a3

q

v

a1

Indices are now consistent within the family, but where do these numbers come from? 36


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To transform 3 indices to 4 indices [u’ v’ w’]

[u v t w]

u = 1/3 2u’ – v’ v = 1/3 (2v’ – u’) t = - (u + v) w = w’

May need to factor to reduce u, v, t, and w to their smallest integers.

e.g. convert q = [1 0 0] to 4-coord. system.

v

u = 1/3 (2x1 – 0) = 2/3 –

2

x3

-

-

x3

t = -(2/3 + (-1/3)) = -1/3 w=0

z

x3

a2

[2 1 1 0]

-1

x3

0

a3

q

v a1 37


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transforming 3 indices to 4 indices r = [1 1 0] to 4-coord. system.

v

convert

u = 1/3 (2x1 – 1) = 1/3 v = 1/3 (2x1 – 1) = 1/3

x3 x3

1 1

x3 t = -(1/3 + 1/3) = -2/3

[1 1 2 0]

-2 x3

w=0

z

0 a2

r

v

a3 © 2007, 2008 Moonsub Shim, University of Illinois

a1 38 MSE280

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Crystallographic Planes To define a plane (i.e. to assign Miller Indices): 1. If the plane passes through the origin, either a) Construct a parallel plane translated or b) choose another origin at the corner of adjacent unit cell

z b a

y

c

x

origin

origin

2. Now, the plane either intersects or parallels each of . 3. Determine length in terms of lattice parameters a, b and c. Parallel to x-axis = length along x Parallel to y-axis = length along y

∞ ∞

Intercepts z-axis at z = c

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Crystallographic Planes 4. Take the reciprocal of the lengths

∞ Along y-axis: length = ∞

0

Along z-axis: length = 1

1

e.g.

ong x-ax s: eng

=

5. If necessary, factor to get the smallest integers. divide through by 3 . .

, ,

, ,

6. Enclose indices in parentheses w/o commas In the example on previous slide, the Miller index would be (0 0 1) 40


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Example 1 Assign Miller indices . z

z’

or g n rom

o

2. Determine lengths along each axis Along x =

b

∞

Intersects y at -1

a

Intersects z at 1/3 y

c

3. Take the reciprocal: 0, -1, 3 O’

x

4. No need to factor. 5. (0 1 3)

x’

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Example 2 Draw the (0 .

1 1) plane.

= The plane runs parallel to x-axis

z

2. along y = -1 (in units of b) take reciprocal: -1 y-intercept at -1 3. along z = 1 (in units of c) take reciprocal: 1 -

|c| |b|

y

x

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Crystallographic planes in FCC z

y

(1 0 0) plane

Look down this direction (perpendicular to the plane)

x © 2007, 2008 Moonsub Shim, University of Illinois

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Crystallographic planes in FCC z

Look down this direction (perpendicular to the plane)

y

x © 2007, 2008 Moonsub Shim, University of Illinois

(1 1 1) plane

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Note: similar to crystallographic directions, planes that are parallel to each other are equivalent

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Crystallographic Planes for Hexagonal Crystals Similar to crystallographic directions for hexagonal crystals, 4-coordinate system is used. i.e. instead of (h k l) for 3-coordinate systems, use (h k i l). For integers h, k, and l , same procedure as 3-coordinate systems is used and i = - (h+k)

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Crystallographic pl anes in HCP z

Name this plane…

a2

a3

(0 0 0 1) plane a1

Parallel to a 1, a2 and a3 -> h = k = i = 0 Intersects at z = 1


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Crystallographic pl anes in HCP z

a2

+1 in a1

a3 -1 in a2

a1 h = 1, k = -1, i = -(1+-1) = 0, l = 0 © 2007, 2008 Moonsub Shim, University of Illinois

(1 1 0 0) plane 48

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z

(1 1 1) plane of FCC

y

x z

(0 0 0 1) plane of HCP

SAME THING!*

a2 a3 a1 © 2007, 2008 Moonsub Shim, University of Illinois

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Close Packi ng

Note: whether the atoms of 2 nd layer fills B-sites or C-sites does not matter. They are equivalent until the 3rd layer is added! 50 © 2007, 2008 Moonsub Shim, University of Illinois

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FCC

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HCP

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Difference between FCC and HCP Looking down (0001) plane

FCC

HCP

ABCABC…

ABABAB…

Looking down (111) plane!

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Densities • rys a s ens y .e. per volume (e.g. g/cm 3).

ens y n un s o mass

• Linear density: number of atoms per unit length (e.g. cm-1). . . cm-2).

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Crystals density (ρ) Mass of unit cell

ρ = M =

Volume of unit cell

V

n = number of atoms in unit cell A = atomic weight N A = Avogadro’s number

nA N A

ρ =

nA N AV 55


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Density example Calculate density of copper given: R = 0.128 nm A = 63.5 g/mol FCC structure Recall for FCC, there are 4 atoms p er unit cell. Express unit cell volume in terms of atomic radius R.

V = a 3

= (2

2 R)3

= 16 2 R 3

Then we have:

=

M

nA

V

N A16 2 R 3 4(65.3 g / mol )

(6.022 x1023 mol −1 )(16 2 )(1.28 x10 −8 cm) 3

= 8.89 g/cm 3

Compare to actual value of 8.94 g/cm 3! © 2007, 2008 Moonsub Shim, University of Illinois

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Linear Density (LD) LD =

Number of atoms centered on a direction vector Length of the direction vector

Example: calculate the linear density of an FCC crystal along [1 1 0]. Length = 4R Effectively 2 atoms alon the [1 1 0] vector. LD = 2/4R = 1/2R

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Planar Density (PD) PD =

Number of atoms centered on a given plane Area of the plane

Example: calculate the planar density on (1 1 0) plane of an FCC crystal.

a = 2 2 R

4 R Count only the parts of the atoms within the plane: 2 atoms 2 1 PD = = area = 2 2 R × 4 R (4 R)( 2 2 R) 4 2 R 2

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Example problem (110)

(101)

(011)

m n 2 . 0

0 . 3 6 n m

0 . 3 2 n m

0.39 nm 0.3 nm

0.253 nm

Given above information answer the following questions. A) What is the crystal structure? B) If atomic radius is 0.08 nm, what is APF? C) If atomic weight is 43 g/mol, calculate density. D) What is the linear density along [210]? E) What is the planar density of (210)? 59


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How to determine crystal structure The two slit experiment Same idea with crystals - Light gets scattered off atoms…

λ d

θ

Incoming light

Constructive interference when n = 2d si n (Bragg’s Law) and d have to be comparable lengths.

But since d (atomic spacing) is on the order of angstroms: x-ray diffraction

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Diffraction Experiment and Signal Scan versus 2x Angle: Polycrystalline Cu

Diffraction Experiment

Measure 2θ

Diffraction collects data in “r eci r ocal s ace” since it is the e uivalent of a Fourier transform of “ real space”, i.e., e ikr , as k ~ 1/r. How can 2θ scans help us determine crystal structure type and distances between Miller Indexed planes ( I.e. structural parameters)?


©D.D. Johnson 2004, 2006-08

Crystal Structure and Planar Distances Bravais Lattice

FCC HCP

Constructive Interference

Destructive Interference

Reflections present

Reflections absent

(h,k,l) All Odd or All Even

(h,k,l) Not All Odd or All Even

Any other (h,k,l)

h+2k=3n, l = Odd

h, k, l are the Miller Indices of the planes

So they determine the important planes of atoms, or symmetry.

n= integer

Distances between Miller Indexed planes For cubic crystals:

For hexagonal crystals: a

d hkl = 4 3


(h 2

a ⎞ + hk + k 2 ) + l 2 ⎛ ⎝ ⎠

2

c

©D.D. Johnson 2004, 2006-08

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Allowed (hkl) in FCC and BCC for principal scattering (n=1) (h k l)

h2+k2+l2

h+k+l

h,k,l all even or odd?

100

1

1

No

110

2

2

No

200

4

2

Yes

210

5

3

No

211

6

4

No

220

8

4

Yes

221

9

5

No

300

9

3

No

311

11

5

Yes

222

12

6

Yes

320

13

5

No

321

14

6

No

Self-Assessment: From what crystal structure is this?

h + k + l was even and gave the labels on graph above, so crystal is BCC.


©D.D. Johnson 2004, 2006-08

Example problem: X-ray crystallography • Given the information below, determine the crystal structure. Consider onl FCC and BCC structures as possibilities. – Lattice parameter: a = 0.4997nm – Powder x-ray: λ = 0.1542 nm

2 (o ) 31 51.8 61.6 64.8


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Diffraction of Single 2D Crystal Grain

Lattice of Atoms: Crystal Grain Diffraction Pattern in (h,k) plane

Powder diffraction figures taken from Th. Proffe and R.B. Neder (2001) http://www.uni-wuerzburg.de/mineralogie/crystal/teaching/pow_a.html


©D.D. Johnson 2004, 2006-08

Diffraction of Multiple Single 2D Crystal Grains (powders) Multiple Crystal Grains: Diffraction Pattern 4 Polycrystals in (h,k) plane (4 grains)

Diffraction Pattern in (h,k) plane (40 grains)

Powder diffraction figures taken from Th. Proffe and R.B. Neder (2001) http://www.uni-wuerzburg.de/mineralogie/crystal/teaching/pow_a.html


©D.D. Johnson 2004, 2006-08

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3.Metal and Ceramic Structures

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