Packing of Atoms in Solids [6]...new atom Size of the largest atom which can fit into the tetrahedral void of FCC CV = r + x e e =r +x 4 6 1 ~ 0.225 2 3 2 ⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝

Packing of Atoms in Solids [6]

1>

Metallic crystals－ are composed of bonded metal atoms. Example: Ni, Cu, Fe, and alloys.

Covalent crystals－consisted of an infinite network of atoms held together by covalent bonds, no individual molecules being present.

Example: diamond, graphite, SiC and SiO2.

Ionic crystals－consisted of an array of positive and negative ions.

Example: NaCl, MgO, CaCl2 and KNO3.

Molecular crystals－are composed of individual molecules. Example: Ar, CO2 and H2O.

Crystalline Solids

2>

Crystalline Solid is the solid form of a substance in which atoms or molecules are arranged in a definite, repeating pattern in three dimension.

Single crystals, ideally have a high degree of order, or regular 3D geometric periodicity, throughout the entire volume of the material.

Crystallography: The branch of science that deals with the geometric description of crystals and their internal arrangement. The word "crystallography" derives from the Greek words crystallon "cold drop, frozen drop", with its meaning extending to all solids with some degree of transparency, and grapho "I write".http://en.wikipedia.org/wiki/Crystallography

Basic Crystallography

3>

An infinite array of points in space,

Each point has identical surroundings to all others.

Arrays are arranged exactly in a periodic manner.

α

a

bCB ED

O A

y

x

Crystal Lattice:

Basic Crystallography

4>

Crystal Lattice: can be obtained by attaching atoms, groups of atoms or molecules which are called motif tothe lattice sides of the crystal lattice (points).

Crystal Structure = Crystal Lattice + Motif

2D Bravais Lattice

5>

Bravais lattice is an infinite array of discrete points with an arrangement and orientation that appears exactly the same, from whichever of the points the array is viewed.

Lattice must be invariant under translation.

All 2D spacemust be filled.

Bravais Latticesin 2D

In three-dimensional space, there are 14 Bravais lattices. These are obtained by combining one of the seven crystalsystems with one of the symmetry operations.

Polyhedral shape: - cell edges (a, b, c) - cell angles (α, β, γ)

3D Unit Cell

6>

A General Crystalline Solid

7>

The unit cell (Bravais Lattice) is the smallest, most symmetrical repeat unit that, when translated in three dimensions, will generate the entire crystal lattice.

http://www.chem.latech.edu/~upali/chem281/notes/C3-metals.htm

A lattice system is generally identified as a set of lattices with the same shape according to the relative lengths of the cell edges (a, b, c) and the angles between them (α, β, γ).

triclinicmonoclinicorthorhombictetragonalrhombohedralhexagonalcubic

The Seven Crystal Systems

8>Symmetry increases

The Seven Crystal Systems

9>

7 Crystal Systems linear and angular distortions

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The Fourteen Bravais Lattices

10>

The lattice types:Primitive (P)Body-Centered (I)Face-Centered (F)Base-Centered (A, B, or C) Rhombohedral (R)

PPP

F

PPRP

CC I

II

F

The Fourteen Bravais Lattices

11>

Body-Centered Cubic (BCC) elements

12>BCC elements: 16/90 (~18%)

Face-Centered Cubic (FCC) elements

13>FCC elements: 21/90 (~23%)

Hexagonal Close-Packing (HCP) elements

14>HCP elements: 31/90 (~34%)

Metallic Crystal Structures

15>

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 directional.Nearest neighbor distances tend to be small in

order to lower bond energy.The “electron cloud” shields cores from each

other.They have the simplest crystal structures.

Metallic Crystal Structures

16>

Cubic Cell Units:

Crystal Structures

17>

Coordinatıon Number (CN): The Bravais lattice points closest to a given point are the nearest neighbours.

Because the Bravais lattice is periodic, all points have the same number of nearest neighbours or coordination number. It is a property of the lattice.

A simple cubic has coordination number 6; a body-centered cubic lattice, 8; and a face-centered cubic lattice,12.

Crystal Structures

18>

Atomic Packing Factor (APF): defined as the volume of atoms (hard spheres) within the unit cell divided by the volume of the unit cell.

Simple Cubic Structure (SC)

19>

Rare due to low packing density (only Po has this structure)Close-packed directions are cube edges.

Coordination Number = 6(nearest neighbors)

Simple Cubic Structure (SC)

20>

APF for a simple cubic structure = 0.52atoms

unit cellatom

volume

unit cellvolumeclose-packed directions

SC contains 8 x 1/8 = 1 atom/unit cell

a

R=0.5a 3

3)5.0(341

aAPF

⎟⎠⎞

⎜⎝⎛⋅

=π

Body-Centered Cubic Structure (BCC)

21>

Coordination Number = 8

Atoms touch each other along cube diagonals.ex: Cr, W, Fe (α), Tantalum, Molybdenum

2 atoms/unit cell:1 center + 8 corners x 1/8

Body-Centered Cubic Structure (BCC)

22>

atomsunit cell

atomvolume

unit cellvolume

APF for a body-centered cubic structure = 0.68

aR

a

length = 4R =Close-packed directions:

3 a

a2

a3

3

3

43

342

a

a

APF⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅⋅

=

π

Face-Centered Cubic Structure (FCC)

23>


Atoms touch each other along face diagonals.ex: Al, Cu, Au, Pb, Ni, Pt, Ag

4 atoms/unit cell:6 face x 1/2 + 8 corners x 1/8

a

2 a

Face-Centered Cubic Structure (FCC)

24>

atomsunit cell

atomvolume

unit cellvolume

APF for a face-centered cubic structure = 0.74

3

3

42

344

a

a

APF⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅⋅

=

π

FCC Close-packed directions: length = 4R = 2 a

FCC Unit cell contains:6 x 1/2 + 8 x 1/8

= 4 atoms/unit cell

Hexagonal Close-Packed Structure (HCP)

25>

HCP formed by compact planes (ABAB...)


ex: Ti, Co, Zr, Zn

2 atoms/unit cell:1 body + 1/2 basal plane x 2

Hexagonal Close-Packed Structure (HCP)

26>

APF for a hexagonal close-packed structure = 0.74In the HCP structure:

r is the atomic radius of the atom.

The volume of the HCP cell:

rc

rba

⋅=

⋅==

324

2

328 rV ⋅=

7405.028342

342

3

33

=×

=×

=r

r

V

rAFP

ππ

FCC – HCP Stacking Sequence

27>

HCP Stacking Sequence: ABABABABA....


28>

FCC Stacking Sequence: ABCABCABCA....


29>

HCP Stacking Sequence: ABABABA....FCC Stacking Sequence: ABCABCA....

https://www.youtube.com/watch?v=tgKC-awk6p4

https://www.youtube.com/watch?v=xyjW59-CYqk

https://www.youtube.com/watch?v=pdFqpDilLwY

Determine the density of BCC iron, which has a lattice parameter of 0.2866 nm.

SOLUTION:

Atoms/cell = 2, a0 = 0.2866 nm = 2.866 × 10-8 cm

Atomic mass = 55.847 g/mol

Volume of unit cell = = (2.866 × 10-8 cm)3 = 23.54 × 10-24

cm3/cell

Avogadro’s number NA = 6.02 × 1023 atoms/mol

30a

Calculation of Theoretical Density (ρ)

30>

32324 /882.7

)1002.6)(1054.23()847.55)(2(

number) sadro'cell)(Avogunit of (volumeiron) of mass )(atomicatoms/cell of(number Density

cmg=××

=

=

−ρ

ρ

Allotropic or Polymorphic Transformations

31>

Allotropy: The characteristic of an element being able to exist in more than one crystal structure, depending on temperature and pressure.

Polymorphism: Compounds exhibiting more than one type of crystal structure.

Titanium Iron


32>


33>

Iron is BCC at 911oC with a lattice parameter of 0.2863 nm. At 913oC, iron is FCC, with a lattice parameter of 0.3591 nm. Determine the % change in volume.

SOLUTION: The volume of a unit cell of BCC iron before transforming is:VBCC = (0.2863 nm)3 = 0.023467 nm3

The volume of the unit cell in FCC iron is:VFCC = (0.3591 nm)3 = 0.046307 nm3

But this is the volume occupied by four iron atoms, as there are four atoms per FCC unit cell. Therefore, we must compare two BCC cells (with a volume of 2 x (0.023467) = 0.046934 nm3) with each FCC cell. The percent volume change during transformation is:

%34.11000.046934

0.046934) - (0.046307 change Volume −=×=

Fe (BCC) Fe (FCC)

Interstitial Voids in FCC Cells

34>

TETRAHEDRAL OCTAHEDRAL

Note: Atoms are coloured differently but are the same

cellntetrahedro VV241

=celloctahedron VV

61

=

¼ way along body diagonal{¼, ¼, ¼}, {¾, ¾, ¾}

+ face centering translations

At body centre{½, ½, ½}

+ face centering translationsFCC


35>

FCC- OCTAHEDRAL

{½, ½, ½} + {½, ½, 0} = {1, 1, ½} ≡ {0, 0, ½}

Face centering translation


Equivalent site for an octahedral void

Site for octahedral void

Interstitial Voids Positions in FCC Cells

36>

Voids / atomVoids / cellPositionFCC voids

14• Body centre: 1 → (½, ½, ½)• Edge centre: (12/4 = 3) → (½, 0, 0)

Octahedral

28¼ way from each vertex of the cube

along body diagonal <111>→ ((¼, ¼, ¼))

Tetrahedral


37>

Radius of the new atom

Size of the largest atom which can fit into the tetrahedral void of FCC

CV = r + xe

xre +=46

225.0~123 2 ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=⇒=

rxre

Size of the largest atom which can fit into the Octahedral void of FCC

2r + 2x = a ra 42 =

( ) 414.0~12 −=rx

** Actually an atom of correct size touches only the top and bottom atoms

Interstitial Voids in BCC Cells

38>

Distorted TETRAHEDRAL Distorted OCTAHEDRAL**

a

a√3/2

a a√3/2

rvoid / ratom = 0.29 rVoid / ratom = 0.155


Coordinates of the void:{½, 0, ¼} (four on each face)

Coordinates of the void:{½, ½, 0} (+ BCC translations: {0, 0, ½})

Interstitial Voids Positions in BCC Cells

39>36• Face centre: (6/2 = 3) → (½, ½, 0)• Edge centre: (12/4 = 3) → (½, 0, 0)

DistortedOctahedral

612• Four on each face: [(4/2) × 6 = 12] → (0, ½, ¼)DistortedTetrahedral

Voids/cell Voids/atom

From the right angled triange OCM: 416

22 aaOC +=5

4a r x= = +

For a BCC structure: 3 4a r= (3

4ra = )

xrr+=

34

45 ⇒ 29.01

35

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

rx


40>

a

a√3/2

BCC: Distorted Tetrahedral Void


41>

2axrOB =+=

324rxr =+ raBCC 43: =

1547.013

32=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

rx

Distorted Octahedral Void

a√3/2

a

aaOB 5.02== aaOA 707.

22

==

As the distance OA > OB the atom in the void touches only the atom at B (body centre).⇒ void is actually a ‘linear’ void

This implies:

Relative Sizes of Voids in Cubic Cells

42>

o

A 292.1=FeFCCr

o

A534.0)( =octxFeFCC

o

A 77.0=Cr

C

N

Void (Oct)

FeFCC

O

FCC

Relative sizes of voids w.r.t to atoms

o

A 258.1=FeBCCr

o

A364.0).( =tetdxFeBCC

o

A195.0).( =octdxFeBCC

BCC

FeBCC

( . ) 0.155FeBCC

FeBCC

x d octr

=

( . ) 0.29FeBCC

FeBCC

x d tetr

=

Interstitial Voids in Hexagonal Cells

43>

TETRAHEDRAL OCTAHEDRAL

These voids are identical to the ones found in FCC


Coordinates: (⅓ ⅔,¼), (⅓,⅔,¾)),,(),,,(),,0,0(),,0,0(: 87

31

32

81

31

32

85

83sCoordinate

Interstitial Voids Positions in Hexagonal Cells

44>

Voids / atomVoids / cellPositionHCP voids

12• (⅓ ⅔,¼), (⅓,⅔,¾)Octahedral

24(0,0,3/8), (0,0,5/8), (⅔, ⅓,1/8), (⅔,⅓,7/8)Tetrahedral

45>

Crystallographic Directions

Lattice directions:

Miller Notation: [hkl] – a specific direction<hkl> – a family of directions

46>


210

X = 1 , Y = ½ , Z = 0[1 ½ 0] [2 1 0]

X = ½ , Y = ½ , Z = 1[½ ½ 1] [1 1 2]

Examples of directions [hkl]:

47>


X = -1 , Y = -1 , Z = 0 [110]X = 1 , Y = 0 , Z = 0 [1 0 0]


48>


X =-1 , Y = 1 , Z = -1/6[-1 1 -1/6] [6 6 1]

We can move vector to the origin.


49>

Crystallographic Planes

Miller notation: (hkl) – a specific plane{hkl} – a family of planes

Miller Indices are a symbolic vector representation for the orientation of an atomic plane in a crystal lattice and are defined as the reciprocals of the fractional intercepts which the plane makes with the crystallographic axes.

To determine Miller indices of a plane, take the following steps:1) Determine the intercepts of the plane along each of the three crystallographic directions;2) Take the reciprocals of the intercepts3) If fractions result, multiply each by the denominator of the smallest fraction

50>


Axis X Y Z

Intercept points 1 ∞ ∞

Reciprocals 1/1 1/ ∞ 1/ ∞Smallest

Ratio 1 0 0

Miller İndices (100)(1,0,0)

51>


Axis X Y Z

Intercept points 1 1 ∞

Reciprocals 1/1 1/ 1 1/ ∞Smallest

Ratio 1 1 0


(0,1,0)

52>


Axis X Y Z

Intercept points 1 1 1

Reciprocals 1/1 1/ 1 1/ 1Smallest

Ratio 1 1 1


(0,1,0)

(0,0,1)

53>


Axis X Y Z

Intercept points 1/2 1 ∞

Reciprocals 1/(½) 1/ 1 1/ ∞Smallest

Ratio 2 1 0

Miller İndices (210)(1/2, 0, 0)

(0,1,0)

54>


Reciprocal numbers are: 21 ,

21 ,

31

Plane intercepts axes at cba 2 ,2 ,3

Indices of the plane (Miller): (2,3,3)

(100)

(200)(110)

(111) (100)

Indices of the direction: [2,3,3]a3

2

2

bc

[2,3,3]

55>

Family of Crystallographic Planes & Directions

)111(),111(),111(),111(),111(),111(),111(),111(}111{)001(),100(),010(),001(),010(),100(}100{

≡

≡

Directions:

Planes:

56>

Atomic Density

a a2 a2

a2a2

aa

a2a2a

(100) (110) (111)

SC

FCC

BCC

ATOMIC DENSITY (atoms/unit area)

(111) < (110) < (100)

(110) < (100) < (111)

(111) < (100) < (110)

57>

Directions in HCP Crystal

Miller notation: [hk.l] or [hkil] Where i = h + k

1. Vector repositioned (if necessary) to pass through origin.2. Read off projections in terms of unit

cell dimensions a1, a2, a3, or c3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvtw]

[ 1120 ]ex: ½, ½, -1, 0 =>

Adapted from Fig. 3.8(a), Callister & Rethwisch 8e.

dashed red lines indicate projections onto a1 and a2 axes a1

a2

a3

-a32

a2

2a1

-a3

a1

a2

zAlgorithm

58>

Crystallographic Planes in HCP Crystal

example a1 a2 a3 c

4. Miller-Bravais Indices (1011)

1. Intercepts 1 ∞ -1 12. Reciprocals 1 1/∞

1 0 -1-1

11

3. Reduction 1 0 -1 1

a2

a3

a1

z

Miller notation: (hk.l) or (hkil) Where i = h + k

59>

Another Important Crystal Structures

Sodium Chloride Structure Na+Cl-

Cesium Chloride Structure Cs+Cl-

Diamond Structure

Zinc Blende (ZnS)

60>

Sodium Chloride Structure (NaCl)

61>

Sodium Chloride Structure (NaCl)

Many common salts have FCC arrangements of anions with cations in OCTAHEDRAL HOLES

Ion Count for the Unit Cell: 4 Na+ and 4 Cl-

Na4Cl4 = NaCl

Similar structures: LiF, NaBr ,KCl, MgO, FeO, etc…

62>

Cesium Chloride Structure (CsCl)

939.0181.0170.0

Cl

Cs ==−

+

r

r

∴ Since 0.732 < 0.939 < 1.0, cubic sites preferred

So each Cs+ has 8 neighbor Cl-

Cesium chloride consists of equal numbers of cesium and chlorine ions, placed at the points of a BCC cubic lattice so that each ion has eight of the other kind as its nearest neighbors.

63>

Diamond Structure (C)

The diamond lattice is consist of two interpenetrating face centered bravais lattices.There are 8 atoms in the structure of diamond.Each atom bonds covalently to 4 others equally spread about atom in 3D.

graphite diamond

142pm

335pm

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

64>

Calcium Fluorite Structure (CaF2) • Cations in cubic sites• UO2, ThO2, ZrO2, CeO2

• Antifluorite structure –positions of cations and anions reversed

65>

Zinc Blende Structure (ZnS)

Zinc Blende is the name given to the mineral ZnS. It has a cubic close packed (face centred) array of S and the Zn(II) sit in tetrahedral (1/2 occupied) sites in the lattice.

66>

Perovskite Structure

ABX3 Crystal Structures

• Barium Titanate,a complex oxide

BaTiO3

67>

X-Rays to Determine Crystal Structure

Bragg equation: nλ=2d.sinθ

68>

X-Rays to Determine Crystal Structure

(110)

(200)

(211)

z

x

ya b

c

Diffraction angle 2θ

Diffraction pattern for polycrystalline α-iron (BCC)

Inte

nsity

(rel

ativ

e)

z

x

ya b

c

z

x

ya b

c

Bragg equation: nλ=2d.sinθ

CALLISTER JR, W. D. AND RETHWISCH, D. G. Materials Science and Engineering: An Introduction, 9th edition.

John Wiley & Sons, Inc. 2014, 988p. ISBN: 978-1-118-32457-8.HOSFORD, W. F.

Elementary Materials Science. ASM International. 2013, 180p. ISBN 978-1-62708-002-6.ASHBY, M. and JONES, D. R. H.

Engineering Materials 1: An Introduction to Properties, Applications and Design. 4th Edition. Elsevier Ltd. 2012, 472p. ISBN 978-0-08-096665-6.

CALLISTER JR, W. D. AND RETHWISCH, D. G. Fundamentals of Materials Science and Engineering: An Integrated Approach, 4th ed.

John Wiley & Sons, Inc. 2012, 910p. ISBN 978-1-118-06160-2.MITTEMEIJER, E. J.

Fundamentals of Materials Science: The Microstructure–Property Relationship Using Metals as Model Systems. Springer-Verlag Berlin. 2010, 594p. ISBN 978-3-642-10499-2.

ASKELAND, D. AND FULAY, P. Essentials of Materials Science & Engineering, 2nd Edition.

Cengage Learning. 2009, 604p. ISBN 978-0-495-24446-2.ABBASCHIAN, R., ABBASCHIAN, L., AND REED-HILL, R. E.

Physical Metallurgy Principles, 4th Ed. Cengage. 2009, 750p. ISBN 978-0-495-08254-5.SMALLMAN, R. E. and NGAN, A.H.W.

Physical Metallurgy and Advanced Materials, 7th Edition. Elsevier Ltd. 2007, 650p. ISBN 978-0-7506-6906-1.

References

69Nota de aula preparada pelo Prof. Juno Gallego para a disciplina Ciência dos Materiais de Engenharia.® 2015. Permitida a impressão e divulgação. http://www.feis.unesp.br/#!/departamentos/engenharia-mecanica/grupos/maprotec/educacional/

Documents

Packing of Atoms in Solids [6]...new atom Size of the largest atom which can fit into the tetrahedral void of FCC CV = r + x e e =r +x 4 6 1 ~ 0.225 2 3 2 ⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝