chapter 3 - chapter 3: structures of metals & ceramics x-ray diffraction photograph x-ray...

$: Chapter 3 - Chapter 3: Structures of Metals & Ceramics X-ray diffraction photograph X-ray diffraction in metals$
Chapter 3 -

Chapter 3: Structures of Metals & Ceramics

X-ray diffraction photograph X-ray diffraction in metals

Chapter 3 - 2

3.1 ISSUES TO ADDRESS...

• How do atoms assemble into solid structures?

• How does the density of a material depend on its structure?

• When do material properties vary with the sample (i.e., part) orientation?

Chapter 3: Structures of Metals & Ceramics

• How do the crystal structures of ceramic materials differ from those for metals?

Chapter 3 - 3

• Non dense, random packing

• Dense, ordered packing

Dense, ordered packed structures tend to have lower energies.

CRYSTAL STRUCTURES3.2 Energy and Packing

Energy

r

typical neighbor bond length

typical neighbor bond energy

Energy

r

typical neighbor bond length

typical neighbor bond energy

Chapter 3 - 4

• atoms pack in periodic, 3D arraysCrystalline materials...

-metals-many ceramics-some polymers

• atoms have no periodic packingNoncrystalline materials...

-complex structures-rapid cooling

crystalline SiO2

noncrystalline SiO2"Amorphous" = NoncrystallineAdapted from Fig. 3.40(b), Callister & Rethwisch 3e.

Adapted from Fig. 3.40(a), Callister & Rethwisch 3e.

Materials and Packing

Si Oxygen

• typical of:

• occurs for:

Chapter 3 - 5

3.4 Metallic Crystal Structures

• How can we stack metal atoms to minimize empty space?

2-dimensions

vs.

Now stack these 2-D layers to make 3-D structures

Chapter 3 - 6

• 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.- Electron cloud shields cores from each other

• Have the simplest crystal structures.

We will examine three such structures...

3.4 Metallic Crystal Structures

Chapter 3 -

FCC = face-centered cubicHCP = hexagonal close-packedBCC = body-centered cubic1 nm = 10-9 m = 10 Ǻ

Chapter 3 - 8

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

• Coordination # = 6 (# nearest neighbors)

(Courtesy P.M. Anderson)

Simple Cubic Structure (SC)

Chapter 3 - 9

• APF for a simple cubic structure = 0.52

APF = a3

4

3(0.5a) 31

atoms

unit cellatom

volume

unit cell

volume

Atomic Packing Factor (APF)

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

Adapted from Fig. 3.42, Callister & Rethwisch 3e.

close-packed directions

a

R=0.5a

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

Chapter 3 -

The body-centered cubic (BCC) crystal structure

Hard sphere unit cell representation

Reduced-sphere unit cell representation

Aggregate of many atoms

Chapter 3 - 11

• Coordination # = 8



• Atoms touch each other along cube diagonals.--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.

Body Centered Cubic Structure (BCC)

ex: Cr, W, Fe (), Tantalum, Molybdenum

QuickTime™ and aCinepak decompressor

are needed to see this picture.

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

Chapter 3 - 12

Atomic Packing Factor: BCC

a

APF =

4

3 ( 3a/4)32

atoms

unit cell atom

volume

a3unit cell

volume

length = 4R =Close-packed directions:

3 a

• APF for a body-centered cubic structure = 0.68

aRAdapted from

Fig. 3.2(a), Callister & Rethwisch 3e.

a 2

a 3

Chapter 3 - 13




• Atoms touch each other along face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.

Face Centered Cubic Structure (FCC)

ex: Al, Cu, Au, Pb, Ni, Pt, Ag

QuickTime™ and aCinepak decompressor

are needed to see this picture.

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

Chapter 3 - 14

• APF for a face-centered cubic structure = 0.74Atomic Packing Factor: FCC

maximum achievable APF

APF =

4

3( 2a/4)34

atoms

unit cell atom

volume

a3unit cell

volume

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

Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

a

2 a

Adapted fromFig. 3.1(a),Callister & Rethwisch 3e.

Chapter 3 - 15

A sites

B B

B

BB

B B

C sites

C C

CA

B

B sites

• ABCABC... Stacking Sequence• 2D Projection

• FCC Unit Cell

FCC Stacking Sequence

B B

B

BB

B B

B sitesC C

CA

C C

CA

AB

C

Chapter 3 -

16


• ABAB... Stacking Sequence

• APF = 0.74

• 3D Projection • 2D Projection


6 atoms/unit cell

ex: Cd, Mg, Ti, Zn

• c/a = 1.633

c

a

A sites

B sites

A sites Bottom layer

Middle layer

Top layer

The Hexagonal Closed-Packed Crystal Structure (HCP)

Chapter 3 - 17

3.5 Theoretical Density,

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

Density = =

VC NA

n A =

Cell Unit of VolumeTotal

Cell Unit in Atomsof Mass

Chapter 3 - 18

• Ex: Cr (BCC)

A = 52.00 g/mol

R = 0.125 nm

n = 2 atoms/unit cell

theoretical

a = 4R/ 3 = 0.2887 nm

actual

aR

= a3

52.002

atoms

unit cellmol

g

unit cell

volume atoms

mol

6.022 x 1023

Theoretical Density,

= 7.18 g/cm3

= 7.19 g/cm3


Chapter 3 - 19

3.6 Ceramic Crystal Structures

Oxide structures– oxygen anions larger than metal cations– close packed oxygen in a lattice (usually FCC)– cations fit into interstitial sites among oxygen ions

Chapter 3 -

c03tf02

Chapter 3 - 21

• Bonding: -- Can be ionic and/or covalent in character. -- % ionic character increases with difference in electronegativity of atoms.

Adapted from Fig. 2.7, Callister & Rethwisch 3e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 byCornell University.

• Degree of ionic character may be large or small:

Atomic Bonding in Ceramics

SiC: small

CaF2: large

Chapter 3 - 22

Factors that Determine Crystal Structure1. Relative sizes of ions – Formation of stable structures: --maximize the # of oppositely charged ion neighbors.


- -

- -+

unstable

- -

- -+

stable

- -

- -+

stable

2. Maintenance of Charge Neutrality : --Net charge in ceramic should be zero. --Reflected in chemical formula:

CaF2: Ca2+cation

F-

F-

anions+

AmXp

m, p values to achieve charge neutrality

Chapter 3 -

c03tf03

Chapter 3 -

c03prob

Example 3.4: Computation of minimum cation-to-anion radius ration for a coordination number of 3

Chapter 3 - 25

• Coordination # increases with

Coordination # and Ionic Radii

Adapted from Table 3.3, Callister & Rethwisch 3e.

2

rcationranion

Coord #

< 0.155

0.155 - 0.225

0.225 - 0.414

0.414 - 0.732

0.732 - 1.0

3

4

6

8

linear

triangular

tetrahedral

octahedral

cubic




ZnS (zinc blende)

NaCl(sodium chloride)

CsCl(cesium chloride)

rcationranion

To form a stable structure, how many anions can surround around a cation?

Chapter 3 -

c03tf04

Chapter 3 - 27

Computation of Minimum Cation-Anion Radius Ratio

• Determine minimum rcation/ranion for an octahedral site (C.N. = 6)

a 2ranion

2ranion 2rcation 2 2ranion

ranion rcation 2ranion

rcation ( 2 1)ranion

arr 222 cationanion

414.012anion

cation r

r

Chapter 3 - 28

Bond Hybridization

Bond Hybridization is possible when there is significant covalent bonding– hybrid electron orbitals form– For example for SiC

• XSi = 1.8 and XC = 2.5

%.)XXionic% 511]}exp[-0.25(-{1 100 character 2CSi

• ~ 89% covalent bonding• Both Si and C prefer sp3 hybridization• Therefore, for SiC, Si atoms occupy tetrahedral sites

Chapter 3 - 29

• On the basis of ionic radii, what crystal structure would you predict for FeO?

• Answer:

5500

1400

0770

anion

cation

.

.

.

r

r

based on this ratio,-- coord # = 6 because

0.414 < 0.550 < 0.732

-- crystal structure is NaCl

Data from Table 3.4, Callister & Rethwisch 3e.

Example Problem: Predicting the Crystal Structure of FeO

Ionic radius (nm)

0.053

0.077

0.069

0.100

0.140

0.181

0.133

Cation

Anion

Al3+

Fe2+

Fe3+

Ca2+

O2-

Cl-

F-

Chapter 3 - 30

Rock Salt StructureSame concepts can be applied to ionic solids in general. Example: NaCl (rock salt) structure

rNa = 0.102 nm

rNa/rCl = 0.564

cations (Na+) prefer octahedral sites


rCl = 0.181 nm

Chapter 3 - 31

MgO and FeO

O2- rO = 0.140 nm

Mg2+ rMg = 0.072 nm

rMg/rO = 0.514

cations prefer octahedral sites

So each Mg2+ (or Fe2+) has 6 neighbor oxygen atoms


MgO and FeO also have the NaCl structure

Chapter 3 - 32

AX Crystal Structures

939.0181.0

170.0

Cl

Cs

r

r


Cesium Chloride structure:

Since 0.732 < 0.939 < 1.0, cubic sites preferred

So each Cs+ has 8 neighbor Cl-

AX–Type Crystal Structures include NaCl, CsCl, and zinc blende

Chapter 3 -

c03f07

Zinc blende or sphalerite or zinc sulfide; highly covalent

AX typeCN = 4All ions tetrahedrally coordinated

Zn ↔ S

Chapter 3 - 34

AX2 Crystal Structures

• Calcium Fluorite (CaF2)

• Cations in cubic sites

• UO2, ThO2, ZrO2, CeO2

• Antifluorite structure –

positions of cations and anions reversed


Fluorite structure

Chapter 3 - 35

ABX3 Crystal Structures


• Perovskite structure

Ex: complex oxide

BaTiO3

Chapter 3 -

c03tf05

Chapter 3 - 37

3.7 Density Computations for Ceramics

A

AC )(

NV

AAn

C

Number of formula units/unit cell

Volume of unit cell

Avogadro’s number

= sum of atomic weights of all anions in formula unit

AA

AC = sum of atomic weights of all cations in formula unit

Chapter 3 -

c03pro

b

Theoretical Density Calculations for Sodium Chloride

Chapter 3 - 39

Densities of Material Classesmetals > ceramics > polymers

Why?

Data from Table B.1, Callister & Rethwisch, 3e.

(g

/cm

)3

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers

Polymers

1

2

20

30Based on data in Table B1, Callister

*GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers

in an epoxy matrix). 10

3

4 5

0.3

0.4 0.5

Magnesium

Aluminum

Steels

Titanium

Cu,Ni

Tin, Zinc

Silver, Mo

Tantalum Gold, W Platinum

Graphite

Silicon

Glass -soda Concrete

Si nitride Diamond Al oxide

Zirconia

HDPE, PS PP, LDPE

PC

PTFE

PET PVC Silicone

Wood

AFRE*

CFRE*

GFRE*

Glass fibers

Carbon fibers

Aramid fibers

Metals have... • close-packing (metallic bonding) • often large atomic masses Ceramics have... • less dense packing • often lighter elements Polymers have... • low packing density (often amorphous) • lighter elements (C,H,O)

Composites have... • intermediate values

In general

Chapter 3 -

c03f10

3.8 Silicate CeramicsSiO4-

4

Chapter 3 - 41

Silicate CeramicsMost common elements on earth are Si & O

• SiO2 (silica) polymorphic forms are quartz, crystobalite, & tridymite

• The strong Si-O bonds lead to a high melting temperature (1710ºC) for this material

Si4+

O2-

Adapted from Figs. 3.10-11, Callister & Rethwisch 3e crystobalite

Chapter 3 -

c03f12

Silicate ion structures formed from Si tetrahedra

Chapter 3 - 43

Bonding of adjacent SiO44- accomplished by the sharing

of common corners, edges, or faces

Silicates

Mg2SiO4 Ca2MgSi2O7


Presence of cations such as Ca2+, Mg2+, & Al3+ 1. maintain charge neutrality, and 2. ionically bond SiO4

4- to one another

Chapter 3 - 44

• Quartz is crystalline SiO2:

• Basic Unit: Glass is noncrystalline (amorphous)• Fused silica is SiO2 to which no impurities have been added • Other common glasses contain impurity ions such as Na+, Ca2+, Al3+, and B3+

(soda glass)


Glass Structure

Si04 tetrahedron4-

Si4+

O2-

Si4+

Na+

O2-

Chapter 3 - 45

Layered Silicates• Layered silicates (e.g., clays, mica,

talc)– SiO4 tetrahedra connected

together to form 2-D plane

• A net negative charge is associated with each (Si2O5)2- unit

• Negative charge balanced by adjacent plane rich in positively charged cations


Silicate sheet structure of (Si2O5)2-

Chapter 3 - 46

• Kaolinite clay alternates (Si2O5)2- layer with Al2(OH)42+

layer

Layered Silicates (cont)

Note: Adjacent sheets of this type are loosely bound to one another by van der Waal’s forces.


Kaolinite ClayTwo layer silicate sheet structure: Al2(Si2O5)(OH)4

15,000x

Electron micrograph of kaolinate crystals

Chapter 3 - 47

3.9 Polymorphic Forms of Carbon

Diamond– tetrahedral bonding of

carbon• hardest material known• very high thermal

conductivity – large single crystals –

gem stones– small crystals – used to

grind/cut other materials – diamond thin films

• hard surface coatings – used for cutting tools, medical devices, etc.


Chapter 3 - 48

Polymorphic Forms of Carbon (cont)

Graphite– layered structure – parallel hexagonal arrays of

carbon atoms

– weak van der Waal’s forces between layers– planes slide easily over one another -- good

lubricant


Chapter 3 - 49

Polymorphic Forms of Carbon (cont) Fullerenes and Nanotubes

• Fullerenes – spherical cluster of 60 carbon atoms, C60

– Like a soccer ball • Carbon nanotubes – sheet of graphite rolled into a tube

– Ends capped with fullerene hemispheres

Adapted from Figs. 3.18 & 3.19, Callister & Rethwisch 3e.

Chapter 3 -

c03impf01

STM

Chapter 3 - 51

• Some engineering applications require single crystals:

• Properties of crystalline materials often related to crystal structure.


-- Ex: Quartz fractures more easily

along some crystal planes than others.

-- diamond single crystals for abrasives

-- turbine blades

Fig. 9.40(c), Callister & Rethwisch 3e. (Fig. 9.40(c) courtesy of Pratt and Whitney).

(Courtesy Martin Deakins,GE Superabrasives, Worthington, OH. Used with permission.)

3.11 Crystals as Building Blocks

Chapter 3 -

c03tf06

Chapter 3 - 53

• Most engineering materials are polycrystals.

• Nb-Hf-W plate with an electron beam weld.• Each "grain" is a single crystal.• If grains are randomly oriented, overall component properties are not directional.• Grain sizes typ. range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers).

Adapted from Fig. K, color inset pages of Callister 5e.(Fig. K is courtesy of Paul E. Danielson, Teledyne Wah Chang Albany)

1 mm

Polycrystals

Isotropic

Anisotropic

Chapter 3 - 54

• Single Crystals-Properties vary with direction: anisotropic.

-Example: the modulus of elasticity (E) in BCC iron:

Data from Table 3.7, Callister & Rethwisch 3e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.)

• Polycrystals

-Properties may/may not vary with direction.-If grains are randomly oriented: isotropic. (Epoly iron = 210 GPa)-If grains are textured, anisotropic.

200 m Adapted from Fig. 5.19(b), Callister & Rethwisch 3e.(Fig. 5.19(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC [now the National Institute of Standards and Technology, Gaithersburg, MD].)

Single vs PolycrystalsE (diagonal) = 273 GPa

E (edge) = 125 GPa

Chapter 3 - 55

Polymorphism

• Two or more distinct crystal structures for the same material (allotropy/polymorphism) titanium

, -Ti

carbon

diamond, graphite

BCC

FCC

BCC

1538ºC

1394ºC

912ºC

-Fe

-Fe

-Fe

liquid

iron system

Chapter 3 -

c03impf02a

Allotropic transformation of tin

Chapter 3 -

c03impf02b

white tin

gray tin

Cooled below 13.2 °C for an extended period of time

Chapter 3 - 58

Fig. 3.20, Callister & Rethwisch 3e.

3.11 Crystal Systems

7 crystal systems

14 crystal lattices

Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal.

a, b, and c are the lattice constants

Chapter 3 - 59

Point CoordinatesPoint coordinates for unit cell

center are

a/2, b/2, c/2 ½ ½ ½

Point coordinates for unit cell corner are 111

Translation: integer multiple of lattice constants identical position in another unit cell

z

x

ya b

c

000

111

y

z

2c

b

b

Chapter 3 -

c03f21

Chapter 3 - 61

3.13 Crystallographic Directions

1. Vector repositioned to pass through origin2. Read off projections in terms of unit cell

dimensions a, b, and c3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]

-1, 1, 1

families of directions <uvw>

z

x

Algorithm

where overbar represents a negative index

[ 111 ]=>

y

Chapter 3 -

c03f22

Chapter 3 -

c03f23

Chapter 3 - 64

HCP Crystallographic Directions

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 =>


dashed red lines indicate projections onto a1 and a2 axes a1

a2

a3

-a3

2

a2

2

a1

-a3

a1

a2

z

Algorithm

Chapter 3 -

c03f24

Chapter 3 - 66

HCP Crystallographic Directions• Hexagonal Crystals

– 4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v'w') as follows.

'ww

t

v

u

)vu( +-

)'u'v2(3

1-

)'v'u2(3

1-

]uvtw[]'w'v'u[

Fig. 3.24(a), Callister & Rethwisch 3e.

-a3

a1

a2

z

Chapter 3 - 67

3.14 Crystallographic Planes


Chapter 3 - 68

Crystallographic Planes• Miller Indices: Reciprocals of the (three) axial

intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices.

• Algorithm 1. Read off intercepts of plane with axes in terms of a, b, c2. Take reciprocals of intercepts3. Reduce to smallest integer values4. Enclose in parentheses, no commas i.e., (hkl)

Chapter 3 - 69

Crystallographic Planesz

x

ya b

c

4. Miller Indices (110)

example a b cz

x

ya b

c


1. Intercepts 1 1 2. Reciprocals 1/1 1/1 1/

1 1 03. Reduction 1 1 0

1. Intercepts 1/2 2. Reciprocals 1/½ 1/ 1/

2 0 03. Reduction 2 0 0

example a b c

Chapter 3 - 70

Crystallographic Planes

z

x

ya b

c


example1. Intercepts 1/2 1 3/4

a b c

2. Reciprocals 1/½ 1/1 1/¾2 1 4/3

3. Reduction 6 3 4

(001)(010),

Family of Planes {hkl}

(100), (010),(001),Ex: {100} = (100),

Chapter 3 - 71

Crystallographic Planes (HCP)

• In hexagonal unit cells the same idea is used

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

Adapted from Fig. 3.24(b), Callister & Rethwisch 3e.

Chapter 3 - 72

Crystallographic Planes

• Atomic packing of crystallographic planes• The atomic packing of the exposed planes

is important in catalysis

• Preparing for the exam:

a) Draw (100) and (111) crystallographic planes

for Fe.

b) Calculate the planar density for each of these planes.

Chapter 3 -

c03f26

Atomic Arrangements in the (100) plane of a FCC

Chapter 3 -

c03f27

Atomic Arrangements in the (100) plane of a BCC

Chapter 3 - 75

ex: linear density of Al in [110] direction

a = 0.405 nm

Linear Density

• Linear Density of Atoms LD =

a

[110]

Unit length of direction vector

Number of atoms

# atoms

length

13.5 nma2

2LD

Chapter 3 - 76

Planar Density of (100) IronSolution: At T < 912C iron has the BCC structure.

(100)

Radius of iron R = 0.1241 nm

R3

34a

Adapted from Fig. 3.2(c), Callister & Rethwisch 3e.

2D repeat unit

= Planar Density = a2

1

atoms

2D repeat unit

= nm2

atoms12.1

m2

atoms= 1.2 x 1019

12

R3

34area

2D repeat unit

Chapter 3 -

c03f28

Chapter 3 - 78

Planar Density of (111) IronSolution (cont): (111) plane 1 atom in plane/ unit surface cell

333 2

2

R3

16R

34

2a3ah2area

atoms in plane

atoms above plane

atoms below plane

ah2

3

a 2

2D re

peat

uni

t

1

= = nm2

atoms7.0m2

atoms0.70 x 1019

3 2R3

16Planar Density =

atoms

2D repeat unit

area

2D repeat unit

Chapter 3 -

c03f29

3.16 Closed-packed crystal structures

3.16 Closed-Packed Crystal Structures

Chapter 3 -

c03f32

Ceramics

Chapter 3 -

c03f33

(111) PLANES OF ROCK SALT

Chapter 3 -c03f34

Crystalline and non crystalline materials3.17 Single Crystals

A garnet single crystalTongbei, Fujian, China

Chapter 3 -

c03f35

3.18 Solidification of a polycrystalline material

Chapter 3 -

3.19 Anisotropy

Chapter 3 -

c03f40

Noncrystalline solids

Chapter 3 - 88

3.20 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.

Chapter 3 -

c03f36

Chapter 3 -

c03f37

Chapter 3 - 91

X-Rays to Determine Crystal Structure

X-ray intensity (from detector)

c

d n

2 sin c

Measurement of critical angle, c, allows computation of planar spacing, d.

• Incoming X-rays diffract from crystal planes.


reflections must be in phase for a detectable signal

spacing between planes

d

incoming

X-rays

outg

oing

X-ra

ys

detector

extra distance travelled by wave “2”

“1”

“2”

“1”

“2”

Chapter 3 -

c03f38

Chapter 3 - 93

X-Ray Diffraction Pattern

Adapted from Fig. 3.20, Callister 5e.

(110)

(200)

(211)

z

x

ya b

c

Diffraction angle 2

Diffraction pattern for polycrystalline -iron (BCC)

Inte

nsity

(re

lativ

e)

z

x

ya b

cz

x

ya b

c

Chapter 3 -

Some hypothetical metal has this simple cubic crystal structure. If its atomic weight is 74.5 g/mol and the atomic radius is 0.145 nm, compute its density.

For the simple cubic crystal structure, the value of n in Equation 3.5 is unity since there is only a single atom associated with each unit cell.

Furthermore, for the unit cell edge length, a = 2R (Figure 3.42).

Therefore, employment of Equation 3.5 yields

= nA

VC N A=

nA

(2 R)3 N A

= (1 atom/unit cell)(74.5 g/mol)

(2)(1.45 10-8

cm)

3

/(unit cell)

(6.02 10

23 atoms/mol)

= 5.07 g/cm3

Problem

Chapter 3 -

problemFirst five peaks, x-ray diffraction, tungsten, BCC monochromatic radiation; wavelength = 0.1542 nm(a) Index (h, k, and l indices) for each of these peaks.(b) Interplanar spacing for each of the peaks.(c) For each peak, determine the atomic radius for W and compare these with the value presented in Table 3.1.

(a) Since W has a BCC crystal structure, only those peaks for which h + k + l are even will appear. Therefore, these five peaks result by diffraction from the following planes: (110), (200), (211), (220), and (310).

Chapter 3 -

First five peaks, x-ray diffraction, tungsten, BCC, monochromatic radiation; wavelength = 0.1542 nm

(b) Interplanar spacing for each of the peaks.

d110 =n

2 sin =

(1)(0.1542 nm)

(2) sin40.2

2

= 0.2244 nm

(b) For each peak, in order to calculate the interplanar spacing we must employ Equation 3.14. For the first peak, which occurs at 40.2°

Chapter 3 -

R110 =(0.2244 nm)( 3) (1)2 + (1)2 + (0)2

4 R110 = 0.1374 nm

(c) Employment of Equations 3.15 and 3.3 is necessary for the computation of R as {a=4R/sqrt(3) and dhkl = a/sqrt(h2+k2+l2)}

R =a 3

4=

(dhkl)( 3) (h)2 + (k)2 + (l)2

4

Similar computations are made for the other peaks:Peak Index 2 dhkl(nm) R (nm)

200 58.4 0.1580 0.1369211 73.3 0.1292 0.1370220 87.0 0.1120 0.1371310 100.7 0.1001 0.1371

Experimental R = 0.1371 nm

Chapter 3 - 98

• Atoms may assemble into crystalline or amorphous 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).

SUMMARY

• Common metallic crystal structures are FCC, BCC, and HCP. Coordination number and atomic packing factor are the same for both FCC and HCP crystal structures.

• Crystallographic points, directions and planes are specified in terms of indexing schemes. Crystallographic directions and planes are related to atomic linear densities and planar densities.

• Ceramic crystal structures are based on: -- maintaining charge neutrality -- cation-anion radii ratios.

• Interatomic bonding in ceramics is ionic and/or covalent.

Chapter 3 - 99

• Some materials can have more than one crystal structure. This is referred to as polymorphism (or allotropy).

SUMMARY

• Materials can be single crystals or polycrystalline. Material properties generally vary with single crystal orientation (i.e., they are anisotropic), but are generally non-directional (i.e., they are isotropic) in polycrystals with randomly oriented grains.

• X-ray diffraction is used for crystal structure and interplanar spacing determinations.

chapter 3 - chapter 3: structures of metals & ceramics x-ray diffraction photograph x-ray...

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