05_mte 271_packing densities (3)

7/29/2019 05_MTE 271_Packing Densities (3)

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Chapter3:PackingDensitiesandChapter3:PackingDensitiesand

CoordinationCoordination

LECTURE #05

earn ng ec ves

Howdoesatomicpackingfactorchangewith

differentatomtypes?

Howdoyoucalculatethedensityofa

material?

2

Pages 46-58.

Relevant Reading for this Lecture...

ReRecap:AtomicPackingFactor(APF)cap:AtomicPackingFactor(APF)

APF=Volumeofatomsinunitcell*

Volumeofunitcell

Describes

Howefficiently

atomsfillspace

withinagiven

unit cellassume ar sp eres

4(2a/4)34

volume

Unitcellcontains:

6x1/2+8x1/8

= 4atoms/unitcell

a

Adaptedfrom

Fig.3.1(a),

Callister7e.

APF= 3

a3

unitcell

volume

Closepackeddirections:

length=4R= 2a

APFforafacecentered

cubicstructure=0.743


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atoms watch out!

Ceramic Crystal Structures!

4

WatchOut!DifferentAtoms. MustModifyAPFEquations.WatchOut!DifferentAtoms. MustModifyAPFEquations.

a

Rblue

a3

aAdaptedfromFig.3.2(a),Callister7e.

a2a

Closepackeddirections:

Mustputincorrectvalue

3adiagonal =2Rblue +2Rred =

APF=

4

3 ( 3a/4) 32

atomsunit cell

a3

unit cell

volume

oreac a om ype

atom

volume

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Whatifionsfromthecrystalstructure?Whatifionsfromthecrystalstructure?

Cl

Cs+

Cesiumchloride(CsCl)unitcellshowing(a)ionpositionsandthetwoionsperlatticepoint

and(b)fullsizeions.NotethattheCs+Clpairassociatedwithagivenlatticepointisnot

amoleculebecausetheionicbondingisnondirectionalandbecauseagivenCs+ isequally

bondedtoeightadjacentCl,andviceversa.

(IPF), similar definition

at APF.

w/respect to crystal

symmetry.

Bravaislattice: simple cubicIons/unit cell: 1 Cs++1 Cl

6

Calculate the following: (1) The CN of CsCl (2) Ionic Packing Factor of CsCl

Recall from the last lecture CN r/R =r+/r

What are the rs for Cs+ and Cl-?

rCs+ = 0.170 nm

rCl- = 0.181 nmThese numbers came from inside cover of text book!

r+/r = 0.939

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Now calculate the Ionic Packing Factor of CsCl

For these ions, they touch along the body diagonal

(make sure you know which direction the hard

spheres are touching)

[# of atoms/unit cell] [volume of a sphere]

volume of unit cell

a3

One atom of Cs + One atom of Cl

4/3(0.170nm)3 + 4/3(0.181nm)3

VCs VCl

=

rCs+ = 0.170 nm

rCl- = 0.181 nm

a2

a

3 a = 2rCs+ + 2rCl-a = 0.405 nm

V = a3 = 0.0664nm3

IPF = 0.683

= . nm

8

For kicks, what if the ions touched along the edge length (not the body diagonal)?


volume of unit cellIPF =

ne a om o s + ne a om o

4/3(0.170nm)3 + 4/3(0.181nm)3

VCs VCl

= 0.0454 nm3

rCs+ = 0.170 nm

rCl- = 0.181 nmwhich r?

average them?

3 a = 2rCs+ + 2rCl-a = 0.405 nm a = 2r

ravg = .V = a3 = 0.0664nm3 a = 0.405 nm 0.352nmV = a3 = 0.0436nm3

Not possible! >100% packing!!!

It is critical that you identify the correct hard sphere touching directions!

IPF = 0.683 1.04

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Lets do another example: Consider NaCl

What is the CN?

What is the IPF?

How do we determine CN?

Correct, CN r+/r

rNa+ = 0.102 nm

rCl- = 0.181 nm

What are the rs for Na+ and Cl-?

Again, got these

numbers from

inside cover of

text book!

r+/r = 0.564

10

Not a very clear representation

lets expand it!

Motif2 ions per lattice point

Do you see how the green spheres form a FCC structure?!

The blue spheres do the same thing, form

a FCC structure

We have two interpenetrating FCC lattices

Commonly called a Rock Salt structure

structure seen in other materials including

TiC, TaC, MgO, CaO, etc.

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Calculate the IPF of NaCl


volume of unit cellIPF =

Two FCC lattices one for Na and one for Cl

4 atoms/Na FCC + 4 atoms/Cl FCC = 8 atoms/unit cell total

Four atoms of Na + Four atoms of Cl

4 x 4/3(0.102 nm)3 + 4 x 4/3(0.181 nm)3= 0.117nm3

What type of unit cell do we have?

12

a = 2rNa + 2rCl

a = 0.566nm0.181nm3

IPF = 0.117nm3 (previous slide)

= 0.65V = a3 = 0.181nm3

What is the volume of the unit cell?

2a

Watc Out D erentIons.Watc Out D erentIons.

CORRECTION!a

Thoughit

looks

like

FCC

symmetry,

theface diagonalatomsdonttouch;

buttheedgeatomsdotouch!

Why? Cationsandanionsdonot

havethesamesize!Beforewe

consideredatomsofthesamesize13


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Unoccupiedspace

incentercan

accommodatesmall

atoms,e.g.Hein

UO2 fuelrods

AtomscanoccupyinterstitialsitesAtomscanoccupyinterstitialsites

uor e a 2

unitcellshowing

(a)ionpositions

(b)fullsizeions.

Not e: useful info on atom

placementFCC interstitial sites

Structure: fluorite (CaF2) type

Bravaislattice: FCCIons/unit cell: 4Ca2++8F-

Typical ceramics: UO2, ThO2, TeO2

14

Calculate

the

ionic

packing

factor

for

UO2,

which

has

the

CaF2 structure

CLASSROOMEXAMPLE:

U and O ions touch along a

portion of the body diagonal

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CLASSROOMEXAMPLE: SOLUTION

This problem is tricky!

The face diagonal has a length of 2a.

CalculatetheionicpackingfactorforUO2,whichhastheCaF2 structure

The body diagonal has a length of 3a.

Along with the cell edge, they form a right

triangle within the unit cell.

The Ca2+ and F- ions touch a short distance

along the body diagonal.

By the principle of similitude, this smaller

3aU2+

O-

a

,

one.2 a

4 2

13 0.105 0.132 0.548

4 U OR R nm nm a

a

Because of this, the length of the bond becomes:

Now you can calculate aand the corresponding unit cell volume.

16

CLASSROOMEXAMPLE: SOLUTION continued

ions in unit cell

unit cell

V

IPF V

Solving for we get: 0.548nma a

4 23 0.105 0.1324 U OR R nm a

33 30.548 0.164

unit cellV a nm nm

3

singleion

4

3V R

4 2 3 33 3 34 8 0.105 0.132 0.09653 3 3 3

ions U OV R R nm

3

3

0.09650.588

0.164ions

unit cell

V nmIPF

V nm

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THEORETICALDENSITY,THEORETICALDENSITY,Density=mass/volume

massofeachatom=atomicweight/Avogadrosnumber

# atoms/unit cell Atomic weight (g/mol)

VcNAVolume/unit cell(cm3/unit cell)Avogadro's number(6.023 x 1023 atoms/mol)

18

n A# atoms/unit cell Atomic weight (g/mol)

THEORETICALDENSITY,

THEORETICAL

DENSITY,

c AVolume/unit cell(cm3/unit cell)

Avogadro's number

(6.023 x 1023 atoms/mol)

Example: Copper

crystalstructure=FCC: 4atoms/unitcell

atomicweight=63.55g/mol (1amu =1g/mol)

a om c

ra us

=

.

nm

nm

=

cm

Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10

-23cm3

Compare to actual: Cu = 8.94 g/cm3

Result: theoretical Cu = 8.89 g/cm3

Why the difference?

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Ex:Cr(BCC)

A = 52.00g/mol

TheoreticalDensity,TheoreticalDensity,

R =0.125nm

n =2

a=4R/3=0.2887nmaR

atoms

= a 3

52.002unitcellmol

unitcell

volume atoms

mol

6.023x1023

theoretical =7.18g/cm3

actual =7.19

g/cm

3

2020

DensitiesofMaterialClassesDensitiesofMaterialClasses

metals >ceramics >polymers

Why?

Graphite/

Ceramics/

Semicond

Metals/

Alloys

Composites/

fibersPolymers

20

30B asedondatainTableB1,Callister

*GFRE,CFRE,&AFREareGlass,

Gold,WPlatinum

Ingeneral

(g/cm

)32

,

Epoxycomposites(valuesbasedon

60%volumefractionofalignedfibers

inanepoxymatrix).10

3

4

5

Magnesium

Aluminum

Steels

Titanium

Cu,Ni

Tin,Zinc

Silver,Mo

an a um

G raphite

Silicon

Glass sodaConcrete

SinitrideDiamondAloxide

Zirconia

PTFE

CFRE *

GFRE*

Glassfibers

Carbon fibers

...

closepacking

(metallicbonding)

oftenlargeatomicmasses

Ceramics have...lessdensepacking

oftenlighterelements

Pol mers have...

DatafromTableB1,Callister7e.

1

0.3

0.4

0.5

HDPE,PSPP,LDPE

PCPETPVC

Wood

AFRE *A ramidfibers

low

packing

density

(oftenamorphous)

lighterelements(C,H,O)

Composites have...intermediatevalues

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Iftherearedifferentionsintheunitcell,

SummarySummary

whichdirectionthehardspherestouch

Theionicpackingfactorcanbecalculatedby

ionsin unit cell

unit cell

VIPF

V

n A

VcNA

# atoms/unit cell Atomic weight (g/mol)

Volume/unit cell

(cm3/unit cell)

Avogadro's number

(6.023 x 1023 atoms/mol)

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05_mte 271_packing densities (3)

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