LECTURE #05. Learning Objectives. How does atomic packing factor change with different atom types? How do you calculate the density of a material?

(1)

Chapter

3:

Packing

Densities

and

Chapter

3:

Packing

Densities

and

Coordination

Learning Objecti es

LECTURE #05

Learning

Objectives

• How

does

atomic

packing

factor

change

with

different

atom

types?

• How

do

you

calculate

the

density

of

a

l?

material?

2

• Pages 46-58.

Relevant Reading for this Lecture...

Re

‐‐

cap:

Atomic

Packing

Factor

(APF)

APF

=

Volume

of

atoms

in

unit

cell*

Volume

of

unit

cell

*

h d

h

Describes

How efficiently

atoms fill space

within a given

unit cell

*assume hard spheres

unit cell

2 a

4 

(

2a/4)

3

4 volume

_atom

Unit

cell

contains:

6 x 1/2

+

8 x 1/8

=

4 atoms/unit

cell

a

2 a

Adapted from

APF

=

3 (

)

atom

a

3 unit

cell

volume

• APF

for

a

face

‐

centered

(2)

If the unit cell has different

atoms – watch out!

Ceramic Crystal Structures!

y

4

Watch

Out!

Different

Atoms.

Must

Modify

APF

Equations.

Watch

Out!

Different

Atoms.

Must

Modify

APF

Equations.

a

R

_blue

a

3 a

Adapted from

Fig. 3.2(a), Callister 7e.

a

2 a

Close

‐

packed

directions:

Must put in correct value for each atom type

3 a

diagonal =

2R

_blue

+

2R

_red

=

APF

=

4

3 

( 3

a

/4)

3

2 atoms

unit cell

a

3 unit cell

volume

for each atom type

atom

volume

(3)

What

if

ions

from

the

crystal

structure?

What

if

ions

from

the

crystal

structure?

Ionic packing factor

Same rules as metals

Structure: CsCl type

Cl

ԟ

Cs

+

Cesium chloride (CsCl) unit cell showing (a) ion positions and the two ions per lattice point and (b) full‐size ions. Note that the Cs+₋_Cl− _pair_associated_with_a_given_lattice_point_is_not a molecule because the ionic bonding is non‐directional and because a given Cs+_is_equally

bonded to eight adjacent Cl−_,_and_vice_versa.

Ionic packing factor

(IPF), similar definition

at APF.

Same rules as metals

w/respect to crystal

symmetry.

Structure: CsCl type

Bravais lattice: simple cubic

Ions/unit cell: 1 Cs

+

_{+ 1 Cl}

ԟ

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 r’s for Cs

+

_{and Cl}

-

_?

rCs+= 0.170 nm

rCl-= 0.181 nm

These numbers came from inside cover of text book!

(4)

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]

IPF

_{volume of unit cell}

a

3

One atom of Cs + One atom of Cl 4/3π(0.170nm)3_{+ 4/3}_π_(0.181nm)3 V_Cs VCl 0 0454 3

IPF =

rCs+= 0.170 nm r_Cl-= 0.181 nm

a

2 a

√3 a= 2r_Cs++ 2r Cl-a= 0.405 nm V = a3= 0.0664nm3 IPF = 0.683 = 0.0454 nm3 8

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

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

O t f C O t f Cl

IPF =

One atom of Cs + One atom of Cl 4/3π(0.170nm)3_{+ 4/3}_π_(0.181nm)3 VCs VCl = 0.0454 nm3 rCs+= 0.170 nm rCl-= 0.181 nm which r? average them? 0 176 √3 a= 2rCs++ 2r Cl-a= 0.405 nm _a_{= 2}_r ravg= 0.176 V = a3_{= 0.0664nm}3 _a_{= 0.405 nm 0.352nm}a V = a3_{= 0.0436nm}3

Not possible! >100% packing!!!

It is critical that you identify the correct ‘hard sphere’ touching directions! IPF = 0.683 1.04

(5)

Let’s do another example: Consider NaCl

What is the CN? What is the IPF?

How do we determine CN?

Correct, CN → r +_/r–

r_Na+= 0.102 nm

r_Cl-= 0.181 nm

What are the r’s for Na+_{and Cl}-_? Again, got these numbers from inside cover of text book!

r +_/r–_{= 0.564}

10

Not a very clear representation – let’s expand it!

Motif 2 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.

(6)

Calculate the IPF of NaCl

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

IPF =

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 4/3π(0.102 nm)3_{+ 4 x 4/3}_π_{(0.181 nm)}3 = 0.117nm3

What type of unit cell do we have?

12

W t h O t! Diff

t I

W t h O t! Diff

t I

a= 2rNa+ 2rCl

a= 0.566nm 0.181nm3

IPF = 0.117nm3_{(previous slide)}

= 0.65 V = a3_{= 0.181nm}3

What is the volume of the unit cell?

2 a

Watch

Out!

Different

Ions.

Watch

Out!

Different

Ions.

CORRECTION!

a

Though it looks like FCC symmetry, the face diagonal atoms don’t touch;

but the edge atoms do touch! Why? Cations and anions do not

have the same size! Before we considered atoms of the same size

(7)

Fl it (C F ) Unoccupied space in center can accommodate small atoms, e.g. He in UO2fuel rods

Atoms

can

occupy

‘interstitial’

sites

Atoms

can

occupy

‘interstitial’

sites

Fluorite (CaF2)

unit cell showing (a) ion positions (b) full‐size ions.

Note: useful info on atom placement

FCC interstitial sites FCC interstitial sites

Structure: fluorite (CaF

2

) type

Bravais lattice: FCC

Ions/unit cell: 4Ca

2+

_{+ 8F}

-Typical ceramics: UO

2

, ThO

2

, TeO

2

14

Calculate the ionic packing factor for UO2, which has the CaF2structure

CLASS ROOM EXAMPLE:

U and O ions touch along a portion of the body diagonal

(8)

CLASS ROOM EXAMPLE: SOLUTION

This problem is tricky!

The face diagonal has a length of 2a.

Calculate the ionic packing factor for UO2, which has the CaF2structure

The body diagonal has a length of 3

a

.

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 triangle measures as ¼ the size of the large

a

2 3a

U

2+

O

-a

triangle, measures as ¼ the size of the large one.

a

2a 4 2 1 3 0.105 0.132 0.548 4 aRURO  nm nm

a

Because of this, the length of the bond becomes:

Now you can calculate

a

and the corresponding unit cell volume.

16

CLASS ROOM EXAMPLE: SOLUTION ‐continued

ions in unit cell

unit cell

V

IPF

V



Solving for we get: a a0.548nm

4 2 3 0.105 0.132 4 RU RO   nm a





3 3 _0.548 _0.164 3 unit cell V a  nm  nm 3 single ion 4 3 V 



R 4 4 16

_

_

32

_

_

4 2 3 3 3 3 3 4 4 16 32 4 8 0.105 0.132 0.0965 3 3 3 3 ions _U _O V  



R   



R  







 nm 3 3 0.0965 0.588 0.164 ions unit cell V nm IPF V nm    17

(9)

THEORETICAL

DENSITY,

THEORETICAL

DENSITY,



Density

=

mass/volume

mass = number of atoms per unit cell



mass of each atom

mass

=

number

of

atoms

per

unit

cell



mass

of

each

atom

mass

of

each

atom

=

atomic

weight

/

Avogadro’s

number

n

A

# atoms/unit cell

Atomic weight (g/mol)

 

n

A

V

_c

N

_A

Volume/unit cell

(cm3/unit cell)

Avogadro's number

(6.023 x 1023 atoms/mol)

18

 

n

A

V

N

# atoms/unit cell

Atomic weight (g/mol)

THEORETICAL

DENSITY,

THEORETICAL

DENSITY,



V

_c

N

_A

Volume/unit cell

(cm3/unit cell)

Avogadro's number

(6.023 x 1023 atoms/mol)

Example:

Copper

• crystal

structure

=

FCC:

4 atoms/unit

cell

• atomic

weight

=

63.55 g/mol

(1

amu =

1 g/mol)

• atomic radius R 0 128 nm (1 nm 10

‐7

_cm)

• atomic

radius

R

=

0.128 nm

(1

nm

=

10

‐7

_cm)

Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10-23cm3

Compare to actual:



_{Cu = 8.94 g/cm3}

(10)

• Ex:

Cr

(BCC)

A

= 52.00

g/mol

Theoretical

Density,

Theoretical

Density,



R

=

0.125 nm

n

=

2 a

=

4R/

3 =

0.2887

nm

a

R

atoms

g



=

a 3

52.00

2 unit

cell

mol

g

unit

cell

volume

atoms

mol

6.023 x 10

23



_theoretical

=

7.18 g/cm

3



_actual

=

7.19 g/cm

3 20 20

Densities

of

Material

Classes

Densities

of

Material

Classes



metals

>



ceramics

>



polymers

Why?

Graphite/ Ceramics/ Semicond Metals/ Alloys Composites/ fibers Polymers 20 30

B ased on data in Table B1, Callister

*GFRE, CFRE, & AFRE are Glass, Carbon & Aramid Fiber‐Reinforced

Tantalum Gold, W Platinum

Metals

have

In

general



(g

/c

m

)

3 2

Carbon, & Aramid Fiber Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers

in an epoxy matrix). 10 3 4 5 Magnesium Aluminum Steels Titanium Cu,Ni Tin, Zinc Silver, Mo Tantalum G raphite Silicon Glass‐soda Concrete Si nitride Diamond Al oxide Zirconia PTFE Silicone CFRE * GFRE* Glass fibers Carbon fibers

Metals

have...

• close

‐

packing

(metallic

bonding)

• often

large

atomic

masses

Ceramics

have...

• less

dense

packing

• often

lighter

elements

Polymers

have...

Data from Table B1, Callister 7e. 1 0.3 0.4 0.5 H DPE, PS PP, LDPE PC PET PVC Silicone Wood AFRE * A ramid fibers