ITNS Lecture 10

7/30/2019 ITNS Lecture 10

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LECTURE 10

Nanomaterials, Surface and

Structure II

Copyright: CRC Press 2009G.L. Hornyak


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Copyright: CRC Press 2009

Crystal Systems

Packing Fraction and Densities

Agenda


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CRYSTAL SYSTEMS



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Bravais

Lattice

Systems

Cubic

Systems


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HCP and FCC


A

B

A

A

B

C

The highest density of packing of spheres follows the rule of C.F. Gauss:

3 2

= 0.74


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The Unit Cell, Coordination and

Packing Fraction


The Unit Cell is an irreducible representation of the crystalstructure of a pure material. It is defined by a set of

parameters called lattice parameters. Lattice parameters

include: cell dimensions a,b,cand crystallographic angles. There are 14 Bravais lattices.

Adding unit cells together result in, eventually, a macroscopic

crystal.

Coordination number(CN or Z) is the number of nearest

neighbors an atom has within the unit cell.

Packing fractionfis the fraction of space occupied by

atoms in the unit cell compared to the whole volume of the

cell.


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Lattice Constants

Lattice constants (or parameters) are characteristics of the

unit cell that define its volume (a,b,c) and the angles betweenthem (,,).

Factors affecting lattice structure and constants:

Crystal type (affects bond length)

Atom size

Temperature and pressure

Chemical composition

Particle size?


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Coordination Number

Coordination numberis:

Large for metals (typically 8 12) Intermediate for ionic compounds (typically 6)

Low for molecular solids (covalent, 4)

Therefore, metals are the most dense covalent solids the least.

Bond length varies with CN:

Coordination Number Relative Radius

12 1.00

8 0.976 0.96

4 0.88

This should serve as an indication of things to come!

Source: D.E. Shriver, P.W. Atkins and C.H. Langford; Inorganic Chemistry, W.H. Freeman & Co., New York

(1990)


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Lattice constants are influenced by

particle size unsaturated

coordination of surface atoms causesdistortions in the lattice.

Nano Phenomenon #38

It takes only a couple unit cells to makea nanometer. Unit cells are at the small

frontier of the nanometer.

Nano Phenomenon #37



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Cubic Systems

Metal atom bonds = non-directional. Hard sphere models

adequately illustrate the packing.

Cubic systems are the simplest to visualizea = b = cand = =

There are three basic kinds: Simple (or primitive) cubic (sc)

Body-centered cubic (bcc)

Face-centered (or closest-packed) cubic (fcc)

Many metals are configured as bccand fcc(others as hcp).Only polonium is sc.

Metals (and ionic compounds) are polymorphic (due to T & P).



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Atomic Radius and Lattice Constants

ao = 2r dbody= ao3 = 4r

dbody

dface

dface = ao2 = 4r

SC BCC FCC

r

Atoms are touching along these vectors.



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Example 10.1

Q: Allotropic transformations: At lower temperatures, iron has

a bccstructure (< 911C). At higher temperatures (> 913C), itmetamorphs into the fccstructure. What is the change in volume

of a bcciron transition to fcciron at temperature given that abcc=

0.2863 nm and afcc= 0.3591 nm.

Iron contracts upon heating.

Vbcc = abcc3 = 0.2863 nm( )3

= 0.02347 nm3

Vfcc = afcc3

= 0.3591 nm( )3

= 0.04631 nm3

However, a bcccell has 2 atoms while an fcchas 4 per unit cell:

V =Vfcc 2 Vbcc

Vfcc

=0.04631 2 0.02347( )

0.04695x100 = 1.34%


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High resolution TEM of Ti and Al multilayers grown by E-beam deposition. The

TiH is hcp but transforms to an an fccatomic arrangement. The Al remains as

an fcclattice. The multilayers have a common [111] growth direction.

Source: Metallurgy Division of the Materials Science and Engineering Laboratory (MSEL), National Institute of

Standards and Testing (NIST) (2008).


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Hexagonal Close Packed Systems

A

A

B

CN = 12

Polytype: ABABABAB

a = bc= = 90

= 120ao = 2r

c= 1.633a (ideally)

Many metals crystallize

as hcp structures.

ao = 2r

co



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Example 10.2Q: Packing fraction ofhcp: Calculate ffor a hexagonal close

packed structure where a = 2r, c= 1.633a, Vhcp

= a2c(sin60).

What is the theoretical density of Mg (r= 0.160 nm)?

fhcp =#Atoms unit cell( ) VAtom

Vhcp

fhcp =

2( ) 43 rhcp3

2rhcp( )2

1.633 2rhcp( ) sin60 =

8

3rhcp

3

1.633 2rhcp( )3

0.866= 0.74

Density of Mg:

VhcpMg = 1.633 2rMg( )

3

sin60 = 1.633 2 0.160 nm( )3

0.866 = 0.04634 nm3

Mg =#Atoms per unit cell( ) MMg

Vhcp NA

=2 24.31g mol1( )

0.04634nm3( ) 6.022x1023 mol-1( )

107 nm

cm

3

= 1.74g cm3



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Tetragonal SystemsMetal bonds are non-directional resulting in simple crystal

structures. However, ionic, covalent and metal compounds

are more complex. Covalent bonds, for example, are

directional.

Tetragonal systems are like cubic systems with one axisextended (a = b c and = = = 90). There are two

kinds of tetragonal lattices: simple and body-centered. Body-centered tetragonal bctis an extended bccsystem.

Diamond cubic (dc) is a special version of an fcccrystal in

which one-half of the tetrahedral holes are filled with

additional atoms. Si, Ge, Sn and diamond are configured inthe dccrystal types.

Zinc Blende (e.g.ZnS) is similar to diamond cubic with Zn2+

filling one-half the tetrahedral holes.


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Example 10.3

Q: Determine the packing fraction ffor a dccell. Visualization is

not easy for this cell. The white circles represent voids.dbody= ao3 = 8r

dbody

ao

There are 8 atoms perdccell:

f =#Atoms VAtom

Vdtc

and 8r= ao 3

f =#Atoms

4

3rSi

3

ao3

=8

4

3rSi

3

8rSi 3( )3= 0.34

The packing fraction is muchlower than for any of the

metals. dc structures are

therefore quite open.



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Example 10.4


Q: Calculate the packing fraction for Indium a metal that

crystallizes in a tetragonal body-centered bctconfiguration:

a = 0.3253 nm, c= 0.4956 nm and r= 0.1626 nm.

aa

c

r

# Atoms are like bcc= 2

f =x VIn

VCell

= 2 4

3r3

a2

c

=2 4.189( ) 0.1626 nm( )

3

0.3253 nm( )2

0.4956 nm( )= 0.687

Atoms touch along the body diagonal where:

dBody = a2+ a2 + c2 = 4r

andc = 1.52a


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Example 10.5Q:What is the theoretical density of indium? of-, -Fe? Atomic mass In =114.818 gmol-1, a and care as before; # atoms = 2 and 4 respectively.

In

=#Atoms M

In

VUnitCell

NA

=2 114.818 g mol1( )

a2c 6.022x1023

107 nm

cm

3

= 7.27 g cm3

The accepted density of indium is 7.31 gcm-3.


Indium:

bccIron ( -Fe):

fcc Iron ( -Fe):

The accepted density of iron is 7.874 gcm-3.

Fe =

#Atoms MFe

Vbcc

NA

=2 55.845 g mol1( )0.2863( )

3

6.022x1023

107 nm

cm

3

= 7.90 g cm3

Since iron contracts upon heating, its density should increase.

Fe =

#Atoms MInVfcc NA

=4 55.845 g mol1( )0.3591( )

3

6.022x1023

107 nm

cm

3

= 8.01 g cm3


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Summary

Once again we are dealing with a concept or characteristic

that exists at the frontier boundary of the nanoscale.

Although this section was fairly descriptive and

fundamental to materials science, it is essential to the

understanding of what is to come later on in the course.

37.Unit cells exist at the frontier of the nanoscale

38.Lattice parameters are influenced by size.

More Nano Phenomena


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