Chapter 3-Crystal Structure ceramic .pdf

material science

1. Crystalline Structure Perfection
Chapter 3
MENG310 LIU-BIUM.B

2. Contents
3.1-Types of solids
3.2-Crystal structures
3.3-Seven systems and fourteen lattices
3.4-Metal structures
3.5-Ceramic structures
3.6-Polymeric Structures
3.7-Semiconductor Structures
3.8-Lattice positions, directions, and planes
2

3. 3
3.1-Types of solids
Solidscanbeclassifieddependingonthearrangementoftheiratomsormolecules.
Crystalline:hashighlydefinedandrepeatablearrangementsofmolecularchains.
Polycrystalline:consistsofmanysmallcrystals(grains)thatareseparatedbygrain
boundariesandnormallyhaverandomcrystallographicorientations.
Amorphous:onekindofnon-equilibriummaterial;itscharacteristicofatomic
arrangementismorelikeliquidandhasnolong-rangeperiodicity.
Thecrystalstructureofmaterialsarepresentedinthefollowingtable:
Material CrystallinePolycrystallineNon-crystalline
Metals
Glasses
Ceramics
Polymers
Semiconductors

4. 4
3.2-Crystal structure
Crystalstructureisauniquearrangementofatomsormoleculesinacrystallineliquid
orsolid
Thecrystalstructureofamaterialcanbedescribedintermsofitsunitcell.
Unitcellisasmallboxcontainingoneormoreatomsarrangedin3D.
Hard sphere unit cell representation
Aggregate (group) of many atoms
Reduced-sphere unit cell

5. 5
3.2-Crystal Structure
Thelengthofunit-celledgesandtheanglesbetweencrystallographicaxesarereferred
toaslatticeconstants,orlatticeparameters.
Thekeyfeatureoftheunitcellisthatitcontainsafulldescriptionofthestructureasa
wholebecausethecompletestructurecanbegeneratedbytherepeatedstackingof
adjacentunitcellsfacetofacethroughoutthree-dimensionalspace.
Geometryoftheunitcell
a,b,andcaretheunitlengthalongx,yandz
axes,respectively.α,β,andγaretheangles
betweenthecrystallographicaxes.

6. 6
3.3-Seven systems and fourteen lattices
Thereareonlysevenuniqueunit-cellshapethatcanbestackedtogethertofillthree-
dimensionalspace.ThesearecalledSevenCrystalSystems.
Thepointswheretheatomsarestackedinagivenunit-cellarecalledlatticepoints.
Thereare14possiblewaysofarrangementofthepoints,called14Bravais’Lattices.

7. 7
3.3-Seven systems and fourteen lattices
Thecrystallinestructuresofmostmetalsbelongtooneofthreerelativelysimpletypes.
Ceramicswhichhavevarietyofchemicalcompositions,exhibitvarietyofcrystalline
structures.
Glassisnon-crystalline,andpolymersmayhaveasmuchas50to100%oftheir
volumenon-crystalline(polycrystalline).

8. 8
3.4-Metal structure
MostMetalshaveoneofthreecrystalstructures:
(HCP)
Hexagonal Compact Packet
Examples:
Be, Ti, Zn, Zr, Mg, Cd
(BCC)
Body Centered Cubic
Examples:
Fe, V, Cr, Mo, W
(FCC)
Face Centered Cubic
Examples:
Al, Ni, Ag, Cu, Au
Metal Crystal lattice examples

9. 9
3.4-Metal Structure
AtomicPackingFactor(APF)isthefractionofvolumeinacrystalstructurethatis
occupiedbyatoms.
APFisdeterminedbyassumingthatatomsarerigidspheres.
APFisrepresentedmathematicallyas:
ThemostdensearrangementofatomshasanAPFofabout0.74
Ingeneral,metalswithahighatomicpackingfactorwillhaveahighermalleability
orductility.
Atomicpackingfactor
Coordinationnumber
CoordinationNumberisthenumberofadjacentatomssurroundingareferenceatom.cellunit
atomsatoms
V
VN
APF



10. 10
3.4-Metal structure
BodyCenteredCubiccrystalstructure
BodyCenteredCubic(BCC)unitcellhasatomsateachoftheeightcornersplusone
atominthecenterofthecube..
Atomsareincontactalongbodydiagonaloftheunitcell.
ThesideofthecubeisaandtheradiusofanatomisR.
1/8 atom1 atom2a 3a a a a 3
4
43
R
aRa  cellunitinsideatomsN
atoms
21
8
1
8  68.0
3
4
3
8
3
42
3
3
3
3












R
R
a
R
V
VN
volumeTotal
volumeAtomic
APF
cellunit
atomatoms


11. 11
3.4-Metal structure
FaceCenteredCubiccrystalstructure
FaceCenteredCubic(FCC)unitcellhasatomsateachoftheeightcornersplusone
atomatthecenterofeachfaceofthecube..
Atomsareincontactalongfacediagonaloftheunitcell.
ThesideofthecubeisaandtheradiusofanatomisR.
1/8 atom1/2 atom2a 3a a a a 2
4
42
R
aRa  cellunitinsideatomsN
atoms
4
2
1
6
8
1
8  74.0
2
4
3
8
3
4
4
3
3
3
3












R
R
a
R
V
VN
volumeTotal
volumeAtomic
APF
cellunit
atomatoms


12. 12
Example: 3.1
Copper Cu has the FCC crystal structurewith an atomic radius of 0.128 nm and
of molar mass equal to 63.55 g/mole.
Determine its atomic packing factor APF and the density of copper
a
Solution
Atoms inside unit cell = 6×1/2 + 8×1/8 = 4 atoms362.0
2
4
42 
R
aRa  
74.0
22
3
16
3
44
3
3
3
3



R
R
a
R
volumeTotal
volumeatomic
APF
  
3
23
3
7
3
/89.8
10023.610362.0
55.634
cmg
Na
MN
V
m
A









The density of FCC unit cell is: AA
N
MN
m
N
N
M
m
n



13. 13
3.4-Metal structure
HexagonalClose-Packedcrystalstructure
HexagonalClose-Packet(HCP)unitcellhasatomsateachofthetwelvecorners,three
atomsinsidetheunitcellandoneoneachbase.
Atomsareincontactalongthesideoftheunitcell.
ThesideoftheunitcellisaandtheradiusofanatomisR.
1/2 atom 1/6 atom
1 atom74.02  APFandRa cellunitinsideatomsN
atoms
6
6
1
12
2
1
23 

14. 14
3.4-Metal Structure
Coordinationnumber
CoordinationNumber(CN)isthenumberofadjacentatomssurroundingareference
atom.
Thecoordinationnumberofthecentralmetalatom/ionmaybeusedtodeducethe
moleculargeometryofthecoordinationcompound.
(BCC)
CN = 8
(FCC)
CN = 12 (HCP)
CN = 12

15. 15
3.5-Ceramic structure
Ceramicsusuallyhaveacombinationofionicbonds.
Ionic bond occurs between a metal and nonmetal with transfer of electrons.
Since the bonding is ionic, the packing factor will be Ionic Packing Factor (IPF).cellunit
ionsions
V
VN
IPF


Ionicpackingfactor
IPFrepresentsthefractionofthevolumeofionsoccupiedinaunitcelloverthetotalunit
cellvolume.
Typesofcrystalstructures
1.MX
2.MX2
3.M2X3
M: Metallic element
X: Non-metallic element

16. 16
3.5-Ceramic structure
MXCeramicStructures:CsCl
CsClcanbethoughtofastwointerpenetratingsimplecubicarrayswherethecornerof
onecellsitsatthebodycenteroftheother.
SimplecubicBravaislattice(8latticepointsatthecornerofthecube).
ThestructureofCsClisnotBCC.ItisasimplecubicBravaislattice.
Thenumberofcompleteionsinsideunitcellis:1CS
+
and1Cl
-
MaterialssimilartoCsClstructureare:CsBr,CsI,CsCN,TiCl,TiBr,andTiCN.

17. 17
3.5-Ceramic structure
MXCeramicStructures:NaCl
NaClhasacubicunitcell.Itisbestthoughtofasaface-centeredcubicarrayofanions
withaninterpenetratingfcccationlattice(orvice-versa).
Whetheronestartswithanionsorcationsonthecorners,thecellhasthesame
structure.
Thenumberofcompleteionsinsideunitcellis:4Na
+
and4Cl
-
MaterialssimilartoNaClstructureare:MgO,CaO,FeO,andNiO.
1/8 ion1/4 ion
1/2 ion

18. Example: 3.2
MgOhas the NaClcrystal structure.
a-Calculate the Ionic packing factor IPF of MgO.
b-Determine the density of MgO
Given: r
Mg
2+
=0.078 nm, r
O
2-
=0.132 nm,
16
O and
24.31
Mg
Solutionnmrra
OMg
42.022
22 
 465.0
3
4
4
3
4
4
3
33
22




a
rr
volumeTotal
volumeIonic
IPF
MgO
  
 
3
23
3
7
3
/61.3
10023.61042.0
16431.244
cmg
Na
MN
V
m
A









18

19. 19
3.5-Ceramic structure
MX2CeramicStructures:CaF2
Thereare12ions(4Ca
2+
and8F
-
)perunitcell.
ThecationsCa
2+
occupythe8cornersoftheunitcellandthecentersofthefaceswhile
theanionsF
-
occupythecornersofacubeinsidetheunitcell.
MaterialssimilartoCaF2structureare:UO2ThO2andTeO2

20. 20
3.5-Ceramic structure
MX2CeramicStructures:CaF2
IonsF
-
(blackspheres)arelocatedonthebodydiagonalofthecubicunitcell.
TheircentersareplacedonthequarterofbodydiagonalsandincontactwithCa
2+
�
�
4 
 
FCaFCa
rrarr
ad
22
3
4
4
3
4
�
�
�

21. 21
3.6-Polymericstructure
Polyethylene unit cell
Polymershavelongchainpolymericmolecules
The arrangement of these long molecules into a regular pattern is difficult.
Plastics tends complex crystalline structure
Polyethylene(C
2H
4)
nand plastics are examples of polymers.

22. 22
3.6-Polymeric structure
CrystallinityofPolymericStructures
Percentageofcrystallinitycanbecomputedbydividing
theamountofthecrystallinephasebythetotalamount
ofthematerialandmultiplyingby100.
Thefractionoftheorderedmoleculesinpolymeris
characterizedbythedegreeofcrystallinity,which
typicallyrangesbetween10%and80%.
StrengthPolymericStructures
Tensilestrengthoftenincreaseswithmolecularweightandpercentageofcrystallinity

23. Example: 3.3
For a density of 0.9979 g/cm
3
Calculate the number of C and H
atoms in the polyethylene unit cell.
Given the volume of unit cell is 0.0933 nm
3
,
1.008
H and
12.01
C.
Solution 
gN
N
m
HC
HC 23
23
1066.4
10023.6
)008.1(4)01.12(2
42
42 



 3
321
23
/997.0
100933.0
1066.4
42
cmg
cm
gN
V
m HC






 atomsNandatomsN
N
HC
HC
84
2
42


23

24. 24
3.7-Semiconductor structure
Elementalsemiconductors:Si,GeandgraySnsharediamonditscubicstructure.
1/8 atom
1/2 atom
1 atom
Thenumberofcompleteatomsinsideunitcellis:4+6×
1
2
+8×
1
8
=8�����
Grayatomsareatomslocatedatthecenteroffacesandcorners.Coloredatomsare
locatedatthebodydiagonalsofthecube.
Thecompleteatomsinsidetheunitcellareplacedonthequarterofthebodydiagonal.
Puresemiconductors

25. 25
ThesemiconductingcompoundsarecomposedofpairsofelementsfromcolumnsIII
andV(e.g.,GaAs)orfromcolumnsIIandVI(e.g.,ZnS).
3.7-Semiconductor structure
Compoundsemiconductors
MaterialssimilartoZnSstructureare:ZnSe,CdS,andHgTe
MaterialssimilartoGaSstructureare:AIP,andInSb
Thenumberofcompleteatomsinsideunitcellis:
4??????+6×
1
2
+8×
1
8
��=4??????+4��

26. Example: 3.4
Silicon Si has a diamond cubic structure.
a-Calculate the Atomic packing factor APF.
b-Determine the density of Si.
Given: r
Si=0.117 nm, and
28.09
Si.
SolutionSiSi
raar
3
8
4
3
2  34.0
3
8
3
4
8
3
4
8
3
3
3
3
3











Si
SiSi
r
r
a
r
volumeTotal
volumeAtomic
APF
  
3
23
3
7
3
/36.2
10023.61054.0
09.288
cmg
Na
MN
V
m
A









26

27. 27
3.8-Lattice positions, directions and planes
Inacrystallattice,eachatom,moleculeorions(constituentparticle)isrepresentedbya
singlepoint.Thesepointsarecalledlatticesiteorlatticepoint.
Latticepositions:sameasthepositioninthecoordinatesystemxyz.Theyareexpressed
asfractionormultiplesoftheunitcelldimensions.
Latticepositions
a,bandcaresidelengthsoftheunitcell
alongx,yandzaxesrespectively.Theyare
theunit-celldimensions.
How to determine the Lattice Position of P:
-Coordinates of P in the coordinate system
xyz is: P(1a, 2b,1c)
-Divide the coordinates by a, b and c
-Lattice Position of P is: 121
P

28. 28
3.8-Lattice positions, directions and planes
Crystaldirectionsareallstraightlinespassingthroughtwonodesinthelatticeusually
oneofthemistheoriginandtheotheroneistheidentifiedpoint.
Adirectionisexpressedassetofintegers,whichareobtainedbyidentifyingthe
smallestintegerpositionsinterceptedbyaparallellinefromtheoriginofthe
crystallographicaxes.
Latticedirections
Notationforlatticedirections.
Notethatparallel[uvw]
directions(e.g.,[111])sharethe
samenotationbecauseonlythe
originisshifted.
Notethatthelinefromtheorigin
ofthecrystallographicaxes
throughthe
1
2
1
2
1
2
bodycentered
positioncanbeextendedto
interceptthe111unitcellcorner
position.

29. 29
3.8-Lattice positions, directions and planes
Afamilyofdirectionsincludesanydirectionsthatareequivalentinlengthandtypesof
atomsencountered.
Latticedirections–Familyofdirections
Familyofdirections,<111>,representingallbodydiagonalsforadjacentunitcellsin
thecubicsystem.111,111,111,111,111,111,111,111111

30. 30
3.8-Lattice positions, directions and planes
Itisimportanttoknowtheanglebetweentwolatticedirectionsandplanessincethat
affectthemechanicalpropertiesofmaterial.
Latticedirections–AnglebetweentwolatticedirectionscosDDDD 
Let??????=&#3627408482;&#3627408462;+&#3627408483;&#3627408463;+&#3627408484;&#3627408464;andሖ??????=ƴ&#3627408482;&#3627408462;+ƴ&#3627408483;&#3627408463;+ƴ&#3627408484;&#3627408464;tobetwolatticedirections.Theangle
θbetweenthemisdefinedas:















222222
1
coscos
wvuwvu
wwvvuu
DD
DD

Example: 3.5
What is the angle between the [110] and [111] directions in the cubic system.
Solution
01
222222
1
5.35
32
011
coscos 


















wvuwvu
wwvvuu


31. 31
3.8-Lattice positions, directions and planes
Alatticeplaneisaplanewhoseintersectionswiththelattice(oranycrystalline
structureofthatlattice)areperiodic(i.e.aredescribedby2dBravaislattices).
MillerIndicesisamethodofdescribingtheorientationofaplaneorsetofplanes
withinalatticeinrelationtotheunitcell.Theyarerepresentedbythreeintegers.
Latticeplanes-MillerIndices(hkl)
Steps for computing Miller Indices
1-Take the intersections of plane with the
Cartesian axis x,y,z
2-Take the reciprocals and make sure that
the three numbers are of the smallest
integers (no fractions or multiples)
3-The obtained set is a miller indices
plane representation (hkl)
Knowledgeofsliporcleavageplanescanfacilitatethefundamentalunderstandingof
mechanicalpropertiesandengineeringoperations,suchastabletingandmilling.

32. 32
3.8-Lattice positions, directions and planes
Belowaresomeexamplesonmillerindices:
Millerplaneexamples

33. 33
Example: 3.6
Find the miller indices of the following planes

34. 34
3.8-Lattice positions, directions and planes
Afamilyofplanescontainsalltheplanesthatarecrystallo-graphicallyequivalent.
EveryfamilyoflatticeplanescanbedescribedbyasetofintegerMillerindicesthat
havenocommondivisors
Millerindices:Familyoflatticeplanes
Family of planes, {100}, representing all faces of unit cells in the cubic system. )100(),010(),001(),001(),010(),100(100

35. 35
3.8-Lattice positions, directions and planes
Thelineardensityisthenumberofatomsoccupiedinaunitlength.
Theintersectionoflatticedirectionwithspherecanbeeitheraradiusoradiameter
whichrepresentshalfandoneatomalongthelinerespectively.
Iftheintersectionisnotaradiusordiameterthennoatomscountalongtheline.
Lineardensitylineoflength
linealongatomsofnb
densityLinear  atom1 atom
2
1

36. 36
3.8-Lattice positions, directions and planes
Theplanardensityisthenumberofatomsoccupiedinaunitarea
Theintersectionoflatticeplanewithspherecanbeeitheradiskorpartofadisk.
Thenumberofatomsinplanecanbecountedasfollows:
PlanardensityplaneofArea
planeinatomsofnb
densityPlaner  atom
4
1 atom
2
1 atom1  atom
360


37. Example: 3.7
Consider FCC Aluminum crystal structure
r
Al= 0.143 nm.
a-Find the linear densities along [110], and [111]
b-Find the planar densities in planes (100), (110), and (111)
Solution
37Length
atomsofnb
densityLinear 
[110]nmatoms
atoms
a
LD /49.3
2404.0
2
2
11
2
1
2
]110[


 nmatoms
atom
a
LD /443.1
3404.0
1
3
2
1
2
]111[



[111]nmara 404.042  2a 3a

38. 38 
2
22)100( /25.12
404.0
2
1
4
14
nmatoms
atoms
a
PD 

 Area
atomsofnb
densityPlanar 
(100)2
2
)110(
/66.8
2404.0
2
2
2
1
2
4
1
4
nmatoms
atoms
aa
PD 



a
a
a2a
(110)2
)111( /14.14
2
60sin22
2
13
6
13
nmatoms
aa
PD 



(111)2a 2a 2a

39. 39
Problems

40. Problem 1
Calculate the IPF for UO
2which has the CaF
2structure. Given r
U
4+
= 0.105 nm and
r
O
2-
= 0.132 nm. Hint the oxygen atoms are located at one-quarter of the body
diagonal.
Solution588.0
3
4
8
3
4
4
3
33
24




a
rr
volumecellUnit
volumeIonic
IPF
OU
 nma
a
rr
OU
548.0
4
3
24 

40

41. Problem 2
Calculate the density of UO
2. Given
238.03
U and
16
O.
Solution
The volume of unit cell is computed in problem 2.
There exist 4 atoms of Uranium and 8 atoms of Oxygen in unit cell. 
 
3
23
3
7
3
/9.10
10023.610548.0
)16(803.2384
cmg
Na
MN
V
m
A









41

42. Problem 3
Calculate the APF for polyethylene. Given the volume of unit cell is 0.09333 nm
3,
r
C
=0.077 nm and r
H=0.046nm. Hint the unit cell contains 4 carbon atoms.
Solution
There exist 4 carbon atoms →8 hydrogen atoms.12.0
0933.0
3
4
8
3
4
4
33



HC
rr
volumeTotal
volumeAtomic
APF

42

43. Problem 4
Calculate the IPF for the zinc blende (ZnS) structure. Given r
Zn
2+
= 0.083 nm and r
S
2-
= 0.174 nm. Hint the sulfur atoms are located at one-quarter of the body diagonal.
Solution
There exist 4 zinc atoms and 4 sulfur atoms. 
468.0
594.0
3
4
4
3
4
4
3
33
3
22




SZn
rr
a
volumeIonic
IPF
 nma
a
rr
SZn
594.0
4
3
22 

43

44. Problem 5
Benefiting from problem 5, calculate the density of zinc blende (ZnS) structure.
Given
65.38
Zn and
32.06
S.
Solution
There exist 4 zinc atoms and 4 sulfur atoms in unit cell. 
 
3
23
3
7
3
/09.3
10023.610594.0
)06.32(438.654
cmg
Na
MN
V
m
Acellunit
ZnS









44

45. x
y
z
Problem 6
(a) Sketch, in a cubic unit cell, a [111] and a [112] lattice direction.
(b) Use trigonometric calculation to find the angle between these two directions.
(c) Use equation 3.3 to determine the angle between these two directions.
[111] 1l
[112] 2l
45
Solution
(b)6,3
21
 ll
Using cosine law
0
21
2
2
2
1
2
5.19cos21  llll
(c)
222222
21
21
cos
wvuwvu
wwvvuu
ll
ll




 18
4
211111
211
cos
222222



 0
5.19
18
4
arccos 







46. Problem 7
(a)Calculate the linear densities along [010], [101], and [111] direction in UO
2.
(b)Calculate the planar densities in the following planes (010), (101), (002), (111),
and (220). Given r
U
4+
= 0.105 nm and r
O
2-
= 0.132 nm.
46
SolutionLength
ionsofnb
densityLinear  nmU
U
a
U
LD /828.1
547.0
1
2
1
2
4
4
4
]010[





 nmarr
a
OU 547.0
4
3
24


[010]a

47. 47 
nmU
U
a
U
LD /585.2
2547.0
2
2
1
2
1
2
4
4
4
]101[






[101]
[111]nmOnmU
a
OU
LD /11.2/055.1
3
2
2
1
2
24
24
]111[




 2a 3a  
 
24
2
4
2
4
)010( /684.6
547.0
2
1
4
14
nmU
U
a
U
PD





 Area
ionsofnb
densityPlanar 
(010)
a
a

48. 2224
24
)101( /45.9/72.4
2
42
nmOnmU
aa
OU
PD





 (101) 
 
24
2
4
4
)002( /684.6
547.0
2
2
14
nmU
U
aa
U
PD







(002)
a
a 
24
4
)111( /718.7
2
60sin22
2
13
6
13
nmU
aa
U
PD






(111)2a 2a 2a  
 
24
2
4
)220( /726.4
547.0
2
2
12
nmU
la
U
PD






(220)
a2
a
l
a2a

49. Problem 8
(a)Calculate the linear densities of ions along [101], and [111] in zinc blende (ZnS).
(b)Calculate the planar densities of ions along the (110) and (020) plane. Given r
Zn
2+
= 0.083 nm and r
S
2-
= 0.174 nm.
49
Solutionnmarr
a
SZn 593.0
4
3
22

 Length
ionsofnb
densityLinear   
nmZn
Zn
a
Zn
LD /38.2
2593.0
2
2
2
1
21
2
2
2
]101[






[101]2a

50. 50
[111]nmSnmZn
a
SZn
LD /97.0/97.0
3
1
2
1
2
22
22
]111[




 3a Area
ionsofnb
densityPlanar  2222
22
)110( /02.4/02.4
2
22
nmSnmZn
aa
SZn
PD






(110) 
 
22
2
2
2
)020( /687.5
593.0
2
2
14
nmZn
Zn
aa
Zn
PD







(020)
a
a
a2a