Primitivelatticevectors
Q:Howcanwedescribetheselatticevectors(thereareaninfinitenumberofthem)?A:Usingprimitivelatticevectors(thereareonlydoftheminad-dimensionalspace).
Fora3Dlattice,wecanfindthree primitivelatticevectors(primitivetranslationvectors),suchthatanytranslationvectorcanbewrittenas
𝑡 = 𝑛%�⃗�% + 𝑛(�⃗�( + 𝑛)�⃗�)where𝑛%, 𝑛( and𝑛) arethreeintegers.
Fora2Dlattice,wecanfindtwo primitivelatticevectors(primitivetranslationvectors),suchthatanytranslationvectorcanbewrittenas
𝑡 = 𝑛%�⃗�% + 𝑛(�⃗�(where𝑛% and𝑛( aretwointegers.
Fora1Dlattice,wecanfindoneprimitivelatticevector(primitivetranslationvector),suchthatanytranslationvectorcanbewrittenas
𝑡 = 𝑛%�⃗�%where𝑛% isaninteger.
Primitivelatticevectors
Red(shorter)vectors:�⃗�% and�⃗�(
Blue(longer)vectors:𝑏% and𝑏(
�⃗�% and�⃗�( areprimitivelatticevectors𝑏% and𝑏( areNOTprimitivelatticevectors
𝑏% = 2�⃗�% + 0�⃗�( �⃗�% =12𝑏% + 0𝑏(
Integercoefficients noninteger coefficients
Primitivecellin2D
Theparallelogramdefinedbytwoprimitivelatticevectorsformaprimitivecell.
Ø theAreaofaprimitivecell:A = |�⃗�%×�⃗�(|Ø Eachprimitivecellcontains1site.
Primitivecellin3D
Theparallelepiped definedbythethreeprimitivelatticevectorsarecalledaprimitivecell.
Ø thevolumeofaprimitivecell:V = |�⃗�%. (�⃗�(×�⃗�))|Ø eachprimitivecellcontains1site.
Aspecialcase:acuboid
Wigner–Seitzcell
Ø thevolumeofaWigner-SeitzcellisthesameasaprimitivecellØ eachWigner-Seitzcellcontains1site(sameasaprimitivecell).
Rotationalsymmetries
Rotationalsymmetries:Ifasystemgoesbacktoitself whenwerotateitalongcertainaxesbysomeangle𝜃,wesaythatthissystemhasarotationalsymmetry.Ø Forthesmallest𝜃,2𝜋/𝜃 isaninteger,whichwewillcall𝑛.Ø Wesaythatthesystemhasa𝑛-foldrotationalsymmetryalongthisaxis.
ForBravaislattices,Ø Itcanbeprovedthat𝑛 canonlytakethefollowingvalues:1, 2, 3, 4 or6.
Cubicsystem
Conventionalcells
Ø Forasimplecubiclattice,aconventionalcell=aprimitivecellØ NOTtrueforbody-centeredorface-centeredcubiclattices
Howcanweseeit?Ø sc:oneconventionalcellhasonesite(sameasaprimitivecell)Ø bcc:oneconventionalcellhastwosites(twiceaslargeasaprimitivecell)Ø fcc:oneconventionalcellhasfourcites(1conventionalcell=4primitivecells)
SimplecubicLatticesites:𝑎(𝑙𝑥? + 𝑚𝑦?+n�̂�)Latticepointperconventionalcell:1 = 8× %
EVolume(conventionalcell):𝑎)Volume(primitivecell):𝑎)Numberofnearestneighbors:6Nearestneighbordistance:𝑎Numberofsecondneighbors:12Secondneighbordistance: 2� 𝑎
Packingfraction:GH≈ 0.524
Coordinatesofthesites:(𝑙, 𝑛, 𝑚)Forthesite 0,0,0 ,6nearestneighbors: ±1,0,0 , 0, ±1,0 and 0,0, ±112nestnearestneighbors: ±1,±1,0 , 0, ±1, ±1 and(±1,0, ±1)
PackingfractionPackingfraction:WetrytopackNspheres(hard,cannotdeform).
Thetotalvolumeofthespheresis𝑁4𝜋 MN
)
ThevolumethesespheresoccupyV > 𝑁4𝜋 MN
)(therearespacing)
Packingfraction=totalvolumeofthespheres/totalvolumethesespheresoccupy
𝑃𝑎𝑐𝑘𝑖𝑛𝑔𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =𝑁4𝜋 𝑅
)
3
𝑉=4𝜋 𝑅
)
3
𝑉/𝑁=
4𝜋 𝑅)
3
𝑉𝑜𝑙𝑢𝑚𝑒𝑝𝑒𝑟𝑠𝑖𝑡𝑒
=4𝜋 𝑅
)
3
𝑉𝑜𝑙𝑢𝑚𝑒𝑜𝑓𝑎𝑝𝑟𝑖𝑚𝑖𝑡𝑖𝑣𝑒𝑐𝑒𝑙𝑙
Highpackingfractionmeansthespaceisusedmoreefficiently
SimplecubicLatticesites:𝑎(𝑙𝑥? + 𝑚𝑦?+n�̂�)Latticepointperconventionalcell:1 = 8× %
EVolume(conventionalcell):𝑎)Volume(primitivecell):𝑎)Numberofnearestneighbors:6Nearestneighbordistance:𝑎Numberofsecondneighbors:12Secondneighbordistance: 2� 𝑎
Packingfraction:GH≈ 0.524
𝑃𝑎𝑐𝑘𝑖𝑛𝑔𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =4𝜋 𝑅
)
3
𝑉𝑜𝑙𝑢𝑚𝑒𝑜𝑓𝑎𝑝𝑟𝑖𝑚𝑖𝑡𝑖𝑣𝑒𝑐𝑒𝑙𝑙
=4𝜋 𝑅
)
3
𝑎)=4𝜋3(𝑅𝑎))=
4𝜋3(𝑎/2𝑎))=
𝜋6≈ 0.524
Ø Abouthalf(0.524=52.4%)ofthespaceisreallyusedbythesphere.Ø Theotherhalf(0.476=47.6%)isempty.
Nearestdistance=2RR= Nearestdistance/2=𝑎/2
bcc
Latticesites𝑎(𝑙𝑥? + 𝑚𝑦?+n�̂�)and𝑎[ 𝑙 + %
(𝑥? + 𝑚 + %
(𝑦? + 𝑛 + %
(�̂�]
Latticepointperconventionalcell:2 = 8× %E+ 1
Volume(conventionalcell):𝑎)Volume(primitivecell):𝑎)/2Numberofnearestneighbors:8
Nearestneighbordistance: (a()(+(a
()(+(a
()(� = )�
(𝑎 ≈ 0.866𝑎
Numberofsecondneighbors:6Secondneighbordistance:𝑎
Packingfraction: )� E𝜋 ≈ 0.680
Coordinatesofthesites:(𝑙, 𝑛, 𝑚)Forthesite 0,0,0 ,8nearestneighbors: ± %
(, ± %
(, ± %
(
6 nestnearestneighbors: ±1,0,0 , 0, ±1,0 and(0,0, ±1)
bccpackingfraction
Volume(primitivecell):𝑎)/2
Nearestneighbordistance: (a()(+(a
()(+(a
()(� = )�
(𝑎 ≈ 0.866𝑎
Packingfraction: )� E𝜋 ≈ 0.680
𝑃𝑎𝑐𝑘𝑖𝑛𝑔𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =4𝜋 𝑅
)
3
𝑉𝑜𝑙𝑢𝑚𝑒𝑜𝑓𝑎𝑝𝑟𝑖𝑚𝑖𝑡𝑖𝑣𝑒𝑐𝑒𝑙𝑙
=4𝜋 𝑅
)
3
𝑎)/2=8𝜋3(𝑅𝑎))=
8𝜋3(3�4 𝑎𝑎
))=3� 𝜋8
≈ 0.680
Ø About68.0%ofthespaceisreallyusedbythesphere.Ø About32.0%ofthespaceisempty.
Nearestdistance=2R
R= Nearestdistance/2= )�
b𝑎
fcc
Latticesites𝑎(𝑙𝑥? + 𝑚𝑦?+n�̂�) 𝑎[ 𝑙 + %
(𝑥? + 𝑚 + %
(𝑦? + 𝑛�̂�]
𝑎[ 𝑙 + %(𝑥? + 𝑚𝑦? + 𝑛 + %
(�̂�]
𝑎[𝑙𝑥? + 𝑚 + %(𝑦? + 𝑛 + %
(�̂�]
Latticepointperconventionalcell:4 = 8× %E+ 6× %
(= 1 + 3
Volume(conventionalcell):𝑎)Volume(primitivecell):𝑎)/4Numberofnearestneighbors:12
Nearestneighbordistance: (a()(+(a
()(+(0)(� = (�
(𝑎 ≈ 0.707𝑎
Numberofsecondneighbors:6Secondneighbordistance:𝑎
Forthesite 0,0,0 ,12nearestneighbors: ± %
(, ± %
(, 0 , ± %
(, 0, ± %
(and 0, ± %
(, ± %
(6 nestnearestneighbors: ±1,0,0 , 0, ±1,0 and(0,0, ±1)
fcc packingfraction
Volume(primitivecell):𝑎)/4
Nearestneighbordistance: (a()(+(a
()(+(0)(� = (�
(𝑎 ≈ 0.707𝑎
𝑃𝑎𝑐𝑘𝑖𝑛𝑔𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =4𝜋 𝑅
)
3
𝑉𝑜𝑙𝑢𝑚𝑒𝑜𝑓𝑎𝑝𝑟𝑖𝑚𝑖𝑡𝑖𝑣𝑒𝑐𝑒𝑙𝑙
=4𝜋 𝑅
)
3
𝑎)/4=16𝜋3
(𝑅𝑎))=
16𝜋3
(2�4 𝑎𝑎
))=2� 𝜋6
≈ 0.740
Ø About74.0%ofthespaceisreallyusedbythesphere.Ø About26.0%ofthespaceisempty.
Nearestdistance=2R
R= Nearestdistance/2= (�
b𝑎
0.740 isthehighestpackingfractiononecaneverreach.Thisstructureiscalled“closepacking”Thereareotherclosepackingstructures(samepackingfraction)
hexagonal
Primitivecell:arightprismbasedonarhombuswithanincludedangleof120degree.
Volume(primitivecell):𝑉𝑜𝑙 = 𝐴𝑟𝑒𝑎×ℎ𝑒𝑖𝑔ℎ𝑡
= 212×
3�
2𝑎%( 𝑎) =
3�
2𝑎(𝑐
2DplanesformedbyequilateraltrianglesStacktheseplanesontopofeachother
HexagonalClose-PackedStructure(hcp)
Ø Let’sstartfrom2D(packingdisks,insteadofspheres).Q:Whatistheclose-packstructurein2D?A:Thehexagonallattice(thesedisksformequilateraltriangles)
Ø Now3D:Q:Howtoget3Dclosepacking?A:Stack2Dclosepackingstructuresontopofeachother.
Atomsinsideaunitcell
Ø WechoosethreelatticevectorsØ ThreelatticevectorsformaprimitiveoraconventionalunitcellØ Lengthofthesevectorsarecalled:thelatticeconstants
Wecanmarkanyunitcellbythreeintegers:𝑙𝑚𝑛𝑡 = 𝑙�⃗�% + 𝑚�⃗�( + 𝑛�⃗�)
Coordinatesofanatom:Wecanmarkanyatominaunitcellbythreerealnumbers:𝑥𝑦𝑧.Thelocationofthisatom:𝑥�⃗�% + 𝑦�⃗�( + 𝑧�⃗�)Noticethat0 ≤ 𝑥 < 1 and0 ≤ 𝑦 < 1 and0 ≤ 𝑧 < 1
Q:Whyxcannotbe1?A:Duetotheperiodicstructure.1isjust0inthenextunitcell
SodiumChloridestructureFace-centeredcubiclatticeNa+ionsformaface-centeredcubiclatticeCl- ionsarelocatedbetweeneachtwoneighboringNa+ions
Equivalently,wecansaythatCl- ionsformaface-centeredcubiclatticeNa+ionsarelocatedbetweeneachtwoneighboringNa+ions
Cesiumchloridestructure
SimplecubiclatticeCs+ionsformacubiclatticeCl- ionsarelocatedatthecenterofeachcube
Equivalently,wecansaythatCl- ionsformacubiclatticeCs+ ionsarelocatedatthecenterofeachcube
Coordinates:Cs:000Cl:%
(%(%(
NoticethatthisisasimplecubiclatticeNOTabodycenteredcubiclatticeØ Forabcclattice,thecentersiteisthe
sameasthecornersitesØ Here,centersitesandcornersitesare
different
Carbonatomscanform4differentcrystalsGraphene(NobelPrizecarbon)
Diamond(moneycarbon/lovecarbon)
Graphite(Pencilcarbon)
Nanotubes
DiamondlatticeisNOTaBravaisLatticeeither
Samestoryasingraphene:Wecandistinguishtwodifferenttypeofcarbonsites(markedbydifferentcolor)Weneedtocombinetwocarbonsites(oneblackandonewhite)togetherasa(primitive)unitcellIfweonlylookattheblack(orwhite)sites,wefoundtheBravaislattice:fcc
CubicZincSulfideStructure
VerysimilartoDiamondlatticeNow,blackandwhitesitesaretwodifferentatomsfcc withtwoatomsineachprimitivecell
Goodchoicesforjunctions
Diode