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Cubic systems. Paul Sundaram University of Puerto Rico at Mayaguez. Review. Seven crystal systems Fourteen Bravais lattices Cubic and Hexagonal systems: 90% of all metals have a cubic or hexagonal structure. Cubic system characteristics. Unit cella=b=c, a = b = g =90˚ - PowerPoint PPT Presentation
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Cubic systemsPaul SundaramUniversity of Puerto Rico at Mayaguez
Review Seven crystal systems Fourteen Bravais lattices Cubic and Hexagonal systems:
90% of all metals have a cubic or hexagonal structure
Cubic system characteristics Unit cell a=b=c, = = =90˚ Face diagonal and body diagonal Number of atoms per unit cell Coordination number:number of nearest neighbor
atoms Close-packed structures Atomic Packing Factor (APF)
APF=(vol.of atoms in unit cell)/(vol. of unit cell) Atom positions, crystallographic directions and
crystallographic planes (Miller indices) Planar atomic density & linear atomic density
Some concepts
Number of atoms per unit cell Corner atom = 1/8 per unit cell Body centered atom = 1 Face centered atom = 1/2
Face diagonal=
Body diagonal= 3a
2a
Simple cubic(P)
Number of atomsper unit cell 1/8 X 8 = 1
Coordinationnumber
Atomic packingfactor
Simple cubic
Number of atomsper unit cell 1/8 X 8 = 1
Coordinationnumber 6
Atomic packingfactor
Simple cubic
Number of atomsper unit cell 1/8 X 8 = 1
Coordinationnumber 6
Atomic packingfactor
0.52
Body centered cubic(I)Number of atoms
per unit cell 1/8 X 8 + 1 = 2
Coordinationnumber
Atomic packingfactor
Real picture
Body centered cubicNumber of atoms
per unit cell 1/8 X 8 + 1 = 2
Coordinationnumber 8
Atomic packingfactor
Body centered cubicNumber of atoms
per unit cell 1/8 X 8 + 1 = 2
Coordinationnumber 8
Atomic packingfactor
0.68
Face centered cubic(F)
Number ofatoms per unit
cell1/8 X 8 + 1/2 X 6 = 4
Coordinationnumber
Atomic packingfactor
Real picture
Face centered cubic
Number ofatoms per unit
cell1/8 X 8 + 1/2 X 6 = 4
Coordinationnumber 12
Atomic packingfactor
Face centered cubic
Number ofatoms per unit
cell1/8 X 8 + 1/2 X 6 = 4
Coordinationnumber 12
Atomic packingfactor
0.74*
*Highest packing possible in real structures
Questions
Atomic Positions
X
Y
Z
(0,0,0)
(1/2,1/2,1/2)
(0,1,1)(1/2,1/2,1)
(1/2,0,1/2)
(0,0,1)
Crystallographic directions
R
R cos()
R cos(90-)
Concept of a vector & components
Examples
ComponentsX:a cos 0=aY:a cos 90=0Z:a cos 90=0Miller index:[100]
ComponentsX:a cos 90=0Y:a cos 90=0Z:a cos 0=aMiller index:[001]
ComponentsX:a cos 90=0Y:a cos 0=aZ:a cos 90=0Miller index:[010]
ComponentsX:a cos 90=0Y:a cos 0=aZ:a cos 90=0Miller index:[010]
Family<100> <010> <001>
Examples
ComponentsX: aY: aZ: 0Miller index:[110]
ComponentsX: 0Y: a Z: aMiller index:[011]
ComponentsX: aY: 0Z: 1Miller index:[101]
Examples
ComponentsX: -aY: -aZ: 0Miller index:[1 1 0]
ComponentsX: 0Y: -a Z: -aMiller index:[0 1 1]
ComponentsX: -aY: 0Z: -aMiller index:[1 0 1]
Family<110> <011> <101>
Examples
ComponentsX: aY: aZ: aMiller index:[111]
ComponentsX: -aY: -aZ: -aMiller index:[111]
Family<111>
Crystallographic planes
X
Y
Z How to determine indices of plane 1.Intersections with X,Y,Z axes
1 2. Take the inverse
1/1 1/ 1/ Miller index(1 0 0)
Family {100}1.Intersections with X,Y,Z axes 1 2. Take the inverse 1/ 1/1 1/ Miller index(0 1 0)
1.Intersections with X,Y,Z axes 12. Take the inverse 1/ 1/ 1/1 Miller index(0 0 1)
Example
X
Y
ZHow to determine indices of plane 1.Intersections with X,Y,Z axes
1 1 2. Take the inverse
1/1 1/1 1/
Miller index(1 1 0)Family {110}
Example
X
Y
ZHow to determine indices of plane 1.Intersections with X,Y,Z axes
1 1 12. Take the inverse
1/1 1/1 1/1
Miller index(1 1 1)Family {111}
Examples
ComponentsX: -1Y: 1Z: 1/2[-1 1 1/2][2 2 1]
ComponentsX: 1/2Y: 1/2Z: 1[1/2 1/2 1][112]
ComponentsX: -1Y: -1/2Z: 1/2[-1 -1/2 1/2][2 1 1]
Examples Intersections1/2,1,1/2Inverse2 1 2(212)
Intersections-1/2,1/2,1Inverse-2 2 1(2 2 1)
Intersections-1,-1,1/2Inverse-1 -1 2(1 1 2)
Intersections1/6,-1/2,1/3Inverse6 -2 3(6 2 3)
