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Atomic packing factor (APF). APF = Volume of atoms in unit cell*. Volume of unit cell. *assume hard spheres. (No.of atoms/unit cell) * volume of each atom.
Typology: Study notes
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•All metals are crystalline solids (special atomic arrangementsthat extend through out the entire material)•Amorphous solids: atoms are arranged at random positions(rubber)•Crystal structure: atoms form a repetitive pattern called lattice.•Crystal lattice: arrays of points (atoms) arranged such that eachpoint has an identical surroundings.
Crystallattice
Unitcell
*Unit cell: The smallest buildingblock of the crystal lattice
Lattice parameters: parameters that completely defines the unit cellgeometry. There are seven possible combinations of a,b,c and
α,β,γ
that give
rise to
crystal systems
Simple cube
Number of atoms per unit cell : 8 (corner atoms) x 1/8 =1 atom / unit cell
Body Center CubeBCC CRYSTALSTRUCTURE Number of atoms per unit cell :
8 (corner atoms) x 1/8 + 1 (interior atom) =
2 atoms / unit cell
Hexagonal close packed (HCP) crystal structure c/a = 1.63 For most HCP metals
Number of atoms per unit cell : 12 (corner atoms) x 1/6 + 3 (interior atoms)
A sitesB sitesA sites
2 a
a
3 a
*BCC (close packed direction are body diagonals ) √
3 a = 4R
a
R=0.5a
a
a
π
( 3a/4)
atoms unit cell
atom volume
volume unit cell
a
π
( 2a/4)
atoms unit cell
atom volume
volume unit cell
ALLOTROPY: Ability of the material to have more than onecrystal structure depending on temperature and pressure. e.g. Pure iron has a BCC crystal structure at room temperaturewhich changes to FCC at 912 C. Example: Determine the volume change of a 1 cm3 cube ironwhen it is heated from 910C, where it is BCC with a latticeparameter of 0.2863 nm, to 915 C, where it is FCC with alattice parameter of 0.3591.
VBcc= a3 = (0.2863)
Vfcc= (0.3591)
On the basis of equivalent number of atoms: 1 fcc unit cell has 4atoms, while 1 bcc unit cell has 2 atoms Volume change = V fcc – 2V Bcc
2 VBcc
= 0.046307 – 2 (0.023467 x
Iron contracts upon heating
Determine the volume change in cobalt when it transforms fromHCP at room temperature to FCC at higher temperatures.
Example:
Directions in unit cell:1.
Subtract the coordinates of the tail point from the coordinatesof the head point.
Clear fractions by multiplying or dividing by a common factorand reduce to lowest integers.
•A direction and its multiple isidentical, [100] is identical to[200], the second was notreduced to lowest integer.•Changing the sign of allindices produces a vector that isopposite in direction.•Parallel vectors have sameindices
