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§2.4 Crystal Structure and Complex Lattice
Ⅰ.Crystal structure is the real arrangement of atom in crystals Crystal structure = Space lattice + Basis or structure unit
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+ =
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The difference between space lattice and crystal structure
Fe : Al = 1 : 1
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2×3 atoms / cell
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Ⅱ.Typical crystal structures of metals
1. BCC Example: α-Fe , V, Nb, Ta, Cr, Mo, W, alkali metals n = 2 atoms/cell CN=8 The number of nearest neighbours around each atom is called — Coordination Number.
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∴ For BCC, Packing fraction = Volume of atoms / cell
To determineξ, The atom is looked as a hard sphere, and the nearest neighbours touch each other. ∴ For BCC, Volume of atoms / cell Volume of unit cell
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Example: γ-Fe , Al , Ni , Pb , Cu ,
2. FCC Example: γ-Fe , Al , Ni , Pb , Cu , Ag , Au , stainless steal n= 8×1/8+6×1/2=4 atoms/cell CN=12
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• 3. HCP Example: Be, Mg, Zn, Cd, Zr, Hf Ti( low temperature) n=
CN=12 ξ=0.74 •
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4. Summary a0 vs. r Structure Atoms per cell Coordination Number
Packing factor Examples SC 1 6 0.52 Polonium (Po),α-Mn BCC 2 8 0.68 Fe,Ti,W,Mo,Nb,Ta,K,Na,V,Zr,Cr FCC 4 12 0.74 Fe,Cu,Au,Pt,Ag,Pb,Ni HCP Ti,Mg,Zn,Be,Co,Zr,Cd
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§2.5 Interstices in typical crystals of metals
Definition: In any of the crystal structures, there are small holes between the usual atoms into which smaller atoms may be placed. These locations are called interstitial sites. Ⅰ. Two types of Interstitials in typical crystals Octahedral interstitial Tetrahedral interstitial
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1. Octahedral interstitial
BCC
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FCC
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HCP
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2. Tetrahedral interstitial
BCC
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FCC
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HCP
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Ⅱ.Determination of the sizes of interstitials
Definition: By size of an interstitial we mean diameter of the maximum hard sphere which can be accommodated in the interstitial without distorting the lattice. di da diameter of interstitial atom diameter of atom in lattice point =
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Octahedral interstitial
condition for touching
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For BCC For FCC
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Tetrahedral interstitial
B L interstitial H D host atom C
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For BCC For FCC
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ξ interstices di/da Summary oct. tete. 6/2=3 12/2=6 4/4=1 8/4=2 6/6=1
CN ξ interstices di/da oct. tete. BCC 2 8 0.68 6 6/2=3 12 12/2=6 0.15 0.29 FCC 4 0.74 4/4=1 8/4=2 0.41 0.22 HCP 6/6=1 12/6=2
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Examples and Discussions
Both FCC and BCC are close-packed structures while BCC is more open? The interstitial atoms most likely occupy the oct. interstitial position in FCC and HCP, while in BCC two types of interstitial can be occupied equally.
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3. The solid solubility in BCC is much lower than in FCC.
4. Diffusion of interstitial atoms in BCC diffusion is much faster than in FCC or HCP at same temperature. 5. Determine the relationship between the atomic radius and the lattice parameter in SC, BCC, and FCC structures when one atom is located at each lattice point.
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6. Determine the density of BCC iron, which has a lattice parameter of 0.2866nm.
Solution: For a BCC cell, Atoms/cell = 2 a0 = nm = 2.866×10-8cm Atomic mass = g/mol Volume of unit cell = a03 = 23.54×10 -24cm3/cell Density
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Exercise Determine the coordinates of centers of both the octahedral and the tetrahedral interstitials in HCP referred to a, b and c. a b c 120o
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Thanks for your attention !
4. Prove that the A-face-centered hexagonal lattice is not a new type of lattice in addition to the 14 space lattices. 5. Draw a primitive cell for BCC lattice. Thanks for your attention !
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