55.85 × 10 −3 kg mol −1 2 −1 23 6.022 × 10 mol 3 (0.2866 × 10 −9 m)
ρ = 7.88 × 103 kg m-3 or 7.88 g cm-3
Atomic concentration is 2 atoms in a cube of volume a3, i.e.
Elementary Crystals ( S. O. Kasap, 1990 - 2001: v.1.0) An e-Booklet
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Elementary Crystals
Safa Kasap Department of Electrical Engineering University of Saskatchewan Canada
Elementary Crystals ( S. O. Kasap, 1990 - 2001: v.1.0) An e-Booklet
2
Solution Consider a face diagonal as shown in Figure 1. Two corner atoms and the central atom are in contact and the length of the face-diagonal is 4R, a2 + a2 = (4R)2 and therefore a= 4 R = 2 2R 2
3. The Diamond and the Zinc Blende Structures
The diamond and the zinc blende crystal structures have similarities. Both are cubic crystals and both have 8 atoms in the unit cell. In the zinc blende unit cell, four are Zn atoms and four are S atoms. Problem: Density of Si and GaAs and atomic concentration The crystal structures for Si and GaAs are shown Figure 3. Given the lattice parameters of Si and GaAs, a = 0.543 nm and a = 0.565 nm respectively, and the atomic masses of each element in the Periodic Table, calculate the density of Si and GaAs. What is the atomic concentration, atoms per unit volume, in each crystal?
a = (4/√3)(0.1241 nm) = 0.2866 nm
There are 2 atoms in the unit cell. There are 8 corners and each corner has 1/8th of an atom within the unit cell. In addition, there is one full atom at the center of the cube. The density is
a = (2√2)(0.1444 nm) = 0.4084 nm
There are 4 atoms in the unit cell. There are 8 corners and each corner has 1/8th of an atom within the unit cell. In addition, there are 6 faces and each face has a 1/2-atom at the center. The density is
Figure 2 Problem: BCC crystal characteristics Iron (below 912 °C) has the BCC crystal structure. The atomic mass of Fe is 55.85 g mol-1. If the radius of the Fe atom is 0.1241 nm find the lattice parameter a, density ρ and the atomic concentration of tungsten. Find also the atomic packing factor APF. Solution Consider a cube diagonal as shown in Figure 2. Two corner atoms and the central atom are in contact and the length of the diagonal is 4R. Since we have a cube a2 + a2 + a2 = (4R)2 and tha) a a (b) Cube diagonal is 4R. 4R (b)
(a) Unit cell of the body centered cubic (BCC) crystal structure. Examples are: Alkali metals (Li, Na, K, Rb), Cr, Mo, W, Mn, -Fe (< 912 C), -Ti (> 882 C).
ρ=
Mass of atoms in unit cell Volume of unit cell
=
( Number of atoms in unit cell) × (Mass of one atom)
Volume of unit cell
that is, i.e.
M 2 at NA ρ= = a3
(Number of atoms in unit cell) × (Volume of one atom)
Volume of unit cell
3 2( 4 3 πR )
a3
=
3 2( 4 3 πR )
4 R 3
3
=
2( 4 3 π) 4 3
3
= 0.68 or 68%
ρ=
Mass of atoms in unit cell Volume of unit cell
=
( Number of atoms in unit cell) × (Mass of one atom)
Volume of unit cell
that is, i.e.
M 4 at NA ρ= = a3
4
107.87 × 10−3 kg mol−1 −1 23 6.022 × 10 mol
(0.4084 × 10
−9
m)
3
ρ = 1.05 × 104 kg m-3 or 10.5 g cm-3
Atomic concentration is 4 atoms in a cube of volume a3, i.e. nat = 4 4 28 m −3 or 5.87 × 10 22 cm −3 = 3 = 5.87 × 10 3 − 9 a (0.4084 × 10 m)
Errors using inadequate data are mush less than those using no data at all Charles Babbage (1792-1871)
1. Unit Cell and FCC Crystals
The unit cell is the most convenient small cell in the crystal structure that carries the properties of the crystal. The repetition of the unit cell in three dimensions generates the whole crystal structure as in Figure 1. Face centered cubic (FCC) unit cell has a cube face with one atom at each face corner and one atom at the center of the face. Only one-eighth of each face-corner atom however belongs to the unit cell as shown in Figure 1. In addition, only half of an atom at the face center belongs to the unit cell. There are therefore 4 atoms in the unit cell. The lattice parameter a is a cube side. Atomic packing factor is the fraction of volume in the crystal actually occupied by atoms.
3
) = 4(
(2
4 3
πR3 )
2R
3
) (2 2 )
=
4( 4 3 π)
3
= 0.74 or 74%
Elementary Crystals ( S. O. Kasap, 1990 - 2001: v.1.0) An e-Booklet
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2. BCC Crystals
Body centered cubic (BCC) unit cell has an atom at each corner of the cube and one atom at the center of the cube. Only one-eighth of each corner atom however belongs to the unit cell as shown in Figure 2. There are therefore 2 atoms in the unit cell.