0.4 Tutorial on chapter 1-crystal and crystal growth.  (Page 8/2)

Tutorial on Chapter 1-Crystal and Crystal Growth.

1. Give the percentage packing of Face-Centered-Cube ? [Ans.74%]

Method:

FCC crystal has 8 atoms at the 8 corners of the cube and 6 atoms on the 6 faces (4 on side faces+2 on the upper and lower faces].

Therefore N = the coordination Number = the number of atoms per unit cell = (8×1/8+6×1/2)= 4 atoms per unit cell.

The atoms are solid spheres and are closely packed so that the atom at the center of the face are touching the 4 atoms on the four corners of the face as shown in the Figure I.

Inspecting the Figure we find that:

4R=diagonal of the face=a√2.

Therefore R/a = √2/4=1/(2√2)

Therefore packing percentage in FCC crystals is 74%.

1. Give the percentage packing of Body-Centered-Cube ? [Ans.68%]

Method:

BCC crystal has 8 atoms at the 8 corners of the cube and 1 atom at the center of the Cube. Therefore N = the coordination Number = the number of atoms per unit cell = (8×1/8+1)= 2 atoms per unit cell.

The atoms are solid spheres and are closely packed so that the atom at the center of the Cube is touching the 2 atoms on the two corners of the cross diagonal of the cube as shown in the Figure 2.

Inspecting the Figure we find that:

4R= cross diagonal of the Cube =a√3.

Therefore R/a = √3/4;

Therefore packing percentage in FCC crystals is 68%.

1. Determine the conducting electrons density in FCC Cu where the lattice parameter is 3.6 Angstrom ? [Ans. 8.5734×10 ²⁸ /m ³ ]

Copper is univalent crystal hence each atom contributes one electron. Therefore atomic density gives the conducting electron density.

Coordination Number of Cu = N = 4 atoms per unit cell.

Unit cell volume is a ³ .

Therefore atomic density = conducting electron density = 4 / (a ³ ) = 8.5734×10 ²⁸ /m ³ = 8.5734×10 ²² /cc.

1. Si, Ge and GaAs are all diamond crystal structure. Diamond structure is two interpenetrating FCC crystals with one sub-lattice displaced w.r.t. the other along the cross diagonal of the cube by a quarter of the cross diagonal length. In case of Ga As it is ZincBlende cubic structure which has one sub-lattice of Ga and the other sub-lattice of As.

Find the weight density of Si,Ge and GaAs. Given Avogadro Number = N _Avo =6.02×10 ²³ atoms/gm-mole, Si, Ge and GaAs have atomic weight of 28.1, 72.6 and 144.63(mean=72.315 respectively and the lattice parameters are 5.43A°, 5.646A° and 5.6533A° respectively.

[Ans. 2.33gm/cc, 5.37gm/cc and 5.32gm/cc respectively]

Method:

Coordination Number =N= 8 corner atoms + 6 face center atoms + 4 body center atoms= (1/8×8 + 1/2×6+ 4)= 8 atoms per unit cell.

Therefore Number density = N ^* = 8/a ³ = 4.9967×10 ²² atoms/cc, 4.445 ×10 ²² atoms/cc, 4.4277 ×10 ²² atoms/cc,of Si, Ge and GaAs respectively.

Weight of one atom= (AW gm/mole)÷ (N _Avo atoms/mole)= 4.66×10 ^-23 gm/atom. 1.157×10 ^-22 gm/atom. 1.244×10 ^-22 gm/atom of Si, Ge, GaAs respectively.

Therefore weight density = Wt of one atom× N ^* = Weight Density = ρ

Therefore density of Si=4.9967×10 ²² atoms/cc×4.6677×10 ^-23 gm/atom=2.33gm/cc.