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Problem Set Solution# 1
ECEN 3320 Fall 2012
Semiconductor Devices
August 27, 2012 – Due September 5, 2012
- Find the Miller indices for the planes shown below.
Figure 1: Planes indicating possible crystal stuctures.
Solution: (a) (3, 3 , 3)! (1/ 3 , 1 / 3 , 1 /3)! (1, 1 , 1). (b) (3, 2 , 2)! (1/ 3 , 1 / 2 , 1 /2)! (2, 3 , 3)
- Consider the three cubic lattices shown below where (a) could be a simple cubic, a face- centered cubic (fcc) or a body-centered cubic (bcc). In the successively labeled panes, the planes a) [100], b) [110] and c) [111] are sketched in on the simple cubic lattice. Suppose all three cubes have the same lattice constant a =0.7nm, that is, the cubic unit cells all have a side length of 0.7 nm.
(a) Suppose one atom is placed at each lattice point of these three lattices. Calculate the volume density of atoms (i. e. the number of atoms per unit volume in units of cm ^3 ) for these three crystals. Solution: The volume of the cubic unit cell is given by
a 3 = (0.7nm) 3 = 0.343nm 3 such that 1 a 3 == 2. 92 ⇥ 1021 /cm 3.
Figure 2: Schematics of three cubic lattices simple cubic, face-centered cubic or body-centered cubic. The planes a) [100], b) [110] and c) [111] are sketched in on the simple cubic lattice in the successively labeled panels.
The simple cubic has one atom per cubic unit cell (1/8 of an atom at each of the 8 corners=1 atom).
n (^) c =
1 atom unit cell volume
0 .343nm 3 = 2. 92 ⇥ 1021 /cm 3
The FCC cubic has four atoms per cubic unit cell (1/8 of an atom at each of the 8 corners+1/2 of an atom on each of 6 facet=4 atoms).
n (^) c = 4 atoms unit cell volume
0 .343nm 3
= 1. 17 ⇥ 1022 /cm 3
The BCC cubic has two atoms per cubic unit cell (1/8 of an atom at each of the 8 corners+1 atom in the middle=2 atoms=1).
n (^) c = 2 atoms unit cell volume
0 .343nm 3
= 5. 83 ⇥ 1021 /cm 3
(b) For the fcc crystal, calculate the surface density of atoms (number of atoms per unit area in units of cm ^2 ) on the (100) plane. Solution: The (100) section of the cubic unit cell of the fcc lattice has 4 spheres at the corners (each having 1/4 inside the unit cell face) and 1 sphere at the center. There are a total of 2 spheres in the area of a 2. The surface density is then
n =
(7 ⇥ 10 ^8 ) 2
= 4. 08 ⇥ 1014 cm ^2
(c) For the fcc crystal, calculate the surface density of atoms on the (111) plane. Solution: The (111) section of the cubic unit cell of the fcc lattice appears as that depicted in figure 3. The 3 spheres at the corners contribute 1/6 each and the ones on the side con- stributes 1/2. So there are a total of 2 spheres in the area of the equilateral triangle with side
p 2 a. The surface density is therefore
n =
p 3 2 a^ 2
p 3 (7 ⇥ 10 ^8 ) 2
= 4. 71 ⇥ 1014 cm ^2
