EEE 352—Spring 2011 Homework 1
1.1
Determine the number of atoms per unit cell in a (a) face-centered cubic, (b) body-centered cubic, and (c) diamond lattice. (a) The body centered cubic has 8 corner atoms, each of which is shared among 8 unit cells, so this gives 1 atom. Then, there are 6 face atoms, each of which is shared among 2 cells, so this gives 3 atoms. Hence, the FCC has 1+3 = 4 atoms per unit cell. (b) The body-centered cubic has 8 corner atoms, each of which is shared among 8 unit cells, so this gives 1 atom. Then, there is 1 atom at the center of the cube. Hence, the BCC has 1+1 = 2 atoms per unit cell. (c) The diamond lattice has the FCC cell with 2 atoms per position. This gives the additional 4 internal atoms. atoms. Hence, the diamond lattice has 2(1+3) = 1+3+4 = 8 atoms per unit cell.
1.2
(a) The lattice constant of GaAs is 5.65 Å. Determine the number of Ga atoms and As atoms per cm3. (b) Determine the volume density of germanium atoms in a germanium semiconductor. The lattice constant of germanium is 5.65 Å. The lattice of Ge is is diamond, while the lattice lattice of GaAs is zincblende. These are both variations of the diamond lattice. In the diamond lattice, lattice, there are 8 atoms per unit cell. Therefore the atomic density is ρ atomic
8 =
(0.565 × 10
7
−
cm)
=
3
4.436 × 10 22 cm − 3 .
(a) In GaAs, one-half of the atoms are Ga and one-half are As. Therefore ρ Ga
=
ρ As
=
2.218 ×10
22
−3
cm
.
(b) In Ge, all of the atoms are Ge. Therefore ρ Ge
1.3
=
4.436 × 10
22
−3
cm
.
Assume that each atom is a hard sphere with the surface of each atom in contact with the surface surface of its nearest neighbor. Determine the percentage of total unit cell volume that is occupied in (a) a simple cubic lattice, (b) a face-centered cubic lattice, (c) a body-centered cubic lattice, and (d) a diamond lattice. (a) In the simple cubic lattice, lattice, the nearest neighbor is along a cube edge. Hence the cube edge a is the diameter of the sphere. There are 8 corner atoms, each shared with eight unit cells, so that there is a net of only 1 atom per unit cell. 3 Hence, the volume of the unit cell is a , while the volume of the atom is 3 4π(a/2) /3. Thus, the percentage of the cell occupied by atom is
per .
1 4 " # a & =
a
3
3
3
"
% ( $ 2'
=
6
=
52.4% .
(b) In the face-centered cube, the nearest neighbor distance is between a corner atom and a face-atom. So, if the edge of the cube is a, the radius of the atom is 2 a / 4 . In the FCC lattice, there are eight corner atoms, each shared with eight unit cells (=1), plus six face atoms, each shared with two unit cells (=3), which gives a total of 4 atoms per unit cell. Hence, the percentage is per .
1 =
a
3
4
4 " # 3
2a &
% $
4
3
( '
2" =
=
6
74% .
(c) In the body-centered cubic, the nearest neighbor distance is along the diagonal of the cube. Hence, the radius of the atom is given by 3a / 4 . In the BCC lattice, there are eight corner atoms, each shared with eight unit cells (=1), plus one body atom which is entirely within the cell, so that there are 2 atoms per cell. Hence, the percentage is
per .
1 =
a
3
2
4 " # 3
3a &
% $
4
3
3"
( '
=
=
8
68%
.
(d) In the diamond lattice, the nearest neighbor distance is 3a / 4 , so that the radius of the atom is one half this, or 3a /8 . In this structure, there are eight corner atoms, each shared with eight unit cells (=1), six face atoms, each shared with two unit cells (=3), and four interior atoms totally within the unit cell, so that there are 8 atoms per cell. Hence, the percentage is
per .
1 =
a
1.5
3
8
4 " # 3
3a &
% $
8
3
( '
3" =
16
=
34%
.
If the lattice constant of silicon is 5.43 Å, calculate (a) the distance from the center of one silicon atom to the center of its nearest neighbor, (b) the number density of silicon atoms, and (c) the mass density (grams per cm3 ) of silicon. (a) If one Si atom is located at (0,0,0), the nearest neighbor is at (1/4,1/4,1/4). Hence, we take ¼ of the edge and multiply by the square root of 3, as d nn
3 =
4
a
3 =
4
5.43
=
2.35 Å.
(b) Here, we use the same procedure as in 1.2, as ρ Si
8 =
(0.543 × 10 − 7 cm) 3
=
5.0 × 10 22 cm − 3 .
(c) In this case, we use Avogadro’s number, which is in units of atoms per mole, and the atomic weight, which is in grams per mole to get
6.02 × 10 28.09
23 =
2.143 × 10
22
atoms / gm .
We now divide this into the atomic density in part (b) to get the mass density of Si to be
ρ M , Si
5.0 × 10
22
=
2.143 × 10
22
=
3 2.33 gm / cm .
