Homework 1 Grading Key

Due: 09/25/03
before class

1. Calculate the
volume density of Si atoms (number of atoms/cm^{3}) given that the
lattice constant of Si is 5.43 angstroms.
Calculate the areal density of atoms (number/cm^{2}) on the
(100) plane. Calculate the distance between two adjacent (111) planes in Si
passing through nearest-neighbor atoms.

*Solution:* Because Si has a Diamond structure, there
are 8 Si atoms per conventional unit cell with a volume of _{}. Hence the volume
density is:

_{} (5 pts)

For the areal density of atoms on
the (100) plane, there are 2 Si atoms per _{}area. Hence the areal
density is:

_{} (10 pts)

The distance formula to calculate the distance between planes is:

_{}

where *h,k,l* are the miller
indices for the plane. For the (111)
plane, this gives a value of 3.14 Angstrom
(10 pts).

2. How many atoms
are found inside a unit cell of a SC, BCC and FCC crystal structure? How far apart in terms of lattice constant *a* are nearest-neighbor atoms in each
case, measured from the center of the atoms.

*Solution:*
There are 1 (5 pts), 2 (5 pts), and 4 (5 pts)
lattice points per unit cell for the SC, BCC, and FCC lattices respectively.

3. Calculate the maximum packing fraction of the unit cell volume that can be filled by hard spheres in the SC, BCC, FCC and Diamond structures. Which structure most efficiently fills space? Can you think of another lattice that is even more efficient at using the available volume?

*SC (10
pts)*: It is easy to
determine that there is 1 lattice point per unit cell. The maximum radius that a hard sphere can
have is _{}, where *a *is the lattice constant. When the radius is this value, then the
spheres on the corners of the unit cell just touch each other. Hence the packing fraction is:

_{}

*B.C.C (10 pts)*: It is easy to determine that there is 2
lattice points per unit cell. The
maximum radius that a hard sphere can have is _{}, where *a *is the lattice constant. When the radius is this value, then the
spheres on the corners of the unit cell just touch the body-centered
sphere. Hence the packing fraction is:

_{}

*F.C.C (10 pts)*: It is easy to determine that there is 4
lattice points per unit cell. The
maximum radius that a hard sphere can have is _{}, where *a *is the lattice constant. When the radius is this value, then the
spheres on the corners of the unit cell just touch the face-centered
sphere. Hence the packing fraction is:

_{}

*Diamond (10 pts)*: It is easy to determine that there is 8
lattice points per unit cell. The
maximum radius that a hard sphere can have is _{}, where *a *is the lattice constant. When the radius is this value, then the
spheres on the corners of the unit cell just touch the nearest neighbor
sphere. Hence the packing fraction is:

_{}

It is seen that the FCC structure most efficiently fills space.

4. Epitaxial
growth of a semiconductor is a method in which a thin semiconductor film is
grown on a semiconductor substrate. Two
common epitaxial growth methods are molecular beam epitaxy (MBE) and
metalorganic chemical vapor deposition (MOCVD). MBE is a process where atoms or molecules are incident on the
substrate and bond to the substrate producing a film. MOCVD is a process in which a flow of gas with different species
passes over a substrate producing a chemical reaction and film growth. If you want a decent thickness for the film
however, the lattice constants of the substrate and film materials have to
match or else defects in the film called misfit dislocations will form. This requirement places a severe restriction
on the available materials for the film.
This question addresses these restrictions.

An engineer wants to grow Ga_{x}In_{1-x}As on an InP
substrate. What value of x (the
fraction of Ga and the fraction of In) is needed so that he can grow a
relatively thick film without defects occurring?

(Hint: A good approximation for the
lattice constant of a ternary semiconductor

is that changes linearly as the mole fraction x is changed:

_{}

where the terms a_{GaAs} = 5.6533 angstroms and a_{InAs}
= 6.0584 angstroms are the lattice constants of the two semiconductors. Also, a_{InP }= 5.8686 angstroms)

*Solution (10 pts):* The problem itself is fairly simple. All you have to do is choose the correct
composition of Ga_{x}In_{1-x}As that is lattice matched to the
InP substrate. Hence we have the
following condition:

_{}

_{}

This gives us _{} and hence we need to
grow _{}.