Homework 1 Grading Key

 

Due:  09/25/03 before class

 

1.  Calculate the volume density of Si atoms (number of atoms/cm3) given that the lattice constant of Si is 5.43 angstroms.  Calculate the areal density of atoms (number/cm2) 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 GaxIn1-xAs 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 aGaAs = 5.6533 angstroms and aInAs = 6.0584 angstroms are the lattice constants of the two semiconductors.  Also, aInP = 5.8686 angstroms)

Solution (10 pts):    The problem itself is fairly simple.  All you have to do is choose the correct composition of GaxIn1-xAs that is lattice matched to the InP substrate.  Hence we have the following condition:

                                                   

                                   

This gives us  and hence we need to grow .