Homework 1
Due: 09/25/03
before class
SC: 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: 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: 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: 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:
_{}
Hexagonal: It is easy to determine that there is 3 lattice points per unit cell. If _{}, then the maximum radius that a hard sphere can have is _{}, where a is the lattice constant. The volume of the hexagonal unit cell if _{} is _{}. Hence the packing fraction is:
_{}
It is seen that the FCC structure most efficiently
fills space.
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:
_{}
For the areal density of atoms on the (100) plane, there are 2 Si atoms per _{}area. Hence the areal density is:
_{}
For the areal density of atoms on the (111) plane, there are 2 Si atoms per _{}area. Hence the areal density is:
_{}
Ratio Table |
S.C. Reciprocal Lattice |
B.C.C. Reciprocal Lattice |
F.C.C. Reciprocal Lattice |
_{} |
1 |
1 |
1 |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
2 |
_{} |
2 |
_{} |
_{} |
2 |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
_{} |
3 |
_{} |
_{} |
a.
Now calculate the set of ratios for the Diamond
structure. You will have to review
the concept of the structure factor in my notes online in order to do this
problem
Note: There are several ways to do this
problem. One way that may be easier
than other ways is to consider the diamond structure as a simple cubic crystal
structure with a basis of eight atoms
Solution: You need to
calculate the structure factor for this case.
I will use the method that considers the Diamond structure a SC lattice
with a basis of eight atoms. The eight
atoms have the following displacement vectors _{}:
Atom |
FCC 1 _{} |
FCC 2 |
1 and 5 |
_{} |
_{} |
2 and 6 |
_{} |
_{} |
3 and 7 |
_{} |
_{} |
4 and 8 |
_{} |
_{} |
The structure factor for _{}:
_{}
The summation over 4 atoms is the structure factor for the FCC and the summation over 2 atoms is an additional structure factor caused by the fact that there for the diamond structure there are two atoms at every FCC lattice point.
For the FCC structure, the structure factor is:
_{}
If all the integers n_{1}, n_{2}, n_{3} are either all even or all odd then the structure factor is _{}. If any integer is odd while the other two are even or if one is even while the other two are odd then _{}.
For the summation over two atoms, the structure factor is:
_{}
If _{}, then _{}. If _{}, then _{}. If _{}, then _{}. Let us compare the FCC and Diamond structure factors as a function of _{}.
_{} |
FCC |
Diamond |
0,0,0 |
4 |
4 |
1,0,0 |
0 |
0 |
1,1,0 |
0 |
0 |
1,1,1 |
4 |
_{} |
2,0,0 |
4 |
0 |
2,1,0 |
0 |
0 |
2,1,1 |
0 |
0 |
2,2,0 |
4 |
8 |
2,2,1 |
0 |
0 |
3,0,0 |
0 |
0 |
3,1,0 |
0 |
0 |
3,1,1 |
4 |
_{} |
2,2,2 |
4 |
0 |
We see that the reciprocal lattice of the FCC is a BCC lattice with a lattice
constant of _{}. The reciprocal
lattice of the diamond structure is similar the BCC structure but with several
of the reciprocal lattice points vanishing including the _{} and the _{} reciprocal lattice
point which are the BCC reciprocal lattice translation vectors that have the
second and fifth shortest translation vectors (excluding _{}). Hence our new
ratio table with the Diamond structure data is:
Ratio Table |
F.C.C. Lattice (i.e. Reciprocal lattice of the BCC) |
B.C.C. Lattice (i.e. Reciprocal lattice of the FCC) |
Diamond Reciprocal Lattice |
_{} |
1 |
1 |
1 |
_{} |
_{}=1.41 |
_{}=1.15 |
Missing |
_{} |
_{}=1.73 |
_{}=1.63 |
_{}=1.63 |
_{} |
2 |
_{}=1.91 |
_{}=1.91 |
_{} |
_{}=2.24 |
2 |
Missing |
_{} |
_{}=2.45 |
_{}=2.30 |
--- |
_{} |
_{}=2.65 |
_{}=2.52 |
--- |
_{} |
_{}=2.82 |
_{}=2.58 |
--- |
b. Now armed with these ratios let us move on to the problem in Ashcroft and Mermin. Identify the F.C.C., B.C.C. and Diamond structure and calculate the lattice constant.
Using the equation for the magnitude of the reciprocal lattice vector _{}, we have:
Sample A _{} |
_{} |
Sample B _{} |
_{} |
Sample C _{} |
_{} |
42.2 |
1 |
28.8 |
1 |
42.8 |
1 |
49.2 |
1.156 |
41.0 |
1.408 |
73.2 |
1.634 |
72.0 |
1.633 |
50.8 |
1.725 |
89.0 |
1.921 |
87.3 |
1.917 |
59.6 |
2 |
115.00 |
2.311 |
We compare the experimental ratios and the theoretical ratios and we see that Sample A has a FCC real space lattice, Sample B has a BCC real space lattice and Sample C has a Diamond real space lattice.
5. The Kronig-Penny model for a periodic potential leads to the concept of energy bandgaps. Let us approach the situation in a slightly different way by looking at a particle in a 1D infinite box of length d that has a small periodic perturbation in the bottom of the well (see figure below).
I want you to investigate a couple of results of this perturbation on the energy and wavefunctions of the electron states in this potential well.