Homework 1
Due: 09/25/03 before class
1. Calculate the maximum packing fraction of the unit cell volume that can be filled
by hard spheres in the SC, BCC, FCC, Diamond and Hexagonal structures.
Which structure most efficiently fills space?
SC: It is easy to determine that there is 1 lattice point per unit cell. The
a
maximum radius that a hard sphere can have is r  , where a is the lattice
2
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:
3
4 a
4
1  r 3 1   

3 2
3
P.F . 

  0.5236
3
3
6
a
a
B.C.C: It is easy to determine that there is 2 lattice points per unit cell. The
3
a , where a is the lattice
maximum radius that a hard sphere can have is r 
4
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:
3
4  3 
4
3
2

 
a 
2  r
3
4

  3  0.6802
3
P.F . 

3
3
8
a
a
F.C.C: It is easy to determine that there is 4 lattice points per unit cell. The
2
a , where a is the lattice
maximum radius that a hard sphere can have is r 
4
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:
3
4  2 
4


4  r3 4 3  4 a

  2  0.7405
3
P.F . 

3
3
6
a
a
Diamond: It is easy to determine that there is 8 lattice points per unit cell. The
3
a , where a is the lattice
maximum radius that a hard sphere can have is r 
8
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:
3
4  3 
4


8  r3 8 3   8 a

  3  0.3401
3
P.F . 

3
3
16
a
a
Hexagonal: It is easy to determine that there is 3 lattice points per unit cell. If
a
c  a , then the maximum radius that a hard sphere can have is r  , where a is
2
3 3 3
a .
the lattice constant. The volume of the hexagonal unit cell if c  a is
2
Hence the packing fraction is:
3
4 a
4
3  r3 3   
3 2
3
P.F . 


3 3 3
3 3 3
a
a
2
2
2
 0.6046
6
It is seen that the FCC structure most efficiently fills space.
2. 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 and (111) plane.
Solution: Because Si has a Diamond structure, there are 8 Si atoms per
conventional unit cell with a volume of a 3  5.433  160.1 Angstrom3 . Hence the
volume density is:
number of Si atoms
Si atoms
8

 0.05
3
volume
160.1 Angstrom
Angstrom 3
For the areal density of atoms on the (100) plane, there are 2 Si atoms per a 2 area.
Hence the areal density is:
number of Si atoms
Si atoms
2

 0.0678
2
area
29.49 Angstrom
Angstrom 2
For the areal density of atoms on the (111) plane, there are 2 Si atoms per
3 2
a area. Hence the areal density is:
2
number of Si atoms
Si atoms
2

 0.0144
2
area
138.65 Angstrom
Angstrom 2
3. 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?
a GaInAs ( x)  x  a GaAs  (1  x)  a InAs
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: 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:
a InP  aGaInAs
5.8686  x  5.6533  (1  x)  6.0584
This gives us x  0.4685 and hence we need to grow Ga0.4685In0.5315 As .
4. The ratio of the lengths of the reciprocal lattice vectors for the S.C., B.C.C.,
F.C.C. structures are given below:
Ratio
Table


Ko Ko


K1 K o


K2 Ko


K3 Ko


K4 Ko


K5 Ko


K6 Ko


K7 Ko
S.C. Reciprocal
Lattice
1
B.C.C. Reciprocal
Lattice
1
2
2
3
2 2
3
3
11
3
2
2
2
2
5
6
3
F.C.C. Reciprocal
Lattice
1
4
5
3
6
2 2
19
3
3
2 5
3
7
2 2
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 d j :
Atom
1 and 5
FCC 1

d1  0
2 and 6

a
d 2   xˆ  yˆ 
2

a
d 3   xˆ  zˆ 
2

a
d 4   yˆ  zˆ 
2
3 and 7
4 and 8
FCC 2

 a
a
d 5  d1  xˆ  yˆ  zˆ    xˆ  yˆ  zˆ 
4
4

 a
a
d 6  d 2  xˆ  yˆ  zˆ   3xˆ  3 yˆ  zˆ 
4
4

 a
a
d 7  d 3  xˆ  yˆ  zˆ   3xˆ  yˆ  3zˆ 
4
4

 a
a
d 8  d 4  xˆ  yˆ  zˆ    xˆ  3 yˆ  3zˆ 
4
4

2
2
2
xˆ  n2
yˆ  n3
zˆ :
The structure factor for K  n1
a
a
a
 
 
8

 2 iK v   4 iK d 
iK d
S K  e j  e j  e j 
j 1
 s 1
  s 1

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:
 
  4 iK d 
S FCC K    e j   1  e i n2  n3   e i n1  n3   e i n1  n3 
 s 1

If all the integers n1, n2, n3 are either all even or all odd then the structure factor is
S FCC n1 , n2 , n3   4 . If any integer is odd while the other two are even or if one is
even while the other two are odd then S FCC n1 , n2 , n3   0 .
For the summation over two atoms, the structure factor is:
 

  2 iK d 
 i  n1  n 2  n3  

S2  atom K    e j   1  e 2
 s 1

 
 

If n1  n2  n3  2  even , then S2  atom K  2 . If n1  n2  n3  2  odd , then


S2  atom K  0 . If n1  n2  n3  odd , then S2  atom K  1  i . Let us compare the
FCC and Diamond structure factors as a function of n1 , n2 , n3 .
 
n1 , n2 , n3
0,0,0
1,0,0
1,1,0
1,1,1
2,0,0
2,1,0
2,1,1
2,2,0
2,2,1
3,0,0
3,1,0
3,1,1
2,2,2
 
FCC
Diamond
4
0
0
4
4
0
0
4
0
0
0
4
4
4
0
0
4  1  i 
0
0
0
8
0
0
0
4  1  i 
0
We see that the reciprocal lattice of the FCC is a BCC lattice with a lattice
4
constant of
. The reciprocal lattice of the diamond structure is similar the
a
BCC structure but with several of the reciprocal lattice points vanishing including
the n1 , n2 , n3  2,0,0 and the n1 , n2 , n3  2,2,2 reciprocal lattice point which are
the BCC reciprocal lattice translation vectors that have the second and fifth

shortest translation vectors (excluding K  0 ). Hence our new ratio table with
the Diamond structure data is:
Ratio
Table

Ko

K1

K2

K3

Ko

Ko

Ko

Ko
F.C.C. Lattice
(i.e. Reciprocal
lattice of the BCC)
1
B.C.C. Lattice
(i.e. Reciprocal
lattice of the FCC)
1
1
Missing
2 =1.41
2
3 =1.73
2 2
3 =1.63
2 2
3 =1.63
11
3 =1.91
11
3 =1.91
2
3 =1.15
Diamond
Reciprocal Lattice

K4

K5

K6

K7

Ko

Ko

Ko

Ko
2
5 =2.24
Missing
---
3 =2.30
6 =2.45
4
7 =2.65
19
3 =2.52
---
2 2 =2.82
2 5
3 =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 K  2k sin   ,
2
we have:
Sample A
n
42.2
49.2
72.0
87.3

sin  n
 2

sin  o
 2
1
1.156
1.633
1.917






Sample B
n
28.8
41.0
50.8
59.6

sin  n
 2

sin  o
 2
1
1.408
1.725
2






Sample C
n
42.8
73.2
89.0
115.00

sin  n
 2

sin  o
 2
1
1.634
1.921
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.
a. First solve for the normalized wavefunctions and their energies for the
unperturbed infinite well of width d. Graph the energies of the states as a function

of the wavevector k .
Solution: Using the Schrödinger equation, the boundary conditions and the
normalization condition it is easy to solve for the electron wavefunctions and
associated quantized energy levels:
2
sin k n x 
d
 2 k n2
En 
2m
n 
kn 
n
d
n = integer
b. Now introduce a small periodic perturbation at the bottom of the potential well
such that the shape of the bottom of the well is now given by the equation:
 
U   cos  x
2N
and  is the strength of the periodic perturbation. We are now
d
going to use time independent perturbation (TIP) theory to evaluate the effects on
the electron states. To do this, you need to note that every state has a two-fold
energy degeneracy because Ek   E k  , hence you have to use the form of TIP
that can handle this degeneracy. I want you to do the following:
where  
1. Calculate the first order energy corrections to the electron states.
Solution: You have to use the form of TIP that can handle degeneracy because
Ek   E k  . To do this, you have to calculate the following:
 k U  k
In the case of two-fold degeneracy that we have in this problem, we have to solve
the following eigenvalue/eigenvector equation for each  k :
  k U  k

  k U   k

  k U  k
 k U  k
  1 

   E  1 
 
  2 
 2

The matix elements  k U  k  are easily calculated to be:
 k U  k  0
  k U  k 
 k U  k 
 k U  k 


 2

 0


 2

 0
0
if k  

2
otherwise
if k  

2
otherwise
Hence the eigenvalue/eigenvector problem becomes:

0

 
2


2   1   E   1 
 
 
 2
0   2 

The solutions to this is:
  1  1 1
  
 
2 1
 2 
E
 1 
1 1
  
 
2   1
 2 
E

2

2


Thus, unless k   , the perturbation does not cause any effect. But at k   ,
2
2

there is a first-order energy correction of E   .
2
2. How are the wavefunctions changed? Are any of the wavefunctions
hybridized?

Solution: When k   , the perturbation does not cause any first-order change
2

in the wavefunctions. But at k   , the wavefunctions are hybridized resulting
2
in the new wavefunctions:
  
1
2
     

3. At the Brillouin zone boundaries of k   , what is the value of the energy
2
gap that is produced from the two initially degenerate energy levels? That is, the

two states with k   initially had the same energy but now the states are
2
changed slightly resulting in two new states with a slightly increased energy for
one and a slightly decreased energy for the other. The difference between these
two new states will be the energy bandgap.
Solution: The bandgap that is opened up is E  E  