ALL 11 CHAPTERS COVERED
SOLUTIONS MANUAL
, Contents
1 Structural Properties of Semiconductors 1
1.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
2 Semiconductor Bandstructure 11
2.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 11
3 Bandstructure Modi cations 25
3.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 25
4 Transport: General Formalism 33
4.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 33
5 Defect and Carrier{Carrier Scattering 37
5.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 37
6 Lattice Vibrations: Phonon Scattering 43
6.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 43
7 Velocity{Field Relations in Semiconductors 52
7.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 52
8 Coherence, Disorder, and Mesoscopic Systems 56
8.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 56
9 Optical Properties of Semiconductors 62
9.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 62
10 Excitonic E ects and Modulation of Optical Properties 74
10.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 74
11 Semiconductors in Magnetic Fields 85
11.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 85
2 Semiconductor Bandstructure 1
2.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
3 Bandstructure Modi cations 1
3.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
i
,ii CONTENTS
4 Transport: General Formalism 1
4.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
5 Defect and Carrier{Carrier Scattering 1
5.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
6 Lattice Vibrations: Phonon Scattering 1
6.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
7 Velocity{Field Relations in Semiconductors 1
7.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
8 Coherence, Disorder, and Mesoscopic Systems 1
8.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
9 Optical Properties of Semiconductors 1
9.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1
10 Excitonic E ects and Modulation of Optical Properties 1
10.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
11 Semiconductors in Magnetic Fields 1
11.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
, CHAPTER
1
STRUCTURAL
PROPERTIES OF
SEMICONDUCTORS
1.1 PROBLEMS AND SOLUTIONS
1.1 Consider the (001) MBE growth of GaAs by MBE. Assuming that the sticking
coecient of Ga is unity, calculate the Ga partial pressure needed if the growth rate
has to be 1 m/hr. The temperature of the Ga cell is 1000 K.
Solution 1.1 A 1.0 m/hr growth rate means that the time taken to grow one
monolayer is
0 1
2:825 2 1008 cm
tm` =
(1:0 2 1004 cm)
2 (60 2 60) s
= 1:017 s
The surface density of Ga atoms on the (001) plane is
2 2
NGa = 2 =
a (5:65 2 1008 cm)2
= 6:265 2 1014 cm02
Thus the ux needed is
0 1
6:265 2 1014 cm02
F =
(1:017 s)
= 6:16 2 1014 cm02 s01
The partial pressure needed is then,
p 0 1
F m(g)T (K ) 6:16 2 1014 (70 2 1000)1=2
P = torr =
3:5 2 10 22 3:5 2 1022
,2 CHAPTER 1. STRUCTURAL PROPERTIES OF SEMICONDUCTORS
= 4:656 2 1006 torr
1.2 In the growth of GaAs/AlAs structures in a particular MBE system, the
background pressure of Ga when the Ga shutter is o is 1007 Torr. If the growth
rate of AlAs is 1 m/hr, what fraction of Ga atoms are incorporated in the AlAs
region? The Ga cell is at 1000 K.
Solution 1.2 The Ga ux is given by
3:5 2 1022 2 1007
F = cm02 s01
(70 2 1000)1=2
= 1:323 2 1013 cm02 s01
This gives a growth rate of GaAs of (surface Ga density = 6:265 2 1014 cm02)
0 1
1:323 2 1013 cm02 s01
R(GaAs) =
(6:265 2 1014 cm02 )
= 2:11 2 1002monolayer per second
This gives a growth rate of (in microns/hr)
0 10 1
R(GaAs) = 2:11 2 1002 2:825 2 1002 (60 2 60)
= 2:15 2 1002 m=hr
The fraction of GaAs incorporated is then
2:15 2 1002
x(GaAs) = = 2:15%
1:0
1.3 A 5.0 m Si epitaxial layer is to be grown. The Si ux is 10
14 cm02 s01 . How
long will it take to grow the lm if the sticking coecient is 0.95?
Solution 1.3 Let us assume the growth is along the (100) direction. The density
of surface atoms is 6:78 2 1014 cm02. The growth time is
(5 2 1004 cm) (6:78 2 1014 cm02s01 ) 1
t(5:0 m) =
(2:715 2 1008 cm)
2 (1014 cm02) 0:95
= 36:5 hours!
1.4 a) Find the angles between the tetrahedral bonds of a diamond lattice.
b) What are the direction cosines of the (111) oriented nearest neighbor bond along
the x,y,z axes.
Solution 1.4 a) To solve this problem we take the unit vectors along the tetra-
hedral bonds and take their dot product. Two unit vectors are
1 1
a1 p (111); a1 = p (11 1)
3 3
Thus
1 1 1
cos = a1 1 a2 = p 1 p (1 0 1 0 1) = 0
3 3 3
or = 109:47
,1.1. PROBLEMS AND SOLUTIONS 3
b) The four nearest neighbors are the points a4 (111); a4 (1 a (1
11); a
4 11); 4 (111). The
corresponding unit vectors are p3 (111); p3 (
1 1 11); p3 (1
1 1 11); p3 (
1 11
1).
1.5 Consider a semiconductor with the zinc blende structure (such as GaAs).
a) Show that the (100) plane is made up of either cation or anion type atoms.
b) Draw the positions of the atoms on a (110) plane assuming no surface recon-
struction.
c) Show that there are two types of (111) surfaces: one where the surface atoms are
bonded to three other atoms in the crystal, and another where the surface atoms are
bonded to only one. These two types of surfaces area called the A and B surfaces,
respectively.
Solution 1.5 a) The zinc-blende crystal can be thought of as two interpenetrating
fcc lattices. On one of these lattices one places, say, Ga atoms and on the other one
places, say, As atoms. On a (100) plane, there are atoms on one fcc lattice only. As
a result, the (100) plane atoms are either cations (Ga) or anions (As).
b) On the (110) plane there are both Ga and As atoms. The arrangements of atoms
is shown.
c) It is easy to see this by examining any (111) plane. If the crystal is on one side
of this plane, the surface atoms are bonded to only one atom in the crystal. If the
crystal in on the other side, one has 3 bonds to the crystal below. See Fig. 1.1 for
the various planes.
1.6 Suppose that identical solid spheres are placed in space so that their centers
lie on the atomic points of a crystal and the spheres on the neighboring sites touch
each other. Assuming that the spheres have unit density, show that density of such
spheres is the following for the various crystal structures:
p
fcc :
p2=6 = 0:74
bcc : 3=8 = 0:68
sc : =6 = 0:52
p
diamond : 3=16 = 0:34
Solution 1.6 To solve this problem we examine the radius of the sphere for each
kind of lattice.
fcc lattice:
The radius of the sphere is a=2 2.
p
The volume of the sphere is Vs = 12ap3 .
2
3
p2 a are 4.
Number of spheres in a volume
The density is, therefore, 6 .
bcc lattice: p
The radius of the sphere is 43a .
p a3
The volume of the sphere is 316 .
The number of spheres in p volume a3 are 2.
The density is, therefore, 83 .
,4 CHAPTER 1. STRUCTURAL PROPERTIES OF SEMICONDUCTORS
ATOMS ON THE (110) PLANE
Each atom has 4 bonds:
• 2 bonds in the (110) plane
• 1 bond connects each atom to
adjacent (110) planes
Cleaving adjacent planes
requires breaking 1 bond per atom
ATOMS ON THE (001) PLANE
2 bonds connect each atom to
adjacent (001) plane
Atoms are either Ga or As in a
GaAs crystal
Cleaving adjacent planes
requires breaking 2 bonds per atom
ATOMS ON THE (111) PLANE
Could be either Ga or As
1 bond connecting an adjacent
plane on one side
3 bonds connecting an adjacent
plane on the other side
Figure 1.1 : Some important planes in the cubic system along with their Miller indices.
This gure also shows how many bonds connect adjacent planes. This number determines
how easy or dicult it is to cleave the crystal along these planes.
,1.1. PROBLEMS AND SOLUTIONS 5
SC lattice:
The radius of the spheres is a2 .
The volume of the sphere is a6 . 3
Number of atoms per volume a3 are 1.
The density is, therefore, =6.
diamond: p
Radius of the sphere is p83a .
Volume of the sphere is 128 3a .
3
Number of spheres per a pare 7 (twice the number in an fcc crystal).
3
The density is, therefore, 163 .
1.7 Calculate the number of cells per unit volume in GaAs (a = 5.65 A). Si has a
4% larger lattice constant. What is the unit cell density for Si? What is the number
of atoms per unit volume in each case?
Solution 1.7 There are 4 unit cells in a volume a3, so that the unit cells per unit
volume are 4=a3(cm03).
1.8 A Si wafer is nominally oriented along the (001) direction, but is found to be
cut 2 o , towards the (110) axis. This o axis cut produces \steps" on the surface
which are 2 monolayers high. What is the lateral spacing between the steps of the
2 o -axis wafer?
Solution 1.8 The lattice constant a of Si is 5.43 A. The Si to Si atomic layer
distance am` in the (001) direction is (one-fourth of a) 1.358 A. Since each step is
two monolayers, the step spacing ds is given by
a
tan 2 = m`
ds
or
ds =
2 2 1:358 A = 77:8 A
0:0349
1.9 Conduct a literature search to nd out what the lattice mismatch is between
GaAs and AlAs at 300 K and 800 K. Calculate the mismatch between GaAs and
Si at the same temperatures.
Solution 1.9 The lattice constants of AlAs, GaAs and Si are (at room tempera-
ture)
AlAs : a = 5:6611 A i1 = 0:14%
GaAs : a = 5:6533 A i1 = 3:93%
Si : a = 5:431 A
The expansion coecients are
AlGaAs : 5:2 2 1006= K
GaAs : 6:86 2 1006= K
Si : 2:6 2 1006= K
1.10 In high purity Si crystals, defect densities can be reduced to levels of
1013 cm03. On an average, what is the spacing between defects in such crystals? In
,6 CHAPTER 1. STRUCTURAL PROPERTIES OF SEMICONDUCTORS
heavily doped Si, the dopant density can approach 1019 cm03. What is the spacing
between defects for such heavily doped semiconductors?
Solution 1.10 For a defect density of 10
13 cm03, the volume per unit defect is
4 3
r = 10013 cm3
3 d
or
rd = 2:88 2 103
A
The spacing between defects on an average is twice this value.
For a doping level of 1019 cm03, we have
4 3
r = 10019 cm3
3 d
or
rd = 29:2
A
The spacing on an average is twice this distance.
1.11 A GaAs crystal which is nominally along (001) direction is cut at an angle
o towards (110) axis. This produces one monolayer high steps. If the step size is
to be no more than 100 A, calculate .
Solution 1.11 The lattice constant of GaAs is 5.65 A and one monolayer distance
(Ga to Ga layer or As to As layer) distance is 2.825
A. The tilt angle is given by
= m`
a
tan
ds
where ds is the step spacing between steps. If ds is to be no more than 100
A, the
angle should be
0 2:825
= tan 1
= 1:65
100
The step spacing is often used to control monolayer by monolayer growth of epitax-
ially grown semiconductors.
1.12 Assume that a GaAs bond in GaAs has a bond energy of 1.0 eV. Calculate
the energy needed to cleave GaAs in the (001) and (110) planes.
Solution 1.12 In the (001) plane there are two GaAs bonds for each atom on the
plane. In the (110) plane there is only one such bond. The number density of Ga
atoms on the (001) plane is
2 2
NGa = = = 6:26 2 1014 cm02
a2 (5:65 2 1008 cm)2
Thus the energy needed to cleave the crystal along (001) planes is
E = 6:26 2 1014 2 2 eV=cm2
= 2 2 1004 J=cm2
, 1.1. PROBLEMS AND SOLUTIONS 7
A
a
c/2
A
A x x
a
y
A A A
a
Figure 1.2 : The geometry used to nd the relation between c and a (Problem 1.13).
On the (110) planes the density of Ga atoms is
NGa = p3 2 = p 3
2a 2 (5:65 2 1008 cm)2
= 6:64 2 1014 cm02
The energy to cleave the planes is
E = 6:64 2 1014 eV=cm02
= 1:06 2 1004 J=cm2
which is about half the value found for the (100) planes.
1.13
pConsider a hcp structure shown in Fig. 1.12. Prove the relation given by
c=a = 8=3 = 1:633.
Solution 1.13 We will nd the distance between the planes formed by the rst
set of atoms A and the second set of atoms B . This distance is c=2. To nd this in
terms of a, we consider the position of the atom B which is placed on top of the
center of a triangle of sides a, as shown in Figure 1.12.
To nd the distance c=2, we rst need to nd the distance x. Now (see Fig.
1.12)
a
y = tan30
2
a a2 a2 a2
x2 = y 2 + ( ) 2 = + =
2 12 4 3
2 2
(c=2)2 = a2 0 x2 = a
c
r 8
3
) a
=
3
= 1:633
1.14 A HgCdTe alloy is to be grown with 10% Hg. The sticking coecient of Hg
is 1002 and that of Cd is 1.0. Te is present in excess and does not limit the growth
of the crystal. Calculate the Hg and Cd uxes and partial pressures needed to grow
the lm at a rate of 1.0 m/hour.
Solution 1.14 This problem is similar to Problem 1.1.
SOLUTIONS MANUAL
, Contents
1 Structural Properties of Semiconductors 1
1.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
2 Semiconductor Bandstructure 11
2.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 11
3 Bandstructure Modi cations 25
3.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 25
4 Transport: General Formalism 33
4.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 33
5 Defect and Carrier{Carrier Scattering 37
5.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 37
6 Lattice Vibrations: Phonon Scattering 43
6.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 43
7 Velocity{Field Relations in Semiconductors 52
7.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 52
8 Coherence, Disorder, and Mesoscopic Systems 56
8.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 56
9 Optical Properties of Semiconductors 62
9.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 62
10 Excitonic E ects and Modulation of Optical Properties 74
10.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 74
11 Semiconductors in Magnetic Fields 85
11.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 85
2 Semiconductor Bandstructure 1
2.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
3 Bandstructure Modi cations 1
3.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
i
,ii CONTENTS
4 Transport: General Formalism 1
4.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
5 Defect and Carrier{Carrier Scattering 1
5.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
6 Lattice Vibrations: Phonon Scattering 1
6.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
7 Velocity{Field Relations in Semiconductors 1
7.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
8 Coherence, Disorder, and Mesoscopic Systems 1
8.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
9 Optical Properties of Semiconductors 1
9.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1
10 Excitonic E ects and Modulation of Optical Properties 1
10.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
11 Semiconductors in Magnetic Fields 1
11.1 Problems and Solutions : : : : : : : : : : : : : : : : : : : : : : : : 1
, CHAPTER
1
STRUCTURAL
PROPERTIES OF
SEMICONDUCTORS
1.1 PROBLEMS AND SOLUTIONS
1.1 Consider the (001) MBE growth of GaAs by MBE. Assuming that the sticking
coecient of Ga is unity, calculate the Ga partial pressure needed if the growth rate
has to be 1 m/hr. The temperature of the Ga cell is 1000 K.
Solution 1.1 A 1.0 m/hr growth rate means that the time taken to grow one
monolayer is
0 1
2:825 2 1008 cm
tm` =
(1:0 2 1004 cm)
2 (60 2 60) s
= 1:017 s
The surface density of Ga atoms on the (001) plane is
2 2
NGa = 2 =
a (5:65 2 1008 cm)2
= 6:265 2 1014 cm02
Thus the ux needed is
0 1
6:265 2 1014 cm02
F =
(1:017 s)
= 6:16 2 1014 cm02 s01
The partial pressure needed is then,
p 0 1
F m(g)T (K ) 6:16 2 1014 (70 2 1000)1=2
P = torr =
3:5 2 10 22 3:5 2 1022
,2 CHAPTER 1. STRUCTURAL PROPERTIES OF SEMICONDUCTORS
= 4:656 2 1006 torr
1.2 In the growth of GaAs/AlAs structures in a particular MBE system, the
background pressure of Ga when the Ga shutter is o is 1007 Torr. If the growth
rate of AlAs is 1 m/hr, what fraction of Ga atoms are incorporated in the AlAs
region? The Ga cell is at 1000 K.
Solution 1.2 The Ga ux is given by
3:5 2 1022 2 1007
F = cm02 s01
(70 2 1000)1=2
= 1:323 2 1013 cm02 s01
This gives a growth rate of GaAs of (surface Ga density = 6:265 2 1014 cm02)
0 1
1:323 2 1013 cm02 s01
R(GaAs) =
(6:265 2 1014 cm02 )
= 2:11 2 1002monolayer per second
This gives a growth rate of (in microns/hr)
0 10 1
R(GaAs) = 2:11 2 1002 2:825 2 1002 (60 2 60)
= 2:15 2 1002 m=hr
The fraction of GaAs incorporated is then
2:15 2 1002
x(GaAs) = = 2:15%
1:0
1.3 A 5.0 m Si epitaxial layer is to be grown. The Si ux is 10
14 cm02 s01 . How
long will it take to grow the lm if the sticking coecient is 0.95?
Solution 1.3 Let us assume the growth is along the (100) direction. The density
of surface atoms is 6:78 2 1014 cm02. The growth time is
(5 2 1004 cm) (6:78 2 1014 cm02s01 ) 1
t(5:0 m) =
(2:715 2 1008 cm)
2 (1014 cm02) 0:95
= 36:5 hours!
1.4 a) Find the angles between the tetrahedral bonds of a diamond lattice.
b) What are the direction cosines of the (111) oriented nearest neighbor bond along
the x,y,z axes.
Solution 1.4 a) To solve this problem we take the unit vectors along the tetra-
hedral bonds and take their dot product. Two unit vectors are
1 1
a1 p (111); a1 = p (11 1)
3 3
Thus
1 1 1
cos = a1 1 a2 = p 1 p (1 0 1 0 1) = 0
3 3 3
or = 109:47
,1.1. PROBLEMS AND SOLUTIONS 3
b) The four nearest neighbors are the points a4 (111); a4 (1 a (1
11); a
4 11); 4 (111). The
corresponding unit vectors are p3 (111); p3 (
1 1 11); p3 (1
1 1 11); p3 (
1 11
1).
1.5 Consider a semiconductor with the zinc blende structure (such as GaAs).
a) Show that the (100) plane is made up of either cation or anion type atoms.
b) Draw the positions of the atoms on a (110) plane assuming no surface recon-
struction.
c) Show that there are two types of (111) surfaces: one where the surface atoms are
bonded to three other atoms in the crystal, and another where the surface atoms are
bonded to only one. These two types of surfaces area called the A and B surfaces,
respectively.
Solution 1.5 a) The zinc-blende crystal can be thought of as two interpenetrating
fcc lattices. On one of these lattices one places, say, Ga atoms and on the other one
places, say, As atoms. On a (100) plane, there are atoms on one fcc lattice only. As
a result, the (100) plane atoms are either cations (Ga) or anions (As).
b) On the (110) plane there are both Ga and As atoms. The arrangements of atoms
is shown.
c) It is easy to see this by examining any (111) plane. If the crystal is on one side
of this plane, the surface atoms are bonded to only one atom in the crystal. If the
crystal in on the other side, one has 3 bonds to the crystal below. See Fig. 1.1 for
the various planes.
1.6 Suppose that identical solid spheres are placed in space so that their centers
lie on the atomic points of a crystal and the spheres on the neighboring sites touch
each other. Assuming that the spheres have unit density, show that density of such
spheres is the following for the various crystal structures:
p
fcc :
p2=6 = 0:74
bcc : 3=8 = 0:68
sc : =6 = 0:52
p
diamond : 3=16 = 0:34
Solution 1.6 To solve this problem we examine the radius of the sphere for each
kind of lattice.
fcc lattice:
The radius of the sphere is a=2 2.
p
The volume of the sphere is Vs = 12ap3 .
2
3
p2 a are 4.
Number of spheres in a volume
The density is, therefore, 6 .
bcc lattice: p
The radius of the sphere is 43a .
p a3
The volume of the sphere is 316 .
The number of spheres in p volume a3 are 2.
The density is, therefore, 83 .
,4 CHAPTER 1. STRUCTURAL PROPERTIES OF SEMICONDUCTORS
ATOMS ON THE (110) PLANE
Each atom has 4 bonds:
• 2 bonds in the (110) plane
• 1 bond connects each atom to
adjacent (110) planes
Cleaving adjacent planes
requires breaking 1 bond per atom
ATOMS ON THE (001) PLANE
2 bonds connect each atom to
adjacent (001) plane
Atoms are either Ga or As in a
GaAs crystal
Cleaving adjacent planes
requires breaking 2 bonds per atom
ATOMS ON THE (111) PLANE
Could be either Ga or As
1 bond connecting an adjacent
plane on one side
3 bonds connecting an adjacent
plane on the other side
Figure 1.1 : Some important planes in the cubic system along with their Miller indices.
This gure also shows how many bonds connect adjacent planes. This number determines
how easy or dicult it is to cleave the crystal along these planes.
,1.1. PROBLEMS AND SOLUTIONS 5
SC lattice:
The radius of the spheres is a2 .
The volume of the sphere is a6 . 3
Number of atoms per volume a3 are 1.
The density is, therefore, =6.
diamond: p
Radius of the sphere is p83a .
Volume of the sphere is 128 3a .
3
Number of spheres per a pare 7 (twice the number in an fcc crystal).
3
The density is, therefore, 163 .
1.7 Calculate the number of cells per unit volume in GaAs (a = 5.65 A). Si has a
4% larger lattice constant. What is the unit cell density for Si? What is the number
of atoms per unit volume in each case?
Solution 1.7 There are 4 unit cells in a volume a3, so that the unit cells per unit
volume are 4=a3(cm03).
1.8 A Si wafer is nominally oriented along the (001) direction, but is found to be
cut 2 o , towards the (110) axis. This o axis cut produces \steps" on the surface
which are 2 monolayers high. What is the lateral spacing between the steps of the
2 o -axis wafer?
Solution 1.8 The lattice constant a of Si is 5.43 A. The Si to Si atomic layer
distance am` in the (001) direction is (one-fourth of a) 1.358 A. Since each step is
two monolayers, the step spacing ds is given by
a
tan 2 = m`
ds
or
ds =
2 2 1:358 A = 77:8 A
0:0349
1.9 Conduct a literature search to nd out what the lattice mismatch is between
GaAs and AlAs at 300 K and 800 K. Calculate the mismatch between GaAs and
Si at the same temperatures.
Solution 1.9 The lattice constants of AlAs, GaAs and Si are (at room tempera-
ture)
AlAs : a = 5:6611 A i1 = 0:14%
GaAs : a = 5:6533 A i1 = 3:93%
Si : a = 5:431 A
The expansion coecients are
AlGaAs : 5:2 2 1006= K
GaAs : 6:86 2 1006= K
Si : 2:6 2 1006= K
1.10 In high purity Si crystals, defect densities can be reduced to levels of
1013 cm03. On an average, what is the spacing between defects in such crystals? In
,6 CHAPTER 1. STRUCTURAL PROPERTIES OF SEMICONDUCTORS
heavily doped Si, the dopant density can approach 1019 cm03. What is the spacing
between defects for such heavily doped semiconductors?
Solution 1.10 For a defect density of 10
13 cm03, the volume per unit defect is
4 3
r = 10013 cm3
3 d
or
rd = 2:88 2 103
A
The spacing between defects on an average is twice this value.
For a doping level of 1019 cm03, we have
4 3
r = 10019 cm3
3 d
or
rd = 29:2
A
The spacing on an average is twice this distance.
1.11 A GaAs crystal which is nominally along (001) direction is cut at an angle
o towards (110) axis. This produces one monolayer high steps. If the step size is
to be no more than 100 A, calculate .
Solution 1.11 The lattice constant of GaAs is 5.65 A and one monolayer distance
(Ga to Ga layer or As to As layer) distance is 2.825
A. The tilt angle is given by
= m`
a
tan
ds
where ds is the step spacing between steps. If ds is to be no more than 100
A, the
angle should be
0 2:825
= tan 1
= 1:65
100
The step spacing is often used to control monolayer by monolayer growth of epitax-
ially grown semiconductors.
1.12 Assume that a GaAs bond in GaAs has a bond energy of 1.0 eV. Calculate
the energy needed to cleave GaAs in the (001) and (110) planes.
Solution 1.12 In the (001) plane there are two GaAs bonds for each atom on the
plane. In the (110) plane there is only one such bond. The number density of Ga
atoms on the (001) plane is
2 2
NGa = = = 6:26 2 1014 cm02
a2 (5:65 2 1008 cm)2
Thus the energy needed to cleave the crystal along (001) planes is
E = 6:26 2 1014 2 2 eV=cm2
= 2 2 1004 J=cm2
, 1.1. PROBLEMS AND SOLUTIONS 7
A
a
c/2
A
A x x
a
y
A A A
a
Figure 1.2 : The geometry used to nd the relation between c and a (Problem 1.13).
On the (110) planes the density of Ga atoms is
NGa = p3 2 = p 3
2a 2 (5:65 2 1008 cm)2
= 6:64 2 1014 cm02
The energy to cleave the planes is
E = 6:64 2 1014 eV=cm02
= 1:06 2 1004 J=cm2
which is about half the value found for the (100) planes.
1.13
pConsider a hcp structure shown in Fig. 1.12. Prove the relation given by
c=a = 8=3 = 1:633.
Solution 1.13 We will nd the distance between the planes formed by the rst
set of atoms A and the second set of atoms B . This distance is c=2. To nd this in
terms of a, we consider the position of the atom B which is placed on top of the
center of a triangle of sides a, as shown in Figure 1.12.
To nd the distance c=2, we rst need to nd the distance x. Now (see Fig.
1.12)
a
y = tan30
2
a a2 a2 a2
x2 = y 2 + ( ) 2 = + =
2 12 4 3
2 2
(c=2)2 = a2 0 x2 = a
c
r 8
3
) a
=
3
= 1:633
1.14 A HgCdTe alloy is to be grown with 10% Hg. The sticking coecient of Hg
is 1002 and that of Cd is 1.0. Te is present in excess and does not limit the growth
of the crystal. Calculate the Hg and Cd uxes and partial pressures needed to grow
the lm at a rate of 1.0 m/hour.
Solution 1.14 This problem is similar to Problem 1.1.
