Lecture 7 January 22, 2014
Silicon crystal surfaces
Nature of the Chemical Bond
with applications to catalysis, materials
science, nanotechnology, surface science,
bioinorganic chemistry, and energy
Course number: Ch120a
Hours: 2-3pm Monday, Wednesday, Friday
William A. Goddard, III, wag@wag.caltech.edu
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry, Materials
Science, and Applied Physics,
California Institute of Technology
Teaching Assistants:Sijia Dong <sdong@caltech.edu>
Samantha Johnson <sjohnson@wag.caltech.edu>
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
1
Diamond
Replacing all H atoms of ethane and with methyls, leads to with a
staggered conformation
Continuing to replace H with methyl groups
forever, leads to the diamond crystal
structure, where all C are bonded
tetrahedrally to four C and all bonds on
adjacent C are staggered
A side view is
This leads to the diamond crystal structure. An expanded view
is Ch120a-Goddard-L07
on the next slide © copyright 2013 William A. Goddard III, all rights reserved
2
Infinite structure from tetrahedral bonding plus staggered bonds
on adjacent centers
2nd layer
3
1
1
1st layer
1
2
02
2
2nd layer
0
0
1c
1st layer
1
1
1
1
2nd layer
1st layer
Chair
configuration
of cylcohexane
Not shown: zero layer just like 2nd layer but above layer 1
st layer but below layer 2
3rd
layer just like the©1copyright
Ch120a-Goddard-L07
2013 William A. Goddard III, all rights reserved
3
c
The unit cell of diamond crystal
An alternative view of the
diamond structure is in terms of
cubes of side a, that can be
translated in the x, y, and z
directions to fill all space.
c
f
c
i
c
i f
f
f
f
i c
i
Note the zig-zag chains c-i-f-i-c
f
and cyclohexane rings (f-i-f)-(i-f-i) c
c
There are atoms at
•all 8 corners (but only 1/8 inside the cube): (0,0,0)
•all 6 faces (each with ½ in the cube): (a/2,a/2,0), (a/2,0,a/2),
(0,a/2,a/2)
•plus 4 internal to the cube: (a/4,a/4,a/4), (3a/4,3a/4,a/4),
(a/4,3a/4,3a/4), (3a/4,a/4,3a/4),
Thus each cube represents 8 atoms.
All other atoms of the infinite crystal are obtained by translating
thisCh120a-Goddard-L07
cube by multiples©of
a in 2013
theWilliam
x,y,zA. directions
copyright
Goddard III, all rights reserved
c
4
Diamond Structure
Now bond one of these
atoms, C2, to 3 new C
so that the bond are
staggered with respect
to those of C1.
5a
3a
1a
4b
2b
5
6
3
4
2
1b
4a
2a
1
Start with C1 and
make 4 bonds to form
a tetrahedron.
5b
3b
1c
7
Continue this process.
Get unique structure:
diamond
Note: Zig-zag chain
1b-1-2-3-4-5-6
Chair cyclohexane
ring: 1-2-3-3b-7-1c
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
5
Properties of diamond crystals
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
6
Properties of group IV molecules (IUPAC group 14)
1.526
There are 4 bonds to each atom, but each bond connects two
atoms.
Thus to obtain the energy per bond we take the total heat of
vaporization and divide by two.
Ch120a-Goddard-L07
© copyright 2013
William
Goddard III,
all rights reserved
Note
for Si, that the average
bond
isA.much
different
than for Si H 7
Comparisons of successive bond energies SiHn and CHn
p
lobe
lobe
lobe
p
lobe
p
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
p
8
Miller indices
A 3D crystal is characterized by a unit cell with axes, a, b, c that
can be translated by integer translations along a, b, c to fill all
space.
The corresponding points in the translated cells are all
equivalent.
Passing a plane through any 3 such equivalent points defines a
plane denoted as (h,k,l). An equally space set of parallel to
(h,k,l) pass through all equivalent points, which the l,m,n
correspond to the reciprocal intersections on the unit cell when
one plane passes through the origin. These are called Miller
c
indices
c/l
Ch120a-Goddard-L07
b/k
b
a/h
a
© copyright 2013 William A. Goddard III, all rights reserved
9
Examples of
special planes
c
c/l
b/k
b
a a/h
To denote all equivalent
planes we use {h,k,l} so
that
From Wikipedia
{1,0,0} for cubic
includes the 3 cases in
the first row)
A number with a bar
Ch120a-Goddard-L07
indicates
negative
© copyright 2013 William A. Goddard III, all rights reserved
10
Crystallographic directions
A lattice vector can be written as
Rmnp = m a + n b + p c
where m,n,p are integers. This is denoted as [m,n,p]
The set of equivalent vectors is denoted as <m,n,p>
Examples are shown here.
From Wikipedia
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
11
The Si Crystal viewed
from the [001] direction
[001]
[010]
[110]
[010
[100]
(001) Surface
[100]
[1,-1,0]
Ch120a-Goddard-L07
1st Layer 
2nd Layer 
3rd Layer 
4th Layer 
RED
GREEN
ORANGE
WHITE
not show bonds
12
© copyright 2013 William A. Goddard III, all rights reserved
to 5th layer
The Si Crystal (100) surface, unreconstructed
Projection of
bulk cubic cell
Surface zig-zag row Every red surface atom
is bonded to two green
2nd layer atoms, but the
other two bonds were to
two Si that are now
removed. This leaves
two non bonding
electrons to distribute
among the two dangling
bond orbitals sticking
out of plane (like the 1A1
state of SiH2)
1st Layer 
2nd Layer 
3rd Layer 
th Layer 
4
Surface
unit
cell
P(1x1)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
RED
GREEN
ORANGE
WHITE 13
Si(100) surface (unreconstructed)
viewed (nearly) along the [110] direction
Ch120a-Goddard-L07
Each surface atom has
two dangling bond orbitals
pointing to left and right,
along
[1,-1,0]
direction
© copyright
2013 William
A. Goddard III, all rights reserved
14
The (100) Surface Reconstruction
viewed (nearly) along the [110] direction
Spin pair dangling bond orbitals of adjacent atoms in
[1,-1,0] direction (originally 2nd near neighbors
Get one strong s bond but leave two dangling bond orbitals on
adjacent now bonded atoms (form weak p bond in plane)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
15
Si(100) surface reconstructed (side view)
Surface atoms now bond to form dimers (move from 3.8 to 2.4A)
Get row of dimes with doubled surface unit cell
One strong s bond, plus weak p bond in plane
Surface
orginal cell New cell
length
Lateral
bond
length
7.6A
displacements
2.4A
3.8A
0.7A 0.7A
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
16
Si(100) surface reconstructed (top view)
Rows of dimer
pairs are parallel
New unit cell
reconstructed
surface
P(2x1)
original unit cell
unreconstructed
surface
P(1x1)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
17
Get 2x2 unit cell but atom at
center is equivalent to atom at
corner, therform c(2x2)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
18
Two simple patterns for (100) Surface
Reconstruction
Dimer rows alternate
C(2x2), high energy
Ch120a-Goddard-L07
Dimer rows parallel
P(2x1), low energy
© copyright 2013 William A. Goddard III, all rights reserved
19
P(2x1) more stable than c(2x2) by ~ 1kcal/mol
The Sisurf-Si2nd-Sisurf bond for c(2x2) opens up to 120º
because the Sisurf move opposite directions
110º
120º
120º
Ch120a-Goddard-L07
110º
For P(2x1) the Sisurf move
the same directions and
Sisurf-Si2nd-Sisurf bond
20
© copyright 2013 William A.
Goddard III, all
reserved
remains
atrights
110º
Construct (111) surface
using cubic unit cell
4
2
2
4
Atom #1 bonded to 3 atoms #2
3
4
4
2
1
Each #2 is bonded straight down
to an atom#3
2
4
1
4
1
4
1
2
2
Ch120a-Goddard-L07
2
2
3c
Go straight down to atom #1
Each #2 is bonded to 3 atoms #1 4
in top layer. Get hexagonal double
layer
0
1 4
3
Start at diagonal atom #0
Each atom #3 is bonded to 3
atom#4.
2
2
2
1
1
2
2
© copyright 2013 William A. Goddard III, all rights reserved
21
Si(111) surface (alternate construction)
Start with red atom on top, bond to 3 green atoms in 2nd layer
Each green atom
is bonded to 2
other 1st layer
atoms plus a 3rd
atom straight
down (not shown)
The 3rd layer
atoms bond to 3
4th layer atoms in
orange
Surface unit cell
P(1x1)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
22
Reconstruction of Si(111) surface
Each surface atom
has a single
dangling bond
electron, might
guess that there
would be some
pairing of this with
an adjacent atom
to form a 2x1 unit
cell.
Ch120a-Goddard-L07
Indeed freshly
cleaved Si(111) at
low temperature
does show 2x1
Surface unit cell
23
P(1x1)
© copyright 2013 William A. Goddard III, all rights reserved
LEED experiments (Schlier and Farnsworth,
1959) observed 7th Order Spots  7x7 unit
cell
This was one of the first LEED experiments, a breakthrough
in surface technology
Very surprizing result, no one expected
From 1959 to 1981 many models proposed to fit various
experiments or calculations, but interpretations from different
calculations and experiments contradicted each other
A 2nd major breakthrough in surface technology occurred in
1982 that showed all previous interpretations of theory and
experiment were WRONG.
What was that?
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
24
LEED experiments (Schlier and Farnsworth,
1959) observed 7th Order Spots  7x7 unit
cell
From 1959 to 1981 many models proposed to fit various
experiments or calculations.
Binnig et al., 1981 did first STM image of Si (7x7) and saw 12
bright spots in 7x7 cell, showed that every previous model
was incorrect
Takayanagi et al.,
1985, proposed the
DAS Model that
explained the
experiments
No one has yet
explained the
reaction
mechanism
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
25
two 7x7 cells
What kind of interactions can go over a
7x7 region, with cell size 26.6 by 26.6 A?
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
26
Origin of complex reconstruction of Si(111)
In 49 surface unit cells have 49 dangling bonds for the
unreconstructed unit cell.
The cohesive energy of Si crystal is 108 kcal/mol
What do you expect the surface energy to be?
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
27
Origin of complex reconstruction of Si(111)
In 49 surface unit cells have 49 dangling bonds for the
unreconstructed unit cell.
Since cohesive energy of Si crystal is 108 kcal/mol
expect average bond energy must be 108/2 = 54
kcal/mol (each atom has 4 bonds, but double count the
bonds)
(H3Si-SiH3 bond energy is 74 kcal/mol)
Thus each dangling bond represents ~ 27 kcal/mol of
surface energy = 1.1 eV per surface atom
Calculated value = 1.224 eV snap and 1.200 ev relaxed.
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
28
Consider bonding an atom on top of 3 dangling bonds
T4
H3
T4
T4
H3
T4
Get 3x2 unit
cell
By adding a cap of one adatom Si per 3 top layer Si, can tie
offCh120a-Goddard-L07
all original dangling
bonds.
ThusA. 4816
© copyright
2013 William
Goddard III, all rights reserved
29
Consider bonding an atom on top of 3 dangling bonds
Two ways to do this. T4 and H3
T4 (observed)
H3 (not observed)
Bond angle strain (H3)
Pauli repulsion (H3)
Bond alignment/linear dependence (T4)
*HOMO delocalization (T4)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
30
T4 versus H3 site bonding to dangling bonds
2nd Nodal plane
orthogonal to
green-orange
bonds
3 for H3 &
1 for T4
Thus T4 better
top nodal plane
orthogonal to
purple-red
bonds
Same for H3 &
T4
SOMO H3
SOMO T4
31
bad
Ch120a-Goddard-L07
good
© copyright 2013 William A. Goddard III, all rights reserved
H3 reconstruction
1
1
0
1
2
1
1
0
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3 Monolayer
to tie up bonds.
Surface energy
increases by 0.13 eV
Because get three linear
0-1-2 links, must
orthogonalize
Unit cell
1
1
0
Ch120a-Goddard-L07
1
0
© copyright 2013 William A. Goddard III, all rights reserved
32
H3 reconstruction, unit cell is √3 x √3
With respect 1x1 unit
cell the new unit cell is
√3 longer
Also the unit cell is
rotated 30º so we
could say new unit cell
is (√3 x√3)R30º
1
1
√3
0
1
1
1
0
1
1
1
1
√3/2
1
0
Ch120a-Goddard-L07
1
0
© copyright 2013 William A. Goddard III, all rights reserved
33
T4 reconstruction √3 x √3
1
1
1
2
1
1
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds.
Surface energy
decreases by 0.10 eV
Because 0-1-2 ~ 100º
Unit cell
1
1
Ch120a-Goddard-L07
1
© copyright 2013 William A. Goddard III, all rights reserved
34
T4 reconstruction 2x2
1
1
1
2
1
1
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds, leave dangling
bond orbital
Surface energy
decreases by 0.12 eV
Per 2x2 cell
Unit cell
1
1
Ch120a-Goddard-L07
1
© copyright 2013 William A. Goddard III, all rights reserved
35
Best Si(111) structure is
DAS7x7 with energy
0.156 eV =3.6 kcal/mol
lower than 1x1
Not much, but per 49
cells this is
176.2 kcal/mol = 7.64 eV
Final structure has 12
adatoms out of 49 
12*0.117*4=5.62 eV
The rest of DAS 
2.02eV/cell=0.04eV/atom
Solares SD, Dasgupta S,
Schultz PA, Kim YH, Musgrave
CB, Goddard WA
Langmuir 21 (26): 1240412414 (2005)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
36
The (111) 7x7 DAS Surface
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
37
The (111) 7x7 DAS Surface Layers
(purple, brown and blue atoms have one dangling bond)
Adatoms on Top layer
These adatoms protrude from
the surface so that they show
up prominently in STM
Which way will this show up,
tunneling from tip to surface
orCh120a-Goddard-L07
surface to tip? © copyright 2013 William A. Goddard III, all rights reserved
38
The (111) 7x7 DAS Surface Layers
(purple, brown and blue atoms have one dangling bond)
1st
2nd
18 + 18 red
atoms, all
bonded to
1st layer
3rd
4th
First unreconstructed layer
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
39
The (111) 7x7 DAS Surface
Ch120a-Goddard-L07
12-membered
ring at corner of
cell
© copyright 2013 William A. Goddard
III, all rights reserved
40
The (111) 7x7 DAS Surface
Side view
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
41
The (111) 7x7 DAS Surface Cornerhole
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
42
Si(111) 7x7
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
43
DAS 7x7 (side view)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
45
DAS 7x7 (top view)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
46
The (111) 3x3
DAS Surface Unit Cell
0.026 eV worse than 7x7
Side view
Ch120a-Goddard-L07
Top view
12-membered rings
© copyright 2013 William A. Goddard III, all rights reserved
47
The (111) 5x5
DAS Surface Unit Cell
Side view
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
48
The (111) 5x5
DAS Surface Unit Cell
0.004 eV worse than 7x7
Top view
12- and 8-membered rings
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
49
The (111) 9x9
DAS Surface Unit Cell
0.011 eV worse than 7x7
Side view
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
50
The (111) 9x9
DAS Surface Unit Cell
Top view
12- and 8-membered rings
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
51
DAS Surface Energies (PBE DFT)
Energy, eV/1x1 Cell
1.09
Regression
Ab Initio
1.08
1.078
1.070
1.07
1.068
1.06
1.055
1.048
1.05
1.044
1.04
3
5
7
9
11
13
DAS Cell Size
Unreconstructed relaxed surface: 1.200 eV/1x1 cell
Infinite© DAS
model: 1.107 eV/1x1 cell
Ch120a-Goddard-L07
copyright 2013 William A. Goddard III, all rights reserved
52
Energy, eV/1x1 Cell
DAS Surface Energies
5x5 Surface (9 dangling bonds)
singlet pairing of remaining dangling bonds
is best (wag does not understand why)
0.04
0.03
0.02
0.01
0.00
0
2
4
6
8
10
Spin Polarization
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
53
DAS Reconstruction Driving Force
• 49 unpaired electrons (1/2 Si-Si bond) per 7x7 cell @ 1.2 eV
= 58.8 eV/cell
• DAS 7x7 Surface energy = 51.2 eV/cell (12+1=13 unpaired
electrons)
• Energy reduction due to reconstruction = 7.6 eV
• Difference is due to strain
• Bond length range = 2.31 – 2.50 Å (equilibrium 2.35 Å)
• Bond angle range = 91 – 117º (Equilibrium 109.4°)
Ch120a-Goddard-L07
© copyright 2013 William A. Goddard III, all rights reserved
54
The (110) plane (outlined in green, layer 1)
[001]
3
1
[-1,1,0]
1
1
2
02
2
0
0
1c
1
1
[010]
1
1
[100]
Ch120a-Goddard-L09-10
[110]
© copyright 2010 William A. Goddard III, all rights reserved
55
Si(110) surface (top view)
Cut through
cubic unit cell
surface unit
cell P(1x1)
Surface
atoms red
Ch120a-Goddard-L09-10
© copyright 2010 William A. Goddard III, all rights reserved
56
Si(110) surface (viewed nearly along [-1,1,0] direction)
One dangling bond
electron per surface atom
Surface atoms red
bulk atoms orange
[1,1,0]
[001]
Ch120a-Goddard-L09-10
© copyright 2010 William A. Goddard III, all rights reserved
57
Reconstruction of (110) surface, surface atoms only
side view
(along [-1,1,0])
Showing just 2 dangling
bond orbitals
54.7º
54.7º
Top view
(from [-1,-1,0])
[001]
[-1,1,0]
Ch120a-Goddard-L09-10
[1,1,0]
[001]
© copyright 2010 William A. Goddard III, all rights reserved
58
Reconstruction of (110) surface, surface atoms only
We have a chain of dangling bond
orbitals along the [-1,1,0] direction,
each tilted by 35.3º from the [110]
(vertical) axis
They will want to tilt toward the
vertical axis, reducing their angle
from 35.3º).
This leads to moving the surface
atoms toward the bulk.
There could be 2 by 2 pairing to
double the surface unit cell in the
[-1,1,0] direction
Ch120a-Goddard-L09-10
side view
(along [-1,1,0])
Showing just 2 dangling
bond orbitals
54.7º
54.7º
54.7º
[110]
[001]
© copyright 2010 William A. Goddard III, all rights reserved
59
The zincblende or sphalerite structure
Replacing each C atom of the diamond structure alternately with
Ga and As so that each Ga is bonded to four As and each As is
bonded to four Ga leads to the zincblende or sphalerite structure
(actually zincblende is the cubic form of ZnS and the mineral
sphalerite is cubic ZnS with some Fe)
•As at corners: (0,0,0)
•As at face centers: (a/2,a/2,0),
(a/2,0,a/2), (0,a/2,a/2)
•Ga 4 internal sites: (a/4,a/4,a/4),
(3a/4,3a/4,a/4), (a/4,3a/4,3a/4),
(3a/4,a/4,3a/4),
Thus each cube has 4 As and 4 Ga.
Ch120a-Goddard-L09-10
© copyright 2010 William A. Goddard III, all rights reserved
60
Bonding in GaAs
Making a covalent bond between to each atoms, one might have
expected tetrahedral As to make 3 bonds with a left over lone pair
pointing away from the 3 bonds, while Ga might be expected to
make 3 covalent bonds, with an empty sp3 orbital point away from
the 3 bonds, as indicated here, where the 3 covalent bonds are
shown with lines, and the donor acceptor (DA) or Lewis acidLewis base bond as an As lone pair coordinated with and empty
orbital on Ga
Of course the four bonds to each
atom will adjust to be equivalent,
but we can still think of the bond as
an average of ¾ covalent and ¼
DA
Ch120a-Goddard-L09-10
© copyright 2010 William A. Goddard III, all rights reserved
61
Other compounds
Similar zincblende or sphalerite compounds can be formed with
Ga replaced by B, Al,In and /or As replaced by N, P, Sb, or Bi.
They are call III-V compounds from the older names of the
columns of the periodic table (new UIPAC name 13-15
compounds).
In addition a hexagonal crystal, called Wurtzite, also with
tetrahedral bonding (but with some eclipsed bonds) is exhibited
by most of these compounds.
In addition there are a variety of similar II-VI systems, ZnS,
ZnSe, CdTe, HgTe, etc
Ch120a-Goddard-L09-10
© copyright 2010 William A. Goddard III, all rights reserved
62
The
(110) plane (outlined in green, layer 1)
[001]
Cut through cubic unit cell
3
1
[-1,1,0]
1
1
2
02
2
0
0
1c
1
1
surface unit
cell P(1x1)
[010]
1
1
[110]
0
As atoms
top layer
Ch120a-Goddard-L09-10
Ga atoms
top layer
[001]
[-1,1,0]
© copyright 2010 William A. Goddard III, all rights reserved
63
Reconstruction of (110) surface, side view along [-1,1,0]
Si has
dangling
bond
electron at
each surface
atom
Surface As has 3 covalent bonds to Ga, with 2 e in
3s lone pair, relaxes upward until average bond
angle is 95º Surface Ga has 3 covalent bonds
leaving 0 e in 4th orbital, relaxes downward until
average bond angle is 119º. GaAs angle 0º 26º
54.7º
54.7º
As
Ga
54.7º
[110]
Si (110)
[001]
Ch120a-Goddard-L09-10
GaAs (110)
© copyright 2010 William A. Goddard III, all rights reserved
64
Reconstruction of GaAs(110) surface
As has 3 covalent bonds,
leaving 2 electrons in 3s
lone pair, Ga has 3
covalent bonds leaving 0
eletrons in 4th orbital
Ga
As 54.7º
54.7º
Top view
(from [-1,-1,0])
[001]
[-1,1,0]
Ch120a-Goddard-L09-10
[1,1,0]
side view
[001] (along [-1,1,0])65
© copyright 2010 William A. Goddard III, all rights reserved
Reconstruction of (110) GaAs
Ch120a-Goddard-L09-10
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66
III-V reconstruction
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67
Ch120a-Goddard-L09-10
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68
Reconstruction of GaAs(110) surface, discussion
We consider that bulk GaAs has an average of 3 covalent bonds
and one donor acceptor (DA) bond. But at the surface can only
make 3 bonds so the weaker DA bond is the one broken to form
the surface.
The result is that GaAs cleaves very easily compared to Si. No
covalent bonds to break.
As has 3 covalent bonds, leaving 2 electrons in 3s lone pair. AsH3
has average bond angle of 92º. At the GaAs surface As relaxes
upward until has average bond angle of 95º
Ga has 3 covalent bonds leaving 0 eletrons in 4th orbital. GaH3
has average bond angle of 120º. At the GaAs surface Ga relaxes
downward until has average bond angle of 119º.
This changes the surface Ga-As bond from 0º (parallel to surface
to 26º. Observed in LEED experiments and QM calculations
Ch120a-Goddard-L09-10
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71
4
Analysis of charges
Bulk structure: each As has 3 covalent bonds and one Donoraccepter bond(Lewis base – Lewis acid). This requires 3+2=5
electrons from As and 3+0=3 electrons from Ga.
We consider that each bulk GaAs bond has 5/4 e from As and ¾
e form Ga.
Each surface As has 5/4+1+1+2 = 5.25e for a net charge of -0.25
each surface Ga has ¾+1+1+0= 2.75 e for a net charge of +0.25
Thus considering both surface Ga and As, the (110) is neutral
0
Ga
3/4
1
5.25e
2.75e
2
As
0
Ga
1
1
3/4
5/4
3/4
g
1
3/4
5/4
5/4
a
1
Net Q =0
2
As
1
1
5/4
3/4
5/4
3/4
3/4
5/4
Ch120a-Goddard-L09-10
5/4
a
1
5/4
3/4
g
3/4
5/4
0
Ga
1
3/4
5/4
a
1
3/4 III, all rights reserved
© copyright 5/4
2010 William A. Goddard
5/4
3/4
5/4
5/4
2
As
3/4
g
3/4
3/4 72
The GaAs (100) surface, unreconstructed
Every red surface atom
is As bonded to two
green 2nd layer Ga
atoms, but the other two
bonds were to two Ga
that are now removed.
This leaves three non
bonding electrons to
distribute among the
two dangling bond
orbitals sticking out of
plane (like AsH2)
Ch120a-Goddard-L09-10
1st Layer 
2nd Layer 
3rd Layer 
th Layer 
© copyright 2010 William A. Goddard4III, all
rights reserved
RED
GREEN
ORANGE
WHITE
73
GaAs(100) surface reconstructed (side view)
For the perfect surface, As in top layer, Ga in 2nd layer, As in 3rd
layer, Ga in 4th layer etc.
For the unreconstructed surface each As has two bonds and
hence three electrons in two nonbonding orbitals.
Expect As atoms to dimerize to form a 3rd bond leaving 2 electrons
in nonbonding orbitals.
Surface As-As bonds
As
Ga
As
Ga
As
Ga
As
Ch120a-Goddard-L09-10
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74
Charges for 2x1 GaAs(100)
2nd layer ga
has 3 e
1
2
1
5/4
2
2nd layer, ga
5/4
3/4
3/4
3/4
3/4
3/4
3/4
3/4
3/4
5/4
2e As-ga
bond
1
2
2e As LP
5/4
3/4
3/4
1st layer As
has 5.5 e
Top layer, As
3/4
Ch120a-Goddard-L09-10
1
5/4
3rd layer, as
2
5/4
Each surface
As has extra
3/4
0.5 e  dimer
3/4
3/4 has extra 1e
2e As-As
3/4
Not stable 75
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2010 William A. Goddard III, all rights reserved
3/4
Now consider a missing row of As for GaAs(100)
1
5/4
1
Top layer, As
2nd layer, ga
5/4
3/4
3/4
3/4
3/4
3/4
0
ga empty
LP
3/4
0
3rd layer, as
2nd layer ga
has 2.25e
0
3/4
1st layer As
has 5.5 e
3/4
Ch120a-Goddard-L09-10
Each 2nd layer ga
next to missing
0
As is deficient by
3/4
0.75e extra 0.5 e
3/4
3/4
 4 ga are
3/4
76
missing
© copyright 2010 William A. Goddard III, all rights
reserved 3e
Consider 1
missing As row
out of 4
Extra 1e
missing 3e
-1-1-1+3=0
net charge
Extra 1e
Thus based on
electron counting
expect simplest
surface
reconstruction to
be 4x2. This is
observed
Ch120a-Goddard-L09-10
Extra 1e
Extra 1e
missing 3e
© copyright 2010 William A. Goddard III, all rights reserved
77
Different views of GaAs(100)4x2 reconstruction
-1.0e
Previous page, 3
As dimer rows
then one missing
Ch120a-Goddard-L09-10
+1.5e
Two missing As row
plus missing Ga row
Exposes 3rd row As
Agrees with experiment
Hashizume
etA.al
PhysIII,Rev
B 51,
4200
© copyright
2010 William
Goddard
all rights
reserved
(1995)
78
Ch120a-Goddard-L09-10
© copyright 2010 William A. Goddard III, all rights reserved
79
summary
Postulate of surface electro-neutrality
Terminating the bulk charges onto the surface layer and
considering the lone pairs and broken bonds on the
surface should lead to:
•the atomic valence configuration on each surface atom.
For example As with 3 covalent bonds and a lone pair
and Ga with 3 covalent bonds and an empty fourth orbital
•A neutral surface
This leads to the permissible surface reconstructions
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Intrinsic semiconductors
+
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Excitation energy
-4.0 eV relative to vacuum=-IP
Energy gap = 1.1 eV
-5.1 eV relative to vacuum = -EA
Ch120a-Goddard-L09-10
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To be added – band states
Ch120a-Goddard-L09-10
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To be added – band states
Ch120a-Goddard-L09-10
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Semiconducting properties
Ch120a-Goddard-L09-10
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Semiconducting properties
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Ch120a-Goddard-L09-10
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Trends: overlaps
between bonded atoms
decrease from 2p to 3p
to 4p etc
Thus bonds are
weaker, but antibonds
are not as band
Thus cohesive energy
and band gaps
decrease as go down
the periodic table
Ch120a-Goddard-L09-10
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Add substitutional impurity, P, to Si
Consider the case in which one Si atom of Si crystal is
replace by a P atom (substitutional impurity)
Main effect is that P has one more electron than Si
Neutral has extra
electron in one bond
Ch120a-Goddard-L09-10
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N-type semiconductor
The substituted P can make covalent bonds to 3 of Si neighbors
but the extra electron is in the way of making the 4th bond.
Thus it is very easy to ionize this extra electron (IP=4.05 eV)
donating it to the conduction band (EA=4.0 eV) leaving behind a P
making covalent bonds to all four Si neighbors. The net excitation
energy is just 4.05-4.00=0.05 eV. Thus as room temperature lots
of electrons in conduction band. Get n type semiconductor and P
is called an n-type dopant
Ionize extra electron
get strong bond
Ch120a-Goddard-L09-10
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To be added – band states
IP(P)=4.05 eV
0.054 eV
Remove e from P, add to conduction band = 4.045-4.0 = 0.045 eV
Thus P leads to donor state just 0.045eV below LUMO or CBM
Ch120a-Goddard-L09-10
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Al substitutional impurity in Si
Consider the case in which one Si atom of Si crystal is
replace by a Al atom (substitutional impurity)
Main effect is that Al has one less electron than Si
The substituted Al can make covalent bonds to 3 of the Si
neighbors but it lacks the electron to make a 4th bond 2-e bond
Thus the EA of add an electron to make the 2 electron bond is
EA=5.033 eV, which is nearly as great as the IP=5.1 eV.
Thus removing an electron from the valence band and adding it to
the Al-Si bond costs only 5.1-5.033=0.067eV. leaving behind an Al
making
covalent bonds
to all four Si neighbors.
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Ch120a-Goddard-L09-10
© copyright 2010 William A. Goddard III, all rights reserved
Next consider Al substitutional impurity in Si
Since the net excitation energy 0.067 eV there are lots of holes
in the valence band at room temperature.
Get p type semiconductor and Al is called a p-type or acceptor
dopant
Ch120a-Goddard-L09-10
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To be added – band states
EA(Al)=5.033 eV
0.067 eV
Add e to Al, from valence band = 5.1 -5.033 = 0.067 eV
Al leads to acceptor state just 0.067eV above HOMO or VBM
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Ch120a-Goddard-L09-10
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III-V Compounds
Energy Gaps for III-V much bigger than
for group IV
Consider GaAs, what happens if we
replace As with Se or Ge
What happens if we replace Ga with
Zn or Ge
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Substitute As for Se or Ge
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Substitute Ga with Zn or Ge
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Dopant levels for GaAs
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Cohesive energies and Bonds for III-V compounds
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Compare IV to III-V same row
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