From Atoms to Solids
Outline
- Atomic and Molecular Wavefunctions
-  Molecular Hydrogen
-  Benzene
1
A Simple Approximation for an Atom
-e
+e
r
Continuum of
free electron
states
n=3
n=2
energy of -e
Lets represent the
atom in space by its
Coulomb potential
centered on the
proton (+e):
n=1
e2
V (r) = −
4πo r
Continuum of
free electron
states
Finite Quantum Well:
2
A Simple Approximation for a Molecule
Continuum of
free electron
Bound states
If this were the
atomic potential,
states
Continuum of
free electron
states
then this would be the
molecular potential:
We do not know exactly what the energy levels are,
although in 1-D we could solve the equation exactly if we had to
3
Molecular Wavefunctions
Atomic Wavefunctions:
L = 1 nm
ΨA
Molecular Wavefunctions:
1.505 eV · nm2
E≈
= 0.4 eV
(2L)2
2 atomic states 2 molecular states
Ψeven
Ψodd
When the wells are far apart, atomic functions dont overlap.
A single electron can be in either well with equal probability, and E = 0.4 eV.
4
Molecular Wavefunctions
d
WELLS FAR APART:
L = 1 nm
Ψeven
Ψodd
WELLS CLOSER TOGETHER:
1.505 eV · nm2
E≈
= 0.4 eV
(2L)2
(Degenerate states)
d
Ψeven
Atomic states are beginning
to overlap and distort.
|Ψeven|2 and |Ψodd|2 are not the
same (note center point).
Energies for these two states
are not equal.
(The degeneracy is broken.)
Ψodd
5
Molecular Wavefunctions
1. Which state has the lower
energy?
d
Ψeven
(a) Ψeven
(b) Ψodd
Ψodd
2. What will happen to the energy of Ψeven as the two wells come together
(i.e., as d is reduced)?
(a) E increases
(b) E decreases
(c) E stays the same
6
Bonding Between Atoms
How can two neutral objects bind together?
H + H H2
Superposition of
Coulomb potentials H2:
r
n=3
n=2
energy of -e
-e
+e
+e
n=1
Continuum of free
electron states
e2
V (r) = −
4πo r
+e
r
energy of -e
Lets represent the
atom in space by its
Coulomb potential
centered on the
proton (+e):
+e
-e
Continuum of free
electron states
e2
e2
V (r) = −
−
4πo |r − r1 | 4πo |r − r2 |
7
Molecular Hydrogen
Energy
σ antibonding molecular orbital
1s atom
orbital
1s atom
orbital
σ bonding
molecular orbital
8
Molecular Hydrogen Binding Energy
4.
3.
2.
1.
Potential Energy [kJ/mol]
0
-100
-200
-300
1.
Energy
released
when bond
forms
(-Bond
Energy)
Energy
absorbed
when bond
breaks
(+Bond
Energy)
4.
-400
-432
-500
2.
3.
74
(H2 bond length)
100
200
Internuclear Distance [pm]
9
Tunneling Between Atoms in Solids
Again start with
simple atomic state:
+e
energy of -e
-e
r
ψA
n=1
energy of -e
Bring N atoms together together forming a 1-d crystal (a periodic lattice)…
-e
10
Let Take a Look at Molecular Orbitals of Benzene
… the simplest conjugated alkene
H
H
C
1.08 Å
C
Typical C-C bond length 1.54 Å
Typical C=C bond length 1.33 Å
H
C
or
1.395 Å
C
H
C
C
H
Length of benzene
carbon-carbon bonds is between
C-C and C=C bond lengths
H
π-electron density
p orbitals
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11
ELECTRONEnergy
ENERGY
Electron
Molecular Orbitals of Benzene
Energy distribution of
Benzene π-molecular orbitals
Carbon atoms are represented by dots
Nodal planes are represented by lines
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Energy
antibonding MOs
energy of isolated
p-orbital
bonding MOs
12
from Loudon
… More Examples – Series of Polyacene Molecules
The lowest bonding MO of anthracene
Benzene
255 nm
Naphthaline
315 nm
Anthracene
380 nm
Tetracene
480 nm
Pentacene
580 nm
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Red: bigger molecules !
Blue: smaller molecules !
20000
25000
Luminescence Spectra
13
35000
30000
ENERGY
cm-1
Tunneling Between Atoms in Solids
energy of -e
Conduction in nanostructured materials
electron
atomic spacing
energy of -e
Conduction in nanostructured materials (with Voltage applied)
TUNNELING IS THE MECHANISM FOR CONDUCTION
14
Coulombs Law and Atomic Hydrogen
-e
Continuum of free
electron states
+e
r
n=3
n=2
energy of -e
Lets represent the
atom in space by its
Coulomb potential
centered on the
proton (+e):
n=1
e2
V (r) = −
4πo r
Time-Independent Schrödinger Equation
2 2 r
e2
r
ψ(r)
Eψ(r) = −
∇ ψ(r) −
4πo r
2m
r
15
Solutions
r
ψ1s (r) = e−r/a0
r
ψ2px (r) = xe
−r/2a0
r
ψ2py (r) = ye−r/2a0
r
Image by Szdori on Wikipedia.
Image in the Public Domain
ψ2pz (r) = ze
−r/2a0
E1s = −IH
E2p = −IH /4
These are some solutions,
without normalization.
16
a0 = 0.0529 nm
IH = 13.606 eV
Solutions
BOUNDARY SURFACES OF
s, p, d atomic orbitals
s
px
dxy
dxz
py
dyz
px
dx2—y2
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17
dz2
1s-np transitions in atomic hydrogen
Image in the Public Domain
The Lyman alpha blob, so
called because of its Lyman
alpha emission line, photons
must be redshifted to be
transmitted through the
atmosphere
Image in the Public Domain
18
Energy Levels of Atomic Lithium
0
l=0
l=1
-1
5s
4s
5p
4p
l=2
5d
4d
3p
3d
l=4
5g
n=5
n=4
n=3
3s
Energy (eV)
-2
l=3
5f
4f
-3
2p
Lithium energy
diagram
n=2
-4
-5
2s
Hydrogen
levels
Ground state
Image in the Public Domain
Ground state is 1s22s, excited states shown are 1s2np
19
Energy Levels of Atomic Sodium
Image in the Public Domain
Image by Orci on Wikipedia.
Image in the Public Domain
Image in the Public Domain
20
Example: Barrier Tunneling
•  Lets consider a tunneling problem:
An electron with a total energy of Eo= 6 eV
approaches a potential barrier with a height of
V0 = 12 eV. If the width of the barrier is
L = 0.18 nm, what is the probability that the
electron will tunnel through the barrier?
V0
Eo
2
F 16Eo (V − Eo ) −2κL
T = ≈
e
2
A
V
κ=
2me
(Vo − E) = 2π
2
T = 4e
2me
(Vo − E) = 2π
2
−2(12.6 nm−1 )(0.18 nm)
metal
metal
0 L air
gap
6 eV
−1
2 ≈ 12.6 nm
1.505 eV-nm
= 4(0.011) = 4.4%
Question: What will T be if we double the width of the gap?
21
x
Leaky Particles
Due to barrier penetration, the electron density of a metal
actually extends outside the surface of the metal !
x
Vo
|ψ|2
Work
function Φ
EF
Occupied levels
x=0
Assume that the work function (i.e., the energy difference between the most energetic
conduction electrons and the potential barrier at the surface) of a certain metal is Φ = 5 eV.
Estimate the distance x outside the surface of the metal at which the electron probability
density drops to 1/1000 of that just inside the metal.
(Note: in previous slides the thickness of the potential barrier was defined as x = 2a)
2
|ψ(x)|
1
−2κx
=e
≈
|ψ(0)|2
1000
using κ =
2me
(Vo − E) = 2π
2
1
x = − ln
2κ
2me
Φ = 2π
h2
22
1
1000
≈ 0.3 nm
5 eV
−1
=
11.5
nm
1.505 eV · nm2
Application: Scanning Tunneling Microscopy
ψ(x)
Vacuum
ψ(x)
Vo
V (x)
Piezoelectric Tube
With Electrodes
E < Vo
Second Metal
V (x)
x
Vo
x
Tunneling
Current Amplifier
Tip
Sample
Tunneling
Voltage
Image originally created by IBM Corporation.
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23
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6.007 Electromagnetic Energy: From Motors to Lasers
Spring 2011
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