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NUCLEAR PHYSICS
Building Atoms, Molecules and solids
U(r)
+e
r
a
r
n=3
n=2
y6
n=1
y5
+e
yeven
+e
r
y4
y3
y2
y1
Last week of Course

Today -- Atoms, Molecules, Solids
States with many electrons – filled
according to the Pauli exclusion principle

Next time – Consequences of quantum
mechanics

Metals, insulators, semiconductors,
superconductors, lasers, . .

Overview
Nuclear Spin and Magnetic Resonance Imaging (MRI)
Atomic Configurations
States in atoms with many electrons –
filled according to the Pauli exclusion principle
Molecular Wavefunctions: origins of covalent bonds
Example: H + H  H2
Electron energy bands in Solids
States in atoms with many electrons –
filled according to the Pauli exclusion principle
Electron Spin and Nuclear Spin


Last time: Electrons have ‘spin’, so we need FOUR quantum
numbers to specify the full electronic state of a hydrogen atom
1
1
 n, l, ml, ms (where ms = - /2 and + /2)
Actually, the proton in the H atom ALSO has a spin, which is
described by an additional quantum number, mp


The energy difference between the two proton spin states in a magnetic
field is 660 times smaller than for electron spin states!
But… There are many more unpaired proton spins than unpaired electron
spins in ordinary matter. Our bodies have many unpaired protons in H2O.
Detect them …...
In order to image tissue of various types,
Magnetic Resonance Imaging detects the
small difference in the numbers of “up” and
“down” hydrogen proton spins generated
when the object studied is placed in a
magnetic field. Nobel Prize (2003):
www.beckman.uiuc.edu/research/mri.html
Lauterbur (UIUC)
Nuclear Spin and MRI: Example
Magnetic resonance imaging (MRI) depends on the absorption of
electromagnetic radiation by the nuclear spin of the hydrogen atoms in
our bodies. The nucleus is a proton with spin ½, so in a magnetic field B
there are two energy states. The energy difference between these states
is DE=2mpB, with mp = 1.41 x 10-26 J /Tesla.
B=0
B0
DE = 2mpB
B
Question 1: The person to be scanned by an MRI
machine is placed in a strong (1 Tesla) magnetic field.
What is the energy difference between
spin-up and spin-down proton states in this field?
Question 2: What photon frequency, f, will be absorbed
by this energy difference?
Nuclear Spin and MRI: Example
Magnetic resonance imaging (MRI) depends on the absorption of
electromagnetic radiation by the nuclear spin of the hydrogen atoms in
our bodies. The nucleus is a proton with spin ½, so in a magnetic field B
there are two energy states. The energy difference between these states
is DE=2mpB, with mp = 1.41 x 10-26 J /Tesla.
B=0
B0
DE = 2mpB
B
Question 1: The person to be scanned by an MRI
machine is placed in a strong (1 Tesla) magnetic field.
What is the energy difference between
spin-up and spin-down proton states in this field?
Solution:
DE = 2mpB = 2 x (1.41 x 10-26 J/T) x 1 T
= 2.82 x 10-26 J x 1 eV/ 1.6 x 10-19 J = 1.76 x 10-7 eV
Nuclear Spin and MRI: Example
Magnetic resonance imaging (MRI) depends on the absorption of
electromagnetic radiation by the nuclear spin of the hydrogen atoms in
our bodies. The nucleus is a proton with spin ½, so in a magnetic field B
there are two energy states. The energy difference between these states
is DE=2mpB, with mp = 1.41 x 10-26 J /Tesla.
B=0
B0
DE = 2mpB
B
Question 2: What photon frequency, f, will be absorbed
by this energy difference?
Solution:
Energy conservation:
so:
Ephoton = hf = DE
f = DE / h
= 2.82 x 10-26 J / 6.626 x 10-34 J-sec = 42 MHz
Radio waves
Act 1
We just saw that radio-wave photons with energy 1.7 x 10-7 eV can cause a
nuclear spin to flip. What is the angular momentum of each photon?
a. 0
b. ħ/2
c. ħ
Act 1 Solution
We just saw that radio-wave photons with energy 1.7 x 10-7 eV can cause a
nuclear spin to flip. What is the angular momentum of each photon?
a. 0
b. ħ/2
c. ħ
The nuclear spin has changed from to (or vice versa).
That is, its z-component has changed by ħ. Conservation
of angular momentum requires that the photon have
carried (at least) this much in.
Pauli Exclusion Principle
Let’s start building more complicated atoms to study the Periodic
Table. For atoms with many electrons (e.g., carbon: 6, iron: 26, etc.) what energies do the electrons have?
“Pauli Exclusion Principle” (1925)
No two electrons can be in the same quantum state.
For example, in a given atom they cannot have the same set of quantum
numbers n, l, ml, ms.
This means that each atomic orbital (n,l,ml) can hold 2 electrons: ms = ±½.
Important consequence:
Electrons do not pile up in the lowest energy state.
It’s more like filling a bucket with water.
They are distributed among the higher energy levels according
to the Exclusion Principle.
Particles that obey this principle are called “fermions”.
Protons and neutrons are also fermions, but photons are not.
Filling Atomic Orbitals
According to the Pauli Principle
n

s
p
d
f
g
l=0
1
2
3
4
4
Energy
En 
 13.6 eV 2
Z
2
n
Lecture 11
In a multi-electron atom, the H-atom
energy level diagram is distorted by
Coulomb repulsion between electrons.
Nevertheless, the H-atom diagram is
useful (with some caveats) for figuring
out the order in which orbitals are filled.
3
2
Example: Na (Z = 11)
1s2 2s2 2p6 3s1
1
Z = atomic number = # protons
l
label
#orbitals (2l+1)
0
s
1
1
p
3
2
d
5
3
f
7
Act 2: Pauli Exclusion Principle
1. Which of the following states (n,l,ml,ms) is/are NOT allowed?
a.
b.
c.
d.
e.
(2, 1, 1, -1/2)
(4, 0, 0, 1/2)
(3, 2, 3, -1/2)
(5, 2, 2, 1/2)
(4, 4, 2, -1/2)
2. Which of the following atomic electron configurations violates the Pauli
Exclusion Principle?
a.
b.
c.
d.
e.
1s2, 2s2, 2p6, 3d10
1s2, 2s2, 2p6, 3d4
1s2, 2s2, 2p8, 3d8
1s1, 2s2, 2p6, 3d5
1s2, 2s2, 2p3, 3d11
Act 2: Solution
1. Which of the following states (n,l,ml,ms) is/are NOT allowed?
a.
b.
c.
d.
e.
(2, 1, 1, -1/2)
(4, 0, 0, 1/2)
(3, 2, 3, -1/2)
(5, 2, 2, 1/2)
(4, 4, 2, -1/2)
ml > l
l=n
2. Which of the following atomic electron configurations violates the Pauli
Exclusion Principle?
a.
b.
c.
d.
e.
1s2, 2s2, 2p6, 3d10
1s2, 2s2, 2p6, 3d4
1s2, 2s2, 2p8, 3d8
1s1, 2s2, 2p6, 3d5
1s2, 2s2, 2p3, 3d11
Act 2: Solution
1. Which of the following states (n,l,ml,ms) is/are NOT allowed?
a.
b.
c.
d.
e.
(2, 1, 1, -1/2)
(4, 0, 0, 1/2)
(3, 2, 3, -1/2)
(5, 2, 2, 1/2)
(4, 4, 2, -1/2)
ml > l
l=n
2. Which of the following atomic electron configurations violates the Pauli
Exclusion Principle?
a.
b.
c.
d.
e.
1s2, 2s2, 2p6, 3d10
1s2, 2s2, 2p6, 3d4
1s2, 2s2, 2p8, 3d8
1s1, 2s2, 2p6, 3d5
1s2, 2s2, 2p3, 3d11
Only 6 p-states.
Only 10 d-states.
Filling Procedure for Atomic Orbitals
Due to electron-electron interactions, the hydrogen levels fail to give us
the correct filling order as we go higher in the periodic table.
The actual filling order is given in the table below. Electrons are added
by proceeding along the arrows shown.
Home exercise:
Bromine is an element with Z = 35.
Find its electronic configuration
(e.g., 1s2 2s2 2p6 …).
Note:
The chemical properties of an atom
are determined by the electrons in
the highest n orbitals, because they
are on the surface of the atom.
Act 3: Pauli Exclusion Principle – Part 2
The Pauli exclusion principle applies to all fermions in all situations
(not just to electrons in atoms). Consider electrons in a 2-dimensional
infinite square well potential.
1. How many electrons can be in the first excited energy level?
a. 1
b. 2
c. 3
d. 4
e. 5
Hint: Remember the (nx,ny) quantum numbers.
2. If there are 4 electrons in the well, what is the energy of the
most energetic one (ignoring e-e interactions, and assuming
the total energy is as low as possible)?
a. (h2/8mL2) x 2
b. (h2/8mL2) x 5
c. (h2/8mL2) x 10
Act 3: Solution
The Pauli exclusion principle applies to all fermions in all situations
(not just to electrons in atoms). Consider electrons in a 2-dimensional
infinite square well potential.
1. How many electrons can be in the first excited energy level?
a. 1
b. 2
c. 3
d. 4
e. 5
The first excited energy level has (nx,ny) = (1,2) or (2,1).
That is, it is degenerate.
Each of these can hold two electrons (spin up and down).
2. If there are 4 electrons in the well, what is the energy of the
most energetic one (ignoring e-e interactions, and assuming
the total energy is as low as possible)?
a. (h2/8mL2) x 2
b. (h2/8mL2) x 5
c. (h2/8mL2) x 10
Act 3: Solution
The Pauli exclusion principle applies to all fermions in all situations
(not just to electrons in atoms). Consider electrons in a 2-dimensional
infinite square well potential.
1. How many electrons can be in the first excited energy level?
a. 1
b. 2
c. 3
d. 4
e. 5
The first excited energy level has (nx,ny) = (1,2) or (2,1).
That is, it is degenerate.
Each of these can hold two electrons (spin up and down).
2. If there are 4 electrons in the well, what is the energy of the
most energetic one (ignoring e-e interactions, and assuming
the total energy is as low as possible)?
a. (h2/8mL2) x 2
b. (h2/8mL2) x 5
c. (h2/8mL2) x 10
Two electrons are in the (1,1) state,
and two are in the (2,1) or (1,2) state.
So, Emax = (12+22)(h2/8mL2).
Bonding Between Atoms
-- How can two neutral objects stick together? H + H  H2
+e
Let’s represent the atom in
space by its Coulomb potential
centered on the proton (+e):
r
n=3
n=2
n=1
+e
The potential energy of
the two protons in an H2
molecule look something
like this:
+e
r
The energy levels for this potential are more complicated, so we
consider a simpler potential that we already know a lot about.
Particle in a Finite Square Well Potential
This has all of the qualitative features of molecular bonding,
but is easier to analyze..
The ‘atomic’ potential:
Bound
states
The ‘molecular’ potential:
Consider what happens when two “atoms” approach one other.
There is one electron, which can be in either well.
‘Molecular’ Wave functions and Energies
“Atomic’ wave functions:
yA
‘Molecular’ Wavefunctions:
L = 1 nm
1.505 eV  nm 2
E
 0.4 eV
2
(2L)
2 ‘atomic’ states  2 ‘molecular’ states
yeven
yodd
When the wells are far apart, the ‘atomic’ functions don’t overlap.
The single electron can be in either well with E = 0.4 eV.
‘Molecular’ Wave Functions and Energies
Wells far apart:
d
Degenerate states:
yeven
1.505 eV  nm2
E
 0.4 eV
(2L)2
yodd
Wells closer together:
yeven
yodd
L = 1 nm
1.505 eV  nm2
E
 0.4 eV
(2L)2
d
‘Atomic’ states are beginning
to overlap and distort. yeven
and yodd are not the same.
The degeneracy is broken:
Eeven < Eodd (why?)
yeven: no nodes,
yodd: one node.
Act 4: Symmetric vs. Antisymmetric states
d
yeven
yodd
What will happen to the energy of yeven as the two wells come
together (i.e., as d is reduced)? [Hint: think of the limit as d  0]
a. Eeven decreases.
b. Eeven stays the same.
c. Eeven increases.
Act 4: Solution
d
yeven
yodd
What will happen to the energy of yeven as the two wells come
together (i.e., as d is reduced)? [Hint: think of the limit as d  0]
a. Eeven decreases.
b. Eeven stays the same.
c. Eeven increases.
As the two wells come together, the
barrier disappears, and the wave
function spreads out over a single
double-width well. Therefore the energy
goes down (by a factor of 4).
Energy as a Function of Well Separation
When the wells just touch (d = 0, becoming one well) we know the
energies:
2L = 2 nm
yeven
1.505 eV  nm2
E1 
 0.1 eV
(4L)2
(n = 1 state)
yodd
1.505 eV  nm2 2
E2 
 2  0.4 eV
(4L)2
(n = 2 state)
As the wells are brought together, the even state always has lower
kinetic energy (smaller curvature). The odd state stays at about the
same energy.
odd
0.4 eV
f
0.1 eV
Splitting between even and
odd states:
even
d
DE = 0.4 – 0.1 eV
= 0.3 eV
Molecular Wave functions and Energies
with the Coulomb Potential
To understand real molecular bonding, we must deal with two issues:
 The atomic potential is not a square well.
 There is more than one electron in the well.
+e
r
Atomic ground state (1s):
yA
n=1
Molecular states:
+e
+e
+e
r
+e
yodd
yeven
Bonding state
Antibonding state
r
Energy as a Function of Atom Separation
The even and odd states behave similarly to the square well, but
there is also repulsion between the nuclei
that prevents them from coming too close.
Schematic picture for the total energy
of the nuclei and one electron:
Binding energy
Let’s consider what happens when
there is more than one electron:
Anti-bonding state
Bonding state
Equilibrium bond length
 2 electrons (two neutral H atoms): Both electrons occupy the bonding
state (with different ms). This is H2.
 4 electrons (two neutral He atoms). Two electrons must be in the
anti-bonding state. The repulsive force cancels the bonding, and
the atoms don’t stick. The He2 molecule does not exist.
d
Electron states in a crystal (1)
+e
Again start with a
simple atomic state:
r
yA
n=1
Bring N atoms together forming a 1-d crystal (a periodic lattice).
N atomic states  N crystal states.
What do these crystal states look like?
Like molecular bonding, the total wave function is (approximately)
just a superposition of 1-atom orbitals.
Electron states in a crystal (2)
The lowest energy combination is just the sum of the atomic states.
This is a generalization of the 2-atom bonding state.
No nodes!
The highest energy state is the one where every adjacent pair of
atoms has a minus sign:
N-1 nodes!
There are N states, with energies lying between these extremes.
Kruse Energy bands
Electron Wavefunctions and Energy Band
Highest energy wavefunction
Energy
y6
y5
y4
y3
y2
Closely spaced energy levels
form a “band” – a continuum
of energies between the max
and min energies
y1
Lowest energy wavefunction
FYI: These states are called “Bloch states” after Felix Bloch who
derived the mathematical form in 1929. They can be written as:
y n (x)  u(x)eik x
n
where u is an atomic-like function and the exponential is a convenient
way to represent both sin and cos functions
Energy Bands and Band Gaps
In a crystal the number of atoms is very large and the states
approach a continuum of energies between the lowest and highest
 a “band” of energies. A band has exactly enough states to hold
2 electron per atom (spin up and down).
Each 1-atom state leads to an energy band. So a crystal has
multiple energy bands. Band gaps (regions of disallowed energies)
lie between the bands.
2s band
Band gap
1s band
Summary
Atomic configurations
States in atoms with many electrons – filled according to the Pauli
exclusion principle
Molecular wave functions: origins of covalent bonds
Example: H + H  H2
Electron energy bands in crystals
Bands and band gaps are properties of waves in periodic systems.
There is a continuous range of energies for “allowed” states of an
electron in a crystal.
A Band Gap is a range of energies where there are no allowed states
Bands are filled according to the Pauli exclusion principle