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17 PROBING DEEP INTO MATTER
Creation and Annihilation
• Describe differences between
matter and antimatter
• Apply conservation laws to
annihilation and
materialisation events
• Explain how antimatter is used
in medical imaging (PET)
Starter: Can you give reasons why
imaging the brain using x rays may
have limitations or drawbacks.
• Conventional x ray imaging
does not show soft tissue well
• Conventional x ray imaging is a
“2 dimensional” technique
• Computerised axial
tomography can give 3D
information but x ray dose is
higher than conventional x ray
imaging
• X ray techniques cannot map
brain activity, only show
structures
Read pages 189-191
Answer the following:
1) What does PET stand for?
2) Try and summarise in you own words how it
works.
3) What is the main difference between matter and
antimatter? What is the one thing that is the same?
Ext:
4) What is meant by creation, annihilation and pair
production?
Energy is conserved
Energy, momentum
Conserved quantities
and Electric Charge
conserved
es
total energy before
= kinetic energy of particles
+ rest energy of particles
minimum value of energy
before
e– is rest energy:
Simplify:
assume head-on
collision with
equal speeds
e–
energy after is
energy of gamma
photons
= 2  0.511 MeV
e+
= 2 mc 2 = 2  0.511 MeV
e
Momentum is conserved
+
total energy before

total linear momentum before
Energy is conserved

total energy after
=
e–

= kinetic energy of particles
+ rest energy of particles
=
total linear momentum after

e+
=
same mass;
equal and
opposite velocities

total energy afterenergy E,
momentum p = E/c
photons identical,
momentums opposite
energy after is
total
momentum
before
=0
total momentum = 0
minimum
value
of
energy
total energy after
=
energy of gamma
before is rest energy:
photons
= 2 mc 2 = 2  0.511 MeV
Electric charge is conserved
= 2  0.511 MeV
total charge before
energy after is
charge
energy of gamma
Momentum is conserved
(–e) + (+e) = 0
photons
=
total charge after
charge
0 +0= 0
Making PET scans
A pair of gamma rays are emitted in opposite
directions as a result of electron/positron
annihilation inside the patient
Scintillator – captures gamma ray
photon and emits lower energy
photons into photomultiplier tubes
signal processing

Photomultiplier –
incoming photon creates
a cascade of electrons,
giving an electrical pulse
output
One pair of detectors will
respond almost
simultaneously. This near
coincidence shows that
the two gamma rays came
from a common source.
The tiny time difference
between the two signals
is then used to work out
where they came from
along the line between the
detectors.



Scintillators are arranged
in a grid on the inside
surface of the scanner. In
any short period of time
many detectors will
respond to gamma rays
from many different
annihilations inside the
body. A computer
produces a slice-by-slice
map of activity in the brain.
Annihilation and Creation
Pair annihilation and creation
Annihilation


gamma energy
= 2  0.511 MeV plus
kinetic energy of electrons
e–
e+
Creation?
e–
e+
extremely rare
(cannot bring two
identical photons
together)




Pair Production
Pair creation
e–
e+
gamma energy
= 2  0.511 MeV (minimum)
nucleus carries away
momentum, to conserve
momentum and energy
close to
nucleus

Starter…
An electron and a positron with negligible
kinetic energy annihilate and produce two
identical gamma ray photons.
(Rest mass of electron =9.11x10-31kg, h=6.63x10-34Js, c
= 3x108 ms-1)
Calculate
a) the energy released (in J and MeV)
b) the frequency of the gamma-photons (in Hz).
Particle interactions
• Describe how charged particles
interact via virtual photon exchange
• Explain how to describe these
processes using Feynman diagrams
and the “try all paths” approach
• Discuss consequences of the
differences between fermions and
bosons
Starter: What are the maximum and
minimum amplitudes that can result
from adding two phasors, each of
length 1 unit, and what are the
phase differences in each case?
Starter: Explain the point this diagram
is making….
“Try all paths” is a quantum rule obeyed by allSource
photons and
electrons.
Detector
The same idea is applied to interactions of particles.
Richard Feynman invented a type of diagram to help
physicists keep track of all the possible ways that particles
can interact. The rule “try all paths” changes to “try
everything allowed” or more technically “everything that is
not forbidden is compulsory”.
Quantum fields create and destroy particles
cannot happen
cannot happen
Electromagnetic quantum
field creates a photon
charge e
photon carries
away energy
and momentum
Electromagnetic quantum field
destroys a photon
photon delivers
energy and
momentum
charge e
energy and momentum not
both conserved
combined process can happen
virtual
(unobservable)
photon
charge e
charge e
photon exchanged:
charges exchange energy and
momentum
energy and momentum conserved overall
Forces between charges arise from the exchange of momentum through
exchange of virtual photons
Feynman diagrams show possibilities to be combined
Propagation and interaction of a pair of electrons
Electrons at A and B arrive at C and D
no photon exchange
C
D
A
A
B
electrons just travel A to C, and
B to D
Electrons are
identical,
so there is no way
to tell these apart:
D
C
add the phasor
arrows for both
diagrams
B
electrons just travel A to D,
and B to C
one photon exchange
D
D
C
C
B
A
electrons exchange a photon
A
B
No way to tell
these apart:
add the phasor
arrows for both
diagrams
electrons exchange a
photon
.... plus diagrams with more photons....
For each diagram, add phasor arrows for all possible space-time locations A, B, C, D.
Add total phasor arrows for each type of diagram
Quantum rule: Try all possible ways to interact
Ways for an electron to scatter a photon
In each diagram one electron and one photon come in, and one electron and one photon go out.
All diagrams represent the same process.
interaction with virtual pair
creation and annihilation
Simple electron–photon interactions
e–
e–
B
B
e–
A
e+
A
e–
Electron:
absorbs photon at A
travels to B
emits photon at B
A
e–
B
e–
Electron:
emits photon at A
travels to B
absorbs photon at B
photon creates e–, e+ pair at B
e+ goes to A
e– and e+ annihilate at A,
emitting photon
Phasor arrows for all diagrams are added to find total amplitude and phase for scattering
plus other
more
complex
diagrams
To find out..........
What is a fermion?
Give an example of a particle
that is a fermion.
What is a boson? Give an
example of a particle that is a
boson.
What happens if you try
to squeeze two fermions
into the same region of
space?
What happens if two identical
bosons encounter each
other?
Can you describe one
important consequence of
this?
Can you give one important
practical application of this?
Electron spin
Photon spin
Fermions and Bosons
Fermions
Bosons
½ integer spin (1/2, 3/2,....)
Electrons, Protons etc.
Integer spin (0,1,...)
Photons
Never occupy same quantum state
(Avoid each other always)
Can occupy same quantum state
(Can “flock” together in step)
Consequently.....
Consequently.....
• Two electrons in same orbital of • In a laser, lots of photons join
an atom must have opposite spin
together to produce beam of
(Pauli exclusion principle)
photons all of identical phase
and polarisation
• Matter is “hard”: atoms are
difficult to squash!
Identical particles in the same state
Identical particles in the same state
Electrons (fermions)
Photons (bosons)
electrons are indistinguishable
these possibilities cannot be distinguished
add the phasors for the two diagrams
D
C
X
C
Y
A
photons are indistinguishable
these possibilities cannot be distinguished
add the phasors for the two diagrams
D
X
Y
X
electron X goes to D
electron Y goes to C
electron X goes to C
electron Y goes to D
C
Y
D
X
Y
B
A
B
D
C
A
B
A
B
photon X goes to D
photon Y goes to C
photon X goes to C
photon Y goes to D
bring points C and D together
bring points C and D together
E
E
Exchanging particles
reverses phase.
Adding phasors gives
zero amplitude
A
B
+
Phasors for the two
diagrams become
the same.
Adding phasors
gives 2  amplitude
4  intensity
=0
A
B
+
=
There is zero probability for two electrons to be at the same space-time
point (be in same state)
This is the Pauli exclusion principle
Photons in a given state increase the chance of others joining them in
the same state
Fermions and Bosons
Fermions
Bosons
½ integer spin (1/2, 3/2,....)
Electrons, Protons etc.
Integer spin (0,1,...)
Photons
Never occupy same quantum state
(Avoid each other always)
Can occupy same quantum state
(Can “flock” together in step)
Consequently.....
Consequently.....
• Two electrons in same orbital of
an atom must have opposite spin
(Pauli exclusion principle)
• Matter is “hard”: atoms are
difficult to squash!
• In a laser, lots of photons join
together to produce beam of
photons all of identical phase
and polarisation
Light Amplification by Stimulated Emission of Radiation
Starter
For each of the following statements, say whether it applies to
FERMIONS or BOSONS:
Q1. A photon is an example of this class of particle.
Q2. These particles have integer spin values.
Q3. Particles which cannot occupy the same quantum state.
Q4. “Matter” particles, like protons, neutrons and electrons
belong to this class.
Q5. Virtual particles which are exchanged between interacting
matter particles belong to this class.
Q6. These particles can have spin values of 1/2, 3/2, 5/2 etc.
Q7. (For chemistry students): Two electrons cannot occupy the
same quantum state. How is it then possible to get two
electrons into the same orbital? Hint: in what way are the two
electrons distinguishable when in the same orbital?
Conservation in nuclear processes
• Explain why a new particle
was needed to account for
the energy spectrum of
beta particles
• Balance nuclear equations:
charge, mass-energy,
baryon number, lepton
number
The Baryon Family
Baryons contain three quarks (we come to them
later).
Protons and neutrons are baryons!
Baryon number must be conserved. Baryon
number is +1 for all protons and -1 for antiprotons.
Note: Protons and neutrons are also described
as nucleons.
Leptons – fundamental particles
Leptons are fundamental particles. As far as we are
aware they are not made up of anything smaller.
Examples are electrons and neutrinos.
Leptons are given a property called lepton number.
Electrons and neutrinos are given lepton number 1.
Where as the antiparticles are given lepton number -1.
All hadrons (non-leptons, which we learn more about
later) have lepton number 0.
‘Missing’ energy in beta decay
Beta decay of strontium–90
rest
energy
strontium–90
–
0.546 MeV
yttrium–90
38 protons
52 neutrons
39 protons
51 neutrons
Energy spectrum of beta decay of strontium–90
fraction of beta
particles per
unit energy
range
0
energy/MeV
0.546
Beta decay of strontium–90, including
antineutrino emission
rest
energy
strontium–90
–
energy E
0.546 MeV

energy 0.546 MeV – E
yttrium–90
Rutherford’s experiment
• Describe how alpha
scattering changed our view
of atomic structure
• Explore effects of changing
alpha particle energy and
nuclear charge on scattering
• Estimate upper limit on
nuclear size
Starter:
Q1. I have a charge of +1 and a lepton number of -1. What could
I be?
Q2. I have a charge of zero and a lepton number of +1. What
could I be?
Q3. I have a baryon number of +1, a lepton number of zero and a
charge of +1. What could I be?
Q4. I have a baryon number of +1 and a charge of zero. What
could I be?
Q5. I have a baryon number of -1 and a charge of -1. What could I be?
Starter:
Q1. I have a charge of +1 and a lepton number of -1. What could
I be? ANTIELECTRON (POSITRON)
Q2. I have a charge of zero and a lepton number of +1. What
could I be?
NEUTRINO
Q3. I have a baryon number of +1, a lepton number of zero and a
charge of +1. What could I be?
PROTON
Q4. I have a baryon number of +1 and a charge of zero. What
could I be?
NEUTRON
Q5. I have a baryon number of -1 and a charge of -1. What could I be?
ANTIPROTON
Gravitational and electric fields compared
Q1. (a) Write down the expression for the force F between two masses, M and
m, separated by a distance R.
Q1. (b) Write down the corresponding expression for the force F between two
charges Q and q, separated by a distance R. (The constant in the equation is
known as the electric force constant, and is denoted by k.)
Q2. (a) Write down the expression for the gravitational potential energy for an
object of mass m at a distance R from the centre of a planet of mass M.
Q2. (b) Write down the corresponding expression for the gravitational
potential energy of a charge +q at a distance R from another charge +Q.
Q3. (a) How much kinetic energy would you need to give the object in Q2. (a)
for it to be able to “climb out” of the potential well of the planet?
Q3. (b) How much kinetic energy would the charge +q in Q3. (a) have if it was
released and allowed to coast far away from +Q?
Rutherford’s
experiment
lead block to select
narrow beam of alpha
particles
radium source of
alpha particles
thin gold
foil
microscope to view zinc
sulphide screen and count
alpha particles
vary angle of
scattering
observed
Note: an alpha
particle
Is a helium
nucleus with the
electrons
removed. So it is
positively
charged!
scattered alpha
particles
alpha particle
beam
zinc sulphide screen;
tiny dots of light where
struck by alpha particle
Copy and complete the table using the statements provided
Rutherford’s
observation
Explanation
Model of the
atom
A. Most alpha particles
passed straight through the
foil, undeflected
B. Some were deflected
off course as they passed
through
nearly all the
mass of the
atom is in the
nucleus
C. Some bounced right
back from the foil
there must be centres of
+ charge in the atom
the centres of + charge
must be much heavier
than the alpha particles
the nucleus is
positively
charged, while
the electrons are
outside it, far
away
the atom is
mostly
empty space
the nucleus is tiny
compared to the
overall size of the
atom
Rutherford’s
observation
Explanation
Model of the atom
A. Most alpha The atom is mostly
particles
passed empty space.
straight through the
foil, undeflected
The nucleus is tiny
compared to the
overall size of the
atom
B.
Some
were There must be
deflected off course centres of + charge
as they passed in the atom.
through
The nucleus is positively
charged, while the electrons
are outside it, far
away
The centres of Nearly all of the mass of the
C. Some bounced +charge must be atom is in the nucleus
right back from the much heavier than
foil
the alpha particles
10 0
0
30 60 90 120 150 180
s c attering angle/degree
F o r calculatio n s
fo rce F =
a lpha particle
s c attered
2Z e 2
4 0d 2
charge +2e

s cattering angle
d
aiming error b
As sump tion s:
a lpha pa rtic le is the He nuc leus , cha rge + 2e
gold nucleus h as charge + Z e, and is muc h m ore
m as siv e than alpha particles
s c attering forc e is inv erse sq uare el ectrical
repuls ion
gold nuc leus
c harge + Ze
e qual forc e F b ut
nuc leus is mas siv e,
s o little r ecoil
TEST:
Are slowed -down alp ha particles
scat tered m ore?
Z
TEST:
Does using nuclei of smaller charge
scatter alpha particles less?
TEST:
Are slowed -down alp ha particles
scat tered m ore?
Z
reduce alpha
energy with
absorber
less energetic alpha
particle turned
around further from
the nucleus
lower speed
Z
alpha
d
er from
TEST:
Does using nuclei of smaller charge
scatter alpha particles less?
Z
r eplace foil by
metal of smaller
atomic number
alpha particle gets closer
to nucleus of smaller
charge and is deflected
less
smaller nucleus with less
charge, e.g. aluminium
Measuring the size of nuclei
• Explain how electron diffraction can be
used to measure nuclear size accurately
and precisely
• Determine nuclear diameter from
scattering curves
• Describe and explain the relationship
between nuclear size and nucleon
number
Starter: From your AS physics waves knowledge, explain the
appearance of all features of the single-slit diffraction curve
shown below:
JJ and GP Thomson: father and son Nobel
physics prize winners
Listen sonny, I got the
Nobel prize for
showing that the
electron is a particle...
Whatever, Dad. I
got it for showing
that the electron is
a wave!
Wave-particle duality
Evidence for
wave-like
character
Light
Electrons
Evidence for
particle-like
character
Electron diffraction
This eerie green glow is
caused by low energy
electrons in a cathode ray
tube striking the
phosphorescent coating
on the inside of the glass
bulb just behind the ruler.
In this case the diffraction is
caused by the electrons
passing through a thin layer
of polycrystalline graphite
(pencil "lead"). The regular
array of carbon atoms in the
crystals is responsible for
the diffraction effects.
How does the electron energy affect what is seen in an electron
diffraction experiment?
Diffraction of low energy electrons
Diffraction by planes of atoms of low energy electrons gives a
diffraction pattern that reveals the inter atomic spacing. Here, the
de Broglie wavelength of the electrons is quite large, as their
momentum is small. The wavelength is comparable to the inter
atomic spacing, so we get a lot of diffraction by the planes of atoms
in the manner of a diffraction grating diffracting light.
Diffraction of high energy electrons
With very high energy electrons, the de Broglie wavelength is
comparable to the size of a nucleus, and the diffraction effects seen
are essentially the same as single slit diffraction. The atomic
nucleus behaves as an obstacle for the electrons to diffract around,
and the diffraction patterns seen is essentially the same as we get
when light passes through a single slit.
Electron diffraction
Q1. Explain why there is a minimum in the curve.
Q2. What would be the effect on the curve of using higher energy electrons?
Q3. What would be the effect on the curve of using a sample of argon-40 in place of
neon-20?
Density of nuclear matter
Volume of nucleus increases linearly with number of nucleons
Electron scattering measures
radius r of nucleus
calculate volume = 43 r3
1500
197
1000
122
59
500
28
16
O
0
C
He
4
1
H
88
Sb
Sr
Si
12
0
Co
Au
50
150
100
number of nucleons
200
Estimate from graph:
100 nucleons in
volume 700  10–45 m3
Data:
mass per nucleon u
= 1.7  10–27 kg
volume per nucleon
= 7  10–45 m3
Calculate density:
density =
density =
mass
volume
1.7  10–27 kg
7  10–45 m3
= 2.4  1017 kg m–3
Density of ‘nuclear matter’ is roughly 2  1017 kg m–3
A matchbox-full of nuclear matter would have a mass of five billion
tonnes
Density of nuclear m atter
Volum e of nucleus increases linearly with number of nucleons
Electron scattering measures
radius r of nucleus
calculate v olume =
4
3
KEY POINT
Volume of nucleus is
proportional to number of
particles it contains.
r 3
1500
1 97
1000
122
59
500
28
16
Co
88
Au
Sb
Sr
Si
O
12
0
0
1
C
4
He
H
50
100
150
number of nucleons
200
Nuclear density questions...
Q1. If nuclear volume is proportional to the number of
nucleons (A) in the nucleus, explain why nuclear
radius r is given by: r = r0 A1/3 where r0 is a constant.
Q2. The nucleon number of a gold nucleus is 197.
a) The radius of the nucleon is 1.2x10-15m. Calculate
the radius r of the gold nucleus.
b) Calculate volume of the gold nucleus.
c) The mass of a nucleon is 1.67x10-27kg, calculate the
nuclear density.
Q3. Silver has a mass number of 108. What is its
nuclear density?
The structure of nucleons
• Explain why electrons are
well suited to be a probe
of nucleon structure
• Deduce the quark
composition of a range of
hadrons (baryons and
mesons)
Starter:
Explain why alpha particles
and protons are not well
suited to probing the size
of nuclei
Deep inelastic scattering
Medium energy: elastic scattering
electron
proton
Quarks move rapidly inside proton.
The interaction time is long enough
for the proton to behave like a blur of
charge.
Deep inelastic scattering
High energy: deep inelastic scattering
electron
scattered
at large angle
u
electron
d
Electron can hit one quark, and be scattered. Exchange
of high energy photons leads to the creation of a jet of
particles and antiparticles.
u
‘jet’ of particles, mainly mesons
Particle Family Tree
All
particles
Fundamental
particles (no
Non-fundamental
Particles Hadrons
internal structure)
(made of quarks)
Leptons
(electron, muon,
tau, neutrinos)
Baryons
Exchange
Particles
(contains 3 quarks)
Gluon, W, Z, photon
e.g. Proton,
neutron
As we currently understand!!
Mesons
(quark + antiquark)
e.g. Pion, kaon
Quarks
The building blocks of protons and neutrons, and other
fundamental particles.
Two flavours of quark…
The up quark (+ 2/3 e) and the down quark (– 1/3 e).
The first direct evidence for quarks was obtained when very
high-energy electrons (approx. 20Gev) in a beam were
scattered from a stationary target as if there were point-like
scattering centres in each proton or neutron.
Quarks do not exist in isolation. They are bound together by
the exchange of gluons (since they are the glue that hold the
quarks together).
Quark–gluon interaction
red quark
blue quark
red–blue gluon
red quark
blue quark
Quarks interact by exchanging gluons, which change the quark colours. Here a red quark and a blue quark
exchange a red–blue gluon. The red quark becomes blue and the blue quark becomes red. The quarks exchange
energy and momentum.
Colour charge
Quarks have a new type of charge
called colour charge.
The charge comes in 3 types.
Red, Green and Blue.
They aren't actually coloured.
Reason for colour charge is you
need one of each colour to make
a proton or neutron just like you
need all colours to make white.
Quarks pulled apart make m ore quarks
quark
quark
two quarks held together by
the gluon field...
gluon field
...pull the quarks apart. The
gluon field increases in
energy...
quark
quark
quark
quark
antiquark
quark
...a quark–antiquark pair
materialises from the gluon
field
You cant
really
separate
quarks!!!
Starter: Copy the table and fill in each box with
example(s) of each phenomenon
WAVES
Light
Electrons
PARTICLES
Chemistry questions that only physics can answer.....
DISCUSS IN PAIRS
o Why are atoms mostly empty space?
o Why don’t the electrons just fall into the nucleus?
o Why are only some energy levels allowed for
electrons in an atom or a molecule?
o Why does delocalisation of electrons stabilise a
molecule?
o Why are carrots orange?
The classical atom dethroned
• Explain the flaws in the
“solar system” model of the
atom
• Investigate an electron-wave
atomic model
• Use an electron-wave model
to explain both the stability
of aromatics and the origin
of colour in chemistry
What is the problem?!?
Maxwell’s theory
which states that an
electron being
accelerated in an
electric field will emit
radiation.
Hint: Think about the Diamond visit!
AS-RECAP
Wave-Particle Duality (for electrons)
What is the evidence?!?
de Broglie
λ = h/p = h/mv
-31
m=9.1x10 kg
-34
h=6.6x10 Js
AS-RECAP
Standing Waves…
at rest
Allowed discrete values of
wavelengths only….
fundamental n = 1
L = λ/2
L= λ
L= 3λ/2
Ie L = n λ /2
harmonic n = 2
L-length of string
n – harmonic
harmonic n = 3
Electrons in atoms… (QM)
Atoms can be thought of as a box in which electrons are trapped.
Results of model:
Electrons can only occupy certain discrete energy levels in
atoms.
Each level has a different associated standing wave of a
particular
λ.
A guitar–string
atom
Sim plify:
energy
C hange the 1/r
potential well
of the nucleus
into a pair of
fixed high walls
energy
d
trapped
electron
nucleus +
trapped
electron
1/ r
potential
well
Putting de Broglie
and standing waves
together….
Beta carotene…..
….gives carrots and pumpkins their orange colour
….in humans, is converted into retinal, which is essential for the vision process
• A conjugated system with alternating single and double bonds
• Average carbon-carbon bond length = 140 pm (1 pm = 10-12 m)
• 2 delocalisable electrons per double bond
Challenge
Can you use the equation for electron delocalisation in one dimension (particle
in a box) to :
(a) determine the pattern of energy levels in beta carotene,
(b) calculate the wavelength at which it absorbs light.
Beta-carotene: a chromophore modelled as a
particle in a box
• A conjugated system with
alternating single and double
bonds
• Average carbon-carbon bond
length = 140 pm
• Calculate the length over
which electrons are
delocalised
• Identify the quantum numbers of the
highest-occupied and lowest
unoccupied energy levels
• Use the formula E=n2h2/8mL2 to
calculate the energies of these levels
• Calculate the energy and wavelength of
the photon that would be absorbed if
an electron was excited from the
highest-occupied to lowest-unoccupied
level
Chemistry questions that only physics can answer.....
o Why are atoms mostly empty space?
o Why don’t the electrons just fall into the nucleus?
o Why are only some energy levels allowed for electrons
in an atom or a molecule? EXPLAINED WITH BASIC
MODEL
o Why does delocalisation of electrons stabilise a
molecule? EXPLAINED WITH BASIC MODEL
o Why are carrots orange? EXPLAINED WITH BASIC
MODEL
Starter
In the last lesson, we modelled the electron in an atom by treating the electron
as a standing wave confined to a potential well.
Q1. Which of the following feature(s) of the atom did this model reproduce?
- Quantised electron energies (electrons can only have certain energies
within the atom)?
- Energy levels becoming closer together with increasing energy, as is
observed in the spectra of atoms?
Q2. The model could be used to describe delocalised electrons in molecules
such as benzene and carotene, and thereby explain why delocalisation
(resonance) stabilises these molecules, and how colour arises in them.
(a) Using the energy level expression derived from the model, explain why
allowing electrons to spread out in a molecule leads to stabilisation.
(b) Explain how to calculate the energy of a photon absorbed by a molecule
with delocalised electrons.
Finally, a model for the atom that fits
experimental data!
• Develop an electron-wave
model that incorporates
potential energy
• Use the model to investigate
atom stability
• Calculate energy levels and
atomic radii with the model,
and see if they match
experiment
Considering Potential energy properly…
How small c ould a hydrogen atom be?
imaginary
box
d = 2r
 = 4r
1/r potential +
nucleus
Replace 1/r potential by a box of
width d = 2r
Calculate kinetic energy for waves
 = 2d = 4r
Calculate potential energy at r
potential energy E p = –
e2
4 0r
standing wave  /2 = d
momentum p = h/
2
kinetic energy = p /2m
h2
kinetic energy E k =
2m 2
Find the m in im um rad iu s of an atom , fo r total energy < 0
s hort wavelength
kinetic
unstable
energy
120
2
potential energy E p = –
nucleus
kinetic energy E k =
4 0r
2m 2
Considering Potential energy properly…
Find the m in im um rad iu s of an atom , fo r total energy < 0
s hort wavelength
kinetic
unstable
energy
120
kinetic energy =
100
+
small radius
Ek + Ep > 0
potential
energy
medium wavelength
just
stable
kinetic
ener gy
h2
2m 
minimum
radius of
bound atom
2
80
total energy > 0
unstable
60
40
total energy = 0
just stable
20
+
potential
energy
medium radius
Ek + Ep = 0
0
0.02
long wavelength
kinetic
energy
stab le
–20
0.04
0.06
radius r /nm
total energy < 0
bound
–40
potential energy = –
potential
energy
large radius
Ek + Ep < 0
+
–60
0.08
e2
40 r
KEY POINT:
If size is too small,
the kinetic
energy is too large
for electrical
potential to bind
the electron.
The Bohr model of the atom
Total Energy = Kinetic Energy + Potential Energy
Q1. Which of the terms in the equation above is always positive, and which is
always negative?
Q2. For a stable atom, one where the electron is bound, to the nucleus, what
must be true about the sum of KE+PE?
Kinetic Energy
KE =
h2
/
2mλ2
where λ is the wavelength of the
standing wave describing the
electron
What will happen to the KE if the
electron is confined in a smaller
space closer to the nucleus?
Potential energy
PE = -Zke2 / r
where r is the nucleus-electron
distance, e is the charge on the
electron and k is a constant (=1/4πε0)
What will happen to the PE if the electron
is confined in a smaller space closer to
the nucleus?
KE + PE for hydrogen atom,
with electron modelled as a "particle on a ring", 2-dimensional standing wave
4E-18
At very small radii, the KE is very large as the electron
is confined to a very small orbit. The result is that KE +
3E-18
Total energy / J
2E-18
1E-18
0
0.00
At r = 0.059 nm, the total energy
is a minimum. This is the
distance at which the electron
settles, and determines the
radius of the atom. The value of r
agrees exactly with experiment in
this case, showing that the model
is a valid one.
0.05
0.10
0.15
At large radii, the KE term is small. The atom is
bound because of the negative PE term, but not as
stable as it could be if the electron approached the
nucleus more closely.
0.20
0.25
-1E-18
-2E-18
-3E-18
r / nm
0.30
0.35
0.40
0.45
Chemistry questions that only physics can answer.....
o Why are atoms mostly empty space?
EXPLAINED WITH BOHR MODEL (electron wave + PE)
o Why don’t the electrons just fall into the nucleus?
EXPLAINED WITH BOHR MODEL (electron wave + PE)
o Why are only some energy levels allowed for electrons in an atom or
a molecule?
EXPLAINED CORRECTLY WITH BOHR MODEL (electron wave + PE)
o Why does delocalisation of electrons stabilise a molecule?
EXPLAINED WITH BASIC MODEL (particle in a box)
o Why are carrots orange?
EXPLAINED WITH BASIC MODEL (particle in a box)
Starter : Explaining atomic structure
Explain the following statements carefully, using the key words in italics
provided if need be.
• An electron in orbit around a nucleus does not radiate energy and fall
into the nucleus as predicted by classical physics.
• Matter is mostly empty space, with the atomic nucleus occupying a
tiny fraction of the volume of the atom.
• The energy levels occupied by an electron in atom can have only
certain values described by a quantum number n, where n = 1, 2, 3
etc.
• Atoms absorb and emit light only at certain specific wavelengths.
wave-particle duality standing wave destructive interference
de Broglie wavelength kinetic energy potential energy
stability photon energy
Energy levels and spectra
• Use atomic energy
level data to
calculate emission
and absorption
spectra
• Interpret results of
the Franck-Hertz
experiment in terms
of atomic energy
levels
Energy levels in
hydrogen
En =
2
-13.6/n
Questions….
Q1. What is the
wavelength of the
Red and blue Balmer
lines?
Q2. If the wavelength
of an emitted photon
was 102 nm, which
transition from the
Lyman series caused
it?
The Franck-Hertz experiment
The Franck-Hertz experiment
• Why does the
current suddenly
drop sharply?
• Why does the drop
occur at different
beam energies for
different elements?
A hydrogen atom has energy levels at -13.6 eV (ground
state, containing one electron), -3.4 eV, -1.5 eV, -0.9 eV. The
energy levels are measured with respect to the ionisation
limit, 0.0 eV.
(a) Calculate the wavelength of the lowest-energy photon
that could be absorbed when an electron in the ground
state of a hydrogen atom is excited to an upper
energy level.
(Data: h = 6.6 x 10-34 Js, c = 3 x 108 ms-1, e = 1.6 x 10-19 C)
[3]
(b) Electrons of energy 9 eV are fired through hydrogen gas in its
ground state. The electrons are scattered without loss of energy.
When the experiment is repeated with electrons of 11 eV
energy, the electrons are scattered inelastically, emerging with
energies of about 1 eV. Explain these observations.
[3]