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5 Quantum Mechanics
G482 Electricity, Waves & Photons
2.5.1 Energy of
A Photon
2.5.2 The
Photoelectric
Effect
2.5.3 WaveParticle Duality
2.5.4 Energy
Levels in Atoms
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Index
Introduction....
The aim of this module is to introduce the concept of quantum behaviour. How do we know
that light is a wave?
The evidence for this comes from diffraction of light. However, this wave-like behaviour
cannot explain how light interacts with electrons in a metal.
A revolutionary model of light (photon model), developed by Max Planck and Albert
Einstein, is needed to describe the interaction of light with matter.
Physicists expect symmetry in nature. If light can have a dual nature, then surely particles
like the electron must also have a dual nature. We study the ideas developed by de Broglie.
The final section looks briefly at the idea that electrons in atoms have discrete bond
energies and they move between energy levels by either absorbing or by emitting photons.
There are many opportunities to discuss how theories and models develop with the history
of wave-particle duality.
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Practical Skills are assessed using OCR set tasks.
The practical work suggested below may be carried out as part of skill development.
Centres are not required to carry out all of these experiments.
This module does not lend itself to many experiments carried by the students. However, it
does contain many revolutionary ideas and engaging students in discussions is vital when
demonstrating some of the experiments.
1. Use a GM tube to ‘count’ gamma ray photons.
2. Determine the wavelength of light from different LEDs
3. Demonstrate the photoelectric effect using a photocell or a negatively charged zinc
plate
4. connected to an electroscope.
5. Observe ‘diffraction rings’ for light passing through a tiny hole.
6. Demonstrate the diffraction of electrons by graphite.
7. Observe emission line spectra from different discharge tubes. (A hand-held optical
spectrometer can be used to observe Fraunhofer lines in daylight. Caution: Do not look
directly at the Sun.)
Index
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Activities...
1. Show how a GM tube clicks when
detecting ‘gamma waves’ from a
radioactive source. Discuss the
implication of the ‘clicks’
2. Discuss historical ideas of light at a
wave not sufficient for all
phenomena. Light then to be seen
as stream of particles or quanta of
energy. Define a photon and give
formula E = hf. Relate with c = fλ
to give E = hc/λ
3. Use eV = hc/λ for different LEDs to
estimate Planck’s constant from
gradient of V-1/λ graph
4. Extend idea of W = eV to other
charged particles and hence eV =
½ mv2
5. Define the electronvolt (eV) as a
useful unit of energy on an atomic
scale
Resources....
Points to Note…
1. Possibly show 12V bulb at different
temperatures from variable supply,
1. Discuss meaning of word
quickly showing
quantum in this context.
red/orange/yellow/white hot and
Historical ideas could
drawing spectra to illustrate black body
include discussion of
radiation
black body radiation and
2. Different ‘coloured’ LEDs of given
cover areas of HSW
wavelength with variable supply and
2. Recap W = VQ from
voltmeter. Measure minimum p.d. to
definition of p.d. in
just give illumination
previous section
3. Use LHC at as an example of measuring
energy in eV and use energy to
calculate speed (non-relativistic) of
protons
http://www.youtube.com/watch?v=HDR7G2NsI6o&list=PLACA85BC3CD7F97FC&inde
x=25
Discovery of the Electron
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2.5.1 Energy of A Photon (p172)
Assessable learning outcomes....
(a) describe the particulate nature (photon model) of electromagnetic radiation;
(b) state that a photon is a quantum of energy of electromagnetic radiation;
(c) select and use the equations for the energy of a photon: E = hf =hc/
(d) define and use the electronvolt (eV) as a unit of energy;
(e) use the transfer equation eV = 0.5mv2 electrons and other charged particles;
(f) describe an experiment using LEDs to estimate the Planck constant h using the
Equation eV = hc/. (no knowledge of semiconductor theory is expected).
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a/b) Which is laser light & why?
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a/b) Photons a General Description...
Under the photon theory of light, a photon is a
discrete bundle, packet or quantum of
electromagnetic or light energy.
Photons are always in motion and, in a vacuum,
have a constant speed of light to all observers of
c = 2.998 x 108 ms-1.
Photons have zero mass but carry both energy and momentum, which are also
related to the frequency f and wavelength  of the electromagnetic wave by
E = hf = hc/
(as c = f )
They can be destroyed/created when radiation is absorbed/ emitted. They can
have particle-like interactions (i.e. collisions) with electrons and other particles.
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a/b) More on Photons...
 The photon is an elementary particle, despite the fact that it has no mass.
 It cannot decay on its own, although the energy of the photon can transfer (or
be created) upon interaction with other particles.
 Photons are electrically neutral and are one of the rare particles that are
identical to their antiparticle, the antiphoton.
 Not needed for AS - Photons are spin-1 particles (making them bosons), which
means that their energy is polarised in a direction. This feature is what allows
for polarisation of light. (i.e. TV aerials) (only need to know the outcome!)
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(b) state that a photon is a quantum of energy of electromagnetic
radiation;
All EM radiation can be thought of “photons” or “packets” or a
“quantum” of energy.
They are like little squiggles of energy!
If we take a photo like this shown using a photon sensor the
results are strange. Instead of a dim pattern getting stronger
we have dots which add to the image.
This experiment is evidence that light is a stream of some type
of particle-like object. (in certain conditions)
In fact many experiments convincingly lead to the
surprising result that electromagnetic waves, although they are
waves, have a particle-like nature. These particle-like
components of electromagnetic waves are called photons.
http://en.wikipedia.org/wiki/Photon
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(a) describe the particulate nature (photon model) of
electromagnetic radiation;
The photon model of electromagnetic waves consists of three
basic postulates:
1.
Electromagnetic waves consist of discrete, massless units
called photons. A photon travels in vacuum at the speed of
light, “c = 3 x 108ms-1
2.
Each photon has energy E = hf where f is the frequency of
the wave and h is a universal constant called Planck’s
constant. The value of Planck’s constant is In other words,
the electromagnetic waves come in discrete “chunks” of
Energy.
3.
The superposition of a sufficiently large number of
photons has the characteristics of a continuous
electromagnetic wave.
http://en.wikipedia.org/wiki/Photon
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(c) select and use the equations for the energy of a photon: E = hf =hc/
The known constants for these calculations are always;
h=6.63
350nm = 5.68 x 10-19J
590nm = 3.37 x 10-19J
700nm = 2.84 x 10-19J
x10-34Js
c=3.00 x 108 ms-1
• Using our formulae of E = hf or since c=f  , f= c/  we
could say for neatness and simplicity that;
E = hc/
• Try working out the energies for different frequencies of
visible light to test out your skills. You should get a range
of answers i.e. 3 x 10-19J. Try 350nm, 590nm, 700nm
High Energy
Low Energy
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(c) select and use the equations for the energy of a photon: E = hf =hc/
The known constants for these calculations are always;
h=6.63 x10-34Js
350nm =
590nm =
c=3.00 x 108 ms-1
• Using our formulae of E = hf or since c=f  , f= c/  we could say for
neatness and simplicity that;
700nm =
E = hc/
• Try working out the energies for different frequencies of visible light
to test out your skills. You should get a range of answers i.e. 3 x 10-19J.
Try 350nm, 590nm, 700nm
High Energy
Low Energy
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(d) define and use the electronvolt (eV) as a unit of energy; (recap)
Charge on the electron is 1e = 1.6x10-19 C
(eq1)
But we also know from electrical circuits; 1V = 1 JC-1
So by multiplying equation 1 by 1V on each side we get:
1e x 1V = 1V x 1.6x10-19 C (eq2)
Then sub in 1JC-1 for the voltage part on the RHS of (eq2) gives us;
1e x 1V = 1JC-1 x 1.6x10-19 C
This leaves us with definition: 1eV = 1.6x10-19 J
1MeV = 1x 106 x 1eV
We can use this a smaller version of the joule (not a smaller version of the volt!)
Convert these photon
energies from Joules
to eV........
5.68 x 10-19J =
3.37 x 10-19J =
2.84 x 10-19J =
eV
eV
eV
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(d) define and use the electronvolt (eV) as a unit of energy; (recap)
Charge on the electron is 1e = 1.6x10-19 C
(eq1)
But we also know from electrical circuits;
So by multiplying equation 1 by 1V on each side we get:
Then sub in 1JC-1 for the voltage part on the RHS of (eq2) gives us;
This leaves us with definition:
We can use this a smaller version of the joule (not a smaller version of the volt!)
Convert these photon
energies from Joules
to eV........
5.68 x 10-19J =
3.37 x 10-19J =
2.84 x 10-19J =
eV
eV
eV
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(e) use the transfer equation eV = 0.5mv2 for electrons and other charged
particles;
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(e) use the transfer equation eV = 0.5mv2 for electrons and other charged
particles;
•
If we imagine that we have two oppositely charged
metal plates in a vacuum with a PD of 5000V between
them.
•
A charged particle such as an electron -1.6 x 10-19C is
accelerated by the field from one plate to another.
•
Electron has a rest mass of 1/1840 of an a.m.u. Or 9.11
× 10–31 kg.
•
We can find the velocity or Kinetic energy that it gains
as.....
Units: (Jkg-1 )0.5 = kgm2s-2kg-1)0.5
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(e) use the transfer equation eV = 0.5mv2 for electrons and other charged
particles;
•
If we imagine that we have two oppositely charged
metal plates in a vacuum with a PD of 5000V between
them.
•
A charged particle such as an electron -1.6 x 10-19C is
accelerated by the field from one plate to another.
•
Electron has a rest mass of 1/1840 of an a.m.u. Or 9.11
× 10–31 kg.
1 2
mv  eV
2
2eV
v
m
or
m v  2meV
2 2
•
We can find the velocity or Kinetic energy that it gains
as.....
19
2 1.6 10 C  5000 JC
v
9.110 31 kg
1
mv  2meV
p  2meV
v  42 106 ms 1
Units: (Jkg-1 )0.5 = kgm2s-2kg-1)0.5
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(e) use the transfer equation eV = 0.5mv2 for electrons and other charged
particles;
What would be the velocity of the following particles
for a similar PD of 5000V?
1.
Proton (mass of 1.673 × 10–27 kg)
2.
Alpha Particle (mass of 6.646 × 10–27 kg)
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(e) use the transfer equation eV = 0.5mv2 for electrons and other charged
particles;
What would be the velocity of the following particles
for a similar PD of 5000V?
1.
Proton (mass of 1.673 × 10–27 kg)
2 1.6 10 19 C  5000 JC 1
v
1.673 1027 kg
v  1106 ms 1
2.
Alpha Particle (mass of 6.646 ×
10–27
19
m v  2meV
2 2
kg)
2  2 1.6 10 C  5000 JC
v
6.646 1027 kg
1 2
mv  eV
2
2eV
v
m
or
1
mv  2meV
p  2meV
v  0.7 106 ms 1
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(f) describe an experiment using LEDs to estimate the Planck constant h using the
Equation eV = hc/. (no knowledge of semiconductor theory is expected).
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LED....
incident light
A photodiode is a circuit component which can
be used to convert a light signal into an
transparent
conducting layer
electrical one.
insulating layer
e-
X
V
conducting layer
Light incident on the thin transparent
conducting surface layer of the diode passes
through it to be absorbed in the insulating
layer.
The energy of each photon is sufficient to
release one electron in the insulating layer. It is
a “quantum” effect based on the idea that that
the light behaves as a “quantum” or “photon”
3 x incident
photons
To the
circuit
The potential difference V applied across the
insulating layer causes these electrons to move
upwards to the upper conducting layer.
This is because it is connected to the positive
terminal on the PSU.
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Visible Light.... (Recap)
The energy E, frequency f, and wavelength λ of a photon are related by the
formula;
where h is Planck's constant and c is the speed of light. For example, the
spectrum of visible light consists of wavelengths ranging from 400 nm to 700
nm. Photons of visible light therefore have energies ranging from
Emin =1.78 eV
Emax = 3.11 eV
An electronvolt is also the energy of an infrared photon with a wavelength of
approximately 1240 nm. Similarly, 10eV would correspond to ultraviolet of
wavelength 124 nm, and so on……
h = 6.63 x 10-34 Js
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Using a ICT Spectrometer
Your teacher will show you some examples of laser light.
Can you use E=hf or hc/ to work out the energies of the
light involved in Joules and eV?
Lastly can you predict an energy that a UV photon might
have with a  of 253.7nm
Source
/ 1x 10-9m
Red
656
Green
532
Joules / J x10-19
eV
UV
h = 6.63 x 10-34 Js & c = 3 x 108ms-1
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Using a Spectrometer
Your teacher will show you some examples of laser light.
Can you use E=hf or hc/ to work out the energies of the
light involved in Joules and eV?
Lastly can you predict an energy that a UV photon might
have with a  of 253.7nm
Source
/ 1x 10-9m
Joules / J x10-19
eV
Red
656
3.0
1.9 ± 0.003
Green
532
3.7
2.3 ± 0.002
UV
254
7.8
4.89 ± 0.01
h = 6.63 x 10-34 Js & c = 3 x 108ms-1
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Revision Question
13. Show that the wavelength of a photon of energy 3.9 eV is 320 nm.
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Revision Question
On average, a student uses a computer of power rating 110 W for 4.0 hours
every day. The computer draws a current of 0.48 A and its screen emits visible
light of average wavelength 5.5 x 10-7m.
1.Calculate the energy of each photon of wavelength 5.5 x 10–7 m emitted from
the computer screen. energy = .............................. J
[3]
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Revision Question
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Revision Question
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Revision Question
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Revision Question
13. Show that the wavelength of a photon of energy 3.9 eV is 320 nm.
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Revision Question
On average, a student uses a computer of power rating 110 W for 4.0 hours
every day. The computer draws a current of 0.48 A and its screen emits visible
light of average wavelength 5.5 x 10-7m.
1.Calculate the energy of each photon of wavelength 5.5 x 10–7 m emitted from
the computer screen. energy = .............................. J
[3]
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Revision Question
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Revision Question
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Revision Question
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Connection
•
•
•
Connect your learning to the
content of the lesson
Share the process by which the
learning will actually take place
Explore the outcomes of the
learning, emphasising why this will
be beneficial for the learner
Demonstration
• Use formative feedback – Assessment for
Learning
• Vary the groupings within the classroom
for the purpose of learning – individual;
pair; group/team; friendship; teacher
selected; single sex; mixed sex
• Offer different ways for the students to
demonstrate their understanding
• Allow the students to “show off” their
learning
Activation
Consolidation
• Construct problem-solving
challenges for the students
• Use a multi-sensory approach – VAK
• Promote a language of learning to
enable the students to talk about
their progress or obstacles to it
• Learning as an active process, so the
students aren’t passive receptors
• Structure active reflection on the lesson
content and the process of learning
• Seek transfer between “subjects”
• Review the learning from this lesson and
preview the learning for the next
• Promote ways in which the students will
remember
• A “news broadcast” approach to learning
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