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G482
Electrons, Photons and Waves
Module 2.5
Quantum Physics
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G482 Module 2.5 Quantum Physics
1.
Energy of a photon
(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 and
E = hc
λ
(d) define and use the electronvolt (eV) as a unit of energy;
(e) use the transfer equation eV = ½ mv2 for electrons and other charged
particles;
(f) describe an experiment using LEDs to estimate the Planck constant h using
the equation
eV = hc
λ
(no knowledge of semiconductor theory is expected).
Photon, Photon
energy, electronvolt,
planck constant,
frequency, kinetic
energy, LED,
semiconductor, quanta
(quantum),
2.
The photoelectric
effect
(a) describe and explain the phenomenon of the photoelectric effect;
(b) explain that the photoelectric effect provides evidence for a particulate
nature of electromagnetic radiation while phenomena such as interference and
diffraction provide evidence for a wave nature;
(c) define and use the terms work function and threshold frequency;
(d) state that energy is conserved when a photon interacts with an electron;
Photon
Photoelectron
Work function
Photon energy
Potential energy
Kinetic energy
Intensity
Planck’s Constant
Electroscope
3.
Einstein’s equation
and the
photoelectric effect
(e)select, explain and use Einstein’s photoelectric equation: hf =φ+KEmax;
(f) explain why the maximum kinetic energy of the electrons is independent of
intensity and why the photoelectric current in a photocell circuit is proportional to
intensity of the incident radiation.
Photoelectric effect,
planck, frequencu,
wprk finction, kinetic
energy, electron,
photon, intensity,
incident radiation
4.
The Photon Model
(a) explain electron diffraction as evidence for the wave nature of particles like
electrons;
(b) explain that electrons travelling through polycrystalline graphite will be
diffracted by the atoms and the spacing between the atoms;
Particulate
Quanta
Corpuscle
Photon
Wave
Interference
5.
Wave particle
duality
(c) select and apply the de Broglie equation
λ=h
mv
(d) explain that the diffraction of electrons by matter can be used to determine
the arrangement of atoms and the size of nuclei.
de Broglie,
momentum, planck
constant, wavelength,
diffraction, nuclei
6.
line spectra
2.5.4 Energy levels in atoms
Candidates should be able to:
(a) explain how spectral lines are evidence for the existence of discrete energy
levels in isolated atoms, i.e. in a gas discharge lamp;
(b) describe the origin of emission and absorption line spectra;
(c) use the relationships hf = E1 – E2 and
hc = E1 – E2
λ
Atom, electrons,
Spectral lines, energy
levels, absorption,
wavelength, frequency,
emission, dispersion
7.
G482 Module 5: 2.5
Quantum Physics
Test
Review quantum physics
8.
G482 review
Review Electrons, Waves and Photons
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Lesson 35 notes – The photoelectric effect and photon
energy.
Objectives
(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 and
E = hc
λ
(d) define and use the electronvolt (eV) as a unit of energy;
(e) use the transfer equation eV = ½ mv2 for electrons and other charged particles;
(f) describe an experiment using LEDs to estimate the Planck constant h using the
equation
eV = hc
λ
(no knowledge of semiconductor theory is expected).
The photoelectric effect
A Gold Leaf Electroscope is used to see the photoelectric effect. An
electroscope’s gold leaf will rise when it is charged and fall when it is
discharged. The photoelectric effect shows that is possible to discharge the
electroscope using light.
But not any sort of light.
The diagram shows 3 different types of light hitting a charged Zinc plate on
top of a gold leaf electroscope.
or
No effect
No effect
With U.V. leaf
falls
immediately
(Diagrams: resourcefulphysics.org)
With red light (from a laser) there is no effect.
With white light from a lamp there was no effect.
But using a UV lamp the electroscope immediately discharges and the gold
leaf falls.
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So what’s happening?
The light falling on the Zinc plates has different frequencies and this
determines whether the plate will discharge.
UV has the highest frequency, lamp light doesn’t produce UV and the red light
hasn’t got a high enough frequency to discharge the Zinc plate.
By discharging we mean getting rid of the electrons. So the light is physically
hitting the electrons from the plate.
The light needs a certain frequency before the electrons will be knocked off
the plate; this is called the Threshold Frequency. The amount of electrons
being knocked off is proportional to the intensity of the light.
Wave particle duality
If light were just a wave then this couldn’t happen.
If light were just a wave then the electrons would absorb some energy no
matter what the frequency.
If light were just a wave then emission of an electron would take longer when
a lower intensity light were used, not instantaneously.
But light is not just a wave.
It can also behave as though it were made of tiny particles or packets. We call
these particles photons. It is one of these photons that will hit one electron on
the plate, the electron will absorb the energy and it will fly off the plate. So if
the intensity is greater, i.e. there are more photons, then more electrons can
be knocked off.
Photon Energy (The Einstein relation)
Einstein assumed that each packet of light had a certain amount of energy.
This energy must be proportional to its frequency.
Energy of a photon, E = hf
Where h is Planck’s constant = 6.63x10-34 J s (Or you could think of it as
Joule per Hertz)
And f is the frequency of the light.
Using c=fλ we get:
E = hc/λ
Where c is the speed of the electromagnetic waves.
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Lesson 36 notes - The Photoelectron
Objectives:
(a) describe and explain the phenomenon of the photoelectric effect;
(b) explain that the photoelectric effect provides evidence for a particulate nature of
electromagnetic radiation while phenomena such as interference and diffraction provide
evidence for a wave nature;
(c) define and use the terms work function and threshold frequency;
(d) state that energy is conserved when a photon interacts with an electron;
.
The Photoelectric effect
The diagram shows photons hitting the surface of a metal and photoelectrons
being ejected.
Photons with their Photon Energy and at least the threshold frequency hit
a metal. If the plate is Zinc, UV will nudge the photoelectrons off, if gamma
rays hit the metal they will be whipped off with more force.
The surface photoelectrons absorb the energy and are emitted out of the
metal with the excess energy in the form of Kinetic energy.
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If the intensity increases so that there are now more photons, more
photoelectrons are emitted.
But each photon arriving at the surface has the same photon energy
therefore each photoelectron emitted has the same kinetic energy.
Quantum Well
Photoelectron
Kinetic energy
Photon
energy
Photon
energy
Φ
electron
The diagram shows a representation of the energy levels involved when
releasing a photoelectron from the surface of a metal.
The Photon has some energy (The Photon Energy equal to hf where h is
Planck’s constant and f is the frequency of the radiation).
The electron that the photon hits absorbs all this energy.
It takes a certain amount of energy to release the electron, this energy is
called the Work Function energy and has the symbol Φ (phi).
If there is any excess energy, then the emitted photoelectron will have some
kinetic energy as it flies off the metal’s surface.
Because of the Law of conservation of energy we can see that:
The Photon
Energy
=
The Work
Function
Energy
+
The
Photoelectron’s
Kinetic Energy.
This idea will be extended in lesson 24.
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Lesson 35 and 36 questions – The Photoelectric effect
and photon energy and photoelectrons.
1)
Fig1.1 shows an electrical circuit including a photocell.
Fig 1.1
The photocell contains a metal plate X that is exposed to electromagnetic radiation.
Photoelectrons emitted from the surface of the metal are accelerated towards the
positive electrode Y. A sensitive ammeter measures the current in the circuit due to
the photoelectrons emitted by the metal plate X.
In this question, one mark is available for the quality of written communication.
Name and describe the process by which the photoelectrons are released from the
plate X by electromagnetic radiation.
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Total [6]
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2)a) State what property of electromagnetic radiation is demonstrated by the
photoelectric effect.
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(1)
b)
Define each of the following terms
i)
photon
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(1)
ii)
threshold frequency
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(1)
3)a) A radioactive material emits photons, each having an energy of 1.6 x 10-13 J.
Calculate the frequency of the electromagnetic radiation emitted by the radioactive
material.
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(4)
b)
Calculate the wavelength of the electromagnetic radiation.
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(2)
c)
State the principal type o electromagnetic radiation emitted by the material.
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(1)
Total [7]
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Lesson 37 notes – Einstein’s equation and the
photoelectric effect
Objectives
(e)select, explain and use Einstein’s photoelectric equation: hf =φ+KEmax;
(f) explain why the maximum kinetic energy of the electrons is independent of intensity
and why the photoelectric current in a photocell circuit is proportional to intensity of the
incident radiation.
Einstein’s equation
Recall from lesson 23 that:
The Photon
Energy
=
The Work
Function
Energy
The
Photoelectron’s
Kinetic Energy.
+
Well let’s look at this in more detail.
The photon energy E is given by E=hf where h is Planck’s constant and f
is the frequency of the incident radiation.
If f equals the threshold frequency, photoelectrons will only just escape the
surface and they have zero kinetic energy, so the photon energy = the work
function energy.
The Work function Φ is the amount of energy needed to just release an
electron from the surface of a material.
And the Photoelectron’s kinetic energy is the excess of energy that has
been absorbed from the incoming photon by the emitted electron and is given
by K.E.max=1/2 mvmax2. (It is the maximum K.E. since it may have lost some
energy if the light came through to lower levels of electrons not at the surface
for instance)
So now we have:
hf
=
Φ
+
1/2 mvmax2
And if f = threshold frequency we get:
hf
=
Φ
The electronvolt
Joules are quite a large unit of energy for the examples that you will look at
here and so another unit for energy is normally used. The electronvolt or eV.
The electronvolt (eV) is a unit of energy equal to the work done when an
electron is moved through a p.d. of 1 V.
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Since W=QV,
Therefore when an electron moves through a potential difference of 1 V, the
work done on that electron is equal to 1.6x10-19J.
So, 1eV = 1.6x10-19J
Extension
You should be familiar with the following but do not need to be able to recall
the experiment detailed.
Experiment to find Planck’s Constant and work function
The diagram shows a vacuum photocell connected to a variable PSU and
galvanometer.
As light strikes the photocell, photoelectrons are emitted and a photocurrent (it
is called a photocurrent since it is the light that makes the current flow) is set
up as long as the photon energy is larger than the work function energy for
the anode. (Typical values of Φ are between 2.2eV for lithium and 6.35 eV for
platinum).
A potential difference is set up across the photocell so that photoelectrons are
just stopped from reaching the cathode of the photocell. (the reading on the
galvanometer is zero). This is called the stopping voltage, Vs. This stopping
voltage happens because each electron leaving the surface of the metal now
has to do extra work because of the potential difference set up, an amount
eVs, (since W=QV, and Q=e and V = Vs).
By changing the filters we can change the frequency of light entering the
photocell and then a graph of Kinetic energy of a photoelectron against
frequency can be drawn.
Einstein’s photoelectric equation can be stated as:
eVs = hf – Φ
which can be compared to:
y = mx + c
So the gradient of the graph gives Planck’s constant h.
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Lesson 37 – Einstein’s equation
1)
a)
hf
Einstein’s photoelectric equation may be written as hf = Φ + ½ mvmax2.
Identify the terms:
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Φ
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½ mvmax2
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(3)
b)
The surface of sodium metal is exposed to electromagnetic radiation of
wavelength 6.5 x 10-7m. This wavelength is the maximum for which photoelectrons
are released.
i)
Calculate the threshold frequency
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(3)
ii)
Show that the work function energy of the metal is 1.9 eV.
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(3)
c)
For a particular wavelength of incident light, sodium releases photoelectrons.
State how the rate of release of photoelectrons changes when the intensity of light is
doubled. Explain your answer.
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Total [11]
2)a) Electrons are emitted from the surface of zinc when it is exposed to ultraviolet
radiation.
i)
Name this phenomenon
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(1)
ii)
State the typical value for the wavelength of ultraviolet radiation in metres.
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(1)
b)
Electromagnetic radiation incident on a metal plate releases energetic
electrons from its surface. The metal plate is placed in an evacuated chamber. The
energy of each photon is 2.8eV. The metal has a work function energy of 1.1eV.
i)
Explain what is meant by the work function energy of the metal.
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(1)
ii)
State the speed of photons.
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(1)
iii)
For an electron emitted from the surface of the metal, calculate
1.
its maximum kinetic energy in joules
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(3)
2.
its maximum speed.
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(2)
iv)
State the change, if any, to your answer for the maximum speed of an electron
emitted from the surface of the metal when the intensity of the incident
electromagnetic radiation is doubled.
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(1)
Total [10]
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Lesson 38 notes – The photon model
Objectives
(a) explain
electron diffraction as evidence for the wave nature of particles like
electrons;
(b) explain that electrons travelling through polycrystalline graphite will be
diffracted by the atoms and the spacing between the atoms;
The photoelectric Effect
The photoelectric effect was the first example of a quantum phenomenon to
be seen at the end of the 19th Century. Light or other forms of electromagnetic
radiation shone onto metals release electrons; the energy supplied by the
radiation frees the electrons from the metal.
The way in which the numbers and energies of electrons released changes
when the frequency and intensity of the radiation changes cannot be
explained using the classical wave model of light.

Increasing the intensity of the radiation does not increase the energy
of the electrons but releases more of them per second.

Increasing the frequency of the light increases the energy of the
electrons.

Below a certain frequency of radiation, f0, no electrons are emitted no
matter how intense the radiation.
These facts are explained using the photon model of light. Light (and all
EM radiation) is emitted and absorbed in little packets or quanta called
photons. The energy of a photon is equal to its frequency multiplied by
Planck's constant, h = 6.63 x 10-34 Js.
E=hf
The photoelectric effect is summed up by Einstein's photoelectric equation for
which he won the Nobel Prize.
KEmax of electrons = hf - Φ where Φ is the energy needed to escape from the
This phenomenon introduces the wave-particle duality of nature: light
behaves as a wave at times (e.g. Young's slits) and as a particle at times.
This duality is central to the way quantum mechanics explains nature as it
applies to everything.
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Young's slits
Thomas Young used this experiment to 'prove' that light was a wave at a time
when light was thought to be a particle. The light going through two slits
interferes and produces a pattern that is easy to explain using a wave model
but which cannot be explained if light acts like particles.
Experimental set-up of Young’s Slits
Observed fringe
pattern at screen
Wave particle duality
If light were just a wave then the electrons would absorb some energy no
matter what the frequency.
If light were just a wave then emission of an electron would take longer when
a lower intensity light were used, not instantaneously.
But light is not just a wave.
It can also behave as though it were made of tiny energy packets or particles.
We call these particles photons. It is one of these photons that will hit one
electron on the plate, the electron will absorb the energy and it will fly off the
plate. So if the intensity is greater, i.e. there are more photons, then more
electrons can be knocked off.
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Lesson 38 – The Photon Model
1)
In this question, 2 marks are awarded for the quality of written
communication.
According to wave-particle duality, electromagnetic radiation can either behave as a
wave or as a photon (which exhibits particle-like behaviour).
Describe the behaviour which supports this dual nature of electromagnetic radiation.
Wave behaviour:
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Particle-like behaviour
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(8)
Quality of written communication (2)
Total [10]
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Lesson 39 notes – Wave-particle duality
Objectives
(c) select
and apply the de Broglie equation
λ=h
mv
(d) explain
that the diffraction of electrons by matter can be used to
determine the arrangement of atoms and the size of nuclei.
Electron Diffraction
You have seen in lesson 24 that light can behave as a wave and a particle so
maybe particles of matter can behave as a wave.
The diagram below shows the equipment needed to view electron diffraction.
What you are seeing are the diffraction patterns of particles (electrons)
through a thin foil. This means that particles behave as waves.
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The way it works is that when the cathode has heated up (you will see it
glowing), the potential difference between the negative cathode (electron gun)
and the positive anode (the graphite or thin metal foil) is increased
accelerating the electrons through the foil. Some go straight through, some
don’t make it through, hitting the particles in the foil, but others are diffracted
out the other side towards the inside surface of the end of the tube which has
a luminescent screen deposited on it that when struck by an electron glows.
There is constructive interference where the tube glows and destructive
interference in between these rings.
If the p.d. between the cathode and anode is increased, the speed of the
electrons increases and the rings become smaller. So there is less diffraction
– their wavelength must have decreased.
The de Broglie equation
All of this was hypothesised 3 years before it was demonstrated in the
Electron diffraction tube by de Broglie in 1923.
He used the idea of photons behaving like matter particles and extended it to
all particles.
He related the wave like behaviour of matter to its momentum:
λ = h/p
where λ is the (de Broglie) wavelength of the particle (m)
h is Planck’s constant (Js)
and p is the momentum of the particle (kgms-1)
It is more conveniently represented as
λ = h/mv
where λ is the (de Broglie) wavelength of the particle (m)
h is Planck’s constant (Js)
m is the mass of the particle (kgms-1)
and v is the velocity (ms-1)
Extension
You will not need to be able to derive this but you may be interested where it
came from.
This equation comes from the photoelectric equation for the energy of a
photon and Einstein’s equation for special relativity.
E=hf = mc2
So
hf=pc (since p=mv (and v=c for light))
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c=hf/p
We know that for all waves:
λ = c/f
So substituting in what we know for c,
λ = hf/pf
Cancelling out f we get:
λ = h/p
or
λ = h/mc (for light)
or for particles not travelling at the speed of light::
λ = h/mv.
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Lesson 39 questions – Wave particle duality
1)a)
State the de Broglie equation. Define any symbols used.
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(2)
b)
Outline the evidence for believing that electrons behave like waves.
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(2)
c)
High speed electrons may be used to probe inside atomic nuclei.
i)
Calculate the de Broglie wavelength for a single electron which has a
momentum (mv) of 2.3 x 10-19kgms-1.
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(2)
ii)
Explain how your answer to (c)(i) would change for
1
a neutron of the same momentum
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2
an electron of half the momentum
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(3)
Total [9]
2)
In this question, 1 mark is available for the quality of written communication.
Describe and interpret the experimental evidence for the wave-like behaviour of
electrons.
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(5)
Written communication (1)
Total [6]
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Lesson 40 notes – Line Spectra
Objectives
(a) explain how spectral lines are evidence for the existence of discrete
energy levels in isolated atoms, i.e. in a gas discharge lamp;
(b) describe the origin of emission and absorption line spectra;
(c) use the relationships hf = E1 – E2 and
hc = E1 – E2
λ
Atomic Line spectra
Electrons orbit about atoms of particular energy levels. If an electron absorbs
energy it will rise into another energy level. As the electron moves back down
the energy that it has absorbed is released as light of a particular wavelength.
The energy absorbed is given by E = hf or E=hc/λ. (hf = E1 – E2 and
hc = E1 – E2)
λ
Electrons can only be in discrete orbits.
A photon can be emitted or absorbed by an atom only when an electron
jumps from one orbit to another.
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Emission and absorption line spectra
A hot solid, liquid or gas at high pressure has a continuous spectrum.
There is energy at all wavelengths.
A gas at low pressure and high temperature will produce emission lines.
There is energy only at specific wavelengths.
A gas at low pressure in front of a hot continuum causes absorption lines.
Dark lines appear on the continuum.
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RAB Plymstock School
So when we see a spectrum we can tell what type of source we are seeing.
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RAB Plymstock School