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GCE TEACHERS’ GUIDE
New Specifications:
for teaching from September 2008
Physics
GCE AS and A PHYSICS Teachers' Guide 1
Contents
GCE AS and A Level Physics
Teachers’ Guide
Page
1.
Introduction
1.1 - Overview of New Specification
1.2 - Changes for teaching from September 2008
2.
Delivering the specification
2.1 - Pathways through the Specification
3.
Support for Teachers
3.1 - Generic Resources
3.2 - NGfL Cymru
3.3 - General Websites
4.
Unit Guides
4.1 - PH1 :
4.2 - PH2 :
4.3 - PH3 :
4.4 - PH4 :
4.5 - PH5 :
4.6 - PH6 :
Motion, Energy & Charge
Waves & Particles
Practical Physics
Matter, Forces and the Universe
Electromagnetism, Nuclei & Options
Experimental & Synoptic Assessment
Contributors to the Teachers’ Guide
Issued January 2008
GCE AS and A PHYSICS Teachers' Guide 3
1
INTRODUCTION
The WJEC AS and A2 PHYSICS specification has been modified and updated for delivery from
September 2008. The first AS awards will be made in Summer 2009 and the first A level awards in
summer 2010. For the first availability of units, see page 2 of the specification. The specification can
be delivered and assessed in centres throughout the UK.
The revised subject criteria for GCE PHYSICS issued by the regulators have necessitated a change in
the course structure from the current 3 externally assessed units at each of AS and A2 to 2 externally
assessed theory units plus 1 internally assessed practical unit per stage.
This Guide is one of a number of ways in which the WJEC provides assistance to teachers delivering
the new specification. Also essential to its introduction are the Specimen Assessment Materials
(question papers and marking schemes) and professional development (INSET) conferences.
Other provision which you will find useful are:
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Examiners’ reports on each examinations series
Free access to past question papers via the WJEC secure website
Easy access to specification and other key documents on main website
Easy access to study materials on the main website
Itemised feedback on outcomes for candidates at question level
Regular INSET delivered by Chief Examiners
Additional materials on the National Grid for Learning Wales (NGfL Cymru)
Easy access to both the Subject Officer (a physics specialist) and to administrative sections
Contact Points for GCE PHYSICS are as follows:
Helen Francis [email protected] 029 2026 5081
(Subject Officer)
Matthew Roberts [email protected] 029 2026 5380
(Administrative Support Officer)
Subject page
www.wjec.co.uk
INSET Section [email protected]
www.wjec.co.uk/professionaldevelopment
GCE AS and A PHYSICS Teachers' Guide 4
1.1
Overview of the Specification
The specification is divided into a total of 6 units: 3 AS units and 3 A2 units. Weightings noted below
are expressed in terms of the full A level qualification.
AS (3 units)
PH1
20% 1¼ hour Written Paper 80 marks[120 UM]
Motion, Energy & Charge
Approx 7 structured questions. No question choice. No sections.
PH2
20% 1¼ hours Written Paper 80 marks [120 UM]
Waves & Particles
Approx 7 structured questions. No question choice. No sections.
PH3
10% Internal Assessment 48 marks [60 UM]
Practical Physics
Experimental tasks, performed under controlled conditions, based upon
experimental techniques developed in the AS course.
A LEVEL (the above plus a further 3 units)
PH4
18% 1¼ hour Written Paper 80 marks [108 UM]
Oscillations & Fields
Approx 7 questions. Includes synoptic assessment. No question choice.
No sections.
PH5
22% 1¾ hour written paper 100 marks[132 UM]
Electromagnetism, Nuclei and Options
Section A: Approximately 5 questions on the compulsory content of the
unit. 60 marks
Section B: Case Study, synoptic in nature, based upon open-source
material distributed by the board. 20 marks
Section C: Options: Alternating Currents, Revolutions, Materials,
Medical Physics, Energy. 20 marks
PH6
10% Internal Assessment [UMS = 60]
Experimental & Synoptic Assessment
An experimental task (25 marks), and a data-analysis task (25 marks)
performed under controlled conditions, both synoptic in nature.



Assessment units PH1, PH2 and PH4 are available in the winter examination series. All units
are available in the summer examination series.
Internally assessed units PH3 and PH6 are timetabled for April and March respectively.
Synoptic assessment is included in PH4 and PH5. It is inherent in the internal assessment
PH6.
GCE AS and A PHYSICS Teachers' Guide 5
1.2
Changes to the specification for delivery in September 2008
60% of the content of the AS and A2 specifications is prescribed in the GCE Physics criteria, or core.
This material is common to all the examination specifications with the title Physics. The content of
the core is little changed from the previous version. The changes mainly affect the positioning of
topics within the specification, i.e. AS or A2. “Energy concepts,” for example, has been moved from
A2 to AS – this allows it to sit nicely against kinematics and electrical power, giving scope for
questions on electrical power generation. There are also changes between units at the two levels for
more logical development of concepts – waves now sits alongside quanta to make PH2 a unit which
deals exclusively with waves and particles.
Significant changes to the theory assessments:
 the introduction of novel and contemporary material in AS – leptons and quarks, lasers and
stellar spectra as a means of understanding the composition of stars.
 the introduction of novel and contemporary material into the compulsory part of A2 – the
application of gravitational and circular motion theory, via the Doppler Effect, to the study of
extra-solar planets and the hypothetical dark matter in galaxies.
 the introduction of options into A2 – as part of the PH5 unit, candidates will study one of:
Further electromagnetism & Alternating Current
Revolutions in Physics
Materials
Biological Measurement & Medical Imaging
Energy Matters
All these options are designed to build upon the compulsory core material in AS and A2 and
lend themselves in part to directed self study.
 The Case Study. Candidates will be issued with an article on a contemporary topic. They will
be expected to study it and to answer questions stimulated by the article as part of the PH5
examination.
Changes to the assessment of practical work
The regulatory authorities required practical work to be assessed externally. Following detailed
consultation with teachers, the assessments at both AS and A2 will take the form of WJEC-set tasks.
These will be timetabled and detailed instructions for supervisors will be sent out in advance to allow
them to assemble the necessary apparatus and materials. These will be marked externally by WJEC
examiners.
As these assessments come under the Internal Assessment criteria there are no inter-board agreed
dates for these activities. In order to separate the AS and A2 assessments, to allow as long as possible
for AS candidates to develop practical skills and to allow centres time to send in samples and marks
by mid May, the tasks will be timetabled as follows:
PH3 Late April
PH6 Mid March
GCE AS and A PHYSICS Teachers' Guide 6
2.
DELIVERING THE SPECIFICATION
The specification was designed for and by physics teachers. The extent and depth of the content in each
of the theory units – PH1, 2, 4 & 5 – was chosen to allow it to be taught in approximately 60 hours. This
time includes individual and demonstration practical work required for the examination of the content of
each unit. Time will also need to be devoted to a systematic development of candidates’ experimental and
investigatory skills in preparation for assessment units PH3 and PH6.
2.1
Pathways through the specification
The specification is unitised and there is no requirement for candidates to take units in a particular
sequence. The following models are supported:
(a)
Fully sequential
Time
January year 12
April year 12
May/June year 12
January year 13
March year 13
June year 13
(b)
Linear AS and unitised A2
Time
April year 12
May/June year 12
January year 13
March year 13
June year 13
(c)
Unit taken
PH1
PH3
PH2
PH4
PH6
PH5
Unit taken
PH3
PH1 & PH2
PH4
PH6
PH5
Linear A level
Time
March year 13
April year 13
May/June year 13
Unit taken
PH6
PH3
PH1, PH2
PH4, PH5
These assessment models are not exhaustive and all units may be retaken on any number of occasions
prior to cashing in.
GCE AS and A PHYSICS Teachers' Guide 7
3,
SUPPORT FOR TEACHERS
The most immediate support for teachers is from the Subject Officer, an experienced teacher of
Physics with 25 years experience of preparing candidates for A level Physics. The contact details are
given on page 2 of this guide. In addition, the physics examining team consists of teachers and
academic with a wide range of subject expertise and experience.
No single text book is recommended for the WJEC physics course. The matter of a textbook is one of
personal taste and most of the contents of the specification are well covered in a range of textbooks.
Some topics, within the specification are not found in other specifications. Others, though generally
covered, often cause confusion for A level Physics students. This guide is intended to cover these
gaps. In particular the following topics are covered:
Topic
Photoelectric
effect
Unit
PH2
Comments
This topic is to be found within the national core. It is however,
conceptually difficult and the notes are intended to provide guidance for
teachers as to the level required.
Lasers
PH2
The concept of atomic energy levels and the spontaneous absorption and
emission of photons is part of the legacy specification and, although
numerical questions are generally well, there is often confusion at a
qualitative level. It is expected that the increased exposure to the
concepts, with the introduction of lasers, will reduce the difficulty. The
notes are intended to introduce teachers to this new topic.
Leptons and
PH2
The coverage of the Standard Model expected is restricted to that of the
quarks
1st generation of particles.
The analysis PH2
Intended to show students how the study of spectra allows astronomers
of
stellar
to develop an understanding of stellar conditions and processes. The
spectra
basic physics of black body radiation and absorption spectra is standard
but the application is one which is not well covered in text books.
Uncertainty
PH3 & The understanding of the limitations of physical measurements is
analysis
PH6
fundamental to an appreciation of the degree of trust which can be
placed on the results of investigations. This is introduced in the AS
course and developed in the A2.
Orbital
PH4
This is the most up-to-date topic in the specification. It deals with the
mechanics
way in which the study of orbital motion, using Doppler measurements
on spectral lines has enabled the masses of binary stars to be
determined, extra-solar planets to be detected and the existence of Dark
Matter within galaxies to be inferred. The treatment is quite
mathematical.
A.C. Theory
PH5
A traditional topic for WJEC specifications, this has been left out of
Option A some other exam boards’ material and is not always covered in text
books.
Revolutions
PH5
An innovative unit which traces the history of scientific understanding
Option B in electromagnetism through the nineteenth century, culminating in the
theory of Special Relativity. In addition to a synopsis for teachers in this
guide, an exhaustive comprehensive pupil reader is available on the
WJEC website.
Materials
PH5
In addition to teacher guidance notes in this document, a student reader
Option C on the development of single crystals for use in turbine blades is
available on the WJEC website.
GCE AS and A PHYSICS Teachers' Guide 8
Topic
Biological
measurement
&
medical
imaging
Energy
Matters
Unit
PH5
Option D
Comments
Although well covered in text books, there is often a degree of
confusion as to the detail required. These notes clarify the depth and
breadth which ill be expected.
PH5
Option E
Many of the areas in this option are covered in the AS part of the
specification. The guidance notes indicate how they will be treated at
A2 and integrated with the A2 topics of thermodynamics and nuclear
energy.
Student Guidance notes and worksheets: As the table indicates, two areas of the specification have
publications for students in addition to those intended for teachers. There are also on the website
several documents which will help students and their teachers in preparing for examinations. These
documents cover, amongst other topics:
Thermodynamics – specifically the 1st Law
Ductile deformation of metals.
It is the intention to add to these documents and an up to date list can be found on the A level Physics
page of the WJEC website.
School Physics
The website www.schoolphysics.co.uk covers most of the areas of the WJEC specifications at both
A level and GCSE. It also provides links from specification topics to the relevant resources. Whilst
this is not a WJEC resource, it is a useful one.
Other websites
There are many websites which are of use in the study of Physics to A level. Unfortunately the URLs
of websites change and some are withdrawn. For a list of suggested websites see the document
Physics Websites of the WJEC website.
Suggested textbooks include:
GCE AS and A PHYSICS Teachers' Guide 9
4.
4.1
UNIT GUIDES AS
PH1 – Motion, Energy & Charge
General
PH1 is intended as a general introduction to the AS/A Physics course. It is designed to
provide:
 a gentle transition from GCSE, assuming that candidates have previously covered the
content of GCSE Science and GCSE Additional Science;
 topics which lend themselves to developing students’ practical and investigatory
skills;
 topics which are drawn upon in the remainder of the course.
PH1.1 – Basic Physics
This section includes the idea of units and dimensions, scalar and vector quantities, forces
and their combination, density and moments. Previously, density was not covered until A2
when it suddenly appeared in the kinetic theory of gases. Placing it in PH1.1 will enable
candidates to develop a feeling for the concept in less conceptually-demanding contexts. It
also provides a useful introduction to the uncertainties associated with individual
measurements and the uncertainties in quantities which are derived from combinations of
measurements.
With the exception of density, previous examination questions can be found in past PH1
papers of which there are over 10.
PH1.2 – Kinematics
This section is essentially unchanged from the previous specification. It is very adequately
covered in standard A-level textbooks and there is no intention to produce guidance notes.
Previous examination questions can be found in past PH1 papers.
PH1.3 – Energy Concepts
This topic has been moved from A2 to its more logical home of AS. The concepts of work,
power and energy form a straightforward development of the concepts in the GCSE
specification. This is especially true of work [extended to include the cos  term], kinetic
energy, gravitational potential energy, power and efficiency. Work as the area under the
force-distance graph is included and elastic potential energy in hookean systems introduced.
Previous examination questions can be found in past PH4 papers but caution should be
exercised as the level of demand of these questions is appropriate for A2. Past WJEC GCSE
Physics papers are a rich seam of source material with a similar warning about level of
demand.
GCE AS and A PHYSICS Teachers' Guide 10
PH1.4 – Conduction of Electricity
PH1.5 – Resistance
PH1.6 – D.C. Circuits
This topic area was previously in PH2. Its position here enables centres to commence early in
the AS year on electrical practical and investigatory work. Some minor adjustments to
content have been made:
 Electric charge has been introduced. This topic is no longer covered in the GCSE
Physics course. Its treatment is restricted to charging by rubbing, positive and
negative charges, attraction and repulsion of charges and current as the rate of flow of
charge.
 Temperature coefficient of resistance is no longer covered, though the qualitative
variation of resistance with temperature for metal wires and the experimental
investigation of variation are on the specification. Examination questions in which
candidates interact with data are still possible.
4.2
PH2 – Waves and Particles
General
PH2 takes the student into new and, we hope, exciting territory. The main ‘blocks’ in PH2 are
Waves and Photons. Having these re-united in the same module means that the nature of light
becomes an important theme of PH2.
The old PH2 section on energy levels and line spectra is now seen to have interesting
applications: it has been extended to include a treatment of lasers, while line absorption
spectra are shown to give us information about the atmosphere of stars. In fact, there is a
small section devoted to the physics of stars. It mainly concerns what we can learn about
them from the radiation they emit.
In an extension to the old PH2 section on nuclear structure, the basic constituents of matter
are acknowledged to be quarks and leptons. The four fundamental forces are also introduced
in this section. One teaching strategy would be to generate interest by starting PH2 here, as
the ideas are quite modern, new to most students, and raise tantalising issues.
PH2.1 – WAVES
Most of the content of the Waves and Light sections of the old PH1 is included here. To make
way for new material elsewhere, some gentle pruning has been done. Beats have disappeared.
So has the relationship between intensity and amplitude. Students are no longer required to
describe experiments showing diffraction or interference of water waves, sound and
microwaves, though they are required to be familiar with such experiments; an examination
question could probe understanding of a set-up given in the question. Students do still have to
be able to describe a version – any they please – of a two-slits experiment for light.
Waves are well covered in standard textbooks.
GCE AS and A PHYSICS Teachers' Guide 11
PH2.2 – REFRACTION OF LIGHT
What is required has been spelt out in the specification in rather more detail than previously.
The Snell’s Law equation is written in such a way as to emphasise symmetry.
The only application of total internal reflection now required to be learnt is the step-index
multimode (‘thick core’) optical fibre. Step-index means simply that the core is glass of one
refractive index and the cladding is glass of a lower index (students need to know why it has
to be lower), with an abrupt change in index at the interface.
While such fibres are fine for conveying light for illumination, students need to know that
they can’t be used for transmitting a rapid sequence of data over a long distance. Multimode
dispersion is the problem. Light travelling at an angle to the axis of the fibre (ray A in the
diagram) will travel further for a given axial length of fibre, than light (such as ray B)
travelling parallel to, or at a smaller angle to, the axis, and so will arrive later. Thus the
arrival time of an element of data encoded in the light is smeared out. The element could start
to arrive (by the shortest route) earlier than the previous element has finished arriving by its
longest route. Even worse confusion can occur.
There are two ways round the problem of multimode dispersion…
• Make fibres with graded index cores. This means cores which have a progressively lower index as
we go out from the axis towards the interface with the cladding. The lower the index the faster the
light travels so, if the grading is correctly calculated, the longer, more zigzaggy, paths cash in on
the ‘faster medium’ and take no longer than the short, axial, route. Clever stuff, but note that
graded index fibres are not in the WJEC specification.
• Make the core very thin. Its diameter must be no more than a few wavelengths of the light
(or infrared) being carried. Such fibres are monomode. Light travels parallel to the axis.
There are no zigzag modes. Students are required to know this. They are not required to
know why very thin fibres are monomode. This is just as well, because it cannot be shown
by ray optics, nor even by simple application of Huygens Principle. Electromagnetic wave
theory is needed.
The website http://www.techoptics.com/pages/Fiber%20Optics%20-%20Optical%20Fiber.html
gives an excellent summary of fibres for data transmission, with some facts and figures.
PH2.3 – PHOTONS
This is a heavily modified, and, we hope, more satisfying, version of the ‘Quantum Physics’
section of the previous specification. Particles behaving as waves, and the electron diffraction
demonstration have disappeared. The electron-volt will not be used in PH2. The origin of the
continuous and line X-ray spectra will not be tested.
GCE AS and A PHYSICS Teachers' Guide 12
The photo-electric effect is still arguably the most direct and easily understood evidence for
photons. Although the wording of our requirements [new PH2.3(a)-(d)] has been re-thought,
the same ground is covered, except that Millikan’s role need not be mentioned. This should
discourage students from spending valuable time in the examination drawing elaborate
diagrams of Millikan’s ‘laboratory in a vacuum’.
Students find the photoelectric effect difficult, especially the experimental determination of
KEmax. We have therefore written a detailed account – below – of the material to be tested, so
that the level of approach required is as clear as we can make it.
The section on atomic energy levels and line spectra now culminates in a treatment of lasers.
Detailed notes follow the photoelectric notes.
PH2.3 (a)-(d) Detailed notes on Photons and the Photo-electric Effect
Photons
Everything in nature seems to come in lumps or quanta (singular: quantum). For example,
ordinary matter is made of atoms, and electric charge comes in units of e. This lumpiness was
only becoming fully accepted a hundred years ago. But in 1905 Einstein made the bold
suggestion that light, too, was ‘lumpy’. Light quanta are now called photons.
A photon is a discrete packet of electromagnetic radiation energy. The energy of a photon is
given by
Ephoton = hf
In which f is the frequency of the light and h is a constant called Planck’s constant.
[h = 6.6 = 10-34 Js] [This is given in the WJEC list of constants.]
[For interest only… The constant h had first arisen in the earlier (1900) work of Max Planck
on the radiation inside a cavity with hot walls. Planck had shown that the energies of
oscillating particles in the wall seemed to be quantised.]
Einstein suggested some experiments in which the quantisation of light should reveal itself.
The simplest to understand involved the photoelectric effect, a phenomenon known about
since the late 1880s.
The Photo-electric Effect
When electromagnetic radiation of high enough frequency falls on a metal surface, electrons are
emitted from the surface.
For most metals, ultraviolet is needed. For some (including sodium, potassium, caesium), light
towards the violet end of the spectrum) will release electrons.
GCE AS and A PHYSICS Teachers' Guide 13
Demonstrating the Photo-electric Effect
EITHER Using a zinc plate with a gold leaf electroscope (or a coulombmeter)…
•
•
Clean a zinc plate with fine emery paper or steel wool.
Attach the plate to the top disc on a gold leaf electroscope, so there is good electrical
contact.
• Charge the zinc plate and inner assembly of the electroscope negatively1, e.g. by rubbing the
zinc plate with a polythene rod which has been rubbed with wool or fur. [Charging by
induction using a perspex rod is more reliable, but might be considered too confusing!]
The leaf should now be raised, because the leaf and the back plate are both charged negatively
and repel each other. The leaf should temporarily rise further if the charged polythene rod is
brought near the zinc plate.
•
Place an ultraviolet lamp near the zinc plate. Switch it on. The leaf should be seen to fall.
[Safety note: Don’t look at the ultraviolet lamp (when it’s turned on!)] Clearly the plate (and
inner assembly of electroscope) is losing charge.
•
Repeat the procedure, but charging the zinc plate and inner assembly of the electroscope
positively, e.g. by rubbing the plate with a charged perspex rod.
This time the ultraviolet does not affect the leaf. Charge is not lost.
The simplest explanation is the correct one… The ultraviolet causes electrons to be emitted from
the zinc plate. If the plate is charged positively, the electrons are attracted back again. If the plate
is charged negatively the emitted electrons are repelled and lost from the plate for ever.
1
Note that it is also possible to charge up the electroscope using an EHT power supply. It is
imperative that the output with the current-limiting resistance [usually several M] is
employed if doing this.
GCE AS and A PHYSICS Teachers' Guide 14
OR Using a vacuum photocell…
The apparatus can be supplied by Philip
Harris (Unilab division).
•
Note the polarity of the power supply.
Any electrons emitted from the caesium
surface will be collected by the
‘collecting electrode’.
•
If the photocell is covered the current is
zero; if light falls on the caesium
electrode there is current.
•
Photo-electric Puzzles
Before 1905, the energy of a beam of light was thought of as distributed uniformly across
broad wavefronts. Calculations showed that it should take some time before an electron in a
metal surface could absorb enough energy from the light to escape from the surface. Yet
emission is observed as soon as the light falls on the surface.
Another puzzle was why, for a given surface, we find that light of frequency below a certain
value (the threshold frequency) causes no electron emission at all.
Einstein’s theory of the photo-electric effect solves both these problems…
Einstein’s Photo-electric Equation
Although the free electrons in a metal have no allegiance to particular atoms, there are forces
‘bonding’ them to the lattice of ions as a whole. In order to escape from the metal an electron
has to do work against these forces. Some have to do more work than others, but there is a
certain minimum quantity of work to be done, so no electron can escape unless it is given a
certain minimum energy.
The work function, , of a metal is the minimum energy needed by an electron in order to
escape from the surface.
Einstein’s key idea was that any electron which leaves the surface is ejected by the action of a
single photon. Photons don’t co-operate in the process.
Recall that a photon of light of frequency f has energy hf.
Suppose that a photon gives its energy hf to an electron, and that the electron is able to
escape. The minimum energy used in escaping is , so the maximum kinetic energy the
escaped electron can have is what’s left over of the photon’s energy. So we have the simple
equation…
KEmax = hf – 
This assumes that the photon energy is greater than (or equal to) the work function; in other
words that hf   , so f   / h.
If f <  / h, the photon energy will be less than the work function so no electrons at all can
escape – a simple explanation of the phenomenon of threshold frequency.
GCE AS and A PHYSICS Teachers' Guide 15
The threshold frequency, f0, for a metal is the minimum frequency of electromagnetic
radiation needed to produce electron emission from the surface.
From the argument just given,
f0 =  / h.
This relationship can also be deduced from Einstein’s equation (without using ‘’ or ‘<’
signs!)… At the threshold frequency even the most energetic electron will only just manage
to escape, so KEmax = 0, and therefore hf 0    0  f0 =  / h.
Provided that the light is above the threshold frequency, as soon as it falls on the metal
surface electrons will start to be emitted, as emission results from individual photon ‘hits’,
and is not a cumulative process as supposed before Einstein.
Experimental test of Einstein’s Equation
We use the arrangement with the vacuum photocell given earlier for demonstrating the
photoelectric effect, but with the power supply polarity reversed.
•
Use white light with a coloured filter, or a light emitting diode, to illuminate the caesium
surface with approximately monochromatic light. [Its wavelength can be found using a
diffraction grating, hence its frequency, using f = c / .]
•
Increase the p.d. between the
collecting electrode and the caesium
surface until the current drops to zero.
At this point the p.d. is called the
stopping voltage, Vstop, because it stops
all emitted electrons, even those with
the most K.E., from reaching the
collector electrode.
•
The maximum K.E. of the emitted
electrons is simply given by
KEmax = e Vstop
[How do we justify this? Because of the applied voltage, emitted electrons are subject to
repulsion by the positive collector electrode and attraction by the emitting surface, hence to a
resultant force towards the emitting surface. The electrons therefore get slower and slower as
they cross the gap. When the stopping voltage is applied, even the most energetic of emitted
electrons have no K.E. left when they have made it across the gap. The K.E. lost is equal to
the P.E. gained for these electrons. That’s what the equation states.
It is just like finding the K.E. of a ball thrown upwards in the Earth’s gravitational field by
measuring the its maximum height, and using the energy conservation equation K.E. lost =
mgh]
•
Repeat the process using two or three more frequencies of light.
•
Plot a graph of KEmax against frequency, f. If Einstein’s equation is correct it should have
a positive slope equal to h and a negative intercept, equal to 
GCE AS and A PHYSICS Teachers' Guide 16
We can see this by comparing Einstein’s equation with y = mx + c.
A sample graph is presented below.
Graph of KEmax against frequency of light for a caesium surface
It is useful practice to find from the graph…
• a value of Planck’s constant,
• the threshold frequency for caesium,
• the threshold wavelength for caesium
• the work function for caesium
[In practice it is difficult to obtain a good value for h or  using a commercially available
vacuum photocell. Slight impurities on the caesium surface (e.g. a thin oxide film), and
unwanted electron emission from the collector electrode, both affect the stopping voltage.]
The first convincing verification of Einstein’s photo-electric equation, leading to an accurate
value of the Planck constant was completed in 1916 by R.A. Millikan, working in the United
States. The secret of his success was a remotely operated knife working in the vacuum to
skim off surface layers from the caesium surface as they became contaminated.]
[For interest only: http://focus.aps.org/story/v3/st23]
GCE AS and A PHYSICS Teachers' Guide 17
Effect of changing the Light Intensity
If we bring a monochromatic light source towards a surface we increase the light energy
falling on the surface, per m2, per s. We are said to be increasing the intensity of the light.
Clearly we can apply the same idea to ultraviolet or any other electromagnetic radiation. We
find that…
(1) For light or ultraviolet of a given frequency, changing the intensity has no effect on the
maximum K.E. of the emitted electrons.
This is exactly what Einstein’s theory predicts. The energy given to individual electrons
comes from individual photons, and a photon’s energy, hf, depends only on the frequency
(or, equivalently, the wavelength) of the radiation. It doesn’t, then, depend on its
intensity (provided we don’t change the frequency).
(2) For light or ultraviolet of a given frequency, increasing the intensity increases the
number of electrons emitted per second.
Again, this is just what we’d expect from Einstein’s theory. Increasing the intensity
means increasing the number of photons arriving at the surface, per m2, per s. Naturally
this means that more electrons will be emitted. [Each identical photon has the same
probability of emitting an electron.]
We can show the effect with the same
vacuum photocell arrangement used
for demonstrating the photoelectric
effect. Note that the polarity of the
power supply is arranged to
encourage electrons to cross the gap.
•
Use a monochromatic light source to
illuminate the caesium surface.
•
Check that increasing the p.d. does not
affect the current, I. This means that
all the electrons emitted per second are
being collected.
•
Bring the light source closer and observe the effect on I.
•
I is the charge flowing per second, so the number of electrons emitted per second is I/e in
which e is the charge on each electron.
GCE AS and A PHYSICS Teachers' Guide 18
PH2.3 ( l)-(t) Detailed notes on Laser Physics
Raison d’être
Lasers, by now, are in nearly every household as they were once in every self-respecting
science fiction film. Although, the death ray mystique will appeal to most A-level students
there is real value in studying the physics which forms the foundation of laser construction.
Unlike other subjects such as relativity or nuclear physics which can only be touched upon
within an A-level course, the essential physics underpinning lasers can be taught reasonably
well in a few lessons. Also, these fundamentals of lasers can be taught at the right level while
not oversimplifying the subject.
Summary
(l)-(q) in the syllabus are related to the theory and construction of a simple laser. The new
concepts here are:
1. Stimulated emission of radiation.
2. Lifetime of energy levels.
3. Population inversion.
4. 3 and 4-level laser systems.
5. Basic construction of a laser.
(r)-(t) are related to semiconductor lasers and their uses.
Note: some of the content has been put in for general interest but will not be examined. Those
details in square brackets or small print will not need to be learned and will not be examined.
LASER is an acronym and stands for Light Amplification by Stimulated Emission of
Radiation. Which leads us nicely onto what is stimulated emission?
The Three Important Atomic Processes
These three processes [two of which are already known from 2.3(j)] are:
1. Absorption of light
2. Spontaneous emission of light
3. Stimulated emission of light
Absorption of light by an atom is shown in the diagram below – a photon of the correct
energy is absorbed by the atom and an electron gains enough energy to move from the
ground state to the excited state (Note: for the moment we are only considering the ground
state and the first excited state only).
Excited state
Electron excited to
higher energy
Photon absorbed
Ground state
GCE AS and A PHYSICS Teachers' Guide 19
Spontaneous emission is the reverse process – an electron drops spontaneously (and
randomly) from the excited state to the ground state and emits a photon of the same energy.
These photons have random phase and random direction.
Excited state
Electron drops to
Lower energy
Photon emitted
Ground state
However, there is also a third process [which was originally proposed by Einstein in 1917].
This process is known as stimulated emission – an electron is ‘stimulated’ to drop from its
excited state by an incoming photon.
Excited state
Electron stimulated
to drop by photon
Incoming photon and emitted photon exit
in phase and in the same direction
Ground state
The reason that the electron is stimulated to drop is that the incoming photon is an
electromagnetic wave and its e-m field will exert an oscillating force on the excited electron.
If the incoming photon is of the correct frequency, this oscillating force will cause the excited
electron to drop and both photons will exit with the same frequency, phase and direction.
Note: again, the incoming photon needs to be of the correct energy.
Inverting The Population
In order to get as much light out of a system as is possible we need to get as many atoms
excited as is possible. Obviously, the more electrons we have in an excited state the more will
drop and emit photons (either spontaneously or through stimulation). However, there is one
serious problem that arises when we produce a lot of light – the very photons that we produce
are the actual photons that can be absorbed (they have the correct energy to produce both
effects). If we have photons being absorbed all the time then our laser beam isn’t getting any
stronger.
Forget, for the moment about spontaneous emission (we are allowed to but we’ll explain why
later). When a photon arrives at an atom one of three things can happen:
1. It can pass by and do nothing.
2. It can be absorbed (if the atom is in the ground state).
3. It can cause stimulated emission (if the atom is in the excited state).
GCE AS and A PHYSICS Teachers' Guide 20
When it comes to producing a laser beam with a high intensity the three options above will
have the following effect on the beam.
1. No change in the beam.
2. Net loss of one photon from the beam.
3. Net gain of one photon in the beam.
We need to arrive at a situation where stimulated emission is more likely than absorption so
that the laser beam increases in intensity. Since stimulated emission occurs if the electrons
are in the upper level and absorption when electrons are in the lower level we need to get
more electrons into the upper, excited level. This is called population inversion (or N2 > N1
as stated in the syllabus, where N2 and N1 are the number of electrons in the excited state and
the ground state respectively).
Unfortunately, this goes against what happens in nature – lower energy levels are always
more heavily populated than higher energy levels when we have thermal equilibrium (as we
go up 1eV to higher energy levels the probability of occupation of the level drops by a factor
of 1017). There’s only one thing for it – get rid of this thermal equilibrium. How do we do
this? We continue to pump energy into exciting electrons to higher energy levels to maintain
a population inversion and to break the conditions of thermal equilibrium.
Population inversion is not usually possible if we only have two energy levels (if pumping is
carried out by light). As we start to pump our system we have the following situation:
Excited state (N2 = 0)
Pumping light
Ground state (N1 = 8)
Many electrons will be promoted to the higher energy and all seems fine. Unfortunately, if
we succeed in exciting half the electrons we are now in the following situation:
Excited state (N2 = 4)
Pumping light
Ground state (N1 = 4)
In this situation the incoming flood of photons is just as likely to cause an electron to drop
(stimulated emission) as it is to cause an electron to rise (absorption). The best we can
achieve here is N2 = N1 which is not quite good enough.
GCE AS and A PHYSICS Teachers' Guide 21
The 3 Energy Level Laser System
E3
2
1
2
3
Pumping. Electrons are promoted from the
ground state (E1) to E3 usually by using an
E2
external light source or by electron collisions.
Electrons drop quickly (because E3 is chosen to 1
have a short lifetime of the order of anoseconds)
3
to the metastable (E2). Calling E2 metastable
means that it has a long lifetime and electrons
stay there for a long time (not that long really
E1
around a millisecond but that’s a very long time
for an electron).
This is the transition that produces the laser photons so we must have N2 > N1. Note that,
although stimulated emission still reduces our population inversion, the pumping is at a
different wavelength. We have to make sure that the pumping [1] exceeds the stimulated
emission [3] to maintain a population inversion.
Other things to note:
 E3 (to E2) has to have a short lifetime because E3 cannot start to fill up – pumping
won’t then be possible. Also, we don’t want the electrons to stay in E3 and have them
stimulated to drop back to E1 by the pumping light – that’s back to the 2-level system
again which wasn’t quite good enough.
 More than half the electrons from E1 must be pumped to E2 (via E3) in order to obtain
a population inversion – that’s a lot of electrons!
The 4 Energy Level Laser System
E4
1
2
3
4
Pumping again.
Quick drop to the metastable state E3.
This is the laser light producing transition so this
time N3 > N2. However, because E1 is the ground
state, E2 is practically empty initially so
obtaining population inversion is far, far easier
(definitely no need to pump half the electrons!).
Another quick transition so E2 has a short
lifetime. This is because we want E2 to be empty
so that we have a population inversion (if N2 is
small it’s easier for N3 to be larger than N2).
2
E3
3
1
E2
4
E1
GCE AS and A PHYSICS Teachers' Guide 22
Laser Construction
Laser beam
Amplifying
medium
100% reflecting
mirror
Mirror: approximately 99% reflecting
(approx. 1% gets transmitted)
In order to ensure that the laser produces light of a high enough intensity, the above set up is
used. The amplifying medium is the region where the population inversion exists. This means
that the conditions are right in the amplifying medium for stimulated emission. Under these
conditions one photon has the potential to produce two photons and these can produce 4
photons, then 8 photons etc. Like a chain reaction, this process will lead to an exponential
increase in output energy. Laser physicists aren’t happy with this, they go even further – they
use mirrors to ensure that this exponential increase happens many times. Because only 1% of
the light exits each time it reflects back and forth between the mirrors, on average, the beam
will pass through the amplifying medium a hundred times before it exits. Now, considering
that each time the beam passes through the amplifying medium it is increasing exponentially,
this factor of 100 makes an enormous difference. [Try calculating e0.1 and then e10 on your
calculator and see the difference!]
This all leads to very high light intensities inside the amplifying medium and this is why (as
was said earlier) we can forget about spontaneous emission. Imagine that you’re an excited
electron sitting happily in your higher energy level. Normally, you’ll just drop down
spontaneously when your time is up. But, inside a laser, there’s so much light that you never
drop spontaneously because before your time’s up you’ve been disturbed by another photon,
stimulated to join in with all the other light and join in coherently as well!
Efficiency
Usually, lasers are very inefficient beasts. Because of the large energies required to maintain
a population inversion, their efficiencies are generally far below 1%. Some reasons for this:
 The pumping energy (see [1] in the 3 and 4 level systems) is considerably larger than
the output photon energy.
 High intensity pumping combined with the high intensity of the laser beam means that
the amplifying medium will get very hot. So, there will be large heat losses. To make
this matter worse, we need to cool the amplifying medium usually so that it, or its
container, doesn’t melt. By cooling the system we just transfer more heat and increase
our losses but better this than destroy a £50 000 laser!
GCE AS and A PHYSICS Teachers' Guide 23
Semiconductor Lasers
The basic structure of a standard ‘edge emitting’ semiconductor laser is shown below. The
whole block shown below is a semiconductor chip with dimensions approximately 0.5mm x
0.5 mm  1 mm.
metal
contact
‘sandwich’ of area with
population inversion i.e.
amplifying medium
light emission
not silicon, usually gallium arsenide.
This surface and its opposite are mirrors due
to the air-solid boundary
metal contact below also
[Although the chip itself has dimensions
approximately 0.5mm x 0.5mm x 1mm,
the beam has an approximate cross
sectional area of 30µm x 5µm and passes
along the ~1 mm length of the chip.]
The above laser fits the basic shape of a normal laser (shown below).
mirror
Amplifying
medium
Mirror
The mirrors, however, are far from the 100% and 99% reflecting ideals discussed earlier. The
mirrors are simply due to the semiconductor-air boundary at the edges of the chip. [This in
fact gives 40% reflection only (at both sides).] This would be disastrous for highly inefficient
gas lasers but not for our semiconductor laser. The reason why:
 The population inversion inside the semiconductor sandwich area is millions of times
higher than in gas lasers [~1025 electrons/m3].
 The exponential increase in light intensity (i.e. 1 photon becoming two, becoming
four etc.) occurs far more quickly because of the higher population inversion.
 So the fact that we lose 60% of the light at each reflection is compensated for by
having huge gains between the mirrors.
GCE AS and A PHYSICS Teachers' Guide 24
How can a population inversion be set up just by applying ~3 V d.c. voltage?
[You don’t need to know this but you might find it interesting]
Some of this next part used to be in the A-level syllabus many years ago. It shows a couple of band
diagrams of a p-n junction in a very highly doped semiconductor.
Electrons
p-type
semiconductor
n-type
semiconductor
Holes
Note: This is a diagram for a horizontal p-n junction but the junction in the laser diode diagram is
vertical.
After a small (~3 V) p.d. has been applied, we get:
Electrons
ve
+ve
Holes
Can you see the area of population inversion? If you look carefully, there’s a small region in the
middle between the p-type and the n-type where we have a high concentration of electrons above
holes. These are the electrons that can be stimulated to drop and provide laser light.
Advantages and Uses of Laser Diodes
These are straightforward and can be summarised as follows:
Advantages:
Some Uses:




Cheaper
Smaller
More efficient
Easy to mass produce





Inside DVD and CD players
Barcode readers
Telecommunications (via optical
fibres)
Image scanning
Laser surgery
The usefulness of laser diodes is ‘reflected’ in the number of them produced annually –
around 1 billion (109) laser diodes are produced worldwide per year!
GCE AS and A PHYSICS Teachers' Guide 25
For more than enough further reading see:
http://members.aol.com/WSRNet/tut/ut1.htm
Physics PH6 written paper 2007, comprehension question.
Wikipedia http://en.wikipedia.org then type ‘laser’ or ‘semiconductor laser’
Google http://www.google.co.uk/ then search ‘laser theory’
PH2.4 – MATTER, FORCES AND THE UNIVERSE
We start in familiar territory. PH2.4(a) deals with the simple model of the atom as nucleus
with orbiting electrons, in which the nucleus is composed of nucleons (protons and neutrons).
A
Z
X notation is required.
The rest of PH2.4 is devoted to the modern quark and lepton picture, with a short review of
the fundamental forces.
PH2.4 ( b)-(g) Detailed notes on Quarks, Leptons and Fundamental Forces
Nucleons have structure
This section, until a third of the way down the next page, is non-testable background
material which could provide a lead into PH2.4 (b).
1.
Although electrons are point-like at scales of 10-18 m, nucleons have diameters of the
order of 10-15 m.
1
Nuclear radius varies approximately as A3 – electron diffraction experiments – so
nuclear volume is proportional to A, showing that the volume of each nucleon is, at least
approximately, the same.
Exercise: Radii of nuclei in fm: 6 C 3.16,
28
Si 3.93, 56 Fe 4.85,
120
Sn 5.99,
208
Pb 7.16
1
3
Show that r  A , and determine the density of nuclear material. [An A2 exercise would be to use
a log-log plot - and then go on to determine g at the surface of a 10 km diameter neutron star!]
2. Free neutrons decay into protons with a half-life of about 10 minutes.
The equation is 01 n  11 p  01 e  ν e and is clearly energetically favourable, as the rest mass
of the neutron is greater than the combined rest masses of the proton and the electron.
This is suggestive of a structure for at least the neutron and therefore the proton. [The
very similar masses of the neutron and proton suggests that the particles are closely
related – hence the use of the generic term nucleon.]
3.




The Stanford/MIT Experiment of 1968
20 GeV Stanford Linear Accelerator (SLAC) in 1968
Electrons fired down 2 km long accelerator tube
Target was a 7 cm container of liquid hydrogen.
Looked at interactions between the incident electrons and the protons in the hydrogen,
in which there was a large momentum transfer, shown by a large scattering angle for the
electrons (looked at 10 and 6 which are quite large in these experiments.)
GCE AS and A PHYSICS Teachers' Guide 26

Results
A large number of interactions was observed at these angles (just like the GeigerMarsden-Rutherford -particle scattering experiment). More technically, the
distribution of scattered electrons was similar to that in the -particle scattering,
suggesting a point charge distribution. In other words – just as in the Geiger-Marsden
experiment – the apparently solid object was in fact a collection of tiny ones.
These large momentum-transfer interactions resulted in the production of new particles
– typically the 0 meson.
The model of nucleon structure which best fits these and other results is the quark model,
proposed by Gell-Mann and Zweig in 1964, in which nucleons are composed of three quarks
bound together by the strong (nuclear) force. The work of Richard Feynman was instrumental
in this interpretation.
Quarks, antiquarks and quark combinations
There are two types of quark in everyday matter – the so-called up quark (u) and down quark
(d). The properties of these quarks are:
quark
charge / e
u
 23
d
 13
Note that there is a potential confusion between the down quark (d) and the deuteron (d) i.e
2
1 H – their symbols are the same. The distinction is usually clear from the context, but care
will be taken over the framing of examination questions.
Every particle has an antiparticle, which is identical to its corresponding particle apart from
having opposite charge. Particles and corresponding antiparticles annihilate. In general the
antiparticle has the symbol of the particle, but with a bar over it. Thus for quarks we have:
antiquark
charge / e
u
 23
d
 13
Quarks are bound together into composite particles called hadrons. These are further
classified…
Combinations of 3 quarks are called baryons [meaning ‘heavy ones’]. Neutrons and protons
are baryons. A proton is composed of uud, and a neutron udd. The order of the letters doesn’t
matter – it carries no information. [Check charges!]
proton
u
u
d
u d
d
neutron
What would be the composition of (i) an antiproton, p (ii) an antineutron, n ?
GCE AS and A PHYSICS Teachers' Guide 27
There does exist a family of four very short-lived baryons also composed of up
and down quarks: the  particles (delta particles). They are designated ++, +, 0, and .
The + and 0 have familiar sets of quarks – these two particles are essentially higher energy
(excited) states of the proton and neutron respectively. What are the quark constituents of the
Δ++ and Δ ?
A combination of a quark and an antiquark is called a meson. A family of mesons that
consists of only 1st generation quarks is the pions: +, 0 and . These contain only up or
down quarks and their antiparticles. So the charged pions must, have these quark
structures…

u
d

u
d
The composition of 0 is ambiguous. It could be either uu or dd .
It is actually – not for examination – a mixture of the two! Why don’t the uu and the dd
self-annihilate? They do! The lifetime of the 0 is much smaller than that of the charged
pions.
No such thing as a free quark
Our evidence for quarks is indirect, in the sense that it appears impossible to obtain a free
quark, i.e. a quark by itself. Within a hadron, indeed within the nucleus, the quarks can move
around, but they are bound together by the so-called strong force which does not decrease if
the separation between the quarks increases. [Contrast the electromagnetic force between
charged particles.] Hence the impossibility of obtaining free quarks.
Instead of freeing the quarks, when the bond between quarks is sufficiently stretched, a
quark-antiquark pair is created using the potential energy in the stretched bond, and a new
particle, a meson is formed. This is what happens in a so-called inelastic collision when a
high [kinetic] energy particle is fired at a nucleon. For example…
[π0]
Key
up quark
down quark
GCE AS and A PHYSICS Teachers' Guide 28
Students will not be expected to recall a host of different hadrons, but could be asked to work
out the quark constituents of specified hadrons using the table below, which will be available
in the examination. For example, a student could be asked to determine the composition at
quark/antiquark level of the +, given that is a meson, or of the Δ , given that it is a baryon.
[Students are expected to know that individually observable particles (i.e. excluding quarks)
have charges in units of e.]
Leptons
The up and down quark are the so-called first generation of quarks. The first generation of
leptons consists of the electron, e- and the electron neutrino, e. ‘Lepton’ means ‘light one’
[The literal meaning of ‘meson’ is now easily guessed!] The table contains the first
generation of quarks and leptons.
particle
(symbol)
Leptons
electron neutrino
electron (e)
(e)
Quarks
up
down
(u)
(d)
charge / e
1
0
 23
 13
Lepton
number
+1
+1
0
0
There is also a second and third generation, each with two quarks, one charged lepton and
one uncharged lepton. No questions will be set involving generations higher than the
first.
The electron’s antiparticle is the positron, e+. [Note that, for historical reasons, we don’t use
the bar notation in this case.]
The electron was the first fundamental particle to be discovered [in 1896-7 by J J Thomson].
A-Level students should need no introduction to the electron. It is probably a different matter
for the electron neutrino – or neutrino for short...
Historically, the evidence for the existence of neutrinos was indirect and arose from the fact
that, unlike -particles, -particles have a range of energies – suggesting that another particle
was present which shared the energy in a random way. This evidence is not available in AS
physics as it relies on the application of both mass/energy and momentum which are not
available until the A2 course.
Nonetheless this cloud chamber photograph of a 6He disintegration (beta decay) – see photo
below – is highly suggestive.
The photograph [from a cloud chamber],
shows paths of charged particles
produced from the decay of a 6He
nucleus at X. The short fat track is that of
the daughter 6Li nucleus. The thin curved
track is the electron (- particle). Clearly
something must have been ejected
upwards and to the left……
Further evidence comes from a
consideration of the - energies [more in
A2!]
Source:
Sang 1995
GCE AS and A PHYSICS Teachers' Guide 29
Lepton number
The electron and the (electron) neutrino are each assigned a lepton number of 1. Their
antiparticles, the positron and the antineutrino, ν e , are each assigned a lepton number of –1.
The point of doing this is that we find lepton number to be conserved. Thus in any interaction
the sum of the lepton numbers before is equal to the sum of lepton numbers afterwards.
For example, the beta decay of the 6He nucleus is
6
2
He  63 Li  e  νe
The lepton numbers add to zero on the right. The lepton number is clearly zero on the left.
[Note also the conservation of charge, another infallible conservation law.]
In this interaction, a neutron in the nucleus is replaced by a proton. So we could write the beta
decay as
n p + e + ν e
Or, at an even more fundamental level, recalling that a proton is uud and a neutron is udd, we
could write the beta decay as
d u + e + ν e
Again, check charge and lepton number conservation.
Although the n, p version of the beta decay equation is fine for demonstrating the
conservation laws, it should not be taken as meaning that beta decay involves one neutron
decaying in isolation. We know that the neutron’s nuclear environment is involved, because
different unstable nuclei have different half lives for beta decay. A similar remark applies
even more strongly to the quark version: an isolated neutron has a half life of about 10
minutes, whereas a proton is to all intents and purposes completely stable – no tendency for
its d to undergo the change shown. So the intra-nucleon environment matters. These
subtleties will not be tested.
Exercise on applying Charge and Lepton number conservation
Over the past 30 years, scientists have detected neutrinos from the Sun by their interactions
with dry-cleaning fluid: specifically their interactions with a neutron in 37Cl nuclei.
37
17
Cl  νe 
38
17
Ar  X [X is unidentified for the moment.]
This equation could be written at the nucleon level:
n + e → p + X.
It could also be written at the quark level:
d + e → u + X.
Candidates could be asked to identify particle X. Solution: In order to balance with the other
particles, it has to have a charge of 1 and a lepton number of +1, which means that it is an
electron. Note that we could also say that it cannot be a baryon, because baryon number is
also conserved, but that is not a requirement of the specification so it will not be asked.
GCE AS and A PHYSICS Teachers' Guide 30
The four forces or interactions experienced by particles
Interaction
Experienced
by
Range
Gravitational
all particles
infinite
Weak
all particles
very short
range
Electromagnetic
all charged
particles
infinite
Strong
quarks
short range
Comments
Extremely weak – negligible except in the context of
large objects such as planets and stars
Not considered here
A very weak interaction – only significant in cases
where the electromagnetic and strong interactions do
not operate. Interaction governed by this are of low
probability [in the case of collision interactions] or of
long life-time [in the case of decays]. Governs any
interactions including both hadrons and leptons, e.g. 
decay and the p-p part of the p-p chain (see PH2.5).
Much stronger [and therefore more likely / shorter
lifetime] than the Weak. Governs interactions
composed entirely of charged particles and photons
[N.B. Also experienced by neutral hadrons because
they are composed of quarks].
Responsible for electric attraction and repulsion.
The strongest interaction – only experienced by quarks
and particles composed of quarks (i.e. hadrons).
Responsible for the production of new particles from
nucleon-nucleon interactions.
Examples of interactions – which interaction is responsible for each?
 Neutron decay: n  p  e  e
Governed by the weak interaction because, (i) includes neutrino, which is not affected
by the strong or electromagnetic interactions, and (ii) it has a long lifetime [~ 10
minutes, which is long in particle terms, so the probability of the interaction is low.]
 Proton – proton collision: p  p  p  p   :
Governed by the strong interaction because all the particles present are hadrons
[consist of quarks]. Given enough energy to create a 0 this interaction is almost
certain to occur, although p  p  p  n   is also a good possibility.
 Electron-electron repulsion: e  e  e  e
Governed by the electromagnetic interaction because (i) the strong doesn’t affect
leptons and (ii) although it would be possible to occur via the weak interaction, this is
so much less likely.
 Proton-proton fusion: p  p  d  e  e
Governed by the weak interaction. Given the presence of the neutrino it is the only
candidate. Also the fact that the average lifetime of a proton in the centre of the Sun is
~ 109 years, means that the process is very unlikely in any individual collision.
 0 decay:      [i.e. decays into two photons]
Governed by the electromagnetic interaction because, (i) the pion is composed of a
charged quark and a charged antiquark, which both feel the electromagnetic
interaction, and (ii) neither the strong nor the weak interactions affect both quarks and
electromagnetic photons. The lifetime of a 0, at 10-12 s, is very short compared to the
charged pions which decay by a slower [less likely] weak process, e.g.
  e  ν e [lifetime ~10-8 s], because there is no electromagnetic process available.
GCE AS and A PHYSICS Teachers' Guide 31
Useful websites
These are written at an appropriate level for AS/A students:
Standard Model: http://en.wikipedia.org/wiki/Standard_Model
from the bbc. : http://www.bbc.co.uk/dna/h2g2/A666173
Particle adventure: http://particleadventure.org/frameless/standard_model.html
The following site is worth it for one memorable sentence in brackets:
http://engr-sci.org/pnu/1990/physnews.004.htm
These sites contain a plethora of information – for self-motivated students only:
http://hypernews.slac.stanford.edu/slacsite/aux/HiPPP/scattering/
http://www.physics.ox.ac.uk/documents/PUS/dis/index.htm
http://physics.nmt.edu/~raymond/classes/ph13xbook/node194.html
GCE AS and A PHYSICS Teachers' Guide 32
PH2.5 – USING RADIATION TO INVESTIGATE STARS
This section of the specification is traditional 19th century physics applied to a current
research area. The emphasis is on the use of spectra to investigate stars – the idea that some,
in principle, very simple observations are enough to pin down important characteristics of
stars including their surface temperature, power output and composition. The line spectra
aspects of this unit are designed to fit in with section PH2.3 and so could be taught as a
continuation of that section. The proton-proton chain section introduces the idea of remote
sensing via subatomic particles and ties in well with section PH2.4. The black-body section is
free standing. Detailed notes follow.
Detailed notes on PH2.5: Using radiation to investigate stars
A star’s spectrum consists of a continuous spectrum, from the dense gas of the surface of a
star, and a line absorption spectrum from the passage of the emitted electromagnetic radiation
through the tenuous atmosphere of the star.
We start with some background material to assist in the interpretation of the continuous
spectrum.
PH2.5 (b), (c) The Black Body Spectrum
Good absorbers of radiation (of a given wavelength) are also good emitters (of that
wavelength). A black body is the name physicists have given to an ideal surface which
absorbs all the radiation that falls on it. Matt black surfaces approach this ideal. No surface at
a given temperature can, purely because it is hot, emit more radiation, at any wavelength,
than a black body.
The power emitted, per unit interval of wavelength, per m2 from a black body depends on the
wavelength as shown by the curves for various temperatures of the body.
GCE AS and A PHYSICS Teachers' Guide 33
Two of the most important laws that we find to apply to black body radiation are…
Stefan’s Law
The total power P of electromagnetic radiation emitted from area A of a black body at
(kelvin) temperature T is given by
P =  A T4
In which  is a constant called the Stefan constant. Its value is 5.67  10-8 W m-2 K-4.
P is the total of the power emitted at all wavelengths. It is proportional to the area under the
curve for the particular temperature.
Wien’s Displacement Law
The wavelength, p, at which the maximum power is emitted by a black body is inversely
proportional to the temperature of the black body.
That is
p 
W
T
In which W is a constant called Wien’s constant. We find W = 2.90  10-3 K m.
Three straightforward exercises
1. Use data from the curves to check Wien’s displacement law.
[The easiest way is to multiply p for each curve by the temperature of the black body.
Similar results, close to the accepted value of Wien’s constant, emerges each time.]
2.
Draw vertical lines on the graph grid to show the extremes of the visible region of the
electromagnetic spectrum. Hence explain why the black body at 3000 K appears
yellowish compared with the body at 5000 K (even though the black bodies at all three
temperatures shown are loosely described as ‘white hot’).
What would be the appearance of a black body at 1000 K?
What would be the appearance of a black body of a body at 8000 K?
3.
A white hot lamp filament approximates to a black body. On this basis, estimate its
temperature if the wavelength for peak emission is 1.16  10-6 m.
[Re-arranging Wien’s displacement law…
T = W / p = 2.90  10-3 K m  1.16  10-6 m = 2500 K.]
Estimate the surface area of the filament if the bulb emits radiation at a power of 70 W.
[To estimate the surface area of the filament we will assume that the filament is a black
body at 2500 K. So, re-arranging Stefan’s Law…
A = P /  T4 = 70 / (5.67  10-8  25004) = 3.2  10-5 m2.]
[Note that most of this radiation will be in the infrared (see curves) and is therefore, from
the point of view of illumination, wasted. What is more, the electrical power input to the
lamp would have to be more than 70W, as some energy is lost as heat conducted along
the wires from the filament and as heat convected from the filament by the inert gas
surrounding it. Hence the phasing out of filament lamps as energy wasters.]
GCE AS and A PHYSICS Teachers' Guide 34
PH2.5 (d) A Star’s Continuous Spectrum
Using a telescope, a special diffraction grating, and a detector sensitive across a wide range
of ultraviolet, visible and infrared wavelengths we can study the way in which a star’s
radiation is distributed among different wavelengths. [Allowances have to be made for
selective absorption by the Earth’s atmosphere.] None of the experimental method is
required, but students need to know the result: a star has a continuous spectrum almost
identical to that of a black body. We can usually assume that a star radiates as a black body.
On this assumption we can find out some key information about a star light years away by
studying its continuous spectrum. We need to extract two figures: (i) the total power (across
all wavelengths) reaching the vicinity of the earth, (ii) p, the wavelength of peak power.
How we could proceed is best shown by an example. This also illustrates the level at which
we feel it would be reasonable to ask examination questions…
Example
Consider the following observational data, which relates to Arcturus, a bright reddish star in
the constellation of Boötes:
Wavelength of peak emission in continuous spectrum = 674 nm
Power received per m2 at the Earth = 309  10-8 W m-2
Distance from Earth = 367 l-y [= 347  1017 m]
a.
b.
c.
d.
Sketch the continuous spectrum – a nice easy intro.
Calculate the Kelvin temperature.
Calculate the total power emitted [This called the star’s luminosity.]
Calculate the radius of Arcturus.
(b) To find the Kelvin temperature: use Wien’s law:
2  898 103 m K
T

 4300 K[3s.f.]
max
674 109 m
(c) To find the total power emitted, consider that, at the distance of the Earth, the power is
spread out over the surface of a sphere of radius 347  1017 m, so that the total power is
given by:
P  4 r 2  3  09 108 W  4  68 1028 W .
W
[Note that the total Solar power output is 390  1026 W, so that Arcturus is ~ 120  as
powerful as the Sun on these data.]
(d) To find the radius use Stefan’s Law: P  4 r 2 T 4 , where r is the stellar radius this time.
P
4  68 1028

 1 38 1010 m
4
8
4
4 T
4  5  67 10  4300
[ i.e. 138 million km, approx 20  that of the Sun.]
r
Additional remarks on PH2.5 (a) – (d)
“How science works” and the 19th century ‘ultra-violet catastrophe’ suggest themselves as discussion
topics. Students will not be tested on the ‘ultra-violet catastrophe’!
The cosmic microwave background radiation is an almost perfect black body spectrum. We should
really have mentioned it in the specification. We didn’t – but that doesn’t mean it couldn’t be included
in questions, given sufficient - background - information. It fits well in PH5 in the comprehension
passage…..
GCE AS and A PHYSICS Teachers' Guide 35
PH2.5 (e) (f) A Star’s Line Absorption Spectrum
Dark lines cross a star’s continuous spectrum. These co-incide in wavelength with the known
emission lines of particular elements. [Allowances may need to be made for Doppler shifts,
but this is not to be tested in PH2 examinations.] The implication is that these elements are
present in the tenuous outer ‘atmosphere’ of the star, and absorb specific wavelengths from
the continuous spectrum of light passing ‘outwards’ through this atmosphere.
This ties in with PH2.3 (h), (j). It would now be fair to test the work on absorption spectra in
this stellar context. For example, students could be given wavelengths of a dark lines, and
simplified atomic energy level diagrams, and asked to make reasoned deductions about
element(s) present in the star’s atmosphere.
http://casswww.ucsd.edu/public/tutorial/Stars.html (up to, but not including, ‘spectral
classification’) presents some useful pictures and diagrams.
http://jersey.uoregon.edu/vlab/elements/Elements.html lets you click on lines to find their
wavelengths.
Almost two hundred years ago Joseph Fraunhofer made a study of dark lines crossing the
Sun’s spectrum, and in particular noted dark lines at exactly the same places in the spectrum
as the yellow [sodium] lines in a candle flame. Although he couldn’t explain what was going
on, he had made an exceedingly important discovery: up to that time no-one had the slightest
idea that the Sun’s composition shared anything in common with that of the Earth. For
interest only, go to
http://astronomy.neatherd.org/Fingerprints%20of%20light.htm
Analysis of stellar spectra reveals that 75% of the universe by mass is Hydrogen, and 24%,
helium, with very small quantities of the other elements.
The question naturally arises: where have the other elements come from? This is not required
to be learnt, but for those interested, a very readable account is given in
http://fire.biol.wwu.edu/trent/alles/Origin_of_Elements.pdf .
PH2.5(g) Source of the Sun’s Energy
In a star like the Sun, energy is transferred from the Sun’s core to its surface layers by a
combination of radiation and convection. But what happens in the core? Energy is ‘produced’
through fusion reactions. The released energy is in the form of gamma ray photons, neutrinos
and kinetic energy of the product nuclei, quickly randomised by collisions to random internal
energy,
We require recall of the main branch (p-pI) of the proton-proton chain, which is the main
energy production mechanism in stars like the Sun. There are 3 steps…
p  p  d  e+  νe
(where d  deuteron 21 H)
p  d  23 He  γ
(where γ  photon)
3
2
He  23 He  24 He  p  p
It helps to remember that at each step the nucleus acquires one more nucleon.
Neutrinos from the first step occurring in the Sun’s core are detected on Earth. Their release
indicates that the weak force is involved. This means that the probability of the step occurring
is very low for a given proton. That’s why the Sun doesn’t emit radiation at a greater rate
than its miserly 390  1026 W, and is expected to survive for thousands of millions of years.
GCE AS and A PHYSICS Teachers' Guide 36
Information on various sub-branches of the p-p chain [ppII and ppIII] might be used as
examples in questions, as might the CNO cycle (important in stars heavier than the Sun).
The ppII chain, which accounts for ~14% of the helium produced takes over at the last step of
the ppI chain. The ppII chain is:
3
4
7
2 He  2 He  4 Be  
7
4
Be  e  73 Li   e
Li  11 He  42 He  42 He
Candidates would not be expected to recall this chain, nor even its existence, but could be
expected to, say, infer that step 1 is governed by the electromagnetic interaction, and step 2
by the weak process, or to write step 2 at the level of quarks  u  e  d  e 
7
3
Useful Websites
Note: Googling “proton-proton chain” gives a plethora of sites, many of which give far too
much detail, but the following are interesting:
P-p chain : http://csep10.phys.utk.edu/astr162/lect/energy/ppchain.html
From Wikipedia http://en.wikipedia.org/wiki/Proton-proton_chain_reaction
Nice – with animations: http://burro.cwru.edu/Academics/Astr221/StarPhys/ppchain.html
Another animation http://www.physics.mun.ca/~jjerrett/protonproton/pp.html
Part of a whole course on stellar physics:
http://www.shef.ac.uk/physics/people/vdhillon/teaching/phy213/phy213_fusion3.html
Dry, but links the production of solar neutrinos to their detection using Chlorine [they put the
dry into dry-cleaning fluid].
http://www.sns.ias.edu/~jnb/Papers/Popular/Scientificamerican69/scientificamerican69.html [nice
diagram of neutrino energies from the different branches of the p-p chain – this then related to the
possibility of detecting using Cl-37]. If it’s not on this site, you didn’t want to know it – that goes for
much of the material that is on the site too!
GCE AS and A PHYSICS Teachers' Guide 37
4.3
PH3 – Practical Physics
The attention of centres is drawn to the specification of this unit and the internal assessment
guidelines on pages 24 and 57 of the GCE Physics specification. The focus of this unit is undertaking
measurements and observations and an appreciation of the uncertainties inherent in these
observations. Reference should be made to the Guidelines on the Treatment of Uncertainties <link to
be inserted> on the WJEC website.
Administration of the Internal Assessment.
Based upon the preliminary entries which centres make in October, centres making entries will
receive in February of each year a document entitled Confidential Instructions for Supervisors. This
will contain general instructions for the administration of the assessment and a detailed description of
the apparatus needed for each task. The questions will not be included with this mailing. Well before
the scheduled time for the assessments, centres should assemble the required apparatus. The subject
officer is available to answer any questions and deal with problems that may arise in doing this.
The scheduled sessions for the assessment will be in two consecutive days in the last week in April.
Centres are expected to enter candidates on the first day and only use the second day if candidates
cannot be conveniently accommodated in one day. As the duration of the assessment is only 1½
hours, there is ample time for 2 or 3 sessions on each day. Centres which need to use more than one
session may opt to use session 1 on day 1 and session 1 on day 2 rather than using two sessions on the
same day.
The assessment will come in two versions: version I is to be used on the first day and version II on the
second.
The structure of the assessment lends itself to accommodating candidates in multiples of six; at any
one time, three of them would be undertaking Section A and three would be working on section B.
For centres with large numbers of candidates, a possible arrangement is to have 18 candidates split
between 2 communicating laboratories. Of course, 12 in each laboratory would also be possible, given
sufficiently large laboratories.
Centres will receive copies of the assessment questions in good time. In addition to the multiple
copies of the assessment papers, centres will receive a single copy (1323/01-E Physics PH3 Practical
– Setting up instructions) of the assessment tasks which may be opened a week in advance. This
version will contain only the sections linked to the actual obtaining of results. The full version of the
assessment paper will not be available until the set date of the examination. Supervisors should work
through the tasks and ensure that the apparatus and questions work as intended.
After the practical assessments have taken place, the completed examination papers must be securely
stored by the exams officer before it is submitted to WJEC. Teachers should not be given access to the
completed examination papers after the actual assessments have taken place.
GCE AS and A PHYSICS Teachers' Guide 38
PH4 – Oscillations & Fields
4.4
PH4.1 – Vibrations
This section includes circular motion, simple harmonic motion, damping and resonance. It is
essentially unchanged from the equivalent sections in the legacy specification. All its
contents are adequately covered in many A-level Physics textbooks and there is no current
intention to publish guidance notes.
For examples of examination questions, see previous PH4 papers.
PH4.2 – Momentum concepts
This short section has been augmented by the introduction of the concept of photon
momentum and hence of radiation pressure. This draws upon the photon ideas in PH2 and
presents the opportunity for synoptic questions. The obvious application is the “light sail”
which is proposed as a method of interplanetary propulsion. No equations in addition to
p
h

hc
will be required. Questions could probe the difference between cases in which

f
photons are absorbed and those in which they are reflected [giving twice the momentum
transfer]. The concept of elastic and inelastic collisions draws upon energy from PH1.
For examples of examination questions, see previous PH4 papers.
PH4.3 – Thermodynamics
Statements PH4.3(a)–(g) deal with the behaviour of ideal gases. They include a simple
treatment of the kinetic theory of gases, including the concept of the mole. It too is essentially
unchanged from the previous specification.
Statements PH4.3(h)–(p) cover the concepts of thermodynamics: heat, work and internal
energy. The 1st Law of Thermodynamics is also included. In spite of its presence in the
current specification, it is a section which many students find obscure and accordingly a set
of notes is provided: go to the WJEC website, www.wjec.co.uk , select Physics and GCE AS/A
under “Find resources” and “view the full list of documents” under Related Information.
For examples of examination questions, see previous PH4 papers.
PH4.4 – Electrostatic and Gravitational Fields of Force
These two fields of force are treated together, in view of their mathematical similarity. The
field line is introduced as indicating the direction of the force upon a test object [charge or
mass, respectively] and leads on to its mathematical expression in the concept of the vector
quantity of field intensity. The scalar potential in a field is defined in terms of the work
required to be done [by an external agent] in bringing a unit test object from a point of zero
potential – infinity for mathematical convenience.
There are many similar equations and candidates will be helped to avoid their misapplication
by the equation sheet included in the question paper.
GCE AS and A PHYSICS Teachers' Guide 39
The only introduced concept in this section is that the gravitational field outside a sphericallysymmetric body is identical to that of an identical point mass situated at the centre of the
body. The point of introducing this statement [essentially Gauss’s Law] is to allow for the
application of Newton’s Law of Gravitation to approximately spherical planets, moons and
stars and, in the next section, to the hypothetical dark matter in which galaxies are supposed
to be embedded. It is not explicitly stated, but the other aspect of Gauss’s Law will be
assumed, i.e. that the net contribution to gravitation field by those parts of a spherically
symmetric mass distribution lying outside the radius of the point in question is zero.
For examples of examination questions, see previous PH5 papers.
PH4.5 – Application to Orbits in the Solar System and the Wider Universe.
This section of the specification contains traditional kinematics and application of Newton’s
Laws of Motion. Much the theory is covered in A level text books. The section on mutual
orbits is an exception. The applications to missing matter in galaxies and the detection of
extra-solar planets (ESOs) require very little additional theoretical input.
Statements PH4.5 (a) – (d) deal with the application of Kepler’s Laws of Planetary Motion
and Newton’s Law of Gravitation to the orbit of objects around a massive central object.
With the exception of the statement of Kepler’s Laws, this work could have been examined
under the legacy specification. Suitable statements of Kepler’s Laws are:
K1: The planets orbit in ellipses with the Sun at one focus.
K2: The radius vector sweeps out equal areas in equal intervals of time.
K3: The square of the period of orbit is directly proportional to the cube of the semimajor axis.
The whole of PH4.5 will concentrate on circular orbits. Very little work will be set on the
elliptical aspects. Candidates should be qualitatively aware of the ellipse. [eccentricity will
not be explored quantitatively] and the meaning of “semi-major axis.” The implication of K3,
that the period of orbit of an object in circular orbit is the same as that of an object in an
elliptical orbit with the same semi-major axis, should be understood. An example of where
this is important is the Transfer Orbit.
Transfer Orbits
Consider a satellite being carried on the upper stage of its launch rocket. It is currently in a
low circular orbit – say an altitude of 500 km [radius of orbit ~ 7000 km]. It needs to be
transferred to a geosynchronous orbit [radius ~42000 km].
low earth
(parking) orbit
A
geosynchronous
orbit
B
transfer orbit
The major axis of the transfer orbit is 7 000 + 42 000 = 49 000 km, so the semi-major axis is
24 500 km. The time taken to transfer can then be worked out because the time taken to
GCE AS and A PHYSICS Teachers' Guide 40
complete half an orbit [the dotted line] is the same as the time for half a circular orbit of
radius 24 500 km. Interestingly, though this is not required knowledge, the energy of the
satellite in the transfer orbit is also the same as if it were in a circular orbit of the same radius,
so we can calculate the additional energy [and therefore the impulse] needed to be given at A
and at injection at B.
“Derivation” of Kepler’s 3rd Law
This follows from Newton’s Law of Gravitation, F  G
m1m2
r2
, and the ideas of centripetal
force developed in section PH4.1.
Consider an object of mass m in a circular orbit of radius r about a much more massive object
of mass M. This could be satellite – natural or artificial - in orbit about a planet, a planet
about a star or a star about the supermassive black hole in the centre of our galaxy.
The centripetal force necessary for the [accelerated] circular motion is given by: F  mr 2 ,
4 2
or equivalently by F  mr 2 , where T is the orbital period.
T
So we can write:
Dividing by m and rearranging, we have:
GMm
4 2
 mr 2
r2
T
2 3
4 r
T2 
.
GM
i.e the orbital period squared is proportional to the radius cubed, which is K3 for a circular
orbit.
Note that, we have assumed that the central body is a point mass, which it will certainly not
be, but it is also correct if the central object is spherically-symmetric [see above]. Note also
that, historically, the derivation was done in the opposite direction, with Kepler’s 3 rd Law
being the evidence for the inverse square relationship.
Weighing the Earth
Experiments to determine G, the universal constant of gravitation, used to be described as
weighing the Earth. This is because a knowledge of G and the orbital radius and period of the
Moon enables us to calculate the mass of the Earth.
Data: Radius of Moon’s orbit = 3844  108 m [380 000 km].
Period of Moon’s orbit = 2732 days = 236  106 s.
G = 6673  1011 N m2 kg2.
4 2 r 3
4 2 r 3
T 
, so M E 
= 604  1024 kg.
G
GM E
2
Note that, in this analysis, we have assumed that the mass of the moon is negligible and that
the moon orbits about the centre of the Earth. In fact ME ~ 81 MM so the assumption leads to
some inaccuracy albeit small [~1%].
GCE AS and A PHYSICS Teachers' Guide 41
This type of analysis is very useful in obtaining information about remote objects in the
universe. For example, we can “weigh” other planets and determine their mean densities,
furnishing data which is useful for developing models of their composition. We can also
weigh stars, black holes and whole galaxies using the same technique, see e.g. “how to
measure the mass of a black hole” on the Physics page of the WJEC website.
The relationship between T and r furnishes data which is useful for developing candidates’
graphical skills. Data on Jupiter’s satellites for example can be used in a log-log plot to
2
establish the power law relationship. Students could also plot, say, T 3 against a and use the
3
gradient to determine MJ. Alternatively, T against a 2 is a possibility. Note that it is
unproductive to plot T2 against a3 as, whereas most of the points are almost at the origin a
couple are a long way out.
Often the speed of an orbiting object is measured directly, e.g. using Doppler shift [see
below] in which case we could use the following analysis to determine the central mass M.
GMm mv 2

r2
r
GM
Dividing by m and simplifying:
.
v2 
r
Dark Matter and he motion of objects in galaxies.
Spiral galaxies are flattened assemblages of stars which all rotate in the plane of the galaxy
around the centre in its gravitational field. In additional to stars, spiral galaxies contain large
quantities of gas and dust from which new stars form. Details of the structure of spiral
galaxies will not be examined. Consider the following observed [simplified] rotational speed
curve for a typical spiral galaxy: [The low radius part of the curve is obtained from
observations of stars and gas clouds in the visible part of the disc. The observed speeds
beyond the visible galactic disc are from orbiting clouds of neutral hydrogen which emit a
characteristic 21 cm line in the microwave region of the spectrum.]
-1
Rotational speed (km s )
observed
200
100
central
galactic bulge
extent of visible disc
calculate
d
50 000
100 000
Distance from centre (light years)
[N.B. the inner ellipse is my crude attempt using “Draw” to represent the central galactic bulge]
The “calculated” curve is that predicted by taking into account the observed normal or
“baryonic” matter in the galaxy. N.B. “Observed” doesn’t only mean “light-emitting” – it
also includes dark gas clouds, whose speeds we can detect by their absorption lines in the
light of more distant object which we view through them.
GCE AS and A PHYSICS Teachers' Guide 42
GM
, where r is the orbital radius, v is
r
the orbital speed and M is the total mass within the orbit [assuming a spherically-symmetric
distribution]
A useful equation in investigating these curves is v 2 
It is worth highlighting two regions of the curves:
(a) The low-radius part of the curves.
Here the curves coincide and the speed is roughly proportional to the orbital radius. In
other words, the wholes central region of the galaxy rotates with roughly the same
angular velocity. This implies that the density of matter is constant within this region:
For a constant density, , the mass within an orbit of radius r = 43  r 3  .
4
3
 r3
 43 G r 2 , so v  r .
r
So we can see that, for the central regions of a galaxy, coinciding with the galactic
bulge, the observed rotational velocity is consistent with the observed constant density
of matter and the value of the matter density is consistent value of the rotational speeds.
So, using the equation above: v 2  G
(b) The high-radius part of the curves.
The approximately constant orbital velocity is explicable if the density of the material
falls off roughly as r 2 :
If   kr 2 , the total mass within the orbit, M   4 r 2  kr 2  dr  4 kr
G  4 kr 
 4 Gk , i.e. v is a constant
r
If we imagined a gas cloud orbiting at, say, 75 000 light years from the centre of the
galaxy, it is doing so in the combined gravitational field of all the matter closer to the
centre. In the case of a spherically-symmetric object we can for such purposes consider
it as a point mass with its whole mass concentrated at its centre. Clearly the visible
galaxy is not spherically symmetric, but it is not a bad approximation to consider it so
for great distances. Beyond the visible disc, where the observed matter density is very
1
low, we’d expect the orbital speed, v, to fall off approximately as r  2 , the same
relationship as we observe for the planets in the Solar System. The observation that,
beyond ~ 50 k l-y, the rotational speed is ~ constant implies that the material of the
galaxy extends well beyond the observed galaxy, i.e. the visible galaxy is embedded in
an unobserved cloud of material and also than the whole galaxy has a much greater
mass [~ 10 times] than that of the observable matter.
So v 2 
N.B. It is worth emphasising that using “Dark Matter” to explain the discrepancy
between the observed and calculated orbital speeds is a hypothesis, albeit one which is
widely supported in the theoretical cosmological community. Some theoretical
cosmologists have proposed modifications to the law of gravitation to account for the
observations. The modifications take into account the fact that, at the scale of the Solar
System, the inverse square law works very well. In one such model, by Milgrom, the
modification takes effect at gravitational accelerations of less than 10-9 m s-2 [i.e. ~10-10
g] and for these accelerations the gravitational force falls of as inverse r rather than
inverse r2]. This is a classic example of “watch this space” or “How Science Works.”
GCE AS and A PHYSICS Teachers' Guide 43
Objects in mutual orbit, leading up to the discovery of Extra-solar Planets.
Centre of Mass
We normally think that planets orbit stars and, to a good approximation, this is true because
the planet is so much less massive than the star. For example, the MEarth = 6  1024 kg and
MSun = 2  1030 kg. For precise work or for situations where the two orbiting bodies are of
similar mass, such as a binary star or the Pluto-Charon system, we need to refer to the Centre
of Mass.
Both bodies orbit around a point which, in the absence of externally-applied forces, is
stationary [or, more strictly, moves with constant velocity]. This point is called the centre of
mass.
Let us consider two spherically-symmetric objects, of comparable mass, orbiting about their
centre of mass. Before we do any algebra, we can infer three things about the system:
1. Symmetry considerations tell us that that the centre of mass must be on the line
joining the centres of the two objects.
2. The centre of mass must be between the objects as the direction of the centripetal
acceleration must be towards it.
3. The angular velocities of the objects must be identical – if this were not the case, the
objects would sometimes be on the same side of the Centre of Mass, which clearly
contradicts point 2.
Time for some algebra: Consider two bodies, of mass m1 and m2 orbiting around their Centre
of Mass, C.
m1
r2
r1
C
m2
d
Each body exerts an attractive force upon the other and, by Newton’s 3rd Law, these are equal
and oppositely directed.
So, we can write
m1r1 2  m2 r2 2
m1r1  m2 r2
So, dividing by  ,
m1r1  m2 (d  r1 )
Substituting for r2
m2
r1 
d

m1  m2
m1
r2 
d
Similarly
m1  m2
The orbits of two massive objects, e.g. a binary star system:
Now that the position of the centre of mass is sorted out, we can use Newton’s Law of
Gravitation to work out the orbital characteristics of the binary system as follows:
Consider the orbit of body 1 about the centre of mass. The centripetal force is provided by the
gravitational attraction of body 2 upon body 1.
GCE AS and A PHYSICS Teachers' Guide 44
So we can write
m1r1 2 
Substituting for r1:
m1
Dividing by m1m2 and rearranging:
2 
as T 
2

Gm1m2
d2
m2
Gm1m2
d 2 
m1  m2
d2
G  m1  m2 
d3
4 2 d 3
d3
or T  2
T2 
G  m1  m2 
G  m1  m2 
Consider the case of the Earth-Sun system. The Earth-Sun distance is 1496 million km. With
the masses given above, the distance of the centre of mass of the two bodies from the centre
of the Sun is given by:
6 1024
r
1 496 108 km  450 km
30
24
2 10  6 10
In this calculation, clearly the mass of the Earth in the denominator is quite insignificant. The
figure of 450 km compares to a radius for the Sun of 700 000 km – so not large!
Aside: In fact, we could come up with this figure without the above analysis, just by using the
idea of Conservation of Momentum. The argument could go as follows.
Let the speed of the Earth in its orbit be vEarth, so its [linear] momentum is given by:
pEarth = 6  1024 kg  vEarth
Assuming the momentum of the Earth-Sun system is zero, it follows that the momentum of
the Sun is the same, in the opposite direction. So the orbital speed of the Sun is given by:
vSun
6 1024  vEarth

 3 106 vEarth .
30
2  0 10
As the two bodies take the same time, T, to orbit the Centre of Mass:
T
2 rEarth 2 rSun

vEarth
vSun
So
2 rSun
2 1 496 1011 m

vEarth
3 106 vEarth
So
rSun = 440 km
which agrees to within the accuracy of the data. Can you spot the approximation?
We’ll return to the idea of using momentum conservation when we analyse extra-solar
planetary systems.
Question 1: If the mass of the Earth were 100 times as great [6  1026 kg], what would be the
effect on:
1. the length of the year;
2. the position of the centre of mass of the Earth-Sun system;
3. the orbital speed of the Earth;
4. the orbital speed of the Sun.
GCE AS and A PHYSICS Teachers' Guide 45
Question 2: The dwarf planet, Pluto, has a
mass of 127  1022 kg. Its moon, Charon has a
mass of 19  1021 kg. The mean separation of
their centres is 19 640 km. Use these data to
determine:
1. the position of their centre of mass;
2. the orbital period of the two bodies;
3. the orbital speeds of the two bodies.
Nice pic of Pluto,
Charon and the
recently discovered
Nix and Hydra
Measuring speeds using the Doppler Effect.
Many objects that we study, including stars and gas clouds, have emission and/or absorption
spectra with identifiable lines. If such an object is moving towards or away from us, the
wavelength of the radiation which we receive is shifted. This shift is towards longer
wavelengths [red shift] if the object is moving away from us and towards shorter wavelengths
if it is moving in our direction.
We shall only use the low-velocity approximation for the Doppler shift,
 v
i.e.
 .
 c
The velocity v in this equation is the component of the objects velocity relative to the
observer along the line joining the observer to the object. This is known as the radial velocity,
which can be slightly confusing, if we are considering an object in orbital motion about
another]. In this low-velocity approximation [“low velocity” is relative to c, the speed of
light, so speeds up to (say) 107 m s-1 would be considered “low”], any motion at right angles
to the line of sight produces a negligible Doppler shift.
Sign convention: In the equation above,  will be positive if v is positive, so we measure v
away from the observer.
Alternative forms of the equation: Because the frequency of radiation, f, is inversely
proportional to , the same equation holds, with the slight complication that there is now a
minus sign:
f
v
 .
f
c
Of course, if you are happy to remember that a positive v produces a smaller f you can forget
about a sign convention.
Information about stars from the Doppler Effect.
Suppose we observe a star which has a massive star in orbit or, more correctly, a star and
massive planet in mutual orbit. Normally, we would not be able to see the planet, but we
would infer its presence from data about the speed of the star.
In questions we would always assume that we see such a system edge-on. In reality, the
situation is more complicated. All candidates might be asked to consider is what effect it
would have on our observations if the system were tilted.
GCE AS and A PHYSICS Teachers' Guide 46
Such a system might look as follows:
star
light from the star to observers
orbit
on the Earth
massive
planet
For such a system the data could be presented graphically, e.g. variation in the received
wavelength of the sodium D2 line which has a laboratory wavelength of 58900 nm.
 / nm
5895
0
58900
58850
0
05
10
15
20 Time /days
From this graph, you can determine:
(a) the mean radial speed of the star system, from the mean wavelength [~5891 nm];
(b) the star’s orbital speed, from the amplitude of the wavelength variation [~02 nm];
(c) the period of the orbit.
Notice here the use of the word “radial”. Question (a) asks you to find the component of
the binary system’s velocity in the direction directly away from the Earth.
N.B. It is not only the wavelength [and frequency] of the radiation itself which
undergoes Doppler shift. Recently astronomers noticed that the period of the pulsations
from a pulsar [a neutron star] vary in a periodic way. This is attributable to the effect of
an orbiting planet or companion star. They used the Doppler equation in the form:
T v
 ,
T
c
where T = period of the pulsations, in the same way as the wavelength in the example
above to work out the orbital parameters and the masses involved.
Where it all comes together:
Most of the information about the masses of stars and the evidence for the existence of extrasolar planets has come from Doppler measurements in orbiting systems.
GCE AS and A PHYSICS Teachers' Guide 47
For the case of Extra-solar planets we can assume that mP mS [where mp is the mass of the
planet and mS is the mass of the star]. With this approximation, the equation for the period of
the mutual orbit reduces to:
d3
(1)
GmS
Algebraic manipulation gives the following approximations for the orbital speeds:
G
vS  mP
(2) and
mS d
T  2
vP 
GmS
(3)
d
For a given star system, normally we would know the mass of the star, mS, from our
knowledge of stellar models and the observations would be the received wavelength of a
spectral line against time. For an edge-on system this variation would be sinusoidal and we
can undertake the following steps:
1.
2.
3.
4.
5.
From the graph, determine the orbital period, T.
From the amplitude of the observed Δ, determine vS .
Use equation (1) to determine the separation of the star and planet, d.
Use equation (2) to determine the planetary mass mP and (3) to determine its speed.
If the planet occults the star [passes in front of it – if it’s a true edge-on system it
should do but many will just miss occulting the star], we can further estimate the
planet’s diameter from the period of occultation and its speed and the ratio of the
stellar to planetary diameters from the fractional decrease in the observed light.
The data needn’t be presented graphically, e.g.:
Example
The wavelength of the H line which, in the laboratory has a value of 4861 nm, in the
radiation emitted from a star is observed to fluctuate with an amplitude of ± 105  10-3 nm
with a period of 125  106 s. The mass of the star is 30  1030 kg. Assuming that this
behaviour is caused by an orbiting planet and that we observe the system edge-on:
1. calculate the distance of the planet from the star;
2. calculate the mass of the planet and its orbital speed.
STOP PRESS: Astronomers studying this star have noticed that its brightness drops by ~1%
once in every orbit of the planet. This dimming lasts for 454 hours. They suggest that this
dimming is caused by the planet blocking of the light from the star as it passes in front as
seen from the Earth.
3. Use this information to estimate the diameter of the star and that of the planet. Calculate
also the planet’s density.
Binary Star Systems
Most information about stellar masses comes from a study of binary star systems, i.e. a pair
of stars in mutual orbit. Because both objects in such a system emit light, the orbital
velocities of the two stars can be found directly and so the masses can be calculated. Consider
the following graphs of the radial speeds to two stars in close mutual orbit:
GCE AS and A PHYSICS Teachers' Guide 48
What information can we glean from this without any algebra?
 The mean radial velocity is + 40 km s-1, i.e. the binary system is receding from
us at this speed.
 Assuming we see the system edge on, the speeds of the two components are
75 km s-1 and 25 km s-1 [these are the amplitudes of the speed variations].
 The period of the orbit is 176 days [152  106 s]
We can do some sums before we need to apply complicated theory:
1. The ratio of the masses is 3:1, i.e. one component [the faster one] has ¼ of the total
mass of the system and the other has ¾ of the total mass. This comes from
momentum considerations: in the frame of reference in which the centre of mass is
at rest, the momentum of each component must be equal and opposite. The speeds
are in the ratio 1:3 so the masses must be in the ratio 3:1.
Another way of looking at this idea is as follows:
The gravitational forces on the components are equal: i.e. m1r112  m2 r22 2 .
Dividing by : m1r11  m2 r22 , i.e. m1v1  m2v2
2. We can work out the circumference [and then the radius] of each of the orbits:
e.g. The slower [more massive] component:
circumference  orbital speed  orbital time
 25 km s -1  1 52 106 s
 38 million km
So the radius of the orbit is calculated at 604 million km [circumference = 2r]
Likewise the orbital radius for the faster component is 1814 million km.
3. From the two orbital radii, we can infer that the separation of the stars, d, is
242 million km [the sum of their orbital radii].
Now the earlier formulae click in:
4. We can apply the formula T  2
d3
to find the total mass, m1 + m2, of
G  m1  m2 
the system – which comes out at 36  1030 kg, so the combined mass of the stars is
roughly twice that of the Sun.
GCE AS and A PHYSICS Teachers' Guide 49
5. From this, we can work out that the masses of the individual stars are 09  1030 kg
and 27  1030 kg [remember the 3:1 ratio].
More or less useful references for gravity, spectra and mutual orbits
Check out the applets on:
http://www.ioncmaste.ca/homepage/resources/web_resources/CSA_Astro9/files/html/applets.
html - more GCSE than GCE for Stars, Spectra and Kepler’s Laws
On http://jersey.uoregon.edu/vlab/elements/Elements.html you can find the wavelengths of
spectral lines [put the mouse cursor on the line and click]
Doppler spectroscopy: http://en.wikipedia.org/wiki/Doppler_spectroscopy
An example of a radial velocity curve: http://www.howstuffworks.com/planet-hunting2.htm
Mutual orbit simulation: http://www.howstuffworks.com/framed.htm?parent=planethunting.htm&url=http://exoplanets.org/doppler.html
This site also has mutual orbit simulation and a plethora of other applets:
http://phet.colorado.edu/new/simulations/sims.php?sim=My_Solar_System
Data for 51Peg: http://zebu.uoregon.edu/51peg.html
Overview of detecting ESOs: http://astro.unl.edu/naap/esp/detection.html
Another overview : http://www.esa.int/esaSC/SEMYZF9YFDD_index_0.html
GCE AS and A PHYSICS Teachers' Guide 50
4.5
PH5 Magnetism, Nuclei & Options
General
PH5 is intended as a terminal unit, though the regulations do not require A level Physics to be
cashed in at the same time as PH5 is taken. The 1¾-hour examination paper has 3 sections:
Section A
This is a 60-mark section based upon the core content of PH5.1 – PH5.5. It is designed to
be answered in about 60 minutes.
Section B
This section carries 20 marks and contains a series of questions relating to the Case Study.
It is designed to be answered in about 20 minutes.
Section C
This section consists of 5 questions, 1 on each of the Optional Topics. The questions each
carry 20 marks and are designed to be answered in about 20 minutes. Candidates will
answer 1 question only.
SECTION A – Core content
In line with the other units of this specification, PH5 is designed to require a teaching time of
approximately 60 hours, of which ¾ should be devoted to the study of the compulsory core
content.
PH5.1 – Capacitance
This topic is part of the national core and follows on from the electrostatic fields section of
PH4. It also draws on energy and electrical circuits concepts. The equation for the decay of a
capacitor is of the same form as that for radioactive decay. It is a good topic for the
introduction of semi-log graphs for the determination of the time constant of the decay and
hence the capacitance.
There are many examination questions in past PH4 papers and the topic is well covered in
most A-level physics textbooks.
PH5.2 – B-fields
This section deals with the concept and definition of magnetic fields, their effect on moving
charges (in wires and in free space), their production and application in particle accelerators.
This traditional topic is well covered in A-level text books and there are many examination
questions to be found in past PH5 papers.
PH5.3 – Electromagnetic Induction
In addition to magnetic flux (linkage) and the laws of electromagnetic induction, this topic is
linked via rotating coil generators to the basic concepts of alternating current electricity –
frequency, period, peak values, r.m.s. values. The relationship between peak and r.m.s. values
for a sinusoidally varying quantity and the use of r.m.s. current and voltage in power
calculations are explored. Candidates will be expected to have used oscilloscopes to measure
voltages and currents [by the p.d. across a resistor] and frequencies. This traditional topic is
well covered in A-level text books and there are many examination questions to be found in
past PH5 papers.
GCE AS and A PHYSICS Teachers' Guide 51
PH5.4 – Radioactivity and Radioisotopes
This core topic is largely unchanged from the previous specification and is well covered in Alevel text books. Candidates will be expected to handle logarithm and exponential functions.
Calculations on decay can be expressed via the exponential function e t or 2 x , where x is
the number of half-lives. This topic area lends itself to synoptic questions which combine
concepts of relative atomic mass and the mole with the decay equations to calculate the
activity of a given mass of material of known decay constant. Past PH5 papers contain many
examination questions of an appropriate level.
PH5.5 – Nuclear Energy
This core topic is largely unchanged from previous specifications. As with radioactivity, the
mole concept will be used in estimating the energy release from macroscopic quantities of
reacting materials – the electron volt is also a concept from AS which is of use here.
Conservation of mass/energy s introduced, using E = mc2. This concept can be applied
generally and not only in particle interactions – e.g. calculate the Sun’s power output, and
hence the Solar Constant, given that its mass loss per second is 4 million tonnes.
A useful concept is the energy equivalence of 1 u. 1 u = 1.6604  1027 kg. The energy of this
mass is 1.6604  1027  c2 = 1491  10-10 J = 931 MeV.
The use of neutrino energy in neutrino detectors is possible. The common reaction used in
detectors is:
37
17
Cl  νe 
38
17
Ar 
1
0
e
The masses of the particles are:
Cl: 36.96590 u
e: 0 [at least negligible]
Ar: 36.96677
e: 0.000548 u
The gain in mass in the interaction Δm = 36.96677 + 0.0005  36.96590 = 0.00142 u  1.32
MeV. This means that only neutrinos with a kinetic energy of more than 1.32 MeV can cause
this interaction. The mean neutrino energy produced by the first step in the proton-proton
chain is only 0.26 MeV, so most of these are not detected.
SECTION B – the Case Study
Centres with candidates for PH5 will receive, in February of the relevant year, multiple
copies of a printed passage based upon a physics topic of contemporary interest. The topic
will be chosen to relate to previously studied areas of the AS and A2 specification. It will be
assumed that candidates have covered the whole of the AS specification and PH4 at least.
The passage should be given to candidates for study and teachers are encouraged to discuss
its contents with them, drawing their attention to the relevant areas of the specification and
considering the sorts of questions which could arise from its contents. Candidates will be
provided with a clean version of this passage in the PH5 examination and will not be
permitted to take notes into the exam.
GCE AS and A PHYSICS Teachers' Guide 52
Previous PH6 examinations contain similar passages with the difference that, in the legacy
specification, candidates had not previously seen the passage and were expected to read it in
the examination. Nevertheless, these passages form a good resource for introducing this
section of the paper. Because candidates will be expected to have studied the passage prior to
the examination, no allowance for reading is built into the duration of PH5, which is 105
minutes.
SECTION C – Options
There are 5 optional topics:
A
B
C
D
E
Further Electromagnetism and Alternating Currents
Revolutions in Physics
Materials
Biological Measurement and Medical Imaging
Energy Matters
Each topic is designed to be studied in approximately 15 hours of teaching time. They could
all be taught at the end of PH5. They fit in with the rest of the specification in different ways,
which suggests that different teaching strategies are appropriate:
Option A follows on immediately from the electromagnetism and A.C. material in PH5. The
filters section relates also to the potential divider ideas in PH1.
The Electromagnetic Revolution aspect of Option B, which will be the setting for questions
for the first 3 years, relates in the early stages to the optics material in PH2, the electrostatics
in PH4 and the electromagnetism in PH5. There is a strong case, if this option is to be
offered, for incorporating its ideas throughout the teaching of the rest of the course,
Option C, materials, consists of ideas which were previously in the compulsory specification
and are now optional. There are few strong links with other sections of the specification.
Option D, Biological Measurement and Medical Imaging, has links to PH2 and PH5.
Option E, Energy Matters, links to PH1, PH2 and PH4 and so a possible approach would be
introduce the content throughout the course.
Guidance notes on each of the options follows:
GCE AS and A PHYSICS Teachers' Guide 53
Unit PH5 Option A – Further Electromagnetism and Alternating Currents
The majority of this option unit is taken from the legacy specification where it was
compulsory content. This historical material divides into 3 parts:
1. Mutual induction and its application to transformers. This treatment is largely qualitative,
except for the treatment of the ideal transformer in terms of turns ratio and the equality of
input and output powers.
2. Self inductance and inductors.
3. Phasor analysis of series RC, RL, LC and RCL circuits.
This material is adequately covered in text books and PH5 papers. Questions on transformers
also appear in past GCSE Physics papers. It is not the current intention to produce teacher
guidance notes on these aspects of Option A.
Statements (p) – (r) deal with the sharpness of a resonance curve (Q- factor) and the
application of the RC potential divider to high pass and low pass filter circuits. These topics
are well dealt with in Electronics text books and past GCE Electronics papers [ET4] of WJEC
but have not been dealt with in previous WJEC GCE Physics specifications.
The Quality factor (Q) of a resonant circuit
The quality (Q) factor of a LCR circuit is related to the sharpness of the resonance curve. A
high Q factor gives a sharp resonance curve while a low Q factor gives a broad resonance
curve (see the diagram below with Q = 8 and Q = 2).
The main component in determining the Q factor of the circuit is the resistance of the circuit
because it is the resistance that dissipates energy away from the circuit. This is similar to
pushing a swing back and forth – if there is a lot of friction taking energy away from the
swing it’s difficult to achieve a high amplitude and ‘sharp’ resonance. The easiest way to
define the Q factor is as follows
Q
r.m.s. pd across inductor at resonance
r.m.s. pd across resistor at resonance
As the capacitor and inductor have equal reactance at resonance, the Q factor can also be
written:
GCE AS and A PHYSICS Teachers' Guide 54
Q
r.m.s. pd across capacitor at resonance
r.m.s. pd across resistor at resonance
These definitions lead to the equations
Q
I 0 L
IR

0 L
R
I
and also Q 
0 C

IR
1
0 CR
If we also incorporate the expression for the resonant frequency, 0 
Q
0 L
R
 1
L


1
LC 


R
R
1
, then
LC
L
C
So we have three expressions for the Q factor.
Q
0 L

R
1
0CR

1
L
R
C
Note that, in the expressions for the Q factor, we can eliminate L, C and ω0 but we cannot
eliminate R – it is in all 3 expressions. Note also that the Q factor is a ratio and it has no units.
Now consider this circuit:
10 V
~
10 nF
10 mH
10 
These values for R, C, L make our arithmetic reasonably easy. They give us the following
figures:
0 
1

LC
1
2
10  10
8

1
10
10
 105 s -1
and
Q
0 L
R

105  102
10
 100
We can also calculate the current flowing at resonance because the whole of the supply p.d. is
across the resistor at resonance (p.d.s across the inductor and capacitance are equal and
opposite, so cancel).
I
V
R

10
10
1 A
GCE AS and A PHYSICS Teachers' Guide 55
All seems nice and straight forward until we look at the p.d. across the capacitor or inductor.
VL  I0 L  1105 102  1000 V   VC 
How can we have 1000 V across the inductor (and capacitor) when the supply voltage is only
10 V? There is no simple answer to this question but a better understanding can be drawn
from considering another type of resonance. Again, consider a swing with very little friction.
You only need to provide a small push regularly in order to obtain a large amplitude – you
might only be pushing the swing for a distance of 30 cm but the amplitude of oscillation
could easily be 2 m.
Understanding how a series LCR circuit can be used to select frequencies
We know now that LCR circuits with high Q factors can increase the p.d. in an a.c. circuit.
This can be used in the design of a simple radio. The circuit below can be used as the
detection part of a simple radio. It consists of an antenna (long wire), inductor, variable
capacitor and earth connection.
A good application of synoptic physics:
antenna
Remember that radio waves are electro-magnetic waves and
have oscillating electric (and magnetic) fields. These oscillating
fields will cause electrons to move in the metal antenna. The
moving electrons will give us an alternating current and an
alternating p.d. (due to the resistance of the antenna).
earth
If you look at the loop in the above circuit, you’ll notice that there is no resistor. It is an LC
circuit without the R. Why is this? Remember that we want a high Q factor and one of the
ways that this is achieved is to keep the resistance low. Does this mean that the resistance in
the LC loop is zero? Obviously the resistance cannot be zero because the connecting wires
aren’t made of superconductors. But the main source of resistance in the LC loop is the
inductor. Remember that an inductor is a long wire wound into a coil. In order to make a
large number of loops we need a thin wire and this increases the resistance of the inductor (a
bit of a Catch 22 situation).
A simplified way of analysing the performance of the detecting circuit above is to consider it
as follows:
~
VOUT
GCE AS and A PHYSICS Teachers' Guide 56
i.e. we have a series LCR circuit and the voltage across the variable capacitor is the output
voltage. Note also that we have redrawn the inductor as a resistor and inductor in series
because of the inherent resistance of the wires of the inductor.
If we have resonance in the LCR circuit we know (from the definition of the Q factor) that the
p.d. across the capacitor will be Q times the supply p.d. Hence, we can amplify the input p.d.
by a factor of Q. Also, because of the shape of the resonance curve we only amplify the
frequencies around the resonance frequency, so we have selectivity.
So why do we use a variable capacitor? This is because we can vary the resonance frequency
by varying the capacitance. We obtain the resonance frequency from the equation below.
f0 
0
1

2 2 LC
So, the above simple circuit does three things:
1. Amplification - it amplifies our signal
2. Tuning - it can tune to a particular resonance frequency (by changing C)
3. Selectivity - it amplifies only those frequencies around the resonance frequency.
If you would like to see this tuning circuit in operation there is a reasonably priced kit
available from Maplins - N51FL crystal radio £5.99. Alternatively, it is possible to design
and build your own radio using instructions available from many internet sites e.g.
http://www.midnightscience.com/cigar.html, http://journeytoforever.org/edu_radio.html,
http://www.electronics-tutorials.com/receivers/crystal-radio-set.htm .
~
Example
This circuit is used in a simple radio.
(i) Calculate the Q factor when C = 6 pF
and when C = 600 pF.
(ii) Calculate the range of frequencies
to which the circuit can tuned.
10 
0.15 mH
6  600 pF
GCE AS and A PHYSICS Teachers' Guide 57
Using CR circuits as Low Pass and High Pass Filters
NOTE:
One thing you must beware is that you cannot simply add the rms p.d.s across the resistor and
capacitor in either of these a.c. filter circuits.
Remember always, for r.m.s. p.d.s ,
VIN  VC 2  VR 2
and that (in general)
VIN  VC  VR
Low pass filter:
R1
Compare with
R
~ Vin
C
R2
Vout
The easiest way to explain how the above circuit behaves as a low pass filter is to compare it
with a voltage divider.
In the circuit on the right, the supply voltage is shared between the two resistors. In the low
pass filter, on the left, the voltage is divided between the capacitor and the resistor.
Remember that the reactance of the capacitor is given by: X C 
1
C
From the above equation, at low frequencies XC will be very large. So at low frequencies we
have a voltage divider with a very large “resistance” in the R2 position. This means that
nearly all the supply voltage will be across the capacitor at low frequencies.
At high frequencies XC will be very small. So at high frequencies we have a voltage divider
with a very low “resistance” in the R2 position. This means that nearly all the supply voltage
will be across the resistor at low frequencies i.e. there will be a very low p.d. across the
capacitor.
If we were to draw a graph of Vout/Vin against frequency we would get:
GCE AS and A PHYSICS Teachers' Guide 58
Low pass filter output
1.0
Vout/Vin
0.8
0.6
0.4
0.2
0 0
10
10
2
10
4
10
6
Frequency/Hz
Note that Vout/Vin is usually called the gain and that it starts at 1 and drops to zero (this is
because Vout = Vin at very low frequencies and Vout = 0 at very high frequencies).
Example
1kΩ
10V ac supply
1nF
~ out
V
Vout
1. Calculate the frequency when the rms p.d.
across the resistor is equal to the rms p.d. across
the capacitor.
2. Calculate the rms p.d. across both the resistor
and the capacitor at the frequency of Q1.
Answers
1
Equating the p.d.s across the capacitor and resistor we get:
IX C  IR . Cancelling I gives us: X C  R
1
1
1
But X C 
, hence
.
 R and rearranging we get  
C
C
CR

1
1
Using   2 f , we get: f 


 159 kHz .
9
2 2 CR 2 10 1000
Beware:
There are 3 pitfalls to avoid if you want to obtain the correct answer even after you’ve
1
obtained the equation f 
.
2 CR
 First, you must remember that kΩ means 1000 Ω.
 Second, you must remember that nF means 10-9 F.
 Third (but this only applies if you have an EXP button on your calculator and if
you’re too lazy to do the powers of 10 in your head!), when putting 10-9 in your
calculator you cannot type 10 exp -9 because this is the same as 1010-9. You must
type (and this might seem strange until you think about it carefully) 1 exp -9 because
this is 110-9.
GCE AS and A PHYSICS Teachers' Guide 59
2
There are many ways of obtaining the correct answer
V
e.g.using Z  X C 2  R 2 and I  etc.
Z
but it is probably more direct and simple to do as follows:
Remember that VS2  VC 2  VR 2 and that VC  VR from question 1.
VS2
V
10
and hence VC  S 
 7  07 V
2
2
2
So the correct answer is that the p.d. across both the capacitor and the resistor is 7.07 V.
So VS2  2VC 2 ,  VC 2 
Beware:
Do not fall into the trap of saying that both rms p.d.s must be 5V so that they add up to 10V.
Although this sort of argument applies to instantaneous p.d.s it is completely wrong for
obtaining rms p.d.s because the p.d. across the capacitor is out of phase with the p.d. across
the resistor.
High pass filter:
In the low pass filter of the previous section we noted that when the p.d. was low across the
capacitor the p.d. was high across the resistor. If we now swap our capacitor and resistor, the
output p.d. will be the p.d. across the resistor instead of the capacitor (see below left). In this
circuit we will have a high output where we previously had a low output and a low output
where we previously had a high output. See the graph at the bottom of the page and compare
it with the previous low pass filter graph. The graph at the bottom of this page is
characteristic of a high pass filter.
C
Compare with
R1
~ Vin
R
Vout
R2
Again, the easiest way to explain how the above circuit behaves as a high pass filter is to
compare it with the voltage divider (on the right).
In the circuit on the right, the supply voltage is shared between the two resistors. In the high
pass filter, on the left, the voltage is divided between the capacitor and the resistor.
1
C
From the above equation, at low frequencies XC will be very large. So at low frequencies we
have a voltage divider with a very large “resistance” in the R1 position. This means that the
output voltage across the resistor at low frequencies will be close to zero.
Again, remember that the reactance of the capacitor is given by: X C 
GCE AS and A PHYSICS Teachers' Guide 60
At high frequencies XC will be very small. So at high frequencies we have a voltage divider
with a very low “resistance” in the R1 position. This means that nearly all the supply voltage
will be across the resistor at high frequencies.
If we were to draw a graph of Vout/Vin against frequency we would now get:
High pass filter output
1.0
Vout / Vin
0.8
0.6
0.4
0.2
0 0
10
10
2
10
4
10
6
Frequency / Hz
Note:
Filters are usually drawn in the following manner:
R
VIN
VOUT
C
0V
This makes it easier to draw higher order filters (i.e. one filter feeding into another to provide
more filtering). This notation has not been used here so that students can compare the circuit
more easily with a potential divider. However, the above notation may well be used in an
examination.
GCE AS and A PHYSICS Teachers' Guide 61
Unit PH5 Option B – Revolutions in Physics
OPTION B: REVOLUTIONS IN PHYICS
ELECTROMAGNETISM AND SPACE-TIME
1.
Lifetime of the Electromagnetism and Space-time material
When a History of Physics option was proposed, two periods of revolutionary change immediately
suggested themselves for study: the century of Kepler, Galileo and Newton, and the century of
Young, Faraday and Maxwell. Rather more interest was expressed in the second of these, and only the
Electromagnetism and Space-time revolution will be examined in 2010, 2011 and 2012. After the new
A-Level has been running for a year or two, teachers will be consulted on whether or not a change
should be made for examinations in 2013 and beyond.
2.
Content
One of the most exciting things in Physics is to discover relationships between phenomena which are
seemingly very different in nature. What happened in electromagnetism in the nineteenth century is a
wonderful example. In the year 1800 there were only the vaguest indications that magnetism had
anything to do with moving electric charges, and no evidence at all that light had anything to do with
electricity or magnetism. By 1900 magnetism and electricity had been firmly linked, and light had
been shown to be an electromagnetic wave.
The seemingly obvious need for electromagnetic waves to have a propagation medium (the ether)
created problems. These were resolved in a very radical way by Einstein’s Special Theory of
Relativity.
The structure of the course is shown in a little more detail in the diagram below. The first main
‘block’ deals with events leading to the acceptance of the wave theory of light, starting with a careful
look at Thomas Young’s description of his two slits experiment. Electromagnetism is the subject of
the next main block, starting with Ørsted’s discovery of the magnetic effect of a current, and
considering at some length the subsequent work of Ampère and Faraday.
Maxwell arrived at the conclusion that light was an electromagnetic wave using what we would now
call a mechanical model of electric and magnetic fields. How he made the synthesis is looked at in
some detail, as are the beautifully simple confirmatory experiments of Hertz. The Michelson-Morley
experiment is then outlined, as are responses to its failure to yield the expected evidence for the ether.
Finally there is a small taste of Special Relativity theory (a simple treatment of time dilation)
A brief survey of light, electricity and magnetism before
1800
(NEWTON, HUYGENS, GILBERT, GALVANI, VOLTA)
is followed by a more detailed study of…..
YOUNG
FRESNEL
ØRSTED
AMPÈRE
FARADAY
synthesis
MAXWELL
HERTZ
MICHELSON
EINSTEIN
GCE AS and A PHYSICS Teachers' Guide 62
3.
Serving Suggestions
All the material to be tested in the examination is contained in the 34 sides of WJEC notes, which are
available in electronic form from the Physics section of the WJEC website <link to be inserted> or as
hard copy from the WJEC subject officer.
The notes contain many self-test questions and could be used by a student for self-study. They are
also peppered with links to websites which help to bring the basic material of the option alive and
make it easier to learn. The sites often contain pictures and diagrams.
The first half of the course (Young, Ørsted, Ampère, Faraday) deals largely with concepts in light and
electromagnetism which are key parts of the non-optional A-level specification, but comes at them
from a different angle, adding ‘human interest’, and (obviously) a historical perspective. The result
should be reinforcement. A possible teaching strategy is to integrate the material of the first part of
this option with the normal teaching of the relevant topics. The second half of the material might lend
itself to self-study with lessons on specific topics, such as time dilation.
4.
Extracts from writings of Young, Faraday, Maxwell, Hertz, Einstein
The extracts contained in the WJEC notes are short but they do give the student something
approaching direct contact with great physicists of the past. They are supported by explanatory notes
and self-test questions to help with understanding. In the examination, part of the Option B question
might present the student with a snippet from one of the extracts and ask him or her to explain certain
points, or to put the extract in its historical context.
Those who associate studying history with the enforced learning of dates need not have too many
fears about this option. Placing discoveries in the right half decade will suffice.
5.
Books
Two thinnish and very readable books which provide good support are…
Michael Faraday and the Royal Institution: by John Meurig Thomas (ISBN 0-7503-0145-7).
Relativity and its Roots: by Banesh Hoffmann (ISBN 0-486-40676-8).
Chapter 4 tells pretty much the same story as this course, but, as the book’s title makes clear,
Hoffmann has a special agenda, and his emphases are different.
Examination questions are restricted, however, to the material in the WJEC notes (though the student
will be assumed to have tackled the embedded self-test questions – which occasionally call for him or
her to find facts elsewhere).
GCE AS and A PHYSICS Teachers' Guide 63
Unit PH5 Option C – Materials
General
The optional unit is designed to provide students with a sound understanding of the behaviour
of materials. Emphasis is placed on understanding the relationship between the physical and
mechanical properties of metals, glasses ceramics, polymers and composites and the
microstructure of these materials.
The unit also touches upon the use of advanced materials such as superalloys and carbon
fibre reinforced plastics and their applications.
It is hoped that the unit will provide students who have an interest in pursuing a career in
engineering or material science with an increased technological awareness of the field of
materials and to alert them to the possibilities that the future holds in this rapidly advancing
and increasingly important field.
Much of the unit is covered in sufficient depth in standard A-level texts. This coverage is
dealt with in Table 1. The text books referred to are detailed below the table. Further useful
material is to be found in the following:
1. Cooke (B) and Seng (D) (1989). Physics of Materials for A-level students (2nd ed).
Leeds. University of Leeds.
2. Advanced Physics project for independent learning (APPIL) – unit: Behaviour of
matter. John Murray.
3. Easterling (K) (1990). Tomorrow’s materials. London Bourne Press.
4. Gordon (J.E.) 1978. Structures, or living things don’t fall down. Great Britain.
Penguin Books.
5. Gordon (J.E.) 1976. The New Science of Strong materials or why you don’t fall
through the floor. Great Britain. Penguin Books.
Useful web pages. The links are active in the electronic version of these guidance notes and
every attempt will be made to ensure their current validity.
1) The Macrogalleria- a cyberwordland of polymer fun.
www.pslc.ws/mactest/index.htm
2) Stress-Strain curves.
www.shodor.org/~jingersoll/weave/tutorial/tutorial.html
3) www.s-cool.co.uk
4) www.antonine-education.co.uk
(Follow links to AS Physics and then to Module 3, Topic 6 – elastic properties of
solid materials).
5) www.schoolphysics.co.uk
GCE AS and A PHYSICS Teachers' Guide 64
Table 1.
Selected
Book
Reference
Specification
Reference
a
Adams & Allday
Duncan
10.3
p. 17 – 19, 24
b
c
d
e
3.8 (p.60 – 61)
10.1
f
10.2 (p. 440 –
441)
10.2 (p. 440 – p. 28 – 29
441)
10.6 (p. 448)
p. 31 – 34
g*
h*
p. 29
p. 30
p. 39
Advanced
Physics
for Muncaster
You
p. 288 – 289,
291
p. 282
p. 284 – 285
p. 286
p. 287
9.10 (p. 147 –
151)
11.2 (p. 182)
p. 183 – 184
11.5 (p. 184 –
185)
11.12, 11.13 (p.
192 – 193)
i*
j
k*
l
m*
n
o*
p*
q
r
10.11 (p. 458
459)
10.6 (p. 448)
10.2 (p. 441)
10.10 (p. 456
457)
10.7 (p. 450)
10.8 (p. 452
453)
10.8 (p. 452
453)
10.9 (p. 454
455)
10.10 (p. 456
457)
– p. 35 – 36
p. 290
p. 33
p. 28 – 29
– p. 34 – 35
p. 289, 290
11.10,
11.11
(p.191)
11.3 (p. 193)
p. 190
p. 290
– p. 41
p. 291
p. 291
– p. 41
p. 291
– p. 37
p. 293
11.8, 11.9 (p.
189 – 190)
p. 189
–
p. 38
p. 294
* Further guidance given in the Teacher Guidance Notes or in associated documents.
Texts referred to in the Table:
1. Duncan G.T. (1987) Physics, A textbook for Advanced Level Students (2nd ed).
London. John Murray Ltd..
2. Adams (S) and Allday (J) (2000). Advanced Physics. Oxford. O. U. Press.
3. Johnson (K) et al (2000). Advanced Physics for You. U. K. Nelson Thornes.
4. Muncaster (R) (1993). A-level Physics. Cheltenham. Stanley Thornes.
GCE AS and A PHYSICS Teachers' Guide 65
Detailed guidance
Specification references (a) – (f), (j). (l), (n), (q) and (r) are treated in sufficient detail in
standard A level text books to obviate the need for guidance in this document.
Elastic and Plastic Strain
The process of deformation of ductile materials, including the movement of edge
dislocations, is treated at the molecular level in the WJEC document:
The plastic behaviour of ductile metals.
SUPERALLOYS
Statement (i) draws upon statements (h) and (j). Candidates should have an understanding of
the effects of dislocations at the molecular level, and the strengthening and stiffening of
materials by the introduction of dislocation barriers such as foreign atoms, other dislocations
and grain boundaries (specification statement (h)). Candidates should also be able to describe
failure mechanisms in ductile materials and have an understanding of creep and fatigue
(specification statement (j)).
Introduction
Aircraft jet engines are required to operate within extreme conditions of temperature and
pressure. Jet engine turbine blades rotate at a typical speed of 10,000 rpm for long periods in
an environment of combustion products at working temperatures of 1250ºC (though the inlet
temperatures of high performance engines can exceed 1650ºC); non aviation gas turbines
operate at approximately 1500˚C. The blades must be able to withstand impact and erosion
from debris drawn in with the air stream. In addition, different parts of the blade may be at
different temperatures and they will be subjected to large and rapid temperature changes
when the engine is started up and turned off.
The following is a list of the properties required of the material from which the blades are
made:
 Creep Resistance
Centripetal forces acting on the blade at high rotational speeds provide a considerable
load along the turbine blade axis. Over prolonged periods of time this can cause creep. It
becomes increasingly pronounced as temperature increases. Creep could cause a turbine
blade to deform sufficiently that it might touch the engine casing.

Corrosion Resistance
Iron corrodes to form rust. At high temperatures, the presence of carbon dioxide, water
vapour and other products of the combustion of fuel constitute a highly corrosive
environment.

Toughness
The blades must resist impact with debris passing through the engine. In addition,
stresses generated by expansion and contraction, between different parts of the blade at
different temperatures, must not give rise to cracking.
GCE AS and A PHYSICS Teachers' Guide 66

Mechanical and Thermal Fatigue Resistance
Variations of gas pressure and temperature on different parts of a blade and mechanical
vibrations may generate cyclical stresses which can cause failure due to fatigue.

Metallurgical Stability
The mechanical properties of metals can be modified by heat treatment. Blade materials
must be resistant to such changes and the microstructure must remain stable at high
temperatures.

Density
The density must be low to keep engine weight as low as possible.
The separate document “Superalloys” <link to be inserted>, which can be found in the GCE
Physics section of the WJEC website, gives details of how engineers have worked to produce
single crystal turbine blades which satisfy these design criteria.
HEAT TREATEMENT OF METALS
Strength and hardness are two mechanical properties of a solid metal which are affected by
crystal grain size. The smaller the grains, the stronger a material is- fracture is more difficult
with small grains because there are more grain boundaries and dislocations (moving within
one grain) have difficulty passing into adjoining grains. The atomic planes of adjoining grains
are in different directions so fractures tend to be halted at grain boundaries. The more
boundaries there are, the stronger the material is. The mechanical properties of metals can be
controlled by the following common treatments:

Work hardening.
This is a process which makes a metal stronger. The metal is worked or deformed (by
hammering or repeated bending) when cold to make it stronger and harder. The effect of
working the metal is to increase the number of dislocations, so increasing its strength. The
effects of work-hardening can be felt by bending the wire of a steel coat hanger
backwards and forwards until it snaps.

Quench hardening.
Suggested experiment: Heat one end of a 20cm (approx) length of steel wire (held with
tongs) in a Bunsen flame until it becomes cherry red in colour- about 800ºC. Then plunge
the hot end of the wire quickly into cold water. When the rest of the wire has cooled try to
bend the quenched end. What do you notice?
Rapid cooling ‘freezes’ a particular grain structure into the metal. The higher the
quenching temperature the smaller the grains and the harder and more brittle the resulting
metal.

Annealing.
Suggested experiment: Use the same sample of wire as above and heat the other end of
until it is red hot, and keep it at red heat for about 15 seconds. Withdraw it from the heat
very slowly so that it cools gradually. When cool, try to bend the annealed end. What do
you notice? This experiment can be carried out using a length of copper instead of steel.
Slow cooling allows grains to grow larger, making the metal softer, more easily bent,
hammered or scratched.
GCE AS and A PHYSICS Teachers' Guide 67
Griffith cracks and brittle failure of amorphous solids
A. A. Griffith, investigating the breaking stress of glass in the 1920s, estimated that its value
should be about 1010 Pa. The glass under study had a breaking stress of only 108 Pa but he
found that very thin fibres had much higher breaking stresses, with fibres of diameter 10 -3
mm having a breaking stress of 3  109 Pa. Thus, the breaking stress varies with the diameter
of the glass rod and approaches the theoretical breaking stress as the diameter of the rod
decreased. The results obtained are sketched in the following graph.
6
Strength/
Arbitrary
units
1
10
20
Diameter/ μm
The glass fractures by a process known as brittle fracture. This is accelerated by the
presence of surface imperfections or cracks.
This is shown in the diagrams on the next page. The stress becomes concentrated around the
tips of a crack. Bonds near the crack will break, increasing the load on neighbouring bonds
which are still intact, causing them to break and the crack propagates rapidly [at
approximately the speed of sound in the glass.
Crack
Key: The pecked
lines like this are
called stress lines.
They represent the
way the tensile load
is transmitted along
the rod from bond to
bond. Notice how the
lines are
concentrated near the
tip of the crack.
GCE AS and A PHYSICS Teachers' Guide 68
Stress lines and stress concentrations can be photographed by making specimens out of
Perspex and stressing them between crossed polaroids. The pictures on the next page show
this. The picture on the left shows stress lines in a uniform bar which is stressed. The bar on
the right has a small crack half way up its left side, resulting in a high concentration of stress
around the tip.
This diagram represents the atoms and bonds around the tip of a crack:
The force in bond A will be large since it has to balance the
forces exerted on molecules X and Y from above and
below.
The top two lines are incomplete because of the crack, so
that the stress they carry is transferred to the line of atoms
below. The bond A at the bottom of the crack is therefore
carrying a much higher stress than the rest of the bonds.
The stress can exceed the breaking stress of the material
only in this region causing the bond to break, increasing the
size of the crack and also the stress concentration. The
crack will therefore propagate quickly through the material
causing it to fracture.
In the case of the glass fibres, surface cracks are caused
among other things by differential cooling at the surface and in the centre. The narrower the
thread, the more uniform the temperature and so the less significant are any cracks that form.
This makes the small diameter rods much stronger. The very narrowest glass threads [~ 1
m] approach the theoretical strength predicted by Griffith. For very narrow threads,
inducing cracks by simply touching the surface brings their tensile strength back to that of the
bulk glass.
This property of brittle materials is exploited by glaziers when “cutting” a piece of glass to
size by putting a scratch in it and then snapping it – similarly with tiles.
Since amorphous solids break by brittle fracture, they will be weak under tension, but under
compression they will be very strong as the stress will cause the cracks to close preventing
propagation. When amorphous solids such as brick are used for building, the structures
produced are strong provided the material is kept under compression.
GCE AS and A PHYSICS Teachers' Guide 69
Experiments to investigate the strength of glass.
1) Heat the end of a glass rod so that it softens enough to bend it into a hook shape to support
weights. When cooled, heat the centre of the rod until soft, then remove the rod from the
flame and quickly draw it out into a fibre. (An alternative method for producing fibres is to
hang a 100g mass from the hook, heat the centre of the glass and as the glass softens and
begins to fall, immediately withdraw the heat. The mass should be allowed to fall through a
distance of about 0.5m to 1.0m). Then clamp the straight end so the fibre hangs vertically.
Load the fibre with weights until it snaps. Note the final weight and, using a micrometer,
measure the diameter of the broken fibre. Collating class results would allow graphs of
breaking force or breaking stress versus diameter or cross-sectional area to be drawn.
2) Support a glass rod horizontally at either end. Load its centre with weights until it snaps.
Repeat the test with the glass marked with a glass cutter underneath at the middle. The cut
rod should snap more easily.
3) Heat a glass rod and pull out a fibre about 0.5 m long (as in experiment 1). When it is cool
bend it into an arc on the bench. Release it, run your fingers over the fibre and bend it again.
The fibre should now snap more easily.
THE BEHAVIOUR OF RUBBER & POLYTHENE
A polymer macromolecule consists of long chain molecules, each containing up to 105 atoms
and these chains are held together by cross-bonds, the structure being similar to that shown in
the diagram.
Diagram 1. A possible arrangement for three long-chain rubber molecules.
The behaviour of a polymer depends on the strength of the cross-bonds and examples of two
types of polymer are considered below.
(i) Rubber – an example of polymer with weak cross bonds. Natural rubber is a polymer
of the molecule iso-prene. It has weak van der Waals cross-bonds and only a few covalent
(strong) cross-bonds. Its behaviour under increasing stress is shown by the following graph.
B
stres
s
A
O
strai
n
GCE AS and A PHYSICS Teachers' Guide 70
Between points O and A, the deformation is elastic and from the slope of the graph it can be
seen that, after an initial stiff region, Young’s modulus is small. At stresses greater than A,
the deformation is still elastic, but the value of Young’s modulus is much greater. If suffi
iently stressed, the material breaks.
Initially, as the rubber is deformed, no bonds are extended; the long chain molecules are
straightened against their thermal motion (which tends to increase the amount of folding in
the molecular chains). The van der Waal’s bonds are responsible for the initial stiff region,
but once they are overcome, the rubber molecules unfold and the material can extend by
several times its original length. Because bonds are not being broken here, the additional
stress needed to do this is small. The structure of the material changes as shown in the
following diagram.
_______ molecule 1
- - - - - - molecule 2
Diagram 2. Stretched rubber
At point A, the molecules the sections of the molecules which are free to unwind are more or
less straight, therefore if any further extension of the material takes place, bonds are
stretched. This is far more difficult to do than straightening the molecules, therefore the
value of Young’s modulus increases at this point.
When the stress is removed the thermal motion of the chain molecules makes the polymer
return to its original dimensions. The value of Young’s modulus for such a polymer
increases with temperature, the opposite to the variation in crystalline and amorphous solids.
This is due to the fact that the chain molecules have to be straightened against their thermal
motion. As the temperature increases, the thermal motion increases the amount of folding, so
that the average end-to-end distance in an individual molecule decreases, with the result that
straightening the molecules becomes more difficult.
These polymers also show elastic hysteresis i.e.
the stress-strain curves for the loading and
unloading do not coincide. This is shown in
this graph:
B
stress
A
OAB is the stretching curve and BCO the
C
contracting. The strain for a given stress is
greater when unloading than for loading. The
strain
unloading strain can be considered to ‘lag
O
behind’ the loading strain.
The area under OAB represents the work done [i.e. the energy supplied] to cause stretching;
similarly the area under BCO represents the energy given up by rubber during contraction.
GCE AS and A PHYSICS Teachers' Guide 71
This closed curve is called a hysteresis loop; its area is the energy per unit volume converted
into internal energy [or, ‘lost as heat’ in common parlance] during one cycle. Thus when a
polymer is repeatedly stressed, its temperature increases. Rubber with a hysteresis loop of
small area is said to have resilience. This is an important property where the rubber
undergoes continual compression and relaxation as does a car tyre when it touches the road
as it rolls on. If the rubber used did not have a high resilience, there would be appreciable
loss of energy resulting in increased petrol consumption and lower maximum speed. Heat
build up could even lead to tyre disintegration.
Not for examination: Chemical engineers alter the properties of natural rubber by the process of
vulcanization, in which strong covalent bonds are deliberately introduced between the long
molecules. This has the effect of making the rubber stiffer and increasing its resilience. This very
stiff form of rubber is useful for applications which involve repeated deformation, e.g. car tyres.
Polymers exhibit a property called creep during which the chain segments slowly disentangle
under a constant stress as a consequence of the thermal motion of the chain segments.
On the release of stress, thermal motion restores the mixing, but slowly, since the segments
get in each other’s way during the shuttling process.
(ii)
Polythene- a semi crystalline polymer.
Polythene is a thermoplastic polymer which contains both crystalline and amorphous
regions. Since these regions co-exist within one structure, polythene can be described as a
semi-crystalline material.
In the crystalline regions polymer chains composed of covalent C-C bonds fold together
many times in an ordered arrangement as shown.
Force
Force
These lamellae, as they are called, (from the word laminate) form small, spherical grains
called spherulites. [Note: The crystalline regions of a thin film of high density polythene
can be observed using polarised light and a microscope]. Between these crystalline grains,
the polymer molecules are tangled together in an amorphous state, where little long range
order exists. The graph below shows how the stress varies with the strain for a strip of
high density polythene when it is stretched.
GCE AS and A PHYSICS Teachers' Guide 72
Between O and A, the strip can regain its original length as its behaviour is elastic. The
parallel parts within the lamellae crystal are held together by weak van-der-Waals bonds
and at low strains these bonds resist the applied stress.
From A to B a ‘neck’ forms in the strip as the tangled molecules in the amorphous
regions start to align with each other.
From B to C the width of the neck remains unchanged as the strip is extended. In this
region the van-der Waals forces between the lamellae are overcome and they begin to
unfold and become parallel to each other.
The strip is now said to be cold-drawn and is very strong along the axis of the applied
stress. Beyond C, the stress is resisted by the strong covalent bonds between the carbon
atoms within each polymer molecule.
Investigating force–extension curves for rubber and polythene.
In these experiments you will investigate how rubber and polythene behave under tension. It
is not intended that you should obtain accurate values for the mechanical properties of these
materials, but basic quantities such as the elongation at fracture and the breaking strength
may be determined from the force-extension or stress-strain curves.
(a)
(i)
Rubber Band (cross-linked polymer).
(1)
Hang a (cut) rubber band of (approximate) cross-section 1mm by 2mm
vertically from a stand, boss and clamp (Hoffmann clips are useful
here to suspend the rubber band). The base of the stand should be
secured using a G-clamp. Attach another Hoffmann clamp to the other
end of the rubber band and use this to hang a 50gramme mass holder
from. Place a metre rule as close as possible to the mass holder. The
extension may be read using an optical pin attached to the base of the
mass holder.
(2)
Measure the length, width and thickness of the rubber when it is
supporting the 50gramme holder. Try to avoid squashing the rubber
with the micrometer screw gauge.
GCE AS and A PHYSICS Teachers' Guide 73
(ii)
(3)
Increase the mass in 50gramme steps. Depending on the thickness of
the rubber, you may need to change the smaller masses for a single
0.5kg mass in order to exceed the elastic limit. Continue to add
50gramme masses until the rubber band breaks.
(4)
Plot the force extension curve and determine the Young Modulus from
the linear section.
Natural Rubber.
A similar experiment can be carried out with a natural rubber strip, (linear
polymer chains with little or no cross-linking) cut so its width is about 5mm.
follow the procedure as for the rubber band, increasing the mass in 5ogramme
steps up to 500grammes. You will probably need to cater for an extension
greater than 1.0 metre. From the force-extension curve you should be able to
identify two regions which are approximately linear. If this is possible,
calculate two values of E for natural rubber.
(b)
(i)
(ii)
Low density Polythene.
(1)
Cut a strip of polythene about 1 metre in length from a thin polythene
bag. You may need to fold the polythene several times in order to get
measurable thickness using a micrometer screw gauge. You can then
calculate the thickness if a single sheet.
(2)
Increase the mass, initially in 50gramme steps, measuring the
extension for each mass added. The polythene, at some stage, will
increase in length without any further increase in load (A to C in above
graph). If you plot a stress-strain graph as you go along it is possible to
find an accurate value for this increase in extension. Using 5gramme
and 10gramme masses will also help you identify the point where this
region begins. When this region ends (beyond C), larger loads will be
needed to produce any further extension.
High density polythene.
Cut a strip of high density polythene from the rings used to hold cans of drink
together (a good excuse here for buying beer!!). Mark the strip at two points
along its length and, using the above procedure, slowly stretch it as much as
possible without breaking it. After the neck forms, observe how it lengthens
without becoming narrower. The strip is now cold-drawn and you can measure
the sample’s breaking stress.
GCE AS and A PHYSICS Teachers' Guide 74
Unit PH5 Option D – Biological Measurement and Medical Imaging
General
The optional unit deals with several different medical imaging and measurement techniques:
 X-rays including production, absorption, use in diagnosis and therapy and CAT
scans.
 Ultrasound including acoustic impedance and Doppler techniques.
 The principles of Magnetic Resonance Imaging including comparison with X-rays
and Ultrasound for imaging.
 ECG including the interpretation of the cardiogram.
 Nuclear imaging including radiation dose and PET scanning.
All these techniques build upon the physics content of PH2, PH4 and PH5. The study of this
option provides an ideal synthesis of many of the ideas developed especially in the A2
course.
X-rays
The nature and properties of X-rays, and their production in an X-ray tube are covered in
PH2.3 (e) and (f). The following diagram describes their production in sufficient detail.
Lead
shielding
Very high
voltage
Copper/
Tungsten
block (anode)
Heater
Curren
t
supply
Heater
filament
Electron
beam
Focussing
anode
X rays
Cooling fins
The heater ‘boils off’ electrons by thermionic emission. These are then accelerated to very
high velocities by the p.d. between the heater filament and anode. They are collimated by the
focussing anode. The tube is evacuated so the electrons travel in straight lines and collide
with a tungsten target (the anode) embedded in a copper block. The resulting deceleration
produces an enormous amount of heat (up to 99% of the energy input) and also X-rays, which
emerge from a window in the lead housing.
A continuous spectrum, then, can be obtained by electrons decelerating rapidly in the target
and transferring their energy to single photons. This radiation is known as ‘Bremsstrahlung’
or braking radiation. Superimposed on the continuous spectrum are several sharp lines. These
result from the bombarding electrons knocking out orbital electrons from the innermost shells
of the target atoms. Electrons from outer shells will then make transitions to fill the gaps in
the inner shells, emitting photons whose energies are characteristic of the target atom.
GCE AS and A PHYSICS Teachers' Guide 75
Transitions into the K shell give rise to K lines, the L shell L lines and so on. For heavy metal
targets the resulting photons are in the X-ray range. A typical intensity spectrum would be:
1.0
0.5
λmin
0.05
0.1
0.15
The intensity of X-rays is defined as the energy per second per unit area passing through a
surface. This can be increased by increasing the voltage of the X-ray tube, or by increasing
the current supplied to the filament. The photon energy is also determined by the tube voltage
with the maximum photon energy being given by:
hc
Emax 
 eV ,
min
where V is the tube voltage. The optimum photon energy for radiography is around 30 keV
which is obtained using a peak tube voltage of 80 – 100 kV.
A narrow beam of X-rays is preferred as this reduces scattering and so leads to a sharper
image. Blurring can also occur due to scattered radiation. This can be reduced by introducing
a grid directly in front of the detector. This grid consists of a large number of lead strips so
that only primary or direct radiation will get through to the film.
X-ray beam
Patient
Scattered
Radiation
Transmitted
Radiation
Grid
photographic
film
GCE AS and A PHYSICS Teachers' Guide 76
When X-rays pass through matter they are absorbed and scattered and therefore the beam is
attenuated. This attenuation can be calculated using the equation:
I  I 0e  x
Where I = intensity at a depth x
Io= intensity at the surface
µ = the attenuation coefficient
Note that the half value thickness, x½ can be calculated using
ln 2
.
x12 

This can be derived in the same way as the half life equation in radioactivity.
The main absorption mechanism of X-rays in the body is the photoelectric effect. The X-ray
photon is absorbed by an electron which then leaves the atom. This is more efficient for
atoms with larger numbers of electrons i.e. higher atomic numbers. Consequently denser
materials such as bone will absorb more X-ray photons than less dense areas such as soft
tissue. This will lead to a large contrast between bones and soft tissue and therefore a sharp
image. If there is not a great contrast between the areas of the body being studied then
sometimes a contrast media is used, e.g. a barium meal when studying the stomach or
intestines.
Computed tomography (CT or CAT scan) also uses X-rays, but in this case the X-ray tube
moves in a circle around the patient taking images of the body at all different angles. A
computer combines these images to produce a cross sectional image of the body. By adding
these slices together a 3-D image can be produced.
CT scans are very quick to produce and show a wide range of different tissue clearly. They
do however subject a patient to a high dose of radiation and the machines are very expensive.
Ultrasound
Ultrasound can be generated using piezoelectric crystals. If you apply an alternating p.d.
across the crystal you cause it to become deformed, with the crystal vibrating at the same
frequency as the applied p.d. This can be used to generate ultrasound. The process also works
in reverse, with the crystal receiving ultrasound and converting it to an alternating p.d. The
crystal, then, can act as both an emitter and a receiver of ultrasound.
Ultrasound can be used in diagnosis in two different ways
1. A-scans, or amplitude scans, where a short pulse of ultrasound is sent into the body and a
detector (usually connected to a C.R.O.) scanning for reflected pulses. Using the time
base, the time the ‘echo’ takes to return can be found and the distance between structures
in the body can calculated. A-scans are usually used when the anatomy of the section is
well known but the precise depth is needed e.g. a delay in measuring a known position in
the brain could indicate the presence of a tumour.
2. B-scans, or brightness scans in which the reflected pulse is displayed by the brightness of
the spot. If an array of transducers is used a 2-D image can be built up. This is widely
used to assess the health and growth of a prenatal foetus.
GCE AS and A PHYSICS Teachers' Guide 77
Acoustic impedance is defined by the equation
Z  v
Where Z = acoustic impedance of the medium
Ρ = Density of the medium
v = Speed of sound in the medium
The acoustic impedance determines how much energy (ultrasound) is reflected at a boundary.
If two materials have a large difference in Z then a lot of the energy will be reflected back,
but if there is little or no difference in Z then there will be little or no reflection.
For soft tissue such as skin the acoustic impedance is very different from that of air, so, if an
ultrasound probe were simply placed upon the exterior of the body, most of the ultrasound
would be reflected rather that entering the body. To prevent this, a coupling medium such as
a gel or oil which has a Z value much closer to that of soft tissue must be first applied to the
patient. This removes the thin layer of air and enables a much greater percentage of the
ultrasound to enter the body.
Blood flow around the body can be studied using a Doppler Ultrasound probe. The Doppler
Effect (PH4.5) is the shift in frequency of a wave when it received by a moving object (either
towards or away from the source). It also occurs when a moving source sends out a wave. A
continuous source of ultrasound is sent out and its echo received back off a moving object
such as blood cells. Because the blood cells are moving, the Doppler shift is doubled: once
for the blood cells receiving the ultrasound and once for the blood cells acting as transmitters
back to the stationary receiver. As the ultrasound is sent out continuously, two transducers are
needed: one to produce the ultrasound; the other to receive it. By analysing the shift in
wavelength of the received wave ∆λ and comparing it to the initial wavelength λ the velocity
can be calculated using the equation:
 2v
,


c
where c = the speed of the ultrasound wave. Note the 2 in the equation.
This technique will show up any changes in the blood flow through a vein or artery and so
can be used to detect clots or thrombosis.
Magnetic Resonance Imaging [MRI]
Nucleons (protons and neutrons) possess spin, which makes them behave like small magnets.
Usually these will cancel each other out. However a hydrogen nucleus only has one proton
and it is this nucleus that is studied using an MRI scan. Under normal conditions the
hydrogen nuclei will be randomly arranged and cancel each other out. However if a strong
magnetic field is applied they tend to align themselves, in almost equal numbers, either with
the field lines (in parallel) or exactly opposite to the field (antiparallel).
The nuclei are in continuous motion, due to thermal energies, and will all wobble or precess
around the field lines at the same frequency (called the Larmor frequency)
If radio waves are directed at the hydrogen nuclei at the same frequency as they are
precessing (Larmor frequency), they will resonate and flip from one alignment to another so
producing a magnetic field. When the radiowaves are switched off the nuclei revert back to
their original state giving off electromagnetic radiation. It is this signal that is detected by the
scanner. The time taken for the nuclei to switch back is called the relaxation time, and
depends on what tissue type the nuclei are in. By measuring the various properties of the MRI
signal along with the relaxation time a detailed image of a cross section of the body can be
built up.
GCE AS and A PHYSICS Teachers' Guide 78
Magnetic resonance imaging is particularly good in obtaining high quality images of soft
tissue such as the brain, but is not as good for harder objects such as bone.
Some of the advantages and disadvantages of X-rays, ultrasound and MRI for examining
internal structures can be summarised by the following table:
Technique
Advantages
Disadvantages
X-rays
X-rays are absorbed by bone
and so produce good shadow
images.
Unlike ultrasound they can
produce images of, e.g. cancer
on the lungs.
High radiation dose for the
patient.
People working with X-rays
need to take care to limit their
annual dosage.
Ultrasound
No known side effects.
Good quality images of soft
tissue.
Moving images can be
obtained.
Machines are relatively cheap
and portable
Doesn’t penetrate bone and
so cannot study the brain.
Cannot pass through air and
so cannot study the lungs.
Low resolution.
MRI
No Known side effects.
High quality images of soft
tissue.
Image can be made for any
part/orientation of the body.
Images of hard tissue such as
bone are poor.
Uncomfortable
for
the
patient,
causes
claustrophobia.
Very expensive.
GCE AS and A PHYSICS Teachers' Guide 79
Electrocardiograph (ECG)
The heart is a large muscle which acts like a double pump. The left hand side receives
oxygenated blood from the lungs and pumps it to the rest of the body. The right hand side
pumps the blood returning from the body, at low pressure, to the lungs.
The heart typically beats at between 60 and 100 times a minute. Each beat is triggered by a
pulse starting in the upper right region by a cluster of cells called the sinoatrial node. This
signal spreads through the atria causing them to contract, forcing blood into the ventricles. A
short time later the electrical pulse reaches the ventricles causing them to contract forcing
blood out of the heart. There is a one way valve between the atrium and the ventricle to
ensure the blood flows the right way.
GCE AS and A PHYSICS Teachers' Guide 80
A typical ECG recording consists of three parts:
1. The P wave which occurs because of the contraction of the atria;
2. The QRS wave which is due to the contraction of the ventricles;
3. The T wave which is due to the relaxation of the ventricles.
It is the signals from the sinoatrial node which are detected by the ECG. This signal is very
weak by the time it reaches the body’s surface. The electrodes ( to detect the signal) need to
be placed on the limbs and chest, where the arteries are close to the surface, and also need to
be attached with conducting gel after all body hair has been removed. The signals also need
to be heavily amplified once received.
Any deviation from a ‘normal’ ECG indicates some form of cardiac disorder. This can be
used to look for muscle damage (heart attacks), irregular pumping (arrhythmia), blockages in
the heart due to disease and the heart going into fibrillation, where the beating is fast and
irregular.
Nuclear Imaging
The effects of α, β and γ radiation on living matter will be examined in the same depth as in
PH5.4.
The unit of absorbed dose for matter is the gray (Gy), where
1Gy = 1 joule per kilogram
Because ,  and  radiation interact differently with living tissue, the gray does not
adequately describe the effects of the different radiations. 1 mJ kg1 of  particles has a much
greater biological effect that X-rays or -rays because the radiation is so heavily ionising and
therefore the range of the  particles is so low. To allow for this, a quality factor Q is used:
For many tissues a value of Q = 20 is used for  particle and Q = 1 for ,  and X-radiation.
Different tissues also respond differently and hence different Q-factors are employed by
medical physicists. With this quality factor, the quantity is referred to as the “dose
equivalent” and its unit, the sievert (Sv).
For  and X-rays the dose is equal to the dose equivalent. In general:
Dose / Sv = Q  Dose equivalent / Gy
In examination questions, the value of the Q-factor will always be given.
A gamma camera uses tracers to produce images. A common nuclide used is technetium 99. This is attached to a molecule that will be taken up by the tissue to be studied. It has a
half life of about 6 hours, which is long enough to allow the nuclide to be transported to the
site of interest and for the radiation to be studied but short enough so that the nuclide not stay
active inside the patient for too long.
Once the isotope has been absorbed by the body the gamma rays are directed towards a
crystal (usually sodium iodide with a small amount of thallium added) through a lead
collimator, which consists of a lead circle which has a regular number of holes drilled in it.
This absorbs all the gamma rays that enter the collimator at an angle. The gamma photons
that pass straight through the collimator hit the crystal and cause it to scintillate. The gamma
ray excites electrons in the crystal causing them to give off visible light. This light is detected
by a bank of photomultiplier tubes which build up an image of the levels of gamma radiation
being emitted from different parts of the tissue.
GCE AS and A PHYSICS Teachers' Guide 81
Positron Emission Tomography (PET) Scanning
This requires the use of an isotope that emits positrons (beta+ particles). These will only
travel very short distances before slowing down and attracting an electron from a nearby
atom. These will annihilate each other giving off two gamma rays, of identical wavelength,
which will move off in opposite directions (thus conserving momentum).
The PET scanner detects these gamma rays when they reach a scintillator in the scanning
device. This produces a small flash of light which is picked up and amplified by a series of
photomultiplier tubes in a similar way to the gamma camera. This technique depends on the
simultaneous detection of a pair of gamma photons. Individual gamma photons are ignored.
In order to produce positron emitting isotopes a cyclotron is needed (spec reference PH5.2) as
they have very short half lives. This is very expensive. PET scans are now often combined
with CT scans which enable both soft and hard tissue to be seen clearly.
REFERENCES
Advanced Physics Adams and Allday Oxford press
Advanced Science Physics P. Fuller Heinemann
Medical Physics M. Hollins University of Bath Science Nelson
Medical Physics J.A.Pope Heinemann
WEB REFERENCES
www.wikpedia.org Medical Physics
www.teachingmedicalphysics.org Institute of Physics
www.s-cool.co.uk
www.antonine-education.co.uk
www.medphys.ucl.ac.uk
GCE AS and A PHYSICS Teachers' Guide 82
Unit PH5 Option E – Energy Matters
The aim of this option is to allow students to explore arguably the most pressing topic of our age in
the context of underlying physical principles. After studying this option, students will be in a position
to absorb and dissect information, often contradictory and misleading which is presented in the
popular media and to make informed decisions. Energy already pervades much of the specification.
The Table below identifies these earlier energy-related topics and gives an overview in the centre box
of how they are developed, extended and linked in this option.
PH2.5
BLACK BODY RADIATION.
STEFAN & WIEN LAWS
PH1
MECHANICS
ENERGY
RENEWABLES
TIDAL
HYDROELECTRIC
WIND
STORAGE
SOLAR ENERGY
GREENHOUSE EFFECT
SAVING
(Thermal Conductivity)
STOCKS
HAZARDS
WIDER ISSUES:
(Economic, Social, Political)
WORK FROM HEAT
SECOND LAW
HEAT PUMP
FUEL CELL
PH4
THERMAL PHYSICS
FISSION
Enrichment
Breeding
PH5.5
NUCLEAR
PHYSICS
FUSION
Problems & Prospects
(JET, ITER)
PH1.5
SUPERCONDUCTIVITY
PH5.2
B FIELDS
GCE AS and A PHYSICS Teachers' Guide 83
Electricity Generation from available environmental kinetic energy and gravitational
potential energy – statements (a) – (c).
The enormous world-wide efforts now being made to increase and develop “renewable”
energy provision are the result of two factors: first, fossil fuels, on which we now chiefly
depend, are running out; secondly, these fuels are causing potentially catastrophic
environmental damage. Renewable in this context simply means “does not run out” within
the life times of human civilisation.
Renewable sources and storage systems of the kind listed in (a), (b) and (c) largely involve
applications of basic physics developed at AS (PH1 Mechanics and Energy). In the case of
wind power, the maximum theoretical power Pmax available is determined by the rate of
kinetic energy transfer through the turbine and candidates should be able to deduce (but not
remember) Pmax = ½Av3 and draw conclusions. The power derived in practice is much
below the theoretical maximum due to not all the KE being transferred, losses in the turbine
etc. For tidal and hydroelectric sources, and also storage systems, estimates can be made
from gravitational potential energy calculations based on simple models.
The importance of projects such as Dinorwig in storing energy at times of low consumption
should be recognised. The importance of renewals is nowhere greater than in the UK where
there are Government targets to increase renewals from the present 4% of our total energy
consumption to 10% by 2010 and 20% by 2020. Of the projects listed in the Specification,
the recently completed Three Gorges (Yangtze) is worth special mention because of its sheer
size, and also because of the social, economic and political implications. This is essentially a
large 1.4 mile dam with 370 miles of headwater flooding 620 square miles and having a
generating capacity of 18.2 GW – originally estimated to supply 10% of China’s energy
needs. More than a million people have been displaced in its construction – an exercise more
easily accomplished in China than in most countries. This is an extreme case; in all major
projects there are negative factors: noise, disruption, use of land, effect on wildlife etc and
almost always there are protest groups.
Though the physics of all the above is straightforward, the handling and conversion of the
various units can be troublesome. The kilowatt-hour conversion 1 kWh = 3.6 MJ is needed
frequently and worth remembering while familiarity with the SI prefixes M, G, T, P and E
will be found helpful. Some National and International data tables give energies in mtoes
(million tonnes of oil equivalent). The conversion is 1 mtoe = 42 TJ approximately and this
will be provided in any question.
GCE AS and A PHYSICS Teachers' Guide 84
Nuclear sources – nuclear equations, fission, breeding and enrichment
Uranium derived from mined ore contains only 0.7% of U235, the isotope which fissions
with thermal neutrons and is the main source of nuclear power. U238, which makes up the
remaining 99.3%, does not fission with thermal neutrons though it does make a small energy
contribution by limited fission with fast neutrons. The ideal fission reactor would be fuelled
by pure U235, but it extremely difficult to separate the isotopes as their masses are so nearly
the same. The current method is by ultracentrifuge. Vessels containing Uranium hexafluoride
gas are spun at enormous angular velocity. Because the two isotopes have different masses
they will experience (slightly) different radial forces with the result that the heavier U238
tend to concentrate on the outside of the vessel and the lighter U235 nearer the axis. The
process is slow and the apparatus complex and costly. In practice reactors running on
“enriched” Uranium, in which the U235 has been increased to just a few percent, are more
efficient and cost effective so that the costs of enrichment are more than recovered. Also,
pure Uranium235, obtained through repeated enrichment, is used in the fission bomb which is
a cause of unease when new nations embark on enrichment programs. The remaining U238
after the U235 has been separated is known as depleted uranium (DU). It is much less
radioactive than the original uranium, because the half-life of U-238 is so long [4.5  109
years] and because of its high density has been used lately in armor penetrating shells amid
some controversy [PH6 resource folder 2003].
U238 can itself be made to produce useful fissionable material by “breeding”. U238 captures
fast (high-energy) neutrons to form unstable U239 which decays to Np239 by  emission
which in turn decays to Pu239 by further  emission . (A useful exercise might be to write
down the equations involved in this sequence given that Z = 92 for Uranium). Pu239 is an
isotope of plutonium which does not occur naturally [or, more strictly, any Pu-239 originally
present in the primordial solar nebula has long since decayed]. It turns out that Pu239 fissions
with thermal neutrons similarly to U235 and can therefore be used as a primary source of
nuclear energy. Reactors which produce Pu239 in this way are known as Fast Breeder
Reactors – fast because fast or high energy neutrons are required to start the process. These
reactors have the materials and moderators so arranged that enough fast neutrons are
absorbed for breeding while sufficient moderation occurs to provide thermal neutrons to
sustain the U235 chain reaction. The eventual separation of plutonium from uranium is
relatively easy (compared with U235 and U238 separation) because they are chemically
different. Note that fission produces radioactive waste which must be safely disposed of (still
a major problem), that there are risks (Chernobyl), that stocks of ore are limited (therefore
non-renewable) but that there are no carbon emissions other than that involved in the initial
building of the necessary infrastructure.
Nuclear fusion
It is important to know the DT (deuterium-tritium) reaction and why it is the most suitable for
terrestrial fusion (timescale and temperature). It is useful to look again at the proton chain in
PH2.5 and to be reminded why this would not work on earth. An excellent account of nuclear
fusion is given on the Joint European Torus (JET) website www.jet.efda.org . The key points
to note are:
 the kinetic energies of the particles must be greater than the Coulomb interaction for
interaction to occur meaning temperatures of 108 K;
 plasma containment at this temperature only possible by a combination of magnetic
fields – the tokamac;
GCE AS and A PHYSICS Teachers' Guide 85


very strong B fields are required to deflect and therefore contain fast particles (see
PH5.2(m) for background);
fields of around 5 T require currents of around 7 MA which are only really achievable
with superconducting coils (see PH1.5 (k) to (p)).
The problems are immense and continuous energy through fusion is still a long way off. But
huge efforts and investments are being made. The International Thermonuclear Experimental
Reactor, ITER ( www.iter.org ), a joint venture by most of the great Nations –including
China, India, USA, Russia, EEC – is underway in France and due to power up in 2016.
Fusion, if it works, will provide all our energy needs. Deuterium is abundant in seawater [~ 1
in 104 hydrogen atom is deuterium] and tritium can be obtained by neutron capture by lithium
in the reactor itself. There are no toxic products.
Heat transfer processes – convection and conduction
These are familiar topics well covered in standard textbooks. Emphasis will be on thermal
insulation and energy-saving, but note that publications on these topics use Thermal
Transmittance, (U value) rather than thermal conductivity K. The U value of a slab of
thickness d is given by
U = K/d.
This will be provided, if required, in any question.
Solar radiation as an energy source
A form of renewable energy quite different from those treated earlier is solar energy. The
background physics has already been developed in PH2.5(a to d). Revision of this material is
a good starting point with emphasis on the laws of Wien and Stefan, the inverse square law
and what is meant by black body.
A key quantity is the Solar Constant – the total radiated power per square metre crossing a
plane perpendicular to the earth-sun radius measured just outside the earth’s atmosphere. The
value is not constant (despite the name), as the earth-sun distance varies over the year, but
averages at 1.35 kW m-2. We can estimate the rate of solar power arriving at earth, ignoring
clouds, atmospheric absorption etc. as of the order of 1017 W and compare this with the rate at
which energy is consumed throughout the world (of the order of 1013 W). So there is
abundant solar energy; the problem is harnessing it effectively. There are two ways:
 solar panels;
 photovoltaic cells.
In the solar panel, water is heated directly from sunlight. The panel contains a flat coil of pipe
connected to the domestic hot water cylinder and is placed, ideally, on a South-facing roof.
As the name suggests, the photovoltaic cell produces electrical energy from solar energy. At
present, photovoltaic cells make very little contribution to our energy because of high
manufacturing cost. Most of the cells currently in use are made of very pure silicon which has
to be doped and cut in a special way – all very expensive.
Not for examination: As semiconductor devices and band theory are not in the specification, this
rough outline may be helpful: the silicon cell consists of n- and p-doped regions forming a p-n
junction. Incident photons excite electrons into the conduction band creating electron-hole pairs
which migrate to form an electric current. Detailed knowledge will not be expected.
GCE AS and A PHYSICS Teachers' Guide 86
Much work is in hand to develop cheaper and more efficient cells using, for example,
composite materials. The Energy Conversion Efficiency of a photovoltaic cell is defined by
the usual efficiency equation, in this case written as:
Conversion Efficiency 
Useful energy extracted
100%
Total energy input
Values range from around 6% for the cheapest commercial cells to around 40% for the most
expensive state-of-art cells. It follows that large areas of cell are required for even moderate
power such as typical domestic consumption, but small cells of around 5 cm2 are sufficient to
power pocket calculators which require less than 1 mW.
Carbon footprint
Candidates can not be expected to remember statistics, but some key figures are worth
bearing in mind. For example, about one fifth of UK electricity is from nuclear reactors and
three quarters from fossil fuels (coal, oil and natural gas). Fossil fuels have two major
drawbacks. First, the stocks are finite. Secondly they produce carbon dioxide gas which is
harmful if allowed to escape into the atmosphere. In a chemical reaction in which carbon is
oxidised (burned), each carbon atom combines with two oxygen atoms from the atmosphere
to form carbon dioxide CO2. Straightforward calculation from the atomic masses shows that
one kilogram of carbon produces 2.66 kg of carbon dioxide.
Carbon dioxide is a “greenhouse” gas and its increased presence in the atmosphere leads to
global warming in the following way:
 The solar radiation spectrum covers a range of wavelengths with maximum power at
around 480nm which is at the blue end of the visible spectrum. This value is
determined by the temperature of the sun’s surface (Wien’s displacement law;
λmax T-1).
 The atmosphere is essentially transparent to this wavelength so the solar energy
passes through and is absorbed at the earth’s surface.
 The earth in turn radiates thermal energy but, because the earth’s surface temperature
is much lower than that of the sun, this peaks at around 10 m which is in the far
infrared region.
 Carbon dioxide absorbs strongly at this wavelength, and re-emits in all directions
including back to earth leading to global warming. Other polyatomic molecules such
as methane and nitrous oxide behave similarly but carbon dioxide is more abundant.
This is how greenhouses heat up – hence the name; glass, like CO2 is transparent in
the visible but absorbs in the IR.
Experiments show that burning 1 kg of carbon produces about 13 kWh of energy which
works out at around 6 eV per carbon atom. It is interesting to compare this with the 200 MeV
produced by the fission of one U235 nucleus.
The consequences of increasing greenhouse gases need to be recognised: global warming,
polar icecap melting, weather changes, flooding, more hurricanes etc. Also important is the
decline of vegetation, particularly the rain forests, which remove CO2 from the atmosphere
through organic growth. Important too is recognition that the worlds population is increasing
as is the industrialisation (and hence energy requirements) of emerging nations: also to be
noted are the measures to counter the ill effects (national and international reduction targets,
Kyoto protocol, “carbon footprints”, Environmental Impact ratings etc).
GCE AS and A PHYSICS Teachers' Guide 87
Fuel cells
Interest is reviving in fuel cells as they offer the possibility of efficient and environmentally
friendly energy production, especially for transport, and could become a replacement for the
internal combustion engine when the oil runs out. Prototype cars powered by fuel cells
already exist. The fuel cell is electrolysis in reverse. As electrolysis has long disappeared
from physics specification, a brief outline is necessary.
When an electric current is passed through water, ionization of the water molecules occurs
through collision with the charge carriers. Avoiding the detailed chemistry, the upshot of this
is that water is broken down into its constituent gases with hydrogen bubbles collecting at the
cathode and oxygen at the anode. The apparatus is simple: a dish of water with two electrodes
each with an inverted jar above it to collect the gases, and a current source. A drop of acid is
needed to make the water conducting and the anode made of platinum to avoid oxidation.
So, in summary, electrical energy breaks water down into oxygen and hydrogen gases. Can
the reverse take place in which hydrogen is recombined with oxygen to provide electrical
energy? Fuel cells do just this. The process is complex but in crude outline the following is
what happens in the Polymer Electrolyte Fuel Cell (PEFC):
 Hydrogen is supplied to the anode where the atoms are ionized by a catalyst.
 A polymer electrolyte then routes the electrons to the cathode via an external circuit
forming a useable electric current.
 The protons continue through the polymer electrolyte to the anode where they
recombine with the electrons, and the hydrogen reacts with oxygen, which is fed
directly into the anode, to form water.
One great advantage is that there are no damaging products – particularly no carbon dioxide.
Another great advantage is that there is no heating – useful energy is not being obtained from
heat; this will be returned to later after Heat Engines. Also the cell can be connected directly
to electric motors on drive wheels of cars, cutting out the heavy and inefficient engines
(cylinders, reciprocating pistons etc) of traditional cars. Hydrogen of course is a hazard and
there are difficulties over its delivery and storage, but its supply will become plentiful if
fusion succeeds – hydrogen and oxygen produced by electrolysis of water from electricity
from turbines powered by fusion reactors – but this is a long way off. At present, the
experimental fuel cell cars are extremely expensive and the electrolyte polymer degrades and
has to be replaced within the lifetime of the car.
A useful website, albeit commercial is:
http://automobiles.honda.com/fcx-clarity/how-fcx-works/v-flow/
GCE AS and A PHYSICS Teachers' Guide 88
Thermodynamics – Carnot cycles, heat pumps and the 2nd Law of Thermodynamics
The groundwork for understanding the central problem of obtaining useful work from heat
has been developed already in PH4.3 Thermal Physics (i) to (d) and a good starting point is to
look again at these. One ideal heat engine consists of a cylinder with a piston and containing
ideal gas (the working substance). The engine is also “ideal” in the sense that the walls and
the piston are perfect heat insulators, the base is a perfect conductor and there are no friction
losses as the piston moves. The process involved is treated in many textbooks but is set out
here for convenience.
1. The engine is first placed on heat reservoir at temperature T1; initially the state of the
gas is shown by point A on the p – V diagram.
Heat Q1 passes from the reservoir into the gas as the state of the gas changes from A
to B. Work is done as the gas expands (piston goes up) but the internal energy does
not change as the curve AB is an isotherm (temperature constant so internal energy
constant). The changes along are given by the First Law of Thermodynamics (in the
form U = Q  W ) and are shown in the first line of the table.
2. The engine is now transferred to a perfectly insulating stand and allowed to expand
further to state C; no heat is transferred but work is done (no need to mention
adiabatic); changes are given in line 2.
3. The engine is now transferred to second reservoir at T2 and the gas is compressed
isothermally to state D; changes are given on line 3 but note particularly that heat is
ejected from the gas into the second reservoir – the sink.
4. Finally the cylinder is placed back on the insulating stand and compressed to return to
state A. The cycle is complete, and we can see from the bookkeeping table that an
amount of work (Q1-Q2) has been obtained from an amount of heat Q1.
Heat
reservoir
T1
T2
BC
AB
CD
Step
Q
U
W
A to B
Q1
0
Q1
B to C
0
(U1U2)
U1U2
C to D
-Q2
0
Q2
D to A
0
U1U2
(U1U2)
Heat
sink
DA
GCE AS and A PHYSICS Teachers' Guide 89
pressure, p
A
Indicator diagram
for a Carnot cycle
Q1
D
B
T1
Q2
C
T2
Volume, V
The procedure can be summarized as shown below:
HOT
T1
HEAT Q1
ENGINE
WORK
W = Q1  Q2
HEAT Q2
COLD
T2
The question arises, “Why does heat have to be ejected?” Or, equivalently, “Why does there
have to be a sink?”
If the process terminated at B the engine would be of no further use. To obtain useful work
the process must be continuous, which requires that the cycle be repeated over and over
again for as long as is necessary. This means repeatedly returning the gas to its original state
– something which can only be achieved by ejecting heat at one stage in each cycle.
The efficiency of the heat energy is defined by:
Useful work output
Efficiency 
100% .
Total energy input
Q  Q2
Q
Efficiency  1
 1 2
So
Q1
Q1
(1)
GCE AS and A PHYSICS Teachers' Guide 90
All that we know, or need to know, about the source and sink is that they have temperatures
of, respectively, T1 and T2. Clearly the amounts of heat transferred will be governed by these
temperatures. In fact, the ratio Q2/Q1 is used to define the ratio of Kelvin temperatures so
that
Q2 T2

Q1 T1
and equation (1) above becomes
T
(2)
Efficiency  1  2
T1
So a Carnot engine working between 100oC and 0oC [373 K and 273 K] can not have an
efficiency of greater than 27%. In practice there will be other factors such as friction and heat
loss which will make the actual efficiency even less. In the ideal engine described here there
are no such losses, and the cycle is therefore reversible in that if the operations described are
performed in the opposite (anticlockwise) sense the system is returned to precisely its initial
state. This would not happen if there were losses.
All this shows a fundamental difference between heat and other forms of energy. While
electrical energy can be transformed entirely into heat (I2R heating in a resistor) only a
fraction of heat can be transformed into useful work. This has been shown here to be true for
the ideal Carnot engine; it is in fact generally true and is formulated in the Second Law of
Thermodynamics which will be looked at later.
Details of actual engines, to which the same principles apply, will not be expected. The
simplest example is probably the steam engine, but this is now of historical rather than
practical interest. The Otto cycle for the internal combustion engine is worth looking at as a
more complex example.
The running of the Carnot cycle in reverse has already been mentioned and this is the basis of
the refrigerator and the heat pump. They operate on the same principle – that of extracting
heat from a cold source and ejecting it at a hot sink – and the Carnot cycle runs anticlockwise around the indicator diagram. Work is required to achieve this and a schematic
diagram for the process is given below. Note that the same diagram applies both to the
refrigerator and the heat pump, but that we do not use the term “efficiency” to describe their
effectiveness – efficiency is reserved for the heat engine. Instead, the figure of merit for these
devices is the Coefficient of Performance. The definition of the CoP is essentially the same as
that for efficiency:
Useful energy transfer
100% .
Work input
The nature of the “Useful energy transfer” differs however and the CoP is defined separately
for each case below.
i.e.
Coefficient of Performance 
GCE AS and A PHYSICS Teachers' Guide 91
For the fridge:
HOT
T1
Heat extracted from cold source
100%
Work input
Q
Q2
 2 
W Q1  Q2
COP 
HEAT Q1
ENGINE
HEAT Q2
COLD
T2
WORK
W = Q1  Q2

T2
T1  T2
For the heat pump:
Heat delivered to hot sink
COP 
100%
Work input
Q
Q1
 1
W Q1  Q2

T1
T1  T2
For refrigeration, the source is the inside of the fridge and the sink is the kitchen or, more accurately,
the cooling grill at the back of the cabinet. In the case of the heat pump the source is the ground,
preferably in ground water or a river for better heat transfer, and the sink is the radiators inside of the
house to be heated. Practical details are not expected but the broad principles should be understood.
For example, the working substance or “refrigerant” might be a liquid which can be made to
evaporate at the cold source, thus absorbing heat, and then to eject heat at the hot sink where the vapor
condenses back to liquid. Work is done in circulating the refrigerant and bringing about the necessary
phase changes.
The COPs are, like the Carnot efficiency, theoretical maximums. In practice, performance is much
poorer owing to the usual losses, but it is still instructive to insert numbers for typical situations. Heat
pumps seem a good proposition, but the capital expenditure is high and the losses great. They are used
in major buildings (Festival Hall, Buckingham Palace, and more recently the Senedd at Cardiff) while
many firms offer domestic appliances (try inserting “Heat Pump” into a search engine).
As already mentioned, the limitations of heat into work are a consequence of the Second Law of
Thermodynamics, the Kelvin statement of which is: “No process is possible the only result of which is
the total conversion of heat into work”. To illustrate this, look again at the Carnot cycle and the step
from A to B. This would seem at first to contradict the law for all the heat absorbed in the step is
converted into work. But this is not the only result because, in the process, the state of the gas
(pressure and volume) has changed. The law is fundamental, far reaching and one of the great
cornerstones of science. For example, an essential stage in obtaining work from a nuclear or any other
power plant is that of transforming heat to drive a turbine, and all the limitations of the Carnot cycle
apply. Suppose steam enters the turbine at 500 oC and then condenses to water at 60 oC; the Carnot
efficiency is 0.57 and the actual efficiency will be much less again. So, most of the energy is wasted.
One way to recover some of this is to pump the hot water from the power station condenser around
local housing in a massive central heating system. This is known as a Combined Heat and Power
(CHP) scheme and was pioneered in Britain at the old Battersea power station.
GCE AS and A PHYSICS Teachers' Guide 92
The overall efficiency obtained in a CHP scheme is much greater than the Carnot efficiency because a
large fraction of the “waste” heat Q2 now becomes useful – though there will be some losses between
the power station and the houses. In practice, CHP generators are run with a higher temperature cold
sink [~ 100C], thus reducing the efficiency of the electricity generation but allowing for more useful
heat energy distribution. It is easy to understand now why electricity should not be used for heating –
most of the original energy has already been wasted as heat. We can also now better understand a
great advantage of fuel cells – no heat is involved so there is no Carnot wastage in the cell itself,
though heat may have been involved at an earlier stage in producing the hydrogen and oxygen.
GCE AS and A PHYSICS Teachers' Guide 93
4.6 – PH6 – Experimental & Synoptic Assessment
The attention of centres is drawn to the specification of this unit and the internal assessment
guidelines on pages 48 and 57 of the GCE Physics specification. The focus of this unit is undertaking
measurements and observations and the an appreciation of the uncertainties inherent in these
observations. Reference should be made to the Guidelines on the Treatment of Uncertainties <link to
be inserted> on the WJEC website.
Administration of the Internal Assessment.
Task A
Based upon the preliminary entries which centres make in October, centres will receive in February
multiple copies of the data analysis task. The scheduled date for this task is in March. Bearing in mind
the need to preserve the integrity of the 45-minute assessment, centres are free to arrange for
candidates to undertake it in several sessions. Those scheduled to take it later on will need to be under
supervision before the first group of candidates completes the task.
Task B
Based upon the preliminary entries which centres make in October, centres making entries will
receive in January of each year a document entitled Confidential Instructions for Supervisors. This
will contain general instructions for the administration of the assessment and a detailed description of
the apparatus needed for the investigation. The question will not be included with this mailing. Well
before the scheduled time for the assessments, centres should assemble the required apparatus. The
subject officer is available to answer any questions and deal with problems that may arise in doing
this.
The scheduled sessions for the assessment will be in two consecutive days in the last week in mid
March. Centres are expected to enter candidates on the first day and only use the second day if
candidates cannot be conveniently accommodated in one day. As the duration of the assessment is
only 1¼ hours, there is ample time for 2 or 3 sessions on each day. Centres which need to use more
than one session may opt to use session 1 on day 1 and session 1 on day 2 rather than using two
sessions on the same day.
The assessment will come in two versions: version I is to be used on the first day and version II on the
second.
Centres will receive copies of the assessment questions in good time. In addition to the multiple
copies of the assessment papers, centres will receive a containing a single copy (1326/01 – G Physics
PH6 Experimental Task – Setting up instructions) of the assessment tasks which may be opened a
week in advance. This version will contain only the sections linked to the actual obtaining of results.
The full version of the assessment paper will not be available until the set date of the examination.
Supervisors should work through the tasks and ensure that the apparatus and questions work as
intended.
After the practical assessments have taken place, the completed examination papers must be
securely stored by the exams officer before it is submitted to WJEC. Teachers should not be
given access to the completed examination papers after the actual assessments have taken
place.
GCE AS and A PHYSICS Teachers' Guide 94
5.
CONTRIBUTORS TO THE TEACHERS' GUIDE
WJEC acknowledges with gratitude the contribution of the following members of the A level Physics
examining team to the production of this guidance material:
Dr J P G Richards
Mr P N Wood
Mr D P Edwards
Mr S W Evans
Dr I L Morris
Chair of Examiners
Chief Examiner
Principal Examiner
Principal Examiner
Principal Examiner
GCE AS and A PHYSICS Teachers' Guide 95
Appendix A
Thermodynamics Notes
GCE AS and A PHYSICS Teachers' Guide 96
Introduction
These notes on Thermodynamics were produced as a result of discussions at the GCE Physics
Inset sessions in autumn 2003 and 2004. The document contains much material, which is
intended to clarify terms and conventions in Thermodynamics, which are often
misunderstood. Examples of this are:



the distinction between heat and internal energy
the use of a sign convention to indicate direction of energy transfer
the use of Work in other than mechanical situations e.g. electrical work
The last section of the notes on the 2nd Law of Thermodynamics and the idea of entropy are
included for interest only and clearly outside the core specification.
GCE AS and A PHYSICS Teachers' Guide 97
Systems
The laws of thermodynamics apply to systems. A system is just about anything which we can
imagine to be defined by a boundary. At A-Level the favourite system – because it’s simple
but not trivial – is a sample of ideal gas in a cylinder fitted with a piston. Other examples of
systems are:
•
•
•
•
•
•
a rubber band
a mixture of oxygen and hydrogen in a sealed vessel
a lamp filament
a star
a black hole
the electromagnetic radiation inside a furnace.
The last example is included because it started a revolution in Physics. By applying
thermodynamics (and its sister science, statistical mechanics), the German physicist Max
Planck was led to conclude that the radiation was emitted from, and absorbed by, the walls of
the furnace in discrete packets of energy. From this idea grew Quantum Theory, which has
guided most of the discoveries in Physics from 1900 to the present day.
The Thermodynamic State of a System
For many systems, including gases, any physical property you
can think of can be expressed in terms of just two variables.
For gases, pressure and volume are often chosen as these two
variables. They are said to determine the state of the system.
So a particular point on a plot of p against V represents a
particular state of a sample of gas.
If the system is n moles of an ideal gas, then we have:
pV = nRT
This is called the equation of state of an ideal gas. It gives the property, temperature, in terms
of pressure and volume.
[We could just as well say that eq.1 gives the pressure in terms of volume and temperature. In
other words we can choose which two variables in terms of which we express the system’s
(other) properties.]
Different systems have different equations of state. For example, the equation of state of a
rubber band, relates tension, length and temperature. This equation is beyond A-Level!
GCE AS and A PHYSICS Teachers' Guide 98
The Internal Energy of a System
The internal energy, U, of a system is the sum of the potential and kinetic energies of its
particles.
[We shall not include gravitational energy or energy due to nuclear forces. This is because
they don’t change in the cases we shall be dealing with, and might just as well be zero.]
The Internal energy of an ideal gas
For an ideal monatomic gas, the forces between molecules are negligible (except during
collisions) and this means that the potential energy can be taken as zero. The internal energy
is, then, simply the sum of the kinetic energies of the molecules. So for N molecules each of
mass m and with mean square speed c 2 :
U  N  12 mc 2
But, using Newton’s Laws, it can be shown that pressure  volume is given by:
pV  13 Nmc 2
So:
U  32 pV
For a given pressure and volume U can, then, have only one value. We say that the system’s
internal energy is a function of the system’s state. This is true of all systems. A special
feature of an ideal gas is that U can be expressed in terms of the number n of moles and the
temperature T alone. Using pV = nRT (eq.1) to define T, we have:
U  32 nRT
eq.2
Internal energy of a real gas
The internal energy of a real gas depends on its volume as well as its temperature. This is
because of forces which act between the molecules. These are fairly negligible for a gas at
lowish density, such as oxygen at room temperature and pressure, which is nearly ideal in
behaviour. A-Level questions on gases will assume ideal behaviour!
Even at low densities the factor of 3/2 in eq.2 is only correct for monatomic gases. For oxygen
and other diatomic molecules (molecules consisting of two atoms bonded together) the factor
is usually close to 5/2. This is because the molecules have kinetic energy of rotation as well as
ordinary translational movement through space.
GCE AS and A PHYSICS Teachers' Guide 99
Internal energy of a rubber band
A molecule of rubber is a long chain of atoms. The bonds
between segments of the chain allow rotation of the segments,
of the type suggested by the top diagram.
This means that the chain is constantly changing shape. The
most frequently occurring shapes are ‘folded up’ tangles (see
lower diagram). Even an approximately straight chain would be
fantastically improbable – unless tension is applied to the
rubber.
The internal energy of a rubber band includes the kinetic
energy of random rotation of the segments. The band’s internal
energy hardly changes at all when it is stretched, provided its
temperature is kept constant. How is this like ideal gas
behaviour?
Changing the internal energy of a system
A system can gain internal energy by having work done on it or by having heat flowing into
it.
A system can lose internal energy by doing work on something external to it, or by having
heat flowing out of it.
Indeed, in accordance with the Law of Conservation of Energy:
Gain in system’s
internal energy
Net heat flowing into
system
=
+
Net work done on
system
eq. 3a
Each of the three terms in this equation can be either positive or negative. For example, the
net heat flowing into a system is negative if, in fact, heat flows out of the system.
Work
The basic definition is:
work done by a
force
=
magnitude of
force

distance moved by force
in direction of force
Here are some examples of systems doing work or having work done on them:
Work done by a rubber band
eq.4
GCE AS and A PHYSICS Teachers' Guide 100
When we increase the length of a rubber band by a small length x by applying a pull F,
then:
Work done on band = Fx
We could equally well write this as
Work done by band = – Fx
To see the point of this, suppose we allow the band to contract, exerting a pull F. In this case
x is negative so the work done by the band is positive, which makes sense.
If we stretch the band a lot, then F will change significantly during the stretching. In this case
we add together all the bits of work, Fx, which means adding together the areas of all the
narrow strips under the graph (see above), from the initial extension x1 up to the final
extension, x2. So:
Work done on (or by) band = area under Force-extension graph
eq.5
A positive amount of work is done on the band if x is increasing; a positive amount of work is
done by the band if x is decreasing.
Work done by a gas
When a gas exerting a pressure p expands by a small volume V,
then:
Work done by gas = pV
[This is, in fact, just a more convenient way of writing Fx, in which F is the force on the
piston and x is the distance it moves.]
We could equally well write
Work done on gas = –pV
If we push the piston in a little way V is negative so a positive amount of work is done on
the gas – as expected, since we’ve had to do the pushing.
If the gas expands a lot, then p will change significantly. In this case we add together all the
bits of work, pV, which means adding together the areas of all the narrow strips under the
graph (see diagram), from the initial volume V1 up to the final volume, V2. So:
Work done by gas = area under pressure - volume graph
eq.6
A positive amount of work is done by the gas if V is increasing; a positive amount of
work is done on the gas if V is decreasing.
GCE AS and A PHYSICS Teachers' Guide 101
Frictional Work
Suppose we push or pull a file across a piece of metal. We are doing work because we are
exerting a force which is moving in the same direction as the force itself. The system on
which we are doing work is the metal block (and the file).
An important feature of frictional work is that it is irreversible – the block won’t push or pull
the file and move it backwards! Compare with the reversible work done on the rubber band or
the gas when we push in the piston.
Electrical Work
If a potential difference is applied across a resistor (e.g. a wire) and a current I flows for a
time t:
Electrical work done on resistor = VIt
eq.7
What has VIt to do with force  distance? It is not very difficult to show that the sum of the
forces on the free electrons caused by the electric field set up by V, multiplied by the mean
distance which they move in time t as a result, is equal to VIt. [See E-fields in PH4]
The electrical work done on a resistor is irreversible. The resistor won’t be able to do work
pushing the electrons back the other way.
The electrical work done on a pure inductor or a capacitor is, however, reversible.
Heat
Heat is energy flowing from a region of higher temperature to a region of lower
temperature because of the temperature difference.
Compare with the flow of charge due to a potential difference…
If there is no temperature difference, no heat will flow. [In the electrical case, if there is no
potential difference, no charge will flow.]
It takes time for a finite amount of heat to flow, though the rate of flow is greater the greater
the temperature difference (other things being the same).
GCE AS and A PHYSICS Teachers' Guide 102
Thermodynamics: Applying the ideas
Example 1: Rapid expansion of an ideal gas
Think about the terms on the right hand side of eq.3a:
Gain in system’s
internal energy
=
Net heat flowing into
system
+
Net work done on
system
What about the heat? If the expansion is really rapid the heat flow will be (almost) zero.
What about the work? As the gas expands work is done by the gas. The last term in the
equation is negative. Another way of seeing this is to write the equation as:
Gain in system’s
internal energy
=
Net heat flowing into
system

Net work done by
system
eq.3b
Whichever way we look at it, because the heat term is zero, the right hand side of the
equation is negative, so the gain in internal energy of the gas is negative, i.e. the internal
energy decreases. So the gas temperature falls (see eq.2). It really does happen…

Try this… Squeeze some air down to a quarter of its volume or less in a large plastic
syringe. Wait a few seconds. Let the piston move outwards quickly, doing work on
your hand (i.e. not simply letting go of it). Feel the coolness of the syringe.

Suppose you let the trapped gas return to room temperature, keeping its volume
constant. Draw a line on the graph above to show what happens to the pressure.

When you squashed the gas in the first place what happened to the temperature and
why? Assume you squashed it very quickly. You notice the effect when using a
bicycle pump. In a diesel engine rapid squashing of air and fuel causes the
temperature of the mixture to rise so high that the fuel ignites (without needing a
spark).
GCE AS and A PHYSICS Teachers' Guide 103
Example 2: Slow expansion of an ideal gas
This time heat will have time to flow. For each small increase in volume the gas temperature
will drop a little (Example 1) and heat will flow in from the surroundings – limiting further
temperature drop. If the expansion is really slow (and the cylinder walls conduct heat well)
the temperature drop is negligible, so the expansion is isothermal.
An isothermal change is a change at constant temperature.
Now look at eq.3b. Re-arranging it slightly we have:
Net heat flowing
into system
=
Gain in system’s
internal energy
+
Net work done by
system
eq.3c
Since, in this case, the temperature does not change, there is no change in the internal energy
of the gas (see eq.2). So
Gain in system’s
internal energy
=
Net work done on
system
This work can, in principle, be used to lift weights, generate electricity, propel a vehicle and
so on. The process has its uses! [Unfortunately the gas pressure drops until no more useful
work can be done. But if we could restore the gas to its original state, we could do the same
useful thing all over again. The problem is that restoring the state simply by pushing the
piston back in again (slowly) requires us to put in the same amount of work the gas gave out
when it expanded! The solution is to use a cycle of changes – see page 10.]
We see from this example that…
•
•
Constant temperature is not at all the same thing as no heat flow.
Heat flowing into a system does not necessarily make it hotter (raise its temperature).
Example 3: A hot gas cools
Suppose we have a sample of hot gas confined in a cylinder by a piston. If we let the gas cool
down at constant volume (that is with the piston held in one place), no work is done. So
adapting eq.3c,
Heat flowing out
of gas
=
Decrease in internal
energy of gas
GCE AS and A PHYSICS Teachers' Guide 104
Can’t we say, then, that the hot gas was ‘storing heat’? No, because the hot gas could have
cooled down (i.e. lost internal energy) without giving out heat. It could have done work
instead, as in Example 1. If we had let the piston move out quickly, doing work, then:
Work done by gas
•
=
Decrease in internal
energy of gas
The energy stored by the gas is internal energy. It isn’t heat and it isn’t work. Its
decrease could result in heat being given out or in work being done (or both).
Heat and work are both energy in transit. Thinking of them as ‘stored’ in a system
leads to confusion in thermodynamics!
Example 4: A spark ignites a mixture of hydrogen and oxygen in a vessel
A chemical reaction occurs and the contents of the vessel turn wholly or partly to steam.
There is a sudden huge rise in the temperature and pressure in the vessel. What has happened
to the internal energy of the system (i.e. the contents of the vessel)?
Has heat flowed into or out of the system? Hardly at all in the short time of the chemical
reaction. The spark itself contributes negligible energy. Afterwards, heat will flow out.
Has work been done or by the system? Not during the reaction. Almost immediately
afterwards, the vessel may shatter and the pieces acquire kinetic energy as the expanding gas
does work on them.
So eq.3 (any version: 3a, 3b or 3c) shows that during the reaction there is no change in the
internal energy. How can this be, when the system gets hotter? The increase in molecular
kinetic energy is at the expense of a decrease in chemical potential energy as the atoms bond
with different partners. The overall internal energy is unchanged – at first.
Example 5: Quick extension of a rubber band
We do work on the band. Hardly any heat will flow out of the band if we do the work
quickly. So the internal energy will rise. Since, as for an ideal gas, this is mainly kinetic in
form, the band’s temperature will rise.

Try it, using a fairly stout band. Feel its temperature immediately (on your lips?)

Wait a few seconds, so that heat can flow out of the band into the air, and the band
regain room temperature. Now let the band contract, doing work on you. Feel it again.
Example 6: A filament lamp
Suppose we connect the lamp to a supply giving a steady p.d. of
12·0V. The filament is initially at room temperature and the
current is 30A. The filament gets hotter, and finally achieves a
steady ‘working’ temperature. It now takes a current of 4.0A.
[The current is lower because the resistance of the filament is
higher.]
GCE AS and A PHYSICS Teachers' Guide 105

How much does the filament’s internal energy increase during the first millisecond?
The filament will still be very nearly at room temperature, so negligible heat will flow
from the system (the filament) during this time.
Using eq.3a (or 3b or 3c – they’re all equivalent):
Gain in filament’s
internal energy

=
Net work done on
filament
 VIt
 12  0 V  30 A  0  0010s
 0  36 J
How much heat does the filament give out in 1.0 ms at its working temperature?
Since the filament’s temperature is steady it is no longer gaining internal energy. So:
Net heat flowing
out of filament
Net work done on
filament
=
 VIt
 12  0 V  4  0 A  0  0010s
 0  048 J
The First Law of Thermodynamics
We’ve already used it – many times – without giving it a name. Equations 3a, 3b and 3c
(where it is understood that internal energy is a function of the system’s state) are all ways of
writing it. In symbols we can use eq.3c and write:
U = Q  W
Gain in system’s
internal energy
=
Net heat flowing
into system
+
Net work done by
system
U is a change in a property of the system, its internal energy, U. A positive value of U
means an increase in U; a negative value means a decrease in U. As we said earlier, for a
given system U is a function of the system’s state. For example, for n moles of an ideal
monatomic gas, U is given by 3/2nRT (= 3/2pV).
Q is heat entering the system from (hotter) surroundings. A negative value of Q means heat
leaving the system (to cooler surroundings). We don’t have a ‘’ in front of Q, because heat
flow is not a change in heat. It’s energy in transit. It is not a function of the system’s state.
W is work done by the system. A negative value of W means a positive amount of work done
on the system. We don’t have a ‘’ in front of W for the same reason as for Q; namely, W is
not a change in work. It’s energy in transit. Work is not a function of the system’s state.
GCE AS and A PHYSICS Teachers' Guide 106
As practice in using the symbols, state whether Q, U, W are positive, negative or zero in the
following cases. The first line of the table is filled in as an example. The empty space is for
your own examples.
Process
A gas expands very quickly, doing work
U
Q
W

0
+
An ideal gas expands into a vacuum very
quickly, doing no work. Note: this is an
irreversible process.[Answers a bit repetitive]
An ideal gas expands isothermally, doing work
Water is being heated in an electric kettle.
Take the system as being the whole kettle
(including heating elements) and contents.
Water is being heated in an electric kettle.
Take the system as being just the water.
Cyclic processes – this theme is taken up in the PH5 Energy option.
Suppose an ideal gas is taken through the cycle of changes ABCDA shown above on the left.
Irrespective of exactly what is going on in the individual stages (AB, BC, CD, DA), or the
exact shapes of the curves, we can draw some general conclusions…

When the gas has undergone one cycle and is back at A, its internal energy is the
same as it was originally, that is U = 0. This is because internal energy is a function
of state (see page 2), and the gas is back in the same state as originally.
GCE AS and A PHYSICS Teachers' Guide 107

Over the cycle as a whole, the gas has done a positive amount of work. This is
because the area under ABC represents the gas doing a positive amount of work. The
area under CDA represents work being done on the gas (see page 4), but this area is
smaller.

Thus applying the First Law, we see that for the cycle as a whole, there has been a net
flow of heat into the gas.
Some heat has been turned into work, and this will happen each time the cycle repeats.
Question: When this was first done on an industrial scale, what (non-ideal) gas was used?
Go through the argument again for the right hand diagram, where the cycle is anticlockwise.
What are the differences?
A Special Cycle
Suppose AB in the left hand diagram represents an isothermal expansion in which an ideal
gas pushes a piston (see page 7, Example 2). Since the gas does work, and U = 0, heat is
taken in. If the compression, CD, is also isothermal, heat is given out, but less than is taken in
along AB. Since pV is less along CD than along AB, the isothermal process CD is at a lower
temperature than the isothermal process AB.
[Note that the surroundings must be slightly hotter than the gas for heat to flow in along AB,
and slightly cooler than the gas for heat to flow out along CD. So different environments (e.g.
an oven and a cold water bath) must be provided for these stages.]
Let us now suppose that BC and DA occur so rapidly that no heat enters or leaves the gas in
these stages. In this case, applying the First Law to the complete cycle:
Net work out over a
cycle
=
heat taken in along AB (at
higher temperature)

heat given out along CD (at
lower temperature)
Refrigerators
If we run the same special cycle backwards (right hand diagram above), heat flows out along
AB, a smaller quantity of heat flows in along CD, and a net quantity of work has to be put in
over the cycle. So…
Net work in over a
cycle
=
heat given out along AB
(at higher temperature)

heat taken in along CD (at
lower temperature)
In an ordinary domestic fridge, you can feel the heat coming out from the pipework at the
back. A fluid capable of evaporating and condensing (not, then, an ideal gas) circulates in the
pipes. The heat is taken in from anything placed in the fridge, (and also comes in through the
walls and door of the fridge). The work is supplied electrically.
GCE AS and A PHYSICS Teachers' Guide 108
The Second Law of Thermodynamics (just for interest)
For the ‘special cycle’ on page 11, not all the heat taken in during the isothermal process AB
could be used to produce work. Some heat had to be ‘excreted’ at a lower temperature
(during the process CD). In fact, no exceptions have ever been found to the rule that…
It is not possible, by any cyclic process using any substance, to take in a quantity of heat at
one temperature and turn it all into work.
This is one version of the Second Law of Thermodynamics.
If we could take in heat at one temperature and turn it all into work, life would be a lot
simpler. For example we might be able to take heat out of the polar ice caps (so slowing
down their melting) and get out electrical work. Of course using (that is converting) the
electrical work would give heat, but we could turn this into work. Dream on!
It is possible to show from the Second Law that a very simple rule relates the quantities of
heat taken in and given out to the temperatures in the special cycle of page 11.
heat taken in along AB TAB

heat given out along CD TCD
This applies for any substance, not just an ideal gas, undergoing the special cycle. The one
proviso is that each stage in the process has to be exactly reversible.
Example: Suppose AB is an isothermal at 400 K, and CD an isothermal at 300 K. If the
substance takes in 100 J of heat along AB, it must give out 75 J of ‘waste’ heat along CD.
The net output of work over the cycle will be 25 J. We say that the efficiency of the cycle is
25%. In practice the efficiency will be lower because of irreversible effects like friction
(page 5).
We can increase the cycle’s efficiency by making TAB larger or TCD smaller. There are clearly
technical limits to how large we can make TAB – components will melt! But what about
making TCD small? The problem here is that we must have some body of large heat capacity
outside the system, at a temperature marginally less than TCD, in order that heat will flow out
of the system during the process CD. That’s why, for large-scale practical purposes, TCD can’t
be much less than 300 K. A modern power station has an efficiency of less than 40%. This
means that for every joule of heat supplied from burning fuel (or from nuclear fission) more
than 0.6 J is given out as low temperature heat. It usually goes to waste. Remember this if
you use an electric heater at home!
Degradation of Energy (for interest)
Could we not use the heat excreted at TCD as the input to another ‘heat engine’? Yes, but note
that it would be even more difficult to get a decent yield of work out of that engine, since
there would be even less scope for a large difference between the TAB and TCD for that engine.
In this sense heat at low temperature is low-grade energy: it is less capable of being
converted to work. That’s why we don’t bother to try to re-use the waste heat from (say) the
backs of fridges and television sets. There is a tendency, in fact, for energy always to degrade
– think about effects of friction and electrical resistance.
GCE AS and A PHYSICS Teachers' Guide 109
There is a deeper, microscopic, interpretation of the degradation of energy. When energy is
delivered to a system as heat it increases the disorderliness of energy distribution among the
particles of the system. This disorderliness (measured by the quantity entropy) makes energy
less available for conversion.
As an example, consider again Example 2, the isothermal expansion of an ideal gas. After the
expansion, the gas has the same amount of internal energy as originally, but the particles are
now moving about all over the place – well, in a larger volume than before! And we know
that it is harder to get the expanded gas to do work.
Work is energy transferred in its least degraded form. The force and the distance moved are
in a single direction; the energy transfer is orderly.
GCE AS and A PHYSICS Teachers' Guide 111
Appendix B
Revolutions in Physics Notes
GCE AS and A PHYSICS Teachers' Guide 112
REVOLUTIONS IN PHYICS
ELECTROMAGNETISM AND SPACE-TIME
1. Introduction
One of the most exciting things in Physics is to discover relationships between observed effects (or
phenomena) that were previously thought to be quite distinct. What happened in electromagnetism in
the nineteenth century is a wonderful example. In the year 1800 there were only the vaguest
indications that magnetism had anything to do with moving electric charges, and no evidence at all
that light had anything to do with electricity or magnetism. By 1900 magnetism and electricity had
been firmly linked, and light had been shown to be an electromagnetic wave. How this came about,
sometimes in small steps, and sometimes by seemingly bizarre lines of reasoning, is the subject of this
option.
In one respect the new theory that linked electricity, magnetism and light seemed not to agree with the
facts, as found in experiments. In 1905 Einstein’s Special Theory of Relativity came to the rescue. At
the same time, it actually simplified the theory of electromagnetism (and light). Included in this option
there is a small taste of Relativity theory.
2. Questions and answers about this option
Q What is the point of studying this option?
A • It re-inforces some of the non-optional A-Level material, coming at it from a
different angle, giving it a wider context, and adding ‘human interest’.
• It brings the student into contact with great minds and great ideas.
• Sheer self-indulgence – it’s a wonderful story.
Q
A
Q
A
Q
A
Q
A
How can the material presented in this option, derived from what others have written, give the
promised ‘contact with great minds’?
A few extracts from some of the key figures (Young, Faraday, Maxwell and Einstein) are
provided. The extracts are not very long, but are to be studied closely. Guidance is given.
Does the student have to learn dates?
No, but having the right half-decade is good. In fact people often ‘absorb’ dates easily when
there’s a chain of events – and when there’s no stress to learn dates!
What has to be left out in order to fit the story into an A-level option?
This is a real problem. Looking back on past events and ideas, it’s easy to see, or to think we see,
which of them led nowhere or were of secondary importance, and to leave them out. But at the
time they may have been considered very important. They may have influenced the way physicists
thought, in ways we cannot now know. By omitting them we distort history. Please be aware that
this option cannot tell the whole story.
Can anything be done to give a more balanced picture?
Websites references are sprinkled throughout this WJEC material. Two thinnish and very
readable books which provide good support are…
Michael Faraday and the Royal Institution by John Meurig Thomas (ISBN 0-7503-0145-7).
Relativity and its Roots by Banesh Hoffmann (ISBN 0-486-40676-8).
Chapter 4 tells pretty much the same story as this course, but, as the book’s title makes clear,
Hoffmann has a special agenda, and his emphases are different.
All the material to be tested in the PH5 examination is contained in this WJEC printed
material, but students are urged to visit the websites, as they help to bring the basic
material of the option alive and make it easier to learn. They often contain pictures and
diagrams.
3. Electricity, Magnetism and Light: What was known in 1800
3.1 Electric Charge
• It had been known from ancient times that objects, in particular lumps of amber, could be
‘charged’ by rubbing, and could sometimes attract attract or repel other objects. [Our word
electricity comes from the greek word for amber.]
GCE AS and A PHYSICS Teachers' Guide 113
• Around 1730, Stephen Grey (www.sparkmuseum.com/BOOK_GRAY.HTM) had found that damp
thread, and metals, would conduct charge from one object to another, whereas many materials
were insulators (when dry). [Charge was often referred to as ‘electricity’ and charging, as
‘electrifying’.]
• Soon after, it emerged that there were two sorts of electric charge, and that these could neutralise
each other. Some years later, the american statesman and scientist, Benjamin Franklin, called them
positive and negative. Amber gains a negative charge when rubbed with fur; glass, a positive,
when rubbed with silk.
• Franklin showed, by extremely dangerous experiments, that thunder clouds contain electric charge,
and
that
lightning
is
an
electrical
phenomenon.
(www.inventors.about.com/cs/inventorsalphabet/a/Ben_Franklin_4.htm )
• In about 1745 Dutch investigators discovered that opposite charges could be stored on conducting
surfaces coating the inside and the outside of a glass bottle, and so separated by the insulator, glass.
The device quickly came to be called a Leyden jar, after Leyden, now Leiden, in the Netherlands. It
was used in demonstrations all over Europe to produce sparks and electric shocks - and much
excitement.
• In the late 1780s, Coulomb (www.en.wikipedia.org/wiki/Charles_Augustin_de_Coulomb ) made the
first quantitative investigation of the forces between charged spheres. These were of small enough
diameter, in relation to their separation, to be considered ‘point charges’. Using a torsion balance of his
own devising, he showed that there was an inverse square law, that is, when the separation of the
centres of the spheres was doubled, the force between the spheres quartered, and so on.
(http://library.thinkquest.org/C001429/electricity/electricity11.htm )
[The reclusive Henry Cavendish had made the same discovery some years earlier, but did not publish
his findings.]
Coulomb and his contemporaries were struck by the similarity between this inverse square law for
charges and Newton’s inverse square law of gravitation for masses.
3. Electricity, Magnetism and Light: What was known in 1800
3.2 Magnetism
In the year 1800, most of the knowledge about magnetism dated from 1600, when William Gilbert
had published his great work De Magnete (‘About the Magnet’). He described his experiments to
magnetise iron bars using a lodestone (naturally occurring magnetised iron ore), reported on the
‘magnets’ having poles at either end (the word ‘poles’ is his), and found that even if you cut a magnet
in half, each of the two halves still had both a North and a South pole. He investigated the effect of the
Earth on a pivoted magnet, and came to the conclusion that the Earth itself was a magnet. He
GCE AS and A PHYSICS Teachers' Guide 114
demolished many superstitions about magnetism, but we would regard his own view as to the cause of
magnetic effects as very odd. (http://galileo.rice.edu/sci/gilbert.html )
Although the attraction and repulsion behaviour of magnetic poles resembles that of electric charges,
Gilbert was very careful to explain that magnetic and electric effects were quite distinct.
3.3 The Battery
This was hot news in the year 1800. Back in the 1780s, Luigi Galvani had observed the twitching of a
leg cut from a dead frog, when a nerve was touched by a piece of metal which was also in contact
with the foot. The effect, he found, was much greater if two different metals were joined together.
There are various versions of how the discovery was made; see for example
www.bioanalytical.com/info/calendar/97/galvani.htm . Galvani attributed the twitching to ‘animal
electricity’, perhaps in the frog’s nerves.
Alessandro Volta took up the investigation and became convinced that it was the different metals
which played the key role. He devised a cell consisting of a strip of zinc and a strip of copper dipping
into a cup of brine or dilute acid, but not touching each other, and then started putting cells in series
(as we would now say). Two forms of ‘battery’ emerged, the ‘crown of cups’
(www.scienceandsociety.co.uk/results.asp?image=10207373 ) and the famous ‘voltaic pile’
(www.en.wikipedia.org/wiki/Voltaic_pile ). [In French the name still survives: une pile or une pile
electrique is a battery.]
News of Volta’s invention spread quickly, and batteries, sometimes very large ones, were built all
over Europe and in America. They were found to melt wires, connected across their terminals, and to
enable the splitting up of water up into oxygen and hydrogen. Some investigators were nearly killed
by
electric
shocks
from
batteries
of
many
cells.
Humphry
Davy
(www.rigb.org/rimain/heritage/ripeople/davy.jsp ), at the recently founded ‘Royal Institution’ in
London, used batteries to perform electrolyses which isolated sodium, potassium and various other
elements for the first time. He also fascinated audiences with demonstrations of what a battery could
do.
Davy’s audiences weren’t made up entirely, or even mainly, of people we would now call ‘scientists’.
Any intelligent person – with the leisure – could contribute to a scientific debate. Davy himself was
quite a gifted poet and was a friend of Wordsworth and Coleridge. There wasn’t really an ‘artsscience divide’. ‘Galvanism’, the term used then for the study of the battery and what it could do, was
much talked about, and we might guess that it was one of the influences on the young Mary Shelley,
when she was writing Frankenstein (published in 1818).
Volta himself had established a connection between batteries and electric charge. He discovered that
the terminals of his batteries were charged positively and negatively. Charge collected from the
terminals could be used to make bodies attract and repel, in specially designed instruments. The
battery provided for the first time the means of producing a continuous flow of charge, or electric
current. [Charge in this context was often referred to as an ‘electric fluid’, and there was controversy
over whether there were really two fluids or just one. We shan’t follow this particular sub-plot.]
GCE AS and A PHYSICS Teachers' Guide 115
3. Electricity, Magnetism and Light: What was known in 1800
3.4 Light
In the 1660s Newton had performed a brilliant series of experiments showing that ‘white light’ was a
mixture of colours. He made other major contributions to optics. Naturally he wondered what light
was.
Newton’s rival, Robert Hooke (of Hooke’s Law fame) believed it to be a wave-like disturbance
travelling through, and by means of, a universal medium (often called the aether or ether). Christiaan
Huygens, a strong supporter of a wave theory of light, showed how to predict where a wavefront will
be, and what its shape will be, if we know its position and shape now. He gave convincing wave
theory accounts of reflection and refraction.
(http://encarta.msn.com/encyclopedia_761567208/Christian_Huygens.html )
For Newton and others, the problem with the wave theory was that light doesn’t seem to bend round
corners, for example when opaque objects are put in its path. Water waves, though, do bend and
spread into the ‘shadow’ behind obstacles, sound travels round corners – and so do Huygens’
wavefronts. For this reason, mainly, Newton could not accept that light was a wave, or, more
accurately, just a wave. He held that it consisted of a stream of corpuscles or particles, coming from
its source. But he knew there were problems with this: if light fell on a sheet of glass, some goes
through and some is reflected. Why should some corpuscles do one thing and others another?
Newton wrote of light as having ‘fits’ of easy reflection and fits of easy refraction, and hinted that
possibly some sort of wave-like disturbance might accompany the corpuscles and determine what
they did.
Such was the awe in which Newton was held for showing how an inverse square law of gravitation
accounted for the motion of the planets, the moon and the tides, that his corpuscular theory of light
was given enormous respect. If you challenged it, even long after Newton’s death, you would have to
defend yourself very convincingly.
3.5 Questions on section 3
(1) It was discovered in the 1700s that metals could be charged up by rubbing with a dry cloth. In
what special way would the metal have to be held?
(2) A leyden jar would now be classed as a sort of …………………….. ?
(3) How, mathematically, do we now write Coulomb’s inverse square law for electric charges?
(4) What, according to William Gilbert, was the ‘soul of the Earth’?
(5) In what you have read, have you come across any pre-1800 evidence for a connection between
electricity and magnetism?
(6) What was ‘galvanism’, and why was it so called?
(7) Is it true that none of the effects of an electric current could have been observed before the work
of Galvani and Volta?
(8) How does the wave theory of light account for refraction?
(9) What political upheaval was shaking Europe in the 1790s?
GCE AS and A PHYSICS Teachers' Guide 116
4. Re-birth of the Wave Theory of Light
4.1 Thomas Young
Thomas Young (born in 1773) was a child prodigy. When four years old, he is said to have read the
bible in its entirety…twice. By the age of
fourteen he had mastered several languages,
ancient and modern.
He lived up to his early promise. As a medical
student he discovered the mechanism by which
the eye focuses (or accommodates), and, at the
age of 21 was elected a Fellow of the Royal
Society. This is Britain’s most prestigious
scientific society, dating from the time of
Newton.
In 1801, when Young had set up as a doctor in
London, he was chosen as Professor of Natural
Philosophy (roughly speaking, Physics) at The
Royal Institution. [He turned out not to be as
charismatic a lecturer as Humphry Davy.]
At about this time Young started his researches
on light – see below.
Later in life he made some headway in
deciphering the ancient Egyptian heiroglyphics
on the Rosetta Stone.
www.whonamedit.com/doctor.cfm/1715.html
Writing about light, Young stated two ‘hypotheses’ ;
“A luminiferous [light-carrying] ether pervades the universe.”
“Undulations [waves!] are excited in this ether whenever a body becomes
luminous.”
He explained that:
“an undulation is supposed to consist in a vibratory motion; transmitted
successively through different parts of a medium without any tendency in each
particle to continue its motion except in consequence of the transmission of
successive undulations from a distinct vibrating body.”
Young’s new idea, apparently not grasped by Huygens, was that light had to be a regular sequence of
undulations. This implied that light from the same source, travelling to the same point by different
routes would interfere either constructively or destructively, according to phase difference. Using the
idea of interference, Young was able to explain ‘Newton’s Rings’ a phenomenon which had puzzled
Newton himself. Visit the website below for pictures – strictly ‘for interest only’!
www.physics.montana.edu/demonstrations/video/6_optics/demos/newtonsrings.html
Note that it did not occur to Young at the time that light could be anything other than a longitudinal
wave, like sound.
Young seems [historians argue about it] first to have shown a version of his famous two slits
experiment in a lecture given to The Royal Society in 1803. Here is the account he gives of such an
experiment…
GCE AS and A PHYSICS Teachers' Guide 117
4. Re-birth of the Wave Theory of Light
4.1 Thomas Young (Continued)
“It has been shown that two equal series of waves, proceeding from centres near
each other, may be seen to destroy each other’s effects at certain points, and at other
points to redouble them; and the beating of two sounds has been explained from a
similar interference. We are now to apply the same principles to the alternate union
and extinction of colours.
“In order that the effects of two portions of light may thus be combined, it is
necessary that they be derived from the same origin, and that they arrive at the
same point by different paths in directions not much deviating from each other.
This deviation may be produced in one or both the portions by diffraction, by
reflection, by refraction, or by any of these effects combined: but the simplest case
appears to be, when a beam of homogeneous light falls on a screen in which there
are two very small holes or slits, which may be considered as centres of divergence,
from whence the light is diffracted in every direction.
“In this case, when the two newly formed beams are received on a surface placed so
as to intercept them, their light is divided by dark stripes into portions nearly equal,
but becoming wider as the surface is more remote from the apertures, so as to
subtend very nearly equal angles from the apertures at all distances, and wider also
in the same proportion as the apertures are closer to each other. The middle of the
two portions is always light, and the brighter stripes on each side are at such
distances, that the light coming to them from one of the apertures, must have
passed through a longer space than that which comes from the other, by an interval
which is equal to the breadth of one, two, three or more of the supposed
undulations, while the intervening dark spaces correspond to a difference of half a
supposed undulation, of one and a half, of two and a half, or more.
“From a comparison of various experiments, it appears that the breadth of the
undulations constituting the extreme red light must be supposed to be, in air, about
one 36 thousandth of an inch, and those of the extreme violet, about one 60
thousandth; the mean of the whole spectrum, being about one 45 thousandth. From
these dimensions it follows, calculating upon the known velocity of light, that
almost 500 millions of millions of the slowest of such undulations must enter the
eye in a single second.”
Young continues with a description of the ‘beautiful diversity of tints’ in the fringes which are seen
when white light is used. The above extract is as Young wrote it, apart from one comma being
removed and one new paragraph created. There were no diagrams (apart from the one below);
readers were supposed to … read. And visualise!
GCE AS and A PHYSICS Teachers' Guide 118
4. Re-birth of the Wave Theory of Light
4.1 Thomas Young (Continued)
Here are some must-do ‘comprehension’ questions on this first-ever description of a now famous
experiment.
(1) What did Young mean by a ‘luminiferous ether’? What purpose did it serve?
(2) Draw the set-up described by Young in the second paragraph and the beginning of the third
paragraph in the long extract. It should be familiar!
(3) What – in a word – does Young mean by ‘the breadth of an undulation’ (near the bottom of the
third paragraph)?
(4) WJEC gives the ‘Young’s fringes formula’ as
(5)
(6)
(7)
(8)
(9)
ay
.
D
(a) Re-arrange it to make the fringe separation the subject.
(b) Pick out the phrase from Young’s third paragraph in which he states the effect on the fringe
separation of altering D.
(c) Pick out the phrase from Young’s third paragraph in which he states the effect on the fringe
separation of altering a.
The bright stripe next the central bright stripe is at such a distance, to use Young’s terminology,
that the light coming to it from one of the apertures must have passed through a longer space
than that which comes from the other, by an interval which is equal to the breadth of one of the
supposed undulations. Put this in modern ‘path difference’ language.
1 inch = 2.54 cm. Hence express in metres Young’s results (fourth paragraph) for the
wavelengths of the extremes of the visible spectrum. Do they agree with what textbooks give?
What is conspicuously missing from this account of a quantitative experiment?
When, at the end of the passage, Young refers to ‘the slowest of such undulations, he means
those of the lowest frequency. What does he give as their approximate frequency?
Young refers near the end to ‘the known velocity of light’. [It had been inferred a long time
previously by two different methods based on two quite different sorts of astronomical
measurements.] Work backwards from Young’s figures for longest wavelength and lowest
frequency to deduce what figure he must have been using for the velocity of light.

GCE AS and A PHYSICS Teachers' Guide 119
4. Re-birth of the Wave Theory of Light (Continued)
4.2 Reactions to Young
Young’s experiment is the classic demonstration that light has wave-like properties. But that is not
how it was seen at the time. Maybe Young’s contemporaries would have been more convinced if
he’d given his actual readings, and explained properly how he’d arrived at his results for
wavelengths. Then there was the long-dead Newton to contend with. How dare this upstart, Young,
challenge the great Newton’s view that light was a stream of particles?
Henry Brougham, a barrister who later rose to become Lord Chancellor, wrote an infamous review
of one of Young’s Royal Society papers. He accused Young of putting forward an (unjustified)
theory, and having to make changes to it …
“A mere theory is in truth destitute of all pretentions to merit of every kind, except
that of a warm and misguided imagination. It demonstrates neither patience of
investigation, nor rich resources of skill, nor vigorous habits of attention, nor
powers of abstracting and comparing, nor extensive acquaintance with nature. It is
the unmanly and unfruitful pleasure of a boyish and prurient imagination, or the
gratification of a corrupted and depraved appetite.
“If, however, we condescend to amuse ourselves in this manner, we have the right
to demand, that the entertainment shall be of the right sort – that the hypothesis
shall be so consistent with itself, and so applicable to the facts, so as not to require
perpetual mending and patching – that the child we stoop to play with shall be
tolerably healthy, and not of the puny, sickly nature of Dr Young’s productions [...]”
Not impressed, then? In another paragraph (which no writer today could expect to get away
with) Brougham accused Young of bringing the Royal Society into disrepute…
“Has the Royal Society degraded its publications into bulletins of news and
fashionable theories for the ladies who attend the Royal Institution? Proh Pudor!
[For shame!] Let the professor continue to amuse his audience with an endless
variety of such harmless trifles; but, in the name of Science, let them not find
admittance into that venerable repository which contains the works of Newton, and
Boyle, and Maskelyne, and Herschell.”
(http://homepages.wmich.edu/~mcgrew/brougham.htm for interest only!)
Brougham’s reaction was extreme, but, even putting it aside, Young’s work on interference and the
wave theory didn’t attract much enthusiasm at the time.
GCE AS and A PHYSICS Teachers' Guide 120
4. Re-birth of the Wave Theory of Light (Continued)
4.3 Transverse waves
Real trouble soon arrived for the wave theory.
In about 1808 Etienne Malus discovered an
astonishing fact about the light reflected from
a transparent sur-face. The effect is observed
to perfection for the light reflected off a glass
plate, A, when the angle of incidence is 57°.
The reflected ray is found to be reflected from
another glass plate, B, when this is as shown
in the left hand diagram, but not when B is
turned about the ray as axis, so that it is as
shown on the right. The light must be
asymmetrical about its direction of travel! [A
related effect involving certain crystals, called
‘double refraction’, had puzzled natural philosophers for well over a century. Polaroid had
not been invented.]
To an A-level student the solution should be
obvious: light is a transverse wave, and A must be polarising it. But it hadn’t occurred to Young that
light could be anything else but a longitudinal wave, like sound. Eventually, though, (c1818) the
penny dropped.
By this time another powerful wave theorist, Augustin Fresnel, was at work in France.
(http://micro.magnet.fsu.edu/optics/timeline/people/fresnel.html ). He came upon the significance of
interference independently of Young, and developed the wave theory mathematically. He showed
convincingly that the reason we don’t normally see light bending round corners is because of its
short wavelength. He accounted for polarisation by reflection, double refraction and the diffraction
patterns caused by various obstacles. For a spherical obstacle his equations made an unlikely
prediction … (www.physics.brown.edu/physics/demopages/Demo/optics/demo/6c2010.htm )
4.4 Problems with the Ether
Fresnel effectively killed off the corpuscular theory. Most natural philosophers were persuaded that
light was a transverse wave. The only sort of wave anyone could imagine was a mechanical wave, in
which a pattern of displacements transmits itself through a medium, the ‘ether’. Try and follow this
crude and sketchy explanation…
In the diagram a transverse wave is travelling to the right. The medium is stiff, so the shaded slice
experiences an upward tangential or ‘shearing’ force from
the upwardly displaced slice to its left. The shaded slice will
accelerate upwards, and the peak displacement, P, will
move to the right – and so on.
There were severe problems with this ‘mechanical’ theory…
• It is difficult to see why the ether shouldn’t transmit
longitudinal waves as well as transverse waves. Yet no
longitudinal waves were observed.
• Transverse waves need a stiff medium, a solid, rather than a liquid or gas. But we receive sunlight
and starlight, so all space must be full of this medium. How, then can the planets move without
obstruction? Indeed, how can anything move freely?
For the next few decades, elaborate attempts were made to devise ether structures which would not
have these problems. We shall return to the ether…
GCE AS and A PHYSICS Teachers' Guide 121
5. Discoveries in Electromagnetism
5.1 Ørsted
Electromagnetism was born in 1820 when Hans Christian Ørsted (or Oersted)
(http://en.wikipedia.org/wiki/Hans_Christian_Ørsted) discovered that a copper wire connected across
the terminals of a battery could deflect a compass needle (in other words, a pivoted magnet). The
effect was just as large if non-magnetic substances other than air were placed between the wire and
the magnet. As long as it was close enough to the wire, the magnet was deflected to be almost at
right angles to the wire. The North-seeking pole pointed in opposite directions according to whether
the magnet was below or above the wire. It seemed as if the magnet directions were tangential to
circles going round the wire.
Quick Checks
• Do the needle directions shown agree with the right hand grip (or screw) rule?
• Why does the needle have to be close to the wire to be deflected almost at right angles to the
wire? What other influence is there on the needle?
• Why won’t the experiment work if the wire runs East-West?
A Historical Puzzle
Twenty years had gone by between the invention of the battery and Ørsted’s discovery, and this is
rather odd. For one thing, there was a sort of ‘galvanism mania’ after Volta announced his invention,
and the powers of the battery were explored with great zeal. For another, there were tantalising clues
that magnetism and electricity were related, such as in stories of cutlery becoming magnetised, and
ship’s compasses suffering reverses in polarity, during thunder-storms.
What is more, many investigators were influenced by a philosophical argument which claimed to
show that the ‘forces of nature’ must have an underlying unity. Ørsted held this view, and seems to
have been searching on and off for years for magnetic effects due to a battery. It wasn’t at all
obvious, though, that the battery had to be in a closed circuit, in other words that there had to be a
current. When the effect was discovered (during one of Ørsted’s lectures, according to a popular
version of the story), it was not as anyone had guessed. Instead of pointing parallel to the wire, or
radially towards or away from the wire, the compass seemed to want to point at right angles to both
these directions.
[Note… Others had found compass needles being ‘affected’ during experiments with batteries. But
Ørsted was the first, as far as we know, to investigate systematically what was happening, and to
publish a clear, detailed description of the phenomenon.]
GCE AS and A PHYSICS Teachers' Guide 122
5. Discoveries in Electromagnetism (Continued)
5.2 Ampère
Ørsted’s experiment was demonstrated at a meeting of the French Académie des Sciences. One of
those present was the mathematician André-Marie Ampère, a friend of Fresnel and a supporter of the
wave theory of light.
(http://www-history.mcs.st-and.ac.uk/history/Biographies/Ampere.html )
Ampère immediately plunged into an intense period of investigations. He reported discoveries at a
rate of around one a week for the next two or three months.
Improved version of Ørsted’s experiment
One of the first things Ampère did was to use magnets to cancel the effect of the Earth on
the compass magnet, over a region around the wire. He then found that even when it was not
very close to the wire, the compass magnet pointed at right angles to the wire and
tangentially to circles around the wire. [Ampère referred to electric current in the wire, and
used this term consistently, with the convention that the current in a wire is from the positive
terminal of the battery to the negative.]
Forces between current-carrying wires
Ampère went on to demonstrate a totally new phenomenon: that wires carrying currents
exert forces on each other. Parallel wires attract each other if carrying currents in the same
direction, and repel if the currents are in opposite directions. On the left is Ampère’s
diagram of his apparatus. The parallel wires are AB (fixed to the base) and CD (able to
swing on pivots E and F).
Coils and
Magnets
Ampère
believed that
the basic
forces
involved
both
between his
parallel
wires, and
between the
wire and the
magnet in Ørsted’s experiment, were forces
between currents.
So Ampère believed there were currents inside magnets? Yes. He strengthened his case by
showing that current-carrying coils and solenoids behaved very much like magnets...
• He showed that the ends of two coils seemed to attract and repel each other like the poles
of two magnets.
• He took the wires from the ends of a solenoid (AB in the right hand diagram) back
through the inside of the solenoid and out through the centre, then taking the wires up and
down to cups of mercury (N and M), connected to the terminals of a battery. Thus the
solenoid could turn freely. He found it to behave like a compass magnet.
GCE AS and A PHYSICS Teachers' Guide 123
5. Discoveries in Electromagnetism (Continued)
5.2 Ampère (Continued)
Ampère’s Theory of Magnetism
What might be the paths of currents inside magnets to
make magnets behave like solenoids? At first Ampère
thought they might be big loops, like the turns of a coil.
He then took up a suggestion of Fresnel, that the loops
were ‘molecular’, in other words on a minute scale. In a
magnet the loops’ axes were supposed to be roughly
parallel (see diagram); in unmagnetised iron they were
supposed to be arranged randomly.
Current elements
Ampère regarded a series circuit as made up of a succession of ‘current elements’, that is very short,
near-enough-straight lengths of current-carrying conductor.
He wanted to find a formula for the force between two current elements which would do the same
for current elements as Coulomb’s inverse square law did for stationary point charges. But it needed
to be more complicated as it had to take account of the angles, (,  and ) between the current
elements, and between them and the line joining them. Take a quick look at the formula Ampère
decided upon, by courtesy of
www.rwgrayprojects.com/energy/VACE/calc/calc01.html (top four lines only)
To find the force that a whole circuit (1) exerts on circuit
2, you would need to add up all the forces that all the
current elements in 1 exert on all the elements in 2. This is
every bit as difficult as it sounds, except for certain
symmetrical cases, like long straight wires. Ampère had to
try various formulae for the force between current
elements until he found one which gave answers for forces
between circuits which agreed with experiment. [There are
other possible formulae which do so.]
Ampère was not the only one in France to be galvanised
into action by Ørsted’s discovery. Jean-Baptiste Biot and Félix Savart discovered that the torque on a
compass magnet due to a long straight current-carrying wire varied inversely with the distance of the
magnet from the wire. Like Ampère, they developed the idea of current-elements.
Self-test questions on Ørsted and Ampère
(1)
(2)
(3)
(4)
(5)
(6)
(7)
If you haven’t already done so, find out Ørsted’s nationality.
In Ørsted’s experiment the tendency of the magnet to point in directions tangential to circles
around the wire was probably the result least expected. Which two directions might have been
considered less strange?
How did Ampère’s ‘improved’ version of Ørsted’s experiment make it more conclusive?
(a) What is the basis of the definition of the SI unit of current?
(b) Discuss the appropriateness of naming it after Ampère.
Do physicists today believe that a magnet’s magnetism has anything to do with small-scale
electric currents inside it?
Explain why the force between current elements cannot be measured directly.
Find, in your list of formulae, the one which contains Biot’s and Savart’s discovery about the
long straight wire.
GCE AS and A PHYSICS Teachers' Guide 124
5. Discoveries in Electromagnetism (Continued)
5.3 Faraday
Michael Faraday is perhaps the best known and most
admired of nineteenth century scientists. His career
began with a fairy-tale ‘elevation’ from bookbinder’s
apprentice to Humphry Davy’s assistant at the Royal
Institution.
(www.rigb.org/rimain/heritage/faradaypage.jsp)
[The first three chapters of Michael Faraday and the
Royal Institution by J Meurig Thomas set the scene. The
whole book is very readable.]
Faraday was more an experimental scientist than a
theorist, but he had extraordinary, almost intuitive,
insight. He had the patience to tease out the details of
the effects he investigat-ed, and the flair to judge which
were important.
His achievements included the discovery of benzene, the
liquefaction of several gases, and the formulation of the
laws of electrolysis. He discovered that materials other
than iron experienced forces (even though weak or very
weak) when placed near the poles of a magnet.
It is his work in electromagnetism for which he is
probably most famous…
Faraday, like the french scientists, was
stimulated by Ørsted’s discov-ery. But,
unlike them, Faraday had no maths
beyond arithmetic, nor was he convinced
that mathematical theories, such as those
using ‘curr-ent elements’, served much
purpose.
Instead, Faraday’s ‘feeling’ for the wireand-magnet phenomenon, led him to
devise set-ups in which rotations took
place – see diagram. On the left hand side,
the uppermost pole of a magnet partially
immersed in mercury rotated about a
current-carrying wire. On the right hand
side a current-carrying wire rotated about
the uppermost pole of a magnet. The wire
was pivoted at its top end, and dipped into
mercury at its lower end. Check that you
can trace the path of the current.
In these experiments, as in those of Ørsted
and Ampère, electric currents produced magnetic effects. But magnets hadn’t been shown to produce
currents. Faraday’s instinct was that there must be such an effect, “magnetism causing electricity”; it
just needed to be found. Over the next ten years, he made several attempts to find it. Success came
on the twenty-ninth of August, 1831…
GCE AS and A PHYSICS Teachers' Guide 125
5. Discoveries in Electromagnetism (Continued)
5.3 Faraday (Continued)
The Discovery of Electromagnetic Induction
Faraday’s famous laboratory diary entry for August 29th 1831 (with a little more punctuation added)
begins as follows:
“Have had an iron ring made (soft iron); iron round and 7/8 inches thick and ring 6
inches in external diameter. Wound many coils of copper round, one half of the
coils being separated by twine and calico – there were 3 lengths of wire each about
24 feet long, and they could be connected as one length or used as separate lengths.
[…] Will call this side of the ring A. On the other side but separated by an interval
was wound wire in two pieces together amounting to about 60 feet in length, the
direction being as with the former coils. This side call B.
Charged a battery of 10 pr plates [10 pairs of plates] 4 inches square. Made the coil
[coils] on B side one coil, and connected its extremities by a copper wire passing to a
distance and just over a magnetic needle (3 feet from iron ring) then connected the
ends of one of the pieces on A side with battery; immediately a sensible effect on
needle. It oscillated and settled at last in original position. On breaking connection
of A side with battery, again a disturbance of the needle.”
Notes and self-test questions on the diary extract
• The coils were insulated from each other and from the ring.
• The two coils on side B were connected in series. [How does Faraday express this?]
• In this first experiment, Faraday used only one of the coils on side A; the other coils on side A
might as well not have been there.
• In the language of transformers, what are coils A and B?
• In magnetic terms, what does coil A do when connected to the battery?
• Faraday used Ørsted’s set-up, with wire and compass needle, as a ‘galvanometer’ to detect any
current in the coil B circuit – pick out the phrase Faraday uses to describe the arrangement. [In
fact his galvanometer wasn’t very sensitive, and he went on to use more sensitive instruments.]
• Why did Faraday place the galvanometer as far as 3 feet away from the ring?
• The current in the B circuit – Faraday soon started calling it the induced current – was only
transient; it was present only when the current in A was turned on and off.
This is probably the main reason why Faraday took so long to find ‘magnetism causing
electricity’. No-one seems to have guessed that the effect would take place only when a change
was occurring.
GCE AS and A PHYSICS Teachers' Guide 126
5. Discoveries in Electromagnetism (Continued)
5.3 Faraday (Continued)
Further Exploration of Electromagnetic Induction
Faraday knew he had made a major discovery, and set about a thorough investigation of the
phenomenon. He soon found, as he had expected, that even without iron, a changing current in one
circuit could induce currents in a nearby circuit, though the effect was much weaker than with iron
present.
If there were any doubts that the induced current was a magnetic effect, Faraday put paid to them by
thrusting one end of a bar-magnet into a solenoid connected to a galvanometer. The needle deflected
in one direction when the pole was thust in, returned to its zero position and stayed there if the
magnet was left stationary inside the solenoid, but deflected in the opposite direction when the pole
was withdrawn. (http://micro.magnet.fsu.edu/electromag/java/faraday2/ - not historical but fun)
Magnetic Lines of Force
Not only did Faraday demonstrate many instances
of electromagnetic induction, he developed a
simple but powerful way of visualising when it
would take place.
He relied on lines of force (now called lines of
flux). These are the curved lines that can be
‘plotted’ with iron filings or a compass magnet. In
Faraday’s diagram (of 1832) they arise from a
magnet, AB.
Faraday explains that if a conductor is part of a
closed circuit, a current will flow in it when it ‘cuts’ lines of force. In the diagram the conductor PN
which he draws is a knife blade – re-inforcing the cutting metaphor. He gives a rule for the direction
of current flow which is equivalent to the (later) Fleming’s Right Hand Generator Rule.
• In Faraday’s diagram A is the North pole of the magnet. If PN is moved upwards
what will be the direction of current flow through it (if the circuit is completed)?
This picture of cutting lines of force doesn’t really seem to apply to Faraday’s original experiment
with the ring of iron. There were no moving conductors (or magnets). But the notion of lines of force
can still be used… At the point of turning on the current in A the number of lines of force going
around inside the ring, and therefore ‘linked’ with coil B suddenly increases. The reverse happens
when the current in A is turned off.
The rule that a current is induced when there is a change in the number of lines of force linking a
circuit changes fits all cases of electromagnetic induction, including…
• plunging one pole of a magnet into a coil – draw the ‘before’ and ‘after’ pictures, including some
lines of the magnet’s lines of flux.
• part of a circuit cutting lines of force: the number of lines linked with the complete circuit will
change as a result of the change in area enclosed by the circuit.
GCE AS and A PHYSICS Teachers' Guide 127
5. Discoveries in Electromagnetism (Continued)
5.3 Faraday (Continued)
A Quantitative Law
Faraday came close to a quantitative rule when he wrote:
“If a wire moves across lines of force slowly, a feeble current is produced in it,
continuing for the time of the motion; if it moves across the same lines quickly, a
stronger current is produced for a shorter time.”
We nowadays sum up electromagnetic induction in the equation:
E

t
We see that Faraday’s insights have been ‘developed’ considerably…
• The equation deals with e.m.f. rather than current, since the induced e.m.f. does not depend on the
resistance of the circuit (whereas the current does).
•  represents Faraday’s idea of the number of lines of force linking a circuit. Check that you can
define  the modern way!
• The minus sign acknowledges the insight of Heinrich Lenz, working in St Petersburg in 1834.
Check you can state Lenz’s Law.
• The proportionality was deduced around 1845 by Franz Neumann (from Ampère’s work!). In the
S.I. the proportionality constant is 1, so we can use ‘=’ rather than ‘’.
Action at a Distance?
How does a current-carrying wire influence a compass magnet, or exert a force on another currentcarrying wire, or, if the current is changing, induce a current in another circuit? How does the one
thing (call it ‘X’) influence the other, ‘Y’, even though there’s empty space in between X and Y?
In general, continental physicists (Ampère and others) saw this ‘action-at-a-distance’ as a thing that
simply happens, not requiring explanation. The work of the physicist, they thought, was to find
mathematical laws for the forces between X and Y.
Faraday, though, was not content with action-at-a-distance. Something had to be going on in the
space between X and Y in order to convey an influence from one to the other. Faraday felt that lines
of force were involved. He knew this was controversial.
[In most of his writings Faraday used the term ‘magnetic lines of force’ uncontrovers-ially to mean
lines (or curves) which tell you which way a compass magnet will point, or iron filings will line up,
if you put them in the vicinity of a wire or magnet.]
Electric Lines of force
Faraday also developed the idea of electric lines of force, starting on positive charges and ending on
negative charges. They can be plotted by using a non-conducting pivoted needle, with a positive
charge at one end and a negative at the other. Although they have some of the properties of magnetic
lines of force, the two sorts of line mustn’t be confused.
Some of the lines of force for a charged capacitor
are sketched in the diagram.
[Faraday found out a great deal about capacitors.
In particular he investigated dielectrics and their
effect
on
capacitance.]
GCE AS and A PHYSICS Teachers' Guide 128
6. Electromagnetic Waves
6.1 Faraday
Faraday drew a clear distinction between his experimental researches and his ‘speculations’ for
which there was little experimental evidence. He talked about one such speculation when he had to
fill in for a Royal Institution guest speaker who had taken fright and run away. Here are some
extracts from a summary Faraday wrote for a friend. Referring to electric, magnetic and gravitational
lines of force…
“[We can] affect these lines of force in a manner which may be conceived as partaking of the nature of a shake or lateral vibration. For suppose two bodies, A, B, distant
from each other and under mutual action, and therefore connected by lines of force,
and let us fix our attention upon one resultant of force having an invariable direction
as regards space; if one of the bodies move in the least degree right or left […] then
an effect equivalent to a lateral disturbance will take place in the resultant […]
My view which I am so bold as to put forth considers, therefore, radiation as a high
species of vibration in the lines of force which are known to connect particles and
also masses of matter together. It endeavours to dismiss the ether, but not the
vibrations. The kind of vibration which, I believe, can alone account for the wonderful, varied, and beautiful phenomena of polarization, is not the same as that which
occurs on the surface of disturbed water, or the waves of sound in gases and
liquids, for the vibrations in these cases are direct, to and from the centre of action,
whereas the former are lateral. It seems to me, that the resultant of two or more
lines of force is an apt condition for that action which may be considered as equivalent to a lateral vibration; whereas a uniform medium, like the ether, does not
appear apt, or more apt than air or water.
The occurrence of a change at one end of a line of force easily suggests a consequent
change at the other. The propagation of light, and therefore probably of all radiant
action, occupies time; and that a vibration of a line of force should account for the
phenomena of radiation it is necessary that such vibration should occupy time also.”
Notes and questions on Faraday’s Speculation
(1) What, in modern wave terminology, does he mean by a ‘shake, or lateral vibration?
(2) What, according to Faraday (first paragraph) would you have to do to send such a vibration
along a line of force? This is crudely illustrated for an electric line …
(3) Faraday says (second paragraph) that (for light) lateral vibrations are needed to account for
polarization effects.
(a) What phrase does he use to describe the vibrations in sound waves?
(b) What is the modern term?
(c) Is he right about the nature of surface water waves?
(4) Faraday’s speculations had implications for the luminiferous ether. [Revise section 4.4 – if
necessary!] These implications are summed up in one short sentence in the second paragraph.
Which sentence?
Faraday’s idea was indeed bold. It was nothing less than an attempt to link light and
electromagnetism. To be considered a successful attempt, it would need developing into a theory
which could make predictions, including quantitative ones.
GCE AS and A PHYSICS Teachers' Guide 129
6. Electromagnetic Waves (continued)
6.2 Maxwell
James Clerk Maxwell was arguably the greatest theoretical
physicist of the nineteenth century. Unlike Faraday, Maxwell
was born to well-to-do parents, and he received a good
education, including a thorough training in mathematics.
Maxwell was brilliant at spotting analogies between different
branches of physics, developing them mathematically – and
knowing when to drop the analogy. His most far-reaching
work was in kinetic theory of gases, and in electromagnetism.
(The following link may not work, but the URL is fine: wwwhistory.mcs.st-and.ac.uk/history/Biographies/Maxwell.html)
On Physical Lines of Force
This is the title of a four-part paper of 1861-2 in which
Maxwell sets out to “examine magnetic phenomena from a mechanical point of view, and determine what
tensions in, or motions of, a medium are capable of producing the mechanical phenomena observed.”
• The ‘mechanical phenomena observed’ are the attractions and repulsions between poles of magnets.
He goes on to hint that his ‘medium’ theory will also account for electromagnetic effects such as
induced currents.
• Since magnets will ‘work’ in a vacuum, Maxwell’s medium must fill even ‘empty’ space. [An
invisible, space-filling medium was not a new idea – revise section 4.4].
Maxwell’s starting point was magnetic lines of force. He writes…
“[If] we strew iron filings on paper near a magnet, each filing will be magnetized by
induction, and the consecutive filings will unite by their opposite poles, so as to form
fibres, and these fibres will indicate the direction of the lines of force. The beautiful
illustration of the presence of magnetic force afforded by this experiment, naturally
tends to make us think of the lines of force as something real, and as indicating
something more than the mere resultant of two forces, whose seat of action is at a
distance and which do not exist there at all until a [compass magnet or iron filing] is
placed in that part of the field. We are dissatisfied with the explanation founded on the
hypothesis of attractive and repellent forces directed towards the magnetic poles […]
and we cannot help thinking that in every place where we find these lines of force, some
physical state or action must exist […]”
• What, then, was Maxwell’s take on action-at-a-distance? [See Section 5.3]
The properties Maxwell gave his space-filling medium allowed it to form into lines of force. The structure
of the medium was machine-like. He showed that the machinery seemed to account for the phenomena of
electromagnetism.
On the next three pages we look in some detail at Maxwell’s ‘machinery’. It may seem weird and quite
different from anything you’ve met in Physics before, but the basic ideas aren’t particularly difficult. If
you do find it a struggle, don’t give up: a general feel for what Maxwell was up to is worth having, even if
you lose some of the details.
GCE AS and A PHYSICS Teachers' Guide 130
6. Electromagnetic Waves (continued)
6.2 Maxwell (continued)
Vortices
Maxwell asks us to suspend disbelief and to suppose that space is
filled with elastic beads. If a bead spins about an axis [diagram (a)],
it will become Smartie-shaped (like the Earth), contracting along its
axis and expanding sideways. He called the spinning beads ‘vortices’
– whirlpools.
Diagram (b) shows some of the lines of force between two opposite
magnetic poles attracting each other. It is as if the lines are under
tension, pulling the poles together, and are pushing out sideways,
pushing each other apart. This is just what would happen if the axes
of the spinning beads lie along the lines of force. So magnetic lines of
force are imagi-nary lines along which lie the axes of spin of the
vortices.
The angular velocity of the vortices was proportional to the field
strength. No field strength meant no spin.
Idlers
If you could look along any line of force going from the North pole
of one magnet to the South pole of another, the vortices would be
spinning in the same sense – anticlock-wise, let us suppose. This
presents a problem if space is chock-a-block with vortices. Between
a North and South pole, they are all rotating in the same sense, so
where vortices on adjacent lines of force touch, the vortex surfaces
will be moving in opposite directions, and will interfere with each
other’s motion [diagram (c)].
Maxwell’s solution was to suppose the vortices to be separated by ball-bearing-like ‘idlers’ [as in (d)]. By
rotating in the opposite direction to the vortices, the idlers enable the vortices to rotate in the same
direction as each other. Note: idlers never slip on vortices.
Maxwell’s (in)famous ‘honeycomb’ diagram of his ‘vortex medium’ is given below. Try not to worry
about the sharp corners and the 2-dimensionality; it is just a stylised way of showing space completely
filled with vortices separated by idlers. But, even so, could Maxwell seriously have believed that space
was full of ‘machinery’ of this sort? He wrote:
“The conception of a particle having its motion
connected with that of a vortex by perfect rolling
contact may appear somewhat awkward. I do not
bring it forward as a mode of connexion existing in
nature, or even as that which I would willingly assent
to as an electrical hypothesis. It is, however, a mode
of connexion which is mechanically con-ceivable, and
easily investigated, and it serves to bring out the
actual mechanical connexions between the known
electro-magnetic phenomena; so that I venture to say
that anyone who understands the provisional and
temporary character of this hypothesis, will find
himself rather helped than hindered by it in his search
after the true interpretation of the phenomena.”
Let us now see how it does help…
GCE AS and A PHYSICS Teachers' Guide 131
6. Electromagnetic Waves (continued)
6.2 Maxwell (continued)
Ørsted revisited
Spinning wasn’t the only motion Maxwell allowed his idlers. They
could also move ‘sideways’. The diagram shows a line, I, of idlers
moving (‘translating’) to the right without spinning. They must exert
tangential forces on the vortices with which they are in contact, making
them rotate as shown. This mot-ion spreads outwards from I via
spinning idlers and vortices.
In this 2-dimensional diagram, the vortices above I are being made to
spin anticlockwise, those below I, clockwise. So we are looking at a
section through lines of force going in circles around I. But we know
that an electric current in a straight wire has circular lines of force
around it. So a line of translating idlers must constitute an electric
current!
• Look again at Maxwell’s ‘honeycomb’ diagram – especially the
arrows – and spot the one (zigzag) line of translating idlers. [Note: in the fourth row down of
vortices, all four should be spinning clockwise!]
Self-induction
This material in this box will not be tested. It should, though, be of interest to anyone
who is also studying the Further Electromagnetism and A.C. Theory option.
The vortices have inertia and will acquire kinetic energy when made to spin. This
energy will have to come from the line of translating idlers that set them in motion,
in other words from the electric current. So the current will experience an opposing
e.m.f., when the current is increasing. Once it reaches a steady value, the line of
idlers will be translating at a constant speed, and the vortices spinning at a constant
angular velocity, so they will not be acquiring KE. If the current decreases, the
vortices will give back energy to the current, opposing its decrease, so there will be
an e.m.f. in the other sense.
This ‘explains’ the phenomenon of self-induction, which had been discovered
independently in the early 1830s by Joseph Henry in America and by Faraday.
• Make sure you know the definition of e.m.f..
• Revise, or look up, the defining equation for self-inductance, L, and
think about how it sums up the phenomenon of self-induction.
[We cannot, it should be said, calculate L for an isolated straight wire;
we have to take into account a ‘return path’ for the current (such a
parallel wire).]
• You should now have some inkling of the capabilities of Maxwell’s ‘machinery’. In fact (with the
help of rather a lot of mathematics) Maxwell showed how it would give rise to all the known
effects of electromagnetism, including forces between currents and the e.m.f. induced in a
conductor cutting lines of force.
• But could the machinery tell us anything about electromagnetism that we didn’t already know? In
other words, could it be used to make predictions?
GCE AS and A PHYSICS Teachers' Guide 132
6. Electromagnetic Waves (continued)
6.2 Maxwell (continued)
Transverse Waves
We now look in more detail at how a current-carrying wire sets up a magnetic field around it.
Maxwell’s machinery predicts that when the current is switched on the field will take time to spread
out. The crude diagram below helps to explain this...
Suppose an idler, i, starts to turn. Because of its inertia, the vortex V ‘above’ i will not turn
immediately, and i will roll to the left. But i will exert a tangential force on V, giving an anticlockwise
torque on V, which will deform as shown, as it is made of elastically deformable material. Soon, the
whole of V will start to turn anticlockwise, the deformation and stress will disappear and i will return
to its original position. As it starts to turn, V will turn the idler ‘above’ it and the same thing will
happen all over again for the ‘next vortex out’, and so on.
Don’t worry if you struggled with the last paragraph. The points to grasp are these...
•
A magnetic field propagates outwards from its source at a finite speed.
• The ‘wavefront’ of the spreading magnetic field is accompanied by temporary stress
on the vortex material. Maxwell interpreted this stress as an electric field. The vortices are
temporarily distorted and the idlers temporarily displaced. Maxwell called their motion a
‘displacement current'. The direction of the electric field is the direction of idler
displacement.
• The magnetic field and the electric field are at right angles to each other, and to the
direction of travel of the disturbance. Maxwell’s machinery is predicting transverse
waves.
GCE AS and A PHYSICS Teachers' Guide 133
6. Electromagnetic Waves (continued)
6.2 Maxwell (continued)
Speed of travel of ‘Vortex’ waves
Maxwell derived a formula for the speed at which the transverse waves would travel, in terms of the
stiffness and the density of the vortex material. But for Maxwell’s ‘machinery’ to reproduce
electromagnetic effects properly, the stiffness and density had to be expressible in terms of constants
which appear in the equations of electromagnet-ism. The wave speed formula could then be written
(using modern notation) as
1
V 
0 0
0 is the permeability of free space, and 0 is the permittivity of free space. V is the speed of the waves
in so-called empty space, where there is nothing (except vortices and idlers!)
Maxwell evaluated the right hand side of this formula using electrical measurements which had
already been made (in Germany). He found:
V = 310 740 000 000 millimetres per second.
He also noted the speed of light, as measured fairly recently in France:
VL = 314 858 000 000 millimetres per second.
He then remarked, in one of the most famous sentences in the history of Physics:
“The velocity of transverse waves in our hypothetical medium, calculated from the
electro-magnetic experiments of MM. Kohlrausch and Weber, agrees so exactly with
the velocity of light calculated from the optical experiments of M. Fizeau, that we can
scarcely avoid the inference that light consists in the transverse undulations of the same
medium which is the cause of electric and magnetic phenomena.”
The Cheshire Cat
The equations relating to Maxwell’s ‘machinery’ could be expressed as relationships between
electromagnetic quantities. Most of them were versions of the known laws of electromagnetism, such
as Coulomb’s Law and Faraday’s Law of electromagnetic induction. One sub-set of the equations,
though, was entirely new. It contained the idea that a changing electric field had lines of magnetic
force curling around it.
Maxwell realised that the equations contained everything that his machinery had to say about
electromagnetism. He kept using the equations and stopped referring to the machinery. [Recalling
Alice’s Adventures in Wonderland, someone later commented that the Cheshire Cat had disappeared,
but its grin remained.] In fact Maxwell still believed that electromagnetic influences did travel by
means of a medium, but he stopped investigating the workings of any particular hypothetical medium.
The equations themselves are enough to predict transverse waves. The waves emitted from a charge
oscillating up and down can be represented as shown, at one instant.
An instant later the ‘profile’ of electric and magnetic fields will have moved to the right. Following the
spirit of Maxwell’s equations, we explain their propagation in this way.. The changing electric field
gives rise to a (changing) magnetic field [Maxwell’s discovery] and the changing magnetic field gives
a (changing) electric field [Faraday’s discovery], and the changing electric field gives a changing
magnetic field and so on. Maxwell was claiming that this was light!
GCE AS and A PHYSICS Teachers' Guide 134
6. Electromagnetic Waves (continued)
6.2 Maxwell (continued)
Self-test Questions
(1) Find out in which country was Maxwell born and brought up.
(2) What was Maxwell trying to do, when he invented his ‘medium’ of vortices and idlers?
(3) (a) What was different in Maxwell’s medium when there was a magnetic field?
(b) What, in terms of vortices, gave the direction of the field?
(c) And what gave the magnitude of the field strength?
(4) What, in terms of vortices and idlers, was an electric field?
(5) What two properties of the vortex medium determined the speed at which waves would
propagate?
(6) Explain the remark about the Cheshire cat. What does its grin represent?
(www.ruthannzaroff.com/wonderland/Cheshire-Cat.htm )
(7) (a) Was anything important lost when Maxwell ‘ditched’ the machinery of the vortex medium
and just kept the equations?
(b) Do equations explain things?
(c) What counts as an explanation in Science?
GCE AS and A PHYSICS Teachers' Guide 135
6. Electromagnetic Waves (continued)
6.3 Hertz
Maxwell’s work commanded great respect, but by no means everyone was convinced it was correct.
What in particular was needed was a direct experimental demonstration that electrical oscillations
could give rise to transverse waves. This, and more, was provided by Heinrich Hertz between 1887
and 1889.
Hertz was working with very high frequency electrical oscillations produced by the apparatus shown
in replica on www.sparkmuseum.com/HERTZ.HTM.
When a spark occurred between the small spheres the air in the gap between them had ‘broken down’
and become a conductor. There was a current in the air gap and the rods either side. This current
dropped to zero, reversed in direction, rose to a maximum, fell to zero, reversed and so on. The
frequency of these electrical oscillations was determined by the system’s inductance (mainly due to
the rods) and its capacitance (mainly due to the large spheres). Hertz estimated the frequency to be 10 8
cycles per second.
When the sparking occurred, Hertz could also see sparks jumping across a narrow gap in a wire ring,
even when the ring was a few metres away. Further investigation strongly suggested that transverse
waves were involved.
Hertz modified his apparatus to improve its range and precision. The ‘transmitter’ lost its large
spheres, and the oscillation frequency increased by about 10 times. Hertz found that the ‘radiation’
could be concentrated into a beam using a concave metal reflector. [See diagram (size of rods and
spheres exaggerated).] To detect the radiation he started using a pair of straight wires with an offset
spark gap. The gap could be adjusted with a micrometer screw. The longer the sparks he could get, the
stronger the electric field.
Stationary Waves
Hertz placed a large flat metal sheet in front of the transmitter and facing it. He moved the detector
between the transmitter and the sheet and reported very distinct maxima and minima. He could
distinguish nodal points at the wall and at 33, 65 and 98 cm distance from it. He concluded that
interference was taking place, leading to a standing wave pattern. Here was clear wave-like behaviour.
• Which two ‘streams’ of waves were interfering?
• What wavelength was Hertz using?
• What was the frequency of the oscillations?
• Which devices in the 21st century use this sort of frequency? ‘v.h.f.’ radios, televisions
with traditional (spiky) aerials, or microwave ovens?
GCE AS and A PHYSICS Teachers' Guide 136
6. Electromagnetic Waves (continued)
6.3 Hertz (continued)
Polarisation
No sparking occurred in the detector when it was turned
so its wires were horizontal, as shown. Hertz deduced that
the waves were polarised, with the electric field direction
parallel to the rods in the transmitter (as predicted by
Maxwell’s equations). Clearly they were transverse
waves.
With the detector wires vertical again, Hertz interposed a
grille of parallel wires between the transmitter and
detector. The detector sparking was unaffected when the
wires were horizontal, but no sparks could be had when
the grille was turned so that the wires were vertical.
• What special material, containing parallel
molecules, can do for light what Hertz’s grille
of wires did for u.h.f. waves?
Refraction
Hertz’s account (translated by D E Jones) began
“In order to find out whether any refraction of the ray takes place in passing from air to another
insulating medium, I had a large prism made of so-called hard pitch, a material like asphalt. The base
was an isosceles triangle 1.2 metres in the side, and with a refracting angle of nearly 30°. The
refracting edge was placed vertical, and the height of the whole prism was 1.5 metres. But since the
prism weighed about 12 cwt [600 kg or 0.6 tonne], and would have been too heavy to move as a
whole, it was built up of three pieces, each 0.5 metres high, placed one above the other.”
What Hertz found is summarised in the plan above. Observe that he was now using a concave reflector
behind his detecting wires as well as behind the transmitting rods.
• Hertz calculated the refractive index of the pitch as 1.69. Check this figure, by drawing relevant
normals and calculating angles. Note the symmetry.
Consequences
Hertz’s findings were soon accepted as establishing the reality of electromagnetic waves. The
possibility of using the waves for communication was taken up by several people, most famously by
Guglielmo Marconi. On 12th December 1901, he reported that signals sent from Cornwall had been
received in Newfoundland. And now we have radio, television and mobile phone technology, all based
on electromagnetic waves.
GCE AS and A PHYSICS Teachers' Guide 137
7. Assault on the Ether
7.1 The Triumph of the Ether?
If light is a wave, surely it has to have a medium to travel in? This was the compelling reason for
belief in the existence of a ‘luminiferous ether’ from about 1820 onwards. [Revise section 4.4] In the
1860’s Maxwell showed that a medium with the right structure might be able to account for
electromagnetic effects – of which light was one. Hertz’s work in the late 1880s seemed to confirm
Maxwell’s ideas. At least in Britain, few physicists doubted the existence of the ether, though its
structure was … debatable.
There was no direct evidence for the ether’s existence. But Maxwell realised that in principle such
evidence was available… Since the time of Galileo people had stopped believing that the Earth was
the stationary centre of the universe, so it would be odd to think of the ether as stationary relative to
the Earth. Stationary relative to the Sun seemed a much better bet. So as the Earth moves round the
Sun it is presumably also moving through the ether, and the motion is in principle detectable.
• The Earth’s orbit is roughly a circle of radius 1.5  1011m. Show that the Earth’s orbital speed is
3.0  104 m s1. What is this as a fraction of the speed of light?
7.2 The Michelson Morley Experiment (1887)
The challenge of detecting the motion of the Earth through the ether was taken up in America by
Albert Michelson.
(http://nobelprize.org/nobel_prizes/physics/laureates/1907/michelson-bio.html )
Michelson designed a piece of apparatus which came to be called an interferometer. The semi-silvered
plate acted as a beam-splitter, so light travelled from source to telescope by two routes: SOAOT and
SOBOT. Interference occurred between the light taking the different routes.
Suppose that the apparatus happened to be orientated so that the interferometer was moving to the left
through the ether. This is equivalent to the ether moving to the right past the apparatus, at velocity v,
say. [Think of a plane in a wind-tunnel.] As a result the observed interference pattern was expected to
change if the apparatus was turned about a vertical axis (see next page).
After an inconclusive first experiment, Michelson, joined by E.W. Morley (who had been making
precision measurements in a quite different area of science), redesigned the apparatus. It was now
mounted on a massive concrete block, floating in mercury, so it could be turned smoothly and was not
affected too badly by vibrations. By using multiple reflections, the effective length, L, of each arm,
OA and OB, was made to be 11 m.
GCE AS and A PHYSICS Teachers' Guide 138
7. Assault on the Ether
7.2 The Michelson Morley Experiment (Continued)
Light taking the route OAO
‘Ordinary’ waves, such as sound waves, travel at a fixed speed relative to their medium, so it was
assumed that light would travel at a fixed speed, c, relative to the ether. If the ether is itself rushing
past the apparatus at velocity v then the light should travel ‘downstream’ (OA) at velocity (c + v) and
‘upstream’ (AO) at velocity (c – v) relative to the apparatus (vector addition). The total time for the
light to travel OAO is therefore
L
L

c v c v
If there were no ‘ether wind’ the total time for AOA would be
2L
c
So the extra time taken to travel AOA because of the ether wind’ is
L
L
2L
tOAO 


cv cv c
8
1
Putting L = 11 m, c = 3.00000  10 m s , and v = 3.0  104 m s1 [Why?] gives
tOAO  7.3 1016 s
• You should check this. Try also using a better figure for c e.g. c= 2.99792  108 m s1.
Light taking the route OBO
In this case the ether wind is at right angles to the ‘forward’ and ‘back’ paths OB and BO. According
to vector addition the velocity of light relative to the apparatus is reduced. However the delay due to
the ether wind turns out to be only half as much as for OAO.
In other words
tOBO  3.6(5) 1016 s
So light will return to O, and from there to the telescope, in a shorter time via B than via A. The
difference in times is (tOAO – tOBO) which is 3.7  10-16 s.
Michelson and Morley took the wavelength of the light as 5.5  10-7 m, corresponding to a frequency
of 5.5  1014 cycles per second (5.5  1014 Hz).
Number of cycles occurring in 3.7  1016 s = 5.5 1014 Hz  3.7  1016 s = 0.20 cycles
Expected results and actual results
The light travelling via A is therefore delayed by 0.2 cycles compared with that via B. Suppose that
the apparatus is turned through 90°. The route OBO will now be the slower one, so the change in
delay will be 0.2 cycles – (–0.2 cycles) = 0.4 cycles. If, in the original orientation, A had been moved a
minute amount towards O so as to give full constructive interference between the light travelling the
two routes, then on turning the apparatus through 90° there would be almost complete destructive
interference.
In fact, Michelson and Morley had the apparatus adjusted so that A and B were not quite at right
angles to each other. This meant that the telescope revealed a pattern of parallel bright and dark fringes
much like Young’s fringes. When the apparatus was rotated through 90°, a 0.4 cycle change in delay
would make the fringe pattern shift by 0.4 of a fringe, so a bright fringe would almost be replaced by a
dark one and vice versa.
In fact hardly any fringe shift was observed. But OA might not have been parallel to the ether wind in
the first place, so Michelson and Morley kept the apparatus slowly turning, and examined the fringe
pattern at 16 orientations of the apparatus. They repeated the observations at different times of day and
night and at different times of year. The maximum shift they found was about 0.01 of a fringe. This
was, to all intents and purposes, negligible.
GCE AS and A PHYSICS Teachers' Guide 139
7. Assault on the Ether
7.3 After Michelson Morley
What do scientists do when a successful theory is contradicted by experimental evidence? Give up the
theory? This is not usually the first response. It certainly wasn’t when the Michelson Morley
experiment gave a null result. Rather than give up the idea of the ether, physicists tried to think up
explanations for why the ether did not show up in the experiment. Here are the two most famous…
• The ether in the neighbourhood of the Earth is dragged along by the Earth, rather as a moving ship
is surrounded with a layer of stationary water. So even though the Earth is moving around the Sun,
and even if the whole solar system is moving through the ether, there will be no ether wind on the
Earth’s surface.
The trouble with this idea was there were other effects which the ether theory could explain, but
only if the ether moved freely past the earth! [For interest only: the main such effect was ‘stellar
aberration’, which is explained in Banesh Hoffmann’s book, Relativity and its Roots – even though
‘aberration’ is not in the index.]
• As well as the ether wind changing the velocity of light relative to the interferometer, it also
changes the shape of the interferometer in a way which exactly neutralises the extra delay on the
upstream-downstream arm due to the velocity changes. This idea was put forward in 1889 by
George Fitzgerald, who was working in Dublin.
A similar explanation was offered independently by the Dutch physicist Hendrik Lorentz some
three years later. He claimed that the only change in shape was a contraction of the ‘upstreamdownstream’ arm. Using Maxwell’s equations, and making various assumptions about electrons
and the role of electromagnetic forces in holding matter together, he argued that all objects should
contract in the direction parallel to the ether wind.
7.4 Einstein
(http://www-groups.dcs.stand.ac.uk/~history/Biographies/Einstein.html)
Albert Einstein was barely known to the world of Science
until, in 1905, at the age of 26, three major papers by him
were published in the prestigious German scientific
journal Annalen der Physik. They have been described as
setting the agenda for Physics for the next hundred years.
The first paper contained the curious notion that light
might sometimes behave as if it consisted of packets of
energy, and the prediction of the photo-electric equation.
The second showed how to demonstrate the existence of
molecules by observations on Brownian motion.
The third paper was called (in translation) On the
Electrodynamics of Moving Bodies. It lays the
foundations of what is now called the Special Theory of
Relativity.
The theory takes two innocent-looking starting points, or ‘postulates’ and builds on them in a
ruthlessly logical fashion, to come to some momentous conclusions. The first postulate is the Principle
of Relativity…
7. Assault on the Ether
7.4 Einstein (Continued)
The Principle of Relativity
The laws of Physics are the same in all inertial frames of reference. All such frames are equivalent.
GCE AS and A PHYSICS Teachers' Guide 140
The Principle generalises the finding that mechanics experiments give exactly the same readings when
performed in a laboratory on a smoothly moving, non-accelerating train or aircraft as they do in a
‘stationary’ laboratory. The train and the plane are (approximately) ‘inertial’ frames of reference –
ones in which a body at rest stays at rest, provided no resultant force acts on it. Inertial frames are the
only ones we use in A-Level Physics.
The generalisation that Einstein makes is that all laws of Physics, including those of electromagnetism
and light, hold good in all inertial frames. There is no privileged ether frame of reference, and in all
inertial frames of reference light travels through empty space at speed c given by
1
c
 0 0
But suppose the light source is moving towards the observer (in a laboratory equipped with rulers and
light-activated clocks accurate to the picosecond!). This is the same thing as saying that the observer is
moving towards the light source. Will the observer then measure a larger speed for the light? Not
according to Einstein…
Second Postulate
The speed of light is independent of the motion of its source.
• The Michelson Morley experiment was repeated in the 1920s using starlight and sunlight, rather
than a lamp in the laboratory. What was the point of doing these further experiments? [Again, no
directional differences were observed.]
The time of an event
To deduce things from the postulates we have to be very precise about what we mean by the time
when an event takes place. Einstein wrote [translated from the German]…
We have to take into account that all our judgements in which time plays a part are always judgements
of simultaneous events. If, for instance, I say, “That train arrives here at 7 o’ clock.”, I mean
something like this: “The pointing of the small hand on my watch to 7 and the arrival of the train are
simultaneous events.”.
It might appear possible to overcome all the difficulties attending the definition of ‘time’ by
substituting ‘the position of the small hand of my watch’ for ‘time’. And in fact such a definition is
satisfactory when we are concerned with defining a time exclusively for the place where the watch is
located, but it is no longer satisfactory when we have to connect in time a series of events occurring in
different places.
The way to deal with the problem is to have a clock at every place where an event might happen – and
to make sure the clocks are synchronised. We can in principle put a whole line of clocks down the
laboratory (or, if needed, a three-dimensional array of clocks).
Suppose we have a line of clocks at intervals of 0.30 m… How do we synchronise them? Before
starting them we could adjust their displays to read 0.0, 1.0 ns, 2.0 ns and so on as we go from left to
right along the line. We then trigger them to start by a light signal originating at the left hand end of
the line and travelling down the line past each clock.
•
How does this work? How long does it take the light signal to travel 0.30 m?
GCE AS and A PHYSICS Teachers' Guide 141
7. Assault on the Ether
7.4 Einstein (Continued)
A though-experiment
We shall look at the consequences of Einstein’s postulates in one
specific case, where we can imagine an experimental set-up and work
out what – according to the postulates – must happen. In other words we
shall do a thought-experiment (or Gedankenexperiment). [The term is
believed to have been invented by Ørsted!]
Suppose a light flashes close to the end, A, of a rod of length L. The
flash triggers a nearby clock to start. The light reflects off a mirror at the
other end, B, of the rod. When it has returned to A it triggers the clock to
stop. If the clock records a time interval  between the events of the
light leaving A and returning to A, then:…
2L = c .
Now suppose that the rod is actually moving through a laboratory at
speed v at right angles to itself. [This won’t affect the previous equation,
since this was written for the rod’s own frame of reference.]
In the laboratory frame of reference the light travels the path PQR (see
diagram below).
Clearly in the laboratory frame the light has had to travel further between its leaving the end A of the
ruler and its arriving back there. But the speed of light is the same in all frames. So the time, t,
between the light leaving A and the light returning to A must be greater than the time, , between the
same two events as measured in the rod’s frame of reference. This effect is called time dilation. Time
intervals are different for different observers!
GCE AS and A PHYSICS Teachers' Guide 142
7. Assault on the Ether
7.4 Einstein (Continued)
A thought experiment (continued)
How different are the time intervals measured in the two frames? It is easy to find out…
In the laboratory frame the rod moves a distance v t in the time t. But the light has travelled distance
c t.
So, applying Pythagoras’s theorem to either of the two right angled triangles in the diagram…
L2   12 ct    12 vt 
Substituting for L from the rod frame equation,
2

2
12 c   21 ct   12 vt 
2
2
2
c    12 ct    12 vt 
Doing the squarings and multiplying through by 4,
1
2
2
2
2
c 2     c 2  t   v 2  t  that is c 2      c 2  v 2   t 
2
2
2
2
2

Dividing both sides by c2 and then taking the square roots of each side,     1  vc2
2
2
  t 
2
so
  1  vc2 t
2
Finally, dividing both sides by the square root,
t 

1  vc2
2
• Try to calculate t if  = 1.000000000 ms and v = 1000 ms-1. Check that your calculator gives
t = 1.000000000 ms. Time dilation doesn’t show up, even to 10 significant figures, for a relative
velocity of three times the speed of sound between the frames of reference. No wonder it is such an
unfamiliar idea!
• But for relative velocities approaching the speed of light the effect is large. Calculate  if  =
1.00 ms and v = 3/5 c.
• The formula doesn’t just apply to this set-up, or just to flashes of light. The time interval between
any two events is greater in a frame of reference in which the events occur in different places than
in the frame in which they occur at the same place. The time interval in this latter frame is called
the proper time interval, .
• The effect has been confirmed by experiment and the formula checked.
(www.youtube.com/watch?v=gdRmCqylsME : a bit incoherent, but gives the idea.)
What causes time dilation?
Are measurements made in the rod’s frame of reference invalid because it is moving? No, all inertial
frames of reference are equivalent; none is favoured. In any case, an exper-imenter on the rod is
perfectly entitled to say that it is the laboratory which is moving!
Has the electronics of the rod’s clock been affected by its motion? No; the laws of Physics are the
same in all inertial frames. Nothing is different about the way the clocks run in the two frames.
Anyway, in the rod’s frame it is the laboratory clocks that are moving.
So what does cause time dilation? It is the non-independence of space and time, as acknowledged in
the term, space-time. The time interval between two events is least when measured in a frame of
reference in which the events occur at the same place. It is greater in a frame of reference in which the
events occur in different places, and there-fore have to have their times recorded by different clocks.
[Note the line of synchronised clocks shown in the laboratory frame diagram.]
[In fact what stays the same for two events if we go from one inertial frame to another is the quantity
c2 ()2 = {c2(t)2 – (x)2– (y)2– (z)2}. This looks a bit like Pythagoras’s theorem in 4 dimensions,
which is why time is sometimes called the fourth dimension.]
GCE AS and A PHYSICS Teachers' Guide 143
7. Assault on the Ether
7.4 Einstein (Continued)
Special Relativity results: not for learning
The time dilation thought-experiment was, as promised in the introduction, a small taste of Special
Relativity theory – but no more than a taste. Many other results, some of them equally extraordinary,
can also be deduced from the two postulates. The results include…
• Events which are simultaneous at different places in one frame of reference are generally not
simultaneous in other frames (moving with respect to the first).
• A ruler stationary in one frame of reference is shorter in any other frame moving parallel to the
ruler.
This might remind the alert reader of the idea of Fitzgerald and Lorentz (section 7.3) that the
upstream-downstream arm in the Michelson Morley experiment was shortened by its motion
through the ether. In fact some of Lorentz’s mathematics was very similar to Einstein’s – and
published before 1905, but his interpretation of the equations was very different. For example,
Lorentz argued that the ruler would be contracted in length compared with its length if stationary in
the ether frame of reference. According to Relativity theory there is no such special frame.
• Lengths at right angles to the relative velocity between frames of reference are the same measured
in either frame. We have already assumed this. Where?
• No object can move faster than the speed of light – in any frame.
• Some of the formulae of Newtonian mechanics have to be altered. The alterations usually only
make any significant difference for bodies moving very fast indeed (say above 1.0  106 m s-1).
• Mass and energy are equivalent. The conversion factor between units is c2.
• Maxwell’s equations are valid in all frames of reference, not just in one special ether frame, as
Maxwell and Hertz believed.
• The strengths of components of magnetic and electric fields vary according to reference frame. We
illustrate this with a thought-experiment…
If you want to learn more, Relativity and its Roots by Banesh Hoffmann is helpful.
Should you want to do some serious study, consider an internet search for a secondhand copy of An
Introduction to Special Relativity by James H Smith.It’s an old book (first published 1965) but it’s
down to earth and the explanations are very clear.
GCE AS and A PHYSICS Teachers' Guide 144
7. Assault on the Ether
7.5 Einstein (Continued)
Electrons moving side-by-side: a thought-experiment
Suppose two electrons emerge simultaneously with the same high velocity from electron guns side by
side. The electrons will repel each other so their paths will diverge.
Consider the ‘top’ electron. Home in on two events: (a) the electron leaves the gun, and
(b) the
electron hits a target, having been displaced upw
repulsive force. In the laboratory frame (left hand diagram) these events are spatially far apart, and the
Now imagine we could ride along with the electrons, keeping pace with their horizontal motion (and
clutching a clock!). In this new frame of reference (the ‘electrons’ frame’), events (a) and (b) are only
slightly separated in space (by distance y). The time interval between the events is to all intents and
purposes the proper time, .
Since t is greater than  we must conclude that the repulsive force between the electrons is less in
the laboratory frame than in the electrons’ frame. The force reduction will be very small unless the
electrons’ speed approaches the speed of light.
Is this some weird new phenomenon? No – we can predict it from A-Level Physics, by a quite
different line of reasoning…
We first note that in the electron’s frame, the only force between the electrons is the repulsive
Coulomb force or electrostatic force. [Gravitational forces are negligible.]
In the laboratory frame the repulsive Coulomb force acts, but there is also an attractive force, because
moving charges constitute electric currents, and like currents attract. We could call this the Ampère
force. [See section 5.2.] We could also call it the magnetic force, because each electron sets up a
magnetic field and the other electron moves through this field and experiences a Motor Effect (BIl)
force.
This Ampère force is much smaller than the Coulomb force unless the electrons’ speed is approaching
that of light. All the same, the resultant force between the moving electrons will be slightly less than
that between stationary electrons.
But we arrived at this conclusion by a time dilation argument, without any appeal to Ampère forces:
that is to magnetic forces. It looks as if the only fundamental force between charges is the Coulomb
force, and magnetic forces are an effect due to measuring the force between charges in certain
reference frames. Quite an insight!
[To tell the whole truth, electric fields strengths also, generally, change according to the frame of
reference. In the thought-experiment, the repulsive Coulomb force between the electrons is actually
slightly increased when they are moving relative to us. But the attractive Ampère force is a greater
effect. If we take account of both effects the reduction in repulsive force turns out to be exactly as
calculated from the time dilation argument.]
GCE AS and A PHYSICS Teachers' Guide 145
7. Assault on the Ether
7.6 Einstein (Continued)
Relativity and the Ether
Clearly, if all inertial frames of reference are equivalent and the speed of light (in empty space) is the
same in all frames, then the Michelson-Morley experiment would have had to give a null result: there
being no difference in the speed of light to detect!
• It would, though, be misleading to say that the Special Theory of Relativity explains the null result
of the experiment. Why?
Einstein was motivated at least as much by wanting to tidy up the theory of electro-magnetism as by
the null result of the Michelson-Morley experiment. In particular he did not like the idea of equations
which were valid only in a special frame of reference.
The Special Theory of Relativity held together well and made predictions which were confirmed in
the years that followed. The theory made no use of the idea of an ether, and took as a starting point the
non-existence of a special ether frame of reference. Physicists came to see the ether as being of no use
in explaining anything. They stopped believing in it.
• Why, in the first place, did the ether become such an important part of nineteenth century thinking
on light? What role was it supposed to play?
• What major discovery arose from using an ether theory of electromagnetic fields?
Some quick Revision
(1) Name the two scientists who did most to establish the wave theory of light between 1800 and
1820.
(2) Name three scientists who made discoveries in electromagnetism between 1820 and 1840, and
make summaries of what they discovered.
(3) Whose concept in (electro)magnetism did James Clerk Maxwell build upon when he started to
develop his idea of a vortex medium? What was the concept?
(4) What was the inference from the calculated value of
1
 0 0
which, according to Maxwell, ‘we
can scarcely avoid’?
(5) What were Einstein’s two postulates on which he based the Special Theory of Relativity?
(6) What is meant by time dilation?
What happened next?
In the 21st century we still believe that light travels like a wave, with oscillating electric and magnetic
fields, and that it doesn’t need a medium.
When it comes to understanding how light interacts with matter we need to think of light as photons.
The most highly developed theory we have in Physics is called Quantum Electrodynamics (QED). It
explains electromagnetic forces in terms of the exchange of photons.
If you’re remotely interested, read: QED: The Strange Story of Light and Matter by Richard
Feynman. He was one of the inventors of the theory, a colourful character, a brilliant teacher and
writer – and one of the greatest physicists of the twentieth century.
GCE AS and A PHYSICS Teachers' Guide 147
Appendix C
Superalloys Notes
GCE AS and A PHYSICS Teachers' Guide 148
INTRODUCTION
Aircraft jet engines are required to operate within extreme conditions of temperature and
pressure. Jet engine turbine blades rotate at a typical speed of 10,000 rpm for long periods in
an environment of combustion products at working temperatures of 1250 ºC (though the inlet
temperatures of high performance engines can exceed 1650 ºC), non aviation gas turbines
operate at approximately 1500 ˚C. The blades must be able to withstand impact and erosion
from debris drawn in with the air stream. In addition, different parts of the blade may be at
different temperatures and they will be subjected to large and rapid temperature changes
when the engine is started up and turned off.
The following is a list of the properties required of the material from which the blades are
made.
CREEP RESISTANCE.
Centripetal forces acting on the blade at high rotational speeds provide a considerable load
along the turbine blade axis. Over prolonged periods of time this can cause creep. It
becomes increasingly pronounced as temperature increases. Creep could cause a turbine
blade to deform sufficiently that it might touch the engine casing.
CORROSION RESISTANCE.
Iron corrodes to form rust. At high temperatures, the presence of carbon dioxide, water
vapour and other products of the combustion of fuel constitute a highly corrosive
environment.
TOUGHNESS.
The blades must resist impact with debris passing through the engine. In addition, stresses
generated by expansion and contraction, between different parts of the blade at different
temperatures, must not give rise to cracking.
MECHANICAL AND THERMAL FATIGUE RESISTANCE.
Variations of gas pressure and temperature on different parts of a blade and mechanical
vibrations, may generate cyclical stresses which can cause failure due to fatigue.
METALLURGICAL STABILITY.
The mechanical properties of metals can be modified by heat treatment. Blade materials
must be resistant to such changes and the microstructure must remain stable at high
temperatures.
DENSITY.
The density must be low to keep engine weight as low as possible.
Metallurgists have developed superalloys to meet these stringent specifications. Nickelbased superalloys are able to withstand stresses of over 250 MPa for 30 hours at 850˚C with
less than 0.1% irreversible creep. The alloy contains 18 constituents and its composition is
shown below:
GCE AS and A PHYSICS Teachers' Guide 149
Table 1: Composition of creep-resistant turbine blade superalloy
Element/wt%
Nickel
Cobalt
Tungsten
Chromium
Aluminium
Tantalum
Titanium
Hafnium
Iron
Ni
Co
W
Cr
Al
Ta
Ti
Hf
Fe
59
10
10
9
5.5
2.5
1.5
1.5
0.25
Element/wt%
Molybdenum
Carbon
Silicon
Manganese
Copper
Zirconium
Boron
Sulphur
Lead
Mo
C
Si
Mn
Cu
Zr
B
S
Pb
0.25
0.15
0.1
0.1
0.05
0.05
0.015
<0.008
<0.005
Why the need for so many elements and what does each do?
Alloying elements (re-introducing ‘foreign’ atoms such as carbon atoms in steel) disturb the
regularity of the lattice and by doing so hinder the movement of dislocations.
Also dislocations have difficulty in moving across grain boundaries.
In superalloys, obstacles in the form of insoluble precipitates are introduced to hinder
dislocation movement. Some elements are also introduced to reinforce grain boundaries.
Some elements form a layer of oxide (Cr2O3 in this case) on the blade surface which greatly
improves resistance to corrosion.
Cobalt, tungsten and chromium are all soluble in Nickel (which crystallizes in a cubic closed
pack or face centred cube structure). Their atoms are different in size from nickel atoms and
they are distributed randomly throughout the nickel matrix – they are in solid solution and
they produce irregularities in the crystal.
Any irregularity in the crystal will hinder the progress of a dislocation. Strains set up around
substitute atoms impede the movement of dislocations. As the temperature of a metal
increases, creep becomes more pronounced. More vacancies are introduced as temperature
rises and diffusion of atoms is easier. Additives which made diffusion more difficult,
therefore, enhance creep resistance.
Hard precipitates are more resistant to deformation. Their presence in a softer matrix makes
it more difficult for planes to slip over one another. We can think of them as acting like
boulders thrown into a stream to dam the flow of water. It is also difficult for dislocations to
pass through regions with closely spaced hard precipitates. A dislocation passing through
such a region has been compared with trying to blow up a balloon in a bird cage. It is
difficult for the balloon membrane to pass through the bars. The dislocation is similarly
confined by the precipitates in superalloys.
GCE AS and A PHYSICS Teachers' Guide 150
Fig. 1 Effect of matrix strengtheners (precipitates) on
dislocation movement
Cobalt, tungsten and chromium thus act as matrix strengtheners.
Aluminium and titanium react with the Nickel to form stable compounds such as Ni3Al and
Ni3Ti.
The crystal structure of Ni3Al and Ni3Ti have a similar packing arrangement to the original
(Nickel based) matrix and do not disrupt the regularity of the original matrix to which they
are bonded. However, these particles are extremely hard and are very resistant to sheer
deformation. Metalurgists sometimes call this process the Gamma prime (’) phase.
[Note: Molybdenum, tantalum, tungsten and titanium also form Carbides MoC, TaC, WC
and TiC. These are very hard materials which can also act as obstructions to matrix
deformation. ]
Other Carbides, mainly based on Chromium (CrC3) accumulate at the grain boundaries in the
matrix. It is believed that they strengthen these regions by reducing the formation of
cracks which lead to failure. The grain boundaries become the source of weakness
following the strengthening of the matrix. If they are filled with hard materials, which are
firmly bonded to the grains, it becomes more difficult to shear the grains apart. Grain
boundary strengtheners act rather like the cement in crazy paving.
Some elements may have a number of functions in strengthening the alloy. Chromium is a
good example. Chromium forms a layer of oxide (Cr2O3) on the blade surface which greatly
improves resistance to corrosion.
Table 2 summarises the purpose of each of the alloying elements.
Table 2: Functions of different elements used in
super-alloys
Purpose
Matrix strengheners
Gamma prime formers
Carbide formers
Oxide scale formers
Grain
boundary
strengtheners
Cr

Al
Co

Mo W
Ti
Ta
Nb
Hf












C
B
Zr





GCE AS and A PHYSICS Teachers' Guide 151
Processing developments
Turbine blades were shaped by forging before the introduction of super-alloys, but the new
materials were so hard that they could not be forged or easily shaped and so had to be
produced by casting into moulds.
Conventional (early) casting processes produced blades with a fire grain structure (fig. 3).
Fig. 3 Conventional as-cast grain structure
The grains were formed during solidification in the casting mould. Each grain had a different
orientation of its crystal lattice from its neighbours.
The weakest parts of the structure were the grain boundaries and blade failure often occurred
at these points due to slippage (under stress and creep).
In the 1960s researchers at the jet engine manufacturer Pratt and Whitney set out to eliminate
grain boundaries from turbine airfoils altogether.
The first development was to produce blades with all the grains aligned parallel to the stress
axis by a process called directional solidification.
A mould of molten metal is enclosed in the hot zone of a furnace and the heat is gradually
removed from the bottom of the mould. As the mould is gradually removed from the furnace,
columnar grains develop along the axis (only) of the blade (fig. 4).
The final result is a turbine airfoil composed of columnar crystals or grains running along the
length of the blade. For the case of a rotating blade subjected to centripetal forces of the
order of 20,000g, the grains are now aligned along the major stress axis. Their alignment
strengthens the blade and eliminates intergranular crack formation in directions normal to the
blade span.
The improved creep properties of directional solidification blades allowed engine
temperatures to be increased by approximately 50˚C with further improvement in efficiency.
GCE AS and A PHYSICS Teachers' Guide 152
Fig. 4 Directionally solidified grain structure
Building upon directional solidification, Pratt and Whitney reached the goal of eliminating
turbine airfoil grain boundaries in the late 1960s.
Complete elimination of grain boundaries has further advantages. A blade without grain
boundaries is a single crystal. Creep resistance is improved and because there are no grain
boundaries, the grain boundary strengtheners such as Carbon, Boron, Zirconium and Hafnium
used in early superalloys are not needed. All these elements contributed to the lowering of
the melting point of the blade. The development of single crystal blades has increased the
melting temperatures of turbine blades by a further 150˚C  200˚C.
Single crystal blades are grown by incorporating a geometrical construction called a ‘pigtail’
into the base of the mould. The pigtail is attached to the ‘starter’ crystal and is helical in
construction. It admits only a few columnar crystals from the starter and only allows one
crystal to emerge into the blade root, to start the single crystal structure of the airfoil itself.
This early pioneering work has been taken over by other manufacturers and improved upon
over the past 30 years. Yields (successfully manufactured single crystals with no defects)
greater than 95% are now commonly achieved in the casting process.
Early on, Pratt and Whitney investigated single-crystal turbine airfoil use in various jet
engines. (One of the first was the J58, which powered the Lockheed SR-71 Blackbird). The
very first actual engine use was in Pratt & Whitney’s JT9D-7R4 which received jet engine
flight certification in 1982. This first single-crystal bladed engine powers the Boeing 767 and
the Airbus A310.
In jet engine use, single-crystal turbine airfoils have proven to have as much as nine times
more relative life in terms of creep strength and thermal fatigue resistance and over three
times more relative life for corrosion resistance, when compared to the original cast, granular
blades. Modern high turbine inlet temperature jet engines with long life (that is, 25,000 hours
of operation between overhauls) would not be possible without the use of single-crystal
turbine airfoils. By eliminating grain boundaries, single-crystal airfoils have longer thermal
and fatigue life, are more corrosion resistant, can be cast with thinner walls – meaning less
material and less weight – and have a higher melting point temperature. These improvements
all contribute to higher efficiencies.
GCE AS and A PHYSICS Teachers' Guide 153
The newest chapter of the story is their recent introduction in new, large land-based gas
turbines. Gas turbines used to produce electric power in the 200 to 400 MW range have
turbine airfoils that can be 10 times larger than jet engine turbine airfoils. These large
castings have had production problems in the industry, causing casting yields to go down,
driving costs up.
As an example, one 1999 study done for the U.S. Department of Energy, found that for a
$6,000, 13.6 kg single-crystal blade, a 90 percent yield would raise the cost to $7,000, while
a 20 percent yield would shoot unit costs up to $30,000 each. Much work has been going on
in the casting industry to increase yields for these large turbine blades.
General Electric’s 9H, a 50 Hz combined-cycle gas turbine, is the world’s largest. The first
model went into service in 2003 at Baglan Bay on the south coast of Wales, feeding as
much as 530 MW into the United Kingdom’s electric grid at a combined-cycle thermal
efficiency just under 60 percent. The 9H, at 367,900 kg, has a first-stage single-crystal
turbine vane with a length of 30 cm and first-stage single-crystal blade of 45 cm (the blade
lengths in the PW JT9D-7R4 are about 8 cm). Each finished casting weighs about 15 kg and
each is a single crystal airfoil.
Paul Edwards. September 2007.
References:
Cooke (B) and Sang (D) (1989). Physics of Materials for A-level students (2nd ed). Leeds
University. Leeds press.
www.memagazine.org. A feature article by Longston (Lee.S); ‘Crown Jewels – these
crystals are the gems of turbine efficiency’. Feb 2006.
GCE AS and A PHYSICS Teachers' Guide 155
Appendix D
Ductile Materials Notes
GCE AS and A PHYSICS Teachers' Guide 156
The plastic behaviour of ductile metals
Recap on terms:
Elastic deformation is a change of size or shape which is reversed when the deforming stress
is removed.
If an object undergoes plastic deformation it doesn’t return to its original size or shape when
the stress is removed, but suffers permanent set.
The key to understanding the behaviour of ductile metals at stresses above the elastic limit
lies in the almost perfectly regular nature of their structure. They are crystalline metals; that is
they possess long-range order, consisting of metal ions arranged in flat planes. Each ion is
surrounded by the same number of ions as all the others, usually 12, in an arrangement
known as close packing.
A two dimensional representation would be:
In this diagram, the black circles represent ions of the lattice, and each is surrounded by 6
neighbours in the plane, and this is just part of one of the planes of ions, with other planes
situated above and below it. The diagram below shows a (grey) plane of ions below.
I’m not going to attempt adding a layer above!
In these diagrams, the size of the ions should not be taken too seriously. It would be better to
think of them as touching one another, but the diagrams are more difficult to cope with. The
ions are bonded into the lattice by bonds between them. [If you’ve covered PH2 already, you
may know that the “glue” that holds them in place comes from free conduction electrons
which move around in the lattice, but that is an issue for another day!]
GCE AS and A PHYSICS Teachers' Guide 157
In the rest of this article, we are going to be thinking more about the bonds between the atoms
than about the atoms themselves, so it will be better to represent the planes of atoms like this:
We are not going to be concerned with the precise atomic arrangement, so we’ll stick to a
square array. The diagrams are easier to draw and to interpret. There are also horizontal
bonds between the ions, but these have been left off to make the diagram simpler.
Polycrystalline materials
This is really a note for the purists. Strictly, we should refer to metals as polycrystalline,
rather than crystalline. They consist of an interlocking mass of individual crystals. The
orientations of the lattice planes in neighbouring crystals are independent of one another.
Plastic ‘slip’
Sometimes textbooks describe plastic strain in terms of crystal planes sliding over one
another.
force
Plane x
Plane y
force
According to this idea, when a force is applied, every atom in plane x has to break bonds with
atoms in plane y and then form new ones with different atoms in plane y. This process does
not happen however because is would requires stresses many times greater than obtained in
practice. Real metals are not as strong as this model suggests. Edge dislocations can account
for this difference.
Edge dislocations
No one puts these crystals together. They grow spontaneously as the molten metal cools
down when it is smelted. Sometimes mistakes happen as the metal ions join the crystal. A
frequent mistake – one every million atomic planes or so – is that half a plane of atoms is
missed out. This is known as an edge dislocation [see next page].
GCE AS and A PHYSICS Teachers' Guide 158
X
Edge dislocations are the key to plastic deformation. The secret is in the bonds around the ion
X. These bonds are all under strain, so they are places of weakness. Suppose forces are
applied to this region as in the diagram below:
F
X
Plane x
Plane
y
F
If the forces, F, are small, the horizontal bonds are stretched reversibly [if the force is
removed, they contract again] and the material behaves elastically.
If the forces become larger, the already strained bonds below and to the right of X [shown in
red] are stretched even more and at some point, the Yield Point of the material, they snap
making the dislocation migrate to the right:
F
X
F
GCE AS and A PHYSICS Teachers' Guide 159
If the same force that causes this movement continues to be applied, the dislocation carries on
moving, through the snapping and reformation of bonds, until it reaches the edge of the
crystal.
In this way, the dislocation has moved from left to right through the crystal. It appears as if
plane x and plane y have slipped over each other. However, it has been achieved more easily
since only one (short) line of bonds, rather than a whole plane, has been broken at a time.
The crystal is now more elongated in the horizontal direction. If the forces are removed, the
dislocation does not move back as there is no force to make it do so, and so the deformation
is permanent.
In the plastic region of deformation, as the bonds are not being stretched, the material does
not increase in volume as it stretches. It contracts laterally. This “necking” can be seen
dramatically if the break point in a ductile fracture is examined. Signs of flow, rather like a
fluid, can be seen around the break point.
In the diagram, the crystal does not look dramatically longer than it did before, but imagine
the effect of a large number of edge dislocations at the top of the crystal migrating to the right
and a similar number of edge dislocation at the bottom moving to the left. The crystal is more
elongated (in the horizontal direction).
And finally [not for examination]
Of course, this isn’t the whole story. With a small amount of increased stress, new edge
dislocations can also be created; especially in locations where there is an atom missing from
the lattice [it happens!] or an impurity atom has a different size from the others. The extra
edge dislocations can produc/EDe more and more plastic strain as they migrate. For a fuller
treatment see Gordon.
Reference: THE NEW SCIENCE OF STRONG MATERIALS
Why You Don't Fall Through the Floor
by J. E. Gordon
London, 1976.
GCE Physics – Teacher Guidance/ED
11 January 2008/HT/11/11/2013