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Exam Techniques

In-exam hints

There are many sources of information on how best to approach exams, but this section examines hints that are specific to the HSC Physics exam.

Read the question first

This is easily the most important thing to remember. Often the question will not be exactly what is written in a syllabus dotpoint. This means that you will have to shift the emphasis of your answer, or even omit irrelevant information, in order to most efficiently get full marks.

Pick out key verbs

Tying in with reading the question first, make sure that when you read the question you look out for the key verbs. It probably isn’t necessary to highlight them, but it can help. This step is vital because it helps you decide the structure of your answer. If you familiarise yourself with all the verbs in the Verb Guide chapter, you will notice that each of them follows a particular structure that can be applied to any question using that verb. Using a memorised structure will help you comprehensively answer the question and help you avoid leaving important things out.

Analyse the question as a whole first, think about what it is asking

Although the syllabus dotpoints are themselves questions, exams will rarely ask questions identical to syllabus dotpoints. This means that any given exam question will be different from its corresponding syllabus dotpoint, and therefore the question you answer the same will almost certainly be different to the answer for the syllabus dotpoint. Therefore it’s important to read through the entire question as a whole, to get a feel for exactly what the question is asking. Often if you skim read the question, identify the relevant dotpoint and then write out the answer as if it was the dotpoint, you will miss key areas and lose marks.

Double check answers

It seems obvious, but it’s a key mistake that is often made. When you’re checking your work, don’t just skim through the question and then carefully go through your answer. You’re better off reading the question thoroughly and then skimming through your answer. This is because you are more likely to have misread the question or missed key information than you are to have made any serious errors in your written answer.

Label your graph

Putting a title on a graph is actually worth marks. All you have to do is think of a name, and write it down. However, many people forget to title graphs or label axes because it seems so trivial. This is another free mark you need to be careful to collect.

Triple check equations

Double check all your answers, but triple check anything with an equation in it. Mathematical errors are easy to make, even in the simplest questions. Extension 2 maths students make careless mistakes in projectile motion questions all the time, even though they know how to do them at a far more advanced level than required for HSC Physics. Common errors include things like forgetting to take into account elevations (eg. A projectile fired off a cliff has the cliff’s extra height). Make sure you literally go through the entire question a second time as if you hadn’t done it before. The maths in HSC Physics isn’t all that difficult and so it shouldn’t take too long to go through all the mathematical questions in this manner.

Write over the lines, but don’t write about too much

In general, there are never enough lines for you to write on if you want to completely answer a question. Think of the lines only as guides to answer length- feel free to exceed them by as much as you like. However, it’s vital to remember that you’re only going to get marks for relevant things. It’s a complete waste of time writing about irrelevant things, and in some cases it can cost you marks (for example, if you accidentally contradict yourself). So write as much as you want, but only about relevant things.

When using the right hand screw rule or palm rule, use your right hand

It seems simple enough, but if like most students you’re right-handed it’s easy to use the hand that isn’t holding a pen. Use your left hand for conventional current, and you’ll lose marks instantly. Don’t give away free marks- use the correct hand. Right hand for conventional current, left hand for electrons.

Choose your jargon carefully

Jargon, otherwise known as terminology, is important to use. But if you find yourself using words like “diamagnetic” you’ve probably gone too far. Using lengthy, complicated words can make it more confusing for you in an exam situation, so it’s often safer to stick to words that are mentioned in the syllabus. Long words don’t mean more marks. Clear, logical explanations with reasonable technical language do.

Draw diagrams in pencil

If you don’t already draw diagrams on your exam paper, you should seriously think about it. Drawing diagrams can be a very effective way of demonstrating your understanding of key concepts- firstly, it conveys a lot of information quickly, but secondly, it will probably help you keep your answer clear and easy to read. Diagrams are especially useful for questions concerning Einstein’s thought experiments, which involve abstract but complex ideas. Similarly, make sure you draw your diagram in pencil first, because it’ll just turn your paper into a mess if you keep making changes to it in pen.

Exam Verb Guide

HSC Exam Verbs

In 2002 the Board of Studies introduced a glossary of verbs to make it easier for students to answer questions. This means that in any physics question there is a directive verb telling you what to do. To answer the question, you need to completely deal with the verb. Note that going beyond the verb doesn’t give more marks. This means that to maximise marks, you need to write as little as possible while ensuring that you cover all the relevant criteria for the question. This necessitates a highly targeted and logical approach to writing answers, a skill that comes through firstly knowing and understanding the meaning of all the verbs in the course, and secondly through plenty of practice.


Identify is the most basic verb encountered, and it requires you to merely state a fact. For example, Identify that moving charged particles in a magnetic field experience a force. All that is required is to state the fact that moving charged particles in a magnetic field experience a force. Identify questions are rarely used by themselves as they are effectively too simple to answer and do not differentiate enough between candidates. This means that if an identify question is encountered, it is most likely to be a 1-mark freebie. However, it’s hard to know how much depth to go into when answering an identify question. Often the identify verb is used to explore key concepts that are built upon by other dotpoints. It can be hard to tell which information is relevant to fully answering an identify question. Generally, you will need to carefully read the question and make a judgement based on how much writing space is available and how much time you have in the exam room. The more you write, the safer you are, but it comes at the expense of other questions and you don’t want to be wasting time on a 1-mark question.


Explain requires you to go into detail about the subject matter. Where identify is simply about the surface, explain is about the processes leading to it. For example, Explain that cathode ray tubes allowed the manipulation of a stream of charged particles. Often when approaching an explain question, it can be helpful to consider where you want to end up first, then structure your answer around that. In this case, the surface or end result is that cathode ray tubes allow the manipulation of a stream of charged particles. This is where the answer ends, and it needs to be clearly stated in your answer. However, explain needs to go into depth about how this manipulation is accomplished- the underlying processes. To answer this dotpoint will require identifying what a cathode ray tube is, especially that it provides a stream of charged particle, identifying that charged particles can be manipulated by obstructions and fields, explaining how this manipulation occurs (deflection and blocking in this case), and linking them together and explaining that cathode ray tubes allow this manipulation.


In a way, define is easier than explain because it’s a more complicated version of identify. Define requires greater depth of knowledge i.e. a longer, more detailed answer, but it still is essentially identify. Define Bohr’s postulates would be the same as identifying each of them with a high level of detail.


Compare is examining similarities and differences between things. Less detail is required in terms of identification and explanation because the focus of the answer is on the comparison. For example, Compare step-up and step-down transformers. The answer should be thought of in two parts- similarities and differences, and structured around these two areas. They are similar because they both have two coils where electron movement in one induces a current in the other. They are different because in a step-up transformer the secondary coil has more turns, whereas in a step-down transformer the secondary coil has fewer turns so that the step-up produces a higher voltage while the step-down produces a lower voltage than the input. “Whereas” is an important word in writing comparisons because it shows to the examiner that you are directly examining differences between two things. Whenever you examine differences, always use the word “whereas”. This will make it clear to both you and the examiner what you are contrasting.


Contrast is exactly half of compare. Where in compare both similarities and differences need to be addressed, with contrast you only need to talk about differences. This means contrast is effectively “show how they’re different”. Contrast is rarely used in the syllabus but can often be found in exam questions. For example, Contrast step-up and step-down transformers. To answer this, you now only need to “show how they’re different”. Step-up has more turns in secondary coil and higher output voltage, whereas step-down has fewer turns in secondary coil and lower output voltage. When answering contrast, to get more marks you show more differences. There were 2 listed in this example, the only way to get more marks is to add more differences. Providing similarities will give NO marks and is therefore a waste of your exam time.


This is probably the most common verb used in tests because it’s comprehensive and really tests depth of knowledge. Broadly, it consists of two parts. The first is the simple verbs identify/explain, and so the first part of a discuss answer will involve identifying or explaining relevant issues. Though you need to go through this first step, think through your answer first because it is a waste of time explaining irrelevant issues. The second part of a discuss answer is provide arguments for and against the issues being discussed. This can only be developed by practice, since it its challenging to learn how to formulate arguments through analysis of issues (although you can memorise points for and against, it is essential that you can also make up points for questions that don’t directly match syllabus dotpoints). For example, Discuss the BCS theory. This answer would be divided into parts. The answer would consist firstly of “What is BCS theory? How does it work? What does it accomplish?”, then arguments for or supporting BCS theory, and finally arguments against BCS theory. Generally, any discuss question can be divided into these parts:

  • Identify what is being discussed, explain it, and identify the key issues
  • Provide arguments for
  • Provide arguments against

In some cases there won’t be arguments for and against. These cases require a structured comparison of all the elements of the question. For example, Discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates. Point charge has a field that radiates outward with field strength obeying inverse square law. Positive charges repel positive, negative attracts. Parallel plates have a constant field strength with the field running in one direction between the plates.


Analyse is a more advanced version of explain. It requires you to not only explain what is happening in a system, but to also use data or process information in order to draw conclusions. For example, Analyse information to explain why a magnet is able to hover above a superconducting material that has reached the temperature at which it is superconducting. This dotpoint requires you to use provided information to explain the Meissner effect. However, even if no information is provided, answering this dotpoint will involve firstly identifying that a magnet can hover over a superconductor, and then explain why the Meissner effect occurs and how this produces hovering. Of course, if information is provided it will need to be integrated into your answer. Generally analyse isn’t used in exams as it is more productive for examiners to use discuss.


Evaluate is essentially a biased version of discuss. To answer an evaluate question, you need to firstly answer the question as if it was a discuss question, and then weigh up the for/against points you’ve provided and use them to draw a conclusion. Generally, less time will be spent on explanation in evaluate compared to discuss because the emphasis is on the conclusion and the issues, not explanation. Evaluation requires a clear, structured approach with a logical conclusion that flows from the arguments you’ve made. For example, Evaluate the relative contributions of electrostatic and gravitational forces between nucleons. As with discuss you first need to identify the two forces and explain their operation. As with a “comparing” discuss question, you need to compare their effects, particularly in terms of strength. But because this is an evaluate question, a judgement must be clearly made e.g. gravitational forces are insignificant in interactions between nucleons compared to electrostatic forces. Ensure that the conclusion you provide directly addresses the question. The essential part is to read the question and decide what needs to be done to completely answer it. Note that evaluate is very similar to assess and they can be considered the same.


Justify is pretty much “prove something”. Though it doesn’t appear in the syllabus explicitly, it can appear in exams, particularly regarding pracs. For example, in the pendulum prac, justify timing the pendulum over 10 swings. This is basically asking you to “prove why it is good to time the pendulum over 10 swings”. The argument you make must be clear and logical to fully answer the question. This would involve firstly stating its effect- timing over 10 swings increases reliability. Then you would need to explain errors that affect reliability- random influences and reaction time error to name two, and finally you would need to explain how timing over 10 swings minimises this error and is therefore why that was the procedure. The last step, explaining how timing over 10 swings minimises error, is the justification but to perform this final step the framework for your justification needs to be set up in your response beforehand.

General Tips

As can be seen, almost all the verbs require some degree of identification and explanation. When answering a question, make sure you set up your response by identifying and explaining relevant concepts. Don’t waste time by writing about irrelevant things, so before you start writing think about the answer in its entirety so you can decide what’s relevant.

Never write an answer that requires more than a basic degree of assumed knowledge- assume the marker only knows the bare minimum (i.e. terminology and basic concepts) and requires everything to be spelt out in terms of interactions and effects. Don’t assume anything.

Finally, don’t write simply anything, be concise and relevant. Don’t waffle- get to the point and show depth of knowledge. You’ll find it easier to think about the science behind your answer this way, and the examiner will find it easier to gauge your knowledge.


Universal Gravitation

F = \frac{Gm_1m_2}{r^2}

Usage Tip: This formula gives the magnitude of gravitational attraction between the masses and is exerted equally on both masses. Be careful when you use it, it may not give you the answer you require.

Gravitational Acceleration

Usage Tip: This formula is derived from Universal Gravitation by F = ma, in this case a repre- senting g. Substituting mg for F in the universal gravitation formula and cancelling the mass of the object from F = mg gives the gravitational acceleration. The radius is required to calculate the acceleration at the surface of the object. Alternatively, the distance from the centre of the object could be used- to calculate the acceleration at a point above the surface, it would be (robject +

altitude)2. Note also that because the bottom of the equation is r2, gravitational acceleration and

therefore field strength obeys the inverse square law.

Weight Formula

W = mg

Weight force = mass of the object × gravitational acceleration

Usage Tip: Weight force is simply the force that a gravitational field exerts on an object. As the formula shows, this force increases as the mass of the object increases so that any object in the field accelerates at a constant rate.

Potential Energy

Ep = −

ttm1m2 r


gravitational constant(6.67 × 10 ) × first mass × second mass

Potential energy = −

distance between the objects

Usage Tip: The reason this formula is negative is actually simple. When you move an object away from a gravitational field, it gains energy. A stone at the top of a building has more Ep than a stone on the ground. However, at an infinite distance from a planet, the field strength is 0 and the object doesn’t have any energy due to the gravitational field. So because Ep increases as the distance increases, and is 0 when distance is infinite, Ep takes a negative value. In this definition, it’s the work done to move an object from an infinite distance to a point in a gravitational field. Note also that the bottom of the formula is simply r, not r2. The equation is similar to Universal Gravitation but with those two key differences.

Velocity Equations

v = u + at v2 = u2 + 2as

s = ut + 1 at2



v = final velocity (m/s) u = initial velocity (m/s) a = acceleration (m/s2) s = displacement (m)

t = time (s)

Usage Tip: Select one based on the unknown required. There are 5 variables between them. Ques- tions will provide 3 variable values, so the choice of equation will be the equation with those 3 values and the variable required to solve for.

Escape Velocity

vescape = Escape velocity =

. 2ttmp


. 2 × gravitational constant(6.67 × 10−11) × mass of the planet

radius of the planet

Usage Tip: This lends itself to two key observations. Firstly, the obvious fact that the greater the mass of the planet, the higher the escape velocity, but further, that the bigger the planet, the lower the escape velocity, regardless of mass. The key reason for this is that the escape velocity depends on the distance not from the planet, but from the centre of the field. Assuming a big planet and small planet have the same mass, a rocket at the surface of the big planet will be further from the centre of the field than a rocket at the surface of the small planet. Because the rocket on the larger planet starts out further from the centre of the field, the escape velocity is lower


Fgravities = Force (in G’s) =

mg + ma mgearth

mass × acceleration due to gravity + mass × acceleration of reference frame

mass × Earth’s gravitational acceleration (9.8m/s2)

Usage Tip: G-Force is essentially the ratio of the force experienced by an object to the force it experiences at rest on Earth. This means that 2 G’s are eqivalent to twice the Earth’s gravitational acceleration (ie. 2 G’s is 19.6m/s2 ). When calculating G Force, make sure that you add on the effect of any gravitational field present- not just the acceleration of a rocket etc.

Centripetal force

Fc = Centripetal force =

mv2 r

mass of the object × the object’s velocity2

radius of its circular motion

Usage Tip: Centripetal force is the force required to keep an object in circular motion at a specified



velocity. From F = ma, it can be seen that centripetal acceleration is simply v . Don’t forget in an

orbit, r is equal to orbital altitude plus the radius of the Earth, orbital radius being measured from the Earth’s centre (i.e. the centre of the field).

Orbital Period

T = 2πr


2 × π × radius of the orbit

Period =

orbital velocity

Usage Tip: As can be seen here, the period is just the time taken for the satellite to travel along the length of a circular orbit i.e. the circumference

Orbital Velocity

v = Orbital velocity =

. ttmp


. gravitational constant(6.67 × 10−11 ) × mass of the planet or central body

orbital radius

Usage Tip: Again, orbital radius is measured to the Earth’s centre and is not the same as orbital altitude

Kepler’s Law of Periods


T 2 =

Orbital radius3 Orbital period2 =




gravitational constant(6.67 × 10 ) × mass of the planet 4 × π2

Usage Tip: This is derived by taking the orbital period formula, switching v and T , then substituting


T 2

into the orbital velocity formula and rearranging. Essentially, it states that in any system, r for any

satellite is a constant, as long as they are orbiting the same central body. Kepler’s law is used for calculating orbital radius or orbital period when the orbit of another satellite of the same system is known, or when the mass of the central body is known and the radius or period is provided

Slingshot Effect

vfinal = vinitial + 2Vinitial

Final probe velocity = initial probe velocity + 2 × planet’s initial velocity

Usage Tip: This formula may not be necessary, but it’s included in the Jacaranda textbook. It’s derived by the assumption that the slingshot effect is an elastic collision. Thus two expressions are written, one equating the total momentum of the system before and after and one equating the total kinetic energy before and after, and then solving simultaneously. It probably isn’t necessary to derive it, but in order to reduce variables, the mass of the planet is expressed as Km where m is the mass of the probe and K is a value such that Km is the mass of the planet. The K’s eventually cancel out, but are necessary to derive the equation

Correction Factor

CF =

. v2

1 − c2

. relative velocity2

Correction factor =

1 − speed of light2

This formula is not explicitly provided by itself nor is it used by itself. But it is a key part of all calculations involving relativity that are encountered in the syllabus. Essentially, the correction factor is the mathematical factor that corrects for relativistic effects. All that needs to be known is the effect relativistic speeds have and the correction factor can be easily applied. Note also that the correction factor is always a value between 0 and 1. The syllabus deals with three scenarios- time, length and mass. Time dilates, so at relativistic speeds the time taken for an event to occur in the eyes of an observer increases. So the original time taken is divided by the correction factor so that it gets bigger, giving the observed time. Conversely, if the observed time is provided, it is multiplied by the correction factor to reduce it to the value in the other frame. Length contracts, so to calculate observed length the length in the other frame is multiplied by the correction factor so that it becomes smaller. Mass dilates at relativistic speeds, so the original mass of an object is divided by the correction factor so that the observed mass is bigger. Simply put, observed time is longer, observed length is smaller, observed mass is bigger. Apply the correction factor to real figures to agree with these ideas, and you will always be correct.

Warning: The relative velocity could exceed the speed of light, for example, if both objects were heading in opposite directions at > 0.5c each. This formula is incapable of dealing with such situations. However, these situations are not considered in the HSC course.

Usage Tip: If instead of providing a specific velocity, a question provides a multiple of c, such as a

spaceship travelling at 0.6c relative to an observer, this can be substituted directly into the formula to give √1 − 0.62, because 0.6 is the ratio of spacecraft velocity to the speed of light (i.e. v ), thus



0.62 = v


Mass and Energy

E = mc2

Energy = mass of the object × speed of light2

Usage Tip: This formula only applies to the rest state. If an object is moving, add its kinetic energy to the right side of the equation

About the Guide

The Student’s Guide to HSC Physics is a brand new form of study guide, modelled on the way many students write their own study notes. Most books such as those published by Jacaranda, Excel and Macquarie are combinations of textbooks and questions. While they’re fine for learning new ideas and concepts for the first time, they’re often difficult to use when studying. This is because they don’t follow the syllabus exactly, mixing and matching content, until it becomes difficult for you to decide what needs to be studied and what doesn’t. The result is that you study irrelevant things, and may omit important things.

This guide is a revision aid, not a textbook. The Board of Studies publishes a syllabus for every course that tells you exactly what you need to know. The guide goes through each of those dotpoints clearly and comprehensively, so that you can revise exactly what you need to know to score highly in exams. Unlike a textbook, the Student’s Guide to HSC Physics sticks to the syllabus. Under each dotpoint you will find only what you need to know to get full marks.  By going through each  of the dotpoints with this book, and by practicing answering questions, you will be prepared for any question in your HSC exam.

This book deals with the syllabus as comprehensively as possible. However, in the 3rd column of the syllabus there are occasionally dot points dealing with the use of formulae. They are usually of the form “solve problems and analyse information using *a formula*”. This book being about content, not questions, these dotpoints aren’t included in the main document. However, the Formulae chapter is an all-inclusive formula guide that summarises all of the formulae encountered in HSC Physics, with some extras from the Preliminary course that are relevant to the HSC, along with detailed explanations and useful hints for using them. Make sure you get familiar with using the formulae by doing practice problems- although you don’t need to memorise them, you do need to know how to apply them quickly in exam conditions.

Also in the 3rd column are dotpoints concerning first-hand experiments that you performed in class. The answers in this guide are examples of experiments that can be performed. Only use them if you didn’t perform the experiment or if your experiment didn’t work, for whatever reason. If you performed a different experiment in class, it’s better for you to write about that, because having done it you will know a great deal more and be able to write about it in far greater detail.

Finally, although this guide is designed to be simpler and more accessible than other guides in order to make it easier to study from, parts of it do get quite advanced. This is necessary to score full marks in all questions. However, the more complicated explanations are always there either so that you properly understand what is happening, or to provide depth of knowledge. Take time to understand everything fully- unlike other books, everything here is relevant and will help you in your exams

Romesh Abeysuriya

Romesh Abeysuriya graduated from Sydney Boys’ High School in 2006 with a final mark of 94 for HSC Physics, and is currently in his 3nd year of a Bachelor of Science (Advanced) at The University of Sydney, majoring in Physics, and is a member of the USYD Talented Student Program

Applications of Nuclear Physics

Explain the basic principles of a fission reactor

A fission reactor uses a nuclear reaction to generate electricity. As with all generators, this involves producing rotation to turn a generator. In a nuclear reactor, heat from the nuclear reaction is used to produce steam which turns a turbine, in the same way that burning coal generates steam in a coal power plant. As outlined before, there are several requirements for a controlled reaction. These must be met in a fission reactor to ensure that firstly a reaction takes place, and secondly that the reaction doesn’t go out of control and produce an explosion. In addition to this, there are several key components to a fission reactor. Fuel rods consisting of enriched uranium are placed inside the reactor to provide the critical mass required. Control rods consisting of cadmium or boron are also placed in the reactor, such that they can be moved in and out to control the reaction. The control rods absorb excess neutrons to prevent the reaction from taking place too quickly. When they are lowered, more neutrons are absorbed and the reaction slows, and when pulled out the reaction rate increases. The entire reactor is immersed or surrounded by a moderator to slow down neutrons and thus increase the rate of reaction. The moderator consists of either heavy water, graphite, or various other organic compounds. A coolant is required to extract heat from the reaction and to prevent the reactor from melting. The coolant flows through the reactor then out into a heat exchanger that takes heat extracted from the coolant and uses it to boil water. Spent fuel rods that have been depleted in the reactor are extracted and processed or stored. They are extremely radioactive, making them very difficult to dispose of. Finally, the reactor is surrounded by multiple layers of shielding. There is a graphite shield that reflects neutrons back into the core, followed by a thermal shield to prevent unwanted heat loss from the core, a pressure vessel surrounding the core to isolate and contain everything inside the core, and lastly a biological shield of about 3 metres of concrete mixed with lead pellets, to absorb gamma rays and neutrons.

Figure 1

Remember- A nuclear reactor has a reactor core with fuel rods, control rods and a moderator. Coolant is heated inside the core and pumped out where it boils water. The steam produced turns a turbine, and is then condensed back into water. Shielding is used inside the reactor to prevent radiation and heat from escaping.

Gather, process and analyse information to assess the significance of the Manhattan Project to society

The Manhattan Project was one of the most significant scientific undertakings of the 20th century because of the dramatic impacts it had on society. It consisted of American efforts to produce nuclear weaponry, which were eventually successful and resulted in the deployment of nuclear weapons over Japan in 1945. In terms of impact on society, there were direct scientific impacts, namely the development of nuclear power offering a possible solution to the depletion of fossil fuels and a way of reducing greenhouse gas emissions from power generation. Much more significant however, were the social impacts that atomic weaponry had on global politics. To begin with, nuclear power had terrified the world with its incredible destructive power as witnessed in Japan. As a result, countries with nuclear weapons, primarily the USA and Russia in the period following WW2, became very reluctant to use them, firstly because of their long term destructive power, but secondly because of fear that retaliation would take the form of nuclear reprisal. As a result, although significant political tension built between Russia and USA, it never broke out into conflict, as either side was concerned that aggressive action would result in nuclear warfare, resulting in mutually assured destruction. Where a conventional war would have broken out previously, peace was maintained due to the development of nuclear weaponry. In modern times however, nuclear power is proving to be a dangerous bargaining chip for rogue states such as North Korea and Iran which are using nuclear weapons as leverage in negotiations with the Western world. It has led to a situation where small nations with comparatively weak conventional forces can use the threat of nuclear warfare to negotiate equally with large nations such as the USA. This has led to significant problems in terms of global politics and the power balance between nations necessary to maintain peace. Arguably however, even in these situations the threat of mutually assured destruction is preventing warfare. As a result of the nuclear threat, the UN and the USA are focussing on a diplomatic, sanctions-based approach to resolving conflict rather than an aggressive military approach. Overall, although the Manhattan Project led to the deaths of many Japanese people in Hiroshima and Nagasaki, and although it resulted in a build-up of nuclear arsenals across many nations providing a constant threat to global security, in the end the resulting nuclear stalemate has prevented several wars and therefore averted many possible deaths.

Remember- Although the Manhattan project led to many deaths at the end of WW2, the threat of nuclear war has prevented conflict in the decades after.

Describe some medical and industrial applications of radioisotopes

There are many applications of radioisotopes in medical and industrial fields. In the medical field, radioisotopes are mainly used for imaging/diagnosis and for treatment. In imaging, the transmission of radiation through the body and the degree to which radiation is absorbed can be used to remotely examine the body. They are often used to examine brain activity (using Positron Emission Tomog- raphy). By injecting radioisotopes into the body and examining where they end up (made possible because the radioisotopes are emitting radiation), the circulatory system can be investigated. Fi- nally, radioisotopes are frequently used to kill cancer cells, the radiation destroying them. In industry, they are used to examine stress fractures in metals such as in aircraft wings (because although the fractures may not be visible, radiation can pass through them), detecting leaks in pipes that may be otherwise difficult to find (since radiation will escape from a leaking pipe), and to irradiate medical supplies and food to kill bacteria.

Remember- Radioisotopes are in medicine used for imaging and cancer treatment, while in industry they are used to examine stress fractures and to sterilise objects and food.

Identify data sources and gather, process and analyse information to describe the use of a named isotope in medicine, agriculture and engineering

Iridium-192 is used in medicine to kill cancerous tumours. Iridium-192 pellets are implanted into the tumour where gamma emissions kill the cancer cells. Because the cancer cells are directly exposed to the radiation since they surround the pellet, damage to healthy cells is minimised (as opposed to external irradiation). Because it has a half-life of around 80 days the iridium must be surgically removed after treatment is complete to prevent over-exposure to radiation. The iridium undergoes beta decay, and transmutates to turns to inert platinum that poses no health risk. This makes iridium implants an extremely effective way to treat cancer.

In engineering, cobalt-60 is used to detect stress fractures in metals, particularly in aircraft. Stress fractures occur when metals are repeatedly exposed to strong forces, such as those experienced by the wings of an aircraft. Small fractures can form in the metal, which can eventually result in a catastrophic failure (e.g. there were several cases where early jet aircraft had fuselage explosions because the metals used to construct the aircraft eventually broke apart due to stress). These fractures are the precursors to actual breaks in the metal, but they are extremely hard to detect. By placing cobalt-60 on one side of the metal, and a gamma detector on the other side (often photographic film), the cracks can be identified easily and non-destructively because the gamma radiation only penetrates in areas where stress fractures have formed.

In agriculture, the elements that plants require can be substituted with radioisotopes of the same element. This allows the path of the material to be tracked through the plant’s structure. For example, replacing the phosphorus in soil with a mix containing radioactive phosphorus-32 will allow the path of phosphorus to be tracked into plants. By measuring how radioactive the plants are, how much phosphorus was used by the plant can be determined, as well as the areas in the plant where the phosphorus is concentrated. This has benefits in terms of better understanding the conditions favourable for plant growth, thereby maximising yield and increasing efficiency of the farming process.

Remember- Iridium-192 is implanted to kill cancer cells, cobalt-60 is used is used to detect fractures in metal, and phosphorus-32 is used to trace element flow in plants.

Describe how neutron scattering is used as a probe by referring to the prop- erties of neutrons

In the same way that an electron microscope uses electrons to probe materials, neutrons too can be used in microscopes. However, unlike electrons, neutrons do not carry charge and are therefore not affected by the nuclei of atoms which would deflect electrons. They are therefore extremely useful for imaging crystal structures, as well as substances containing light atoms such as hydrogen. While electrons do pass through crystals and diffract, they are deflected by charges in the crystal, resulting in errors. Neutrons penetrate crystals very effectively, allowing for a clearer and more accurate interference pattern to be produced. According to the de Broglie equation, neutrons have a wavelength shorter that light, similar to the wavelength of an electron.

Remember- Neutrons can be used as effective probes because they have wavelengths similar to electrons. However, because they are not charged they can image objects where charge interferes with electrons.

Identify ways in which physicists continue to develop their understanding of matter, using accelerators as a probe to investigate the structure of matter

Physicists now develop their understanding of matter by examining the component of atoms to better understand them. This necessitates separating the atom into its components, requiring large inputs of energy and sophisticated equipment that was previously unavailable. However, modern particle accelerators are used to break atoms into their components, which are then examined, developing our understanding of matter.

All particle accelerators use magnetic fields to accelerate charged atoms or particles to very high velocities. The three most common types are the linear accelerator, the cyclotron and the synchrotron. A linear accelerator is simply a very long track down which an atom is propelled. It is simple, but its main constraint is size, and energy input is dependant on the length of the accelerator. A cyclotron uses high-frequency AC current to generate a magnetic field that causes the electron to accelerate in a spiral. This reduces the size of the reactor, while at the same time using relatively simple equipment. A synchrotron is a complete circle in which a particle travels. While the particle can be accelerated indefinitely, powerful computers are required to manipulate the magnetic field in the synchrotron in order to propel the particle. Particle accelerators also usually have the ability to generate collisions between high-energy accelerated particles, allowing scientists to examine the properties of matter from the collision. So through using particle accelerators, scientists are able to develop their understanding of matter.

Remember- Scientists are now trying to understand the components of atoms. Accelerators provide the high energies required to break atoms into their components.

Discuss the key features and components of the standard model of matter, including quarks and leptons

The standard model of matter is a theory that states all matter is composed of small elementary particles that exist by themselves or group together to form subatomic particles and to transmit force (because under quantum theory, forces that result due to a field are caused by particles travelling between the objects). There are broadly 3 types of particles. Bosons are force-carrying particles, examples of which include photons that carry electric and magnetic force, gluons that carry the strong nuclear force, and gravitons that cause gravity. Leptons are single elementary particles that exist by themselves and are not affected by the strong nuclear force. They include electrons, muons and taus as well as their neutrino subsidiaries (the electron, muon and tau neutrinos respectively). Quarks are the building blocks of hadrons, which are groups of quarks.

There are 6 types of quarks- up, down, top, bottom, strange and charm, each with different properties. Baryons are groups with 3 quarks, such as protons and neutrons, while mesons are pairings of a quark and an antiquark. Because quarks each have a half-integer spin value (spin being one of Pauli’s quantum numbers), combining two gives a whole integer, while combining three gives a half-integer. Therefore, baryons have half-integer spin values while mesons have whole integer spin values. Thus baryons are also fermions, as fermions are particles that have half-integer spins and therefore obey Pauli’s exclusion principle. Bosons are not fermions.

Figure 2


Nuclear Physics and Nuclear Energy

Define the components of the nucleus (protons and neutrons) as nucleons and contrast their properties

Protons and neutrons are both nucleons- particles found in the nucleus, and are slightly different. Both have masses on the same order (measured in amu) but the neutron is slightly heavier than the proton. In terms of charge, the proton has the same charge as an electron only positive, while the neutron has no charge at all. Protons are therefore affected by magnetic and electric fields, while neutrons are not.

Discuss the importance of conservation laws to Chadwick’s discovery of the neutron

Don’t forget that the focus of this dotpoint is the use of conservation laws in regard to discovering the neutron. In an exam make sure you don’t waste time by going into too much detail about the experiment itself.

Chadwick predicted the existence of the neutron based on an experiment that otherwise had no other explanation. When a beryllium atom was bombarded by alpha particles, it emitted a form of radiation. This radiation could not be detected in a cloud chamber and didn’t appear to be a particle- in fact it was initially thought to be gamma radiation. The radiation was capable of knocking protons out of a block of paraffin wax, with the protons travelling away with considerable momentum. In terms of conservation laws there were two applicable to this- conservation of atomic mass and number, and conservation of momentum/energy. Chadwick found that the energy required to eject the proton with the observed momentum could not have been produced by EMR as the energy required would be insufficient (and conservation of momentum would be violated as the photon would not contain enough momentum). However, he realised that a neutral particle would be capable of colliding with a proton and imparting the observed momentum without violating conservation laws. So conservation of momentum was vital in terms of discovery of the neutron. Secondly, the nuclear reaction was \frac{9}{4} Be + \frac{4}{2} He \frac {12}{6} C + ? . By adding mass numbers, according to conservation of atomic mass there would have to be an unknown particle with \frac{1}{0} ? to explain the reaction- so through conservation of mass Chadwick was able to prove the existence of the neutron (and show that the initially observed radiation was in fact a particle).

Figure 1

Remember- Chadwick used conservation of energy to determine the radiation was a particle, and conservation of mass to determine its mass and charge.

Define the term transmutation

Transmutations are nuclear reactions where one element is transformed into another because the number of protons in the nucleus changes- this can occur either due to alpha or beta decay.

Describe nuclear transmutations due to natural radioactivity

Some atoms are inherently unstable because their nuclei exist outside the zone of stability in terms of proton-neutron ratio, or because they have too many protons. This can cause natural radioactive decay to occur, resulting in nuclear transmutation. There are two forms of natural radioactive decay that result in transmutations- alpha and beta decay. In alpha decay, the nucleus emits an alpha particle consisting of two protons and two neutrons, in the process reducing its mass by 4 and its atomic number by 2. A common example is the alpha decay of uranium-238, which occurs according to


\frac{238}{92}U -> \frac {234}{90}Th + \frac{4}{2}He

More accurately, the alpha particle doesn’t have any electrons, thus a more complete equation would be

\frac{238}{92}U -> \frac {234}{90}Th^2- + \frac{4}{2}He^2+

However, the alpha particle rapidly gains electrons from surrounding atoms (hence why alpha radiation is the least penetrative form of nuclear radiation) and the Thorium ion formed rapidly loses its extra electrons. Therefore, the charges are typically omitted since they are so short lived.

In beta decay, a neutron decays into a proton (which stays in the nucleus raising the atomic number by one), an electron (which is emitted), and an antineutrino (which is also emitted). In the typical case of beta decay

\frac{1}{0}n -> \frac {1}{1}p + \frac{0}{-1}e+ \overline{v}

This form of beta decay is known as ‘beta-minus’. There is another form of beta decay, ‘beta-plus’, where the proton decays into a neutron, a neutrino, and a positron (antielectron). However, beta-plus decay is not included in the HSC- only beta-minus is considered.

Remember- Alpha decay releases 2 protons and 2 neutrons (an alpha particle) while beta decay releases an electron (a beta particle), an antineutrino, and converts a neutron into a proton.

Describe Fermi’s initial experimental observation of nuclear fission

Fermi initially joined the many other scientists who were using neutron bombardment of heavy nuclei in order to investigate their properties. Fermi was trying to cause uranium to undergo beta decay to produce transuranic elements heavier than uranium. What he initially found was that slow neutrons (slowed by a paraffin wax block) were far more effective than fast neutrons, because they had a greater chance of being captured by the nuclei (since slow neutrons spend more time close to the nucleus, because they travel slower). But most importantly, what he observed was that when he bombarded the nuclei with neutrons, instead of producing a single heavy radioisotope he found 4 separate products each with different half lives. This was his first observation of fission, although he did not realise what was happening in his experiment.

Remember- Fermi was the first to observe nuclear fission when he realised that following a nuclear reaction there was more than one product.

Perform a first-hand investigation or gather secondary information to observe radiation emitted from a nucleus using a Wilson Cloud Chamber or similar detection device

The shape of the trails can be directly linked to the properties of alpha and beta particles. Alpha particles form strong trails because they are more highly charged, and therefore ionise more air as they travel (resulting in more condensation), while beta particles form less intense trails because their ionisation strength is not as great. However, alpha particles trails are shorter, because their strong charge causes them to attract electrons rapidly, so before they travel a long distance they get converted to neutral helium, and can therefore no longer ionise the air. In fact, this is the same reason that alpha radiation is both not very penetrative, yet highly dangerous inside the body. Beta particles react less with their surroundings, which is why they travel a longer distance. Finally, the alpha trails are relatively straight, because the large mass of the alpha particle means that it is deflected less by other particles as it travels. On the other hand, beta particles have a very low mass, and so are very susceptible to having their path changed through interactions with other particles. However, their path is still relatively straight because their high velocity and low charge means that they are less likely to have their path changed.

In this experiment, we constructed a cloud chamber by filling a transparent glass container with a supersaturated vapour. We did this by placing filter paper soaked in methylated spirits inside the container, then cooling the container with dry ice placed below the container. When we placed a radiation source next to the chamber, we were able to observe trails left by alpha and beta particles. This is because when the charged particles travel through the chamber, they ionise the surrounding air. The ions created served as points for the vapour to condense, leaving a trail. The trails from alpha particles were straight, relatively short, and thick/well-defined, while the trails from beta particles were longer and thinner, though they too were straight. As the gamma radiation did not create a stream of ions for condensation, gamma emissions were not visible.

Figure 2

Remember- Alpha particles produce short, straight trails, beta particles produce longer, less straight trails, and gamma radiation doesn’t produce any trail.

Discuss Pauli’s suggestion of the existence of the neutrino and relate it to the need to account for the energy distribution of electrons emitted in beta decay

During beta decay, initially scientists thought only beta particles were emitted. When they evaluated the energies involved, they came up with a figure for the maximum kinetic energy that a beta particle should have. All beta particles should have been emitted with this velocity, but this wasn’t the case. Instead, almost none were emitted with the full amount of kinetic energy, and most of them were emitted with significantly less. This meant that the slow beta particles were missing kinetic energy, leading to a violation of conservation of energy. Also, the sum of the momentums before and after beta decay was not equal- assuming the nucleus starts off stationary, the sum of momentums should be zero. However, when the momentums of the beta particle and the remainder of the nucleus were added, it was not zero, so conservation of momentum was being violated.

Pauli realised that conservation of energy (and momentum) could be resolved if there was an addi- tional particle being emitted in beta decay- the neutrino, so named as it means little neutral one. The sum of all 3 momentums was equal, thereby maintaining conservation of momentum, although the neutrino couldn’t be detected at the time. So the proposal of the neutrino explained the variable kinetic energies of beta particles, and resolved conservation of momentum. It was an excellent idea in this regard, but critically it lacked evidence because scientists at the time could not detect neutrinos. They were only detected 20 years later using more advanced techniques (Later they were shown to be antineutrinos, rather than neutrinos).


Figure 3

Remember- Pauli suggested the existence of the neutrino to account for variable beta particle veloc- ities and to fulfil conservation of momentum.

Evaluate the relative contributions of electrostatic and gravitational forces between nucleons

Electrostatic repulsion between like-charged positive protons and gravitation attraction between masses in the nucleus are two of the forces that act between nucleons. However, in terms of relative contributions i.e. relative strength, electrostatic repulsion is far stronger than gravitational attrac- tion. Indeed, the force of gravitational attraction is so insignificant that it can be disregarded in most calculations regarding forces acting between nucleons. The end result of this is that if electrostatic repulsion is forcing the nucleus apart, and gravitational attraction cannot hold it together, then there must be another strong force acting to hold the nucleus together, to prevent it from disintegrating.

Remember- Electrostatic repulsion is far stronger than gravitational attraction within the nucleus.


Account for the need for the strong nuclear force and describe its properties

The strong nuclear force is required to hold the nucleus together (as observed countless times because nuclei don’t just fly apart) given that the only significant other force is electrostatic repulsion in the nucleus. Therefore, the strong nuclear force is needed to be an attractive force that opposes electrostatic repulsion and holds the nucleus together. The strong nuclear force is experienced only over very short distances- at extreme short distances it is repulsive, then it becomes attractive as distance increases, then increasingly weaker at large distances (while electrostatic repulsion remains relatively strong). This means that there is a balance of separation where at a particular point, the two forces are balanced and the nucleus is stable. Incidentally, the strong nuclear force is only repulsive at extremely small distances- at reasonably small distances it is attractive, and far stronger than the electrostatic force in terms of magnitude. In fact, the strong nuclear force is the strongest known force in the universe. The force is independent of charge and only acts on neighbouring nucleons, not on the entire nucleus.

Figure 4

Remember- The strong nuclear force is required to hold together the nucleus given that gravitational attraction is so much weaker than electrostatic repulsion. It repels at extremely short distances, then attracts with decreasing strength at increasing distances.


Explain the concept of a mass defect using Einstein’s equivalence between mass and energy

See the Extra Content chapter for an explanation of the apparent contradiction described in this dotpoint, where both fusion and fission release energy. Also note that the correct unit for ‘atomic mass unit’ is simply ‘u’, not ‘amu’.

The actual mass of a nucleus is always less then the sum of the masses of the constituents of the nucleus. This means that a helium nucleus with 2 neutrons and 2 protons has less mass than the combined mass of 2 neutrons and 2 protons measured separately. This implies that there is missing mass- this missing mass is called mass defect. It is related to the need for nucleons to lose energy in order to bind together, a stable bond being representative of a low energy state. In order for the nucleons to bond together, they need to lose energy. They do this by losing mass, as according to Einstein mass and energy are equivalent. The mass defect is calculated by simply taking the difference between the mass of the nucleus and the sum of its constituents, usually all carried out in amu. Further, this mass loss can be expressed in energy terms, as MeV, according to 1u = 931.5MeV. The mass defect in terms of energy is also known as the binding energy- the energy required to completely separate out all the parts of the nucleus by breaking bonds. Binding energy is the energy input required to restore the nucleons to their original energy states, thereby breaking the bonds that hold them together in the nucleus. This is also related to the release of energy in fission- the binding energy of a single atom is less than the binding energy of the two atoms produced when the single atom is split. If the total binding energy has increased, that means that more energy is now required to break the bonds, and therefore some energy must have been emitted in the splitting process- this is the energy release from fission (similarly, when split the total mass defect of two smaller nuclei is more than the mass defect of just one nucleus)


Figure 5

Remember- Binding energy is the energy required to break a nucleus into its constituents, and mass defect is binding energy expressed in amu by using Einstein’s equivalence between mass and energy.

Describe Fermi’s demonstration of a controlled nuclear chain reaction in 1942

Fermi realised that since the fission of a uranium atom released 3 neutrons, and that since only 1 neutron is required to cause fission in a uranium nucleus, a chain reaction of nuclear fission could be produced that would release a great deal of energy. If neutrons were absorbed such that not all of them produced additional fission, a controlled chain reaction could be produced to release power. This is exactly what Fermi demonstrated in 1942 in a squash court in Chicago at Stagg Field, when he took 50 tonnes of natural uranium in 20000 slugs, in a reactor with 400 tonnes of graphite as a moderator. He used cadmium control rods to prevent the reaction from going out of control. His reaction was successful and was able to generate 0.5 watts in a self-sustaining reaction.

Remember- Fermi built a nuclear reactor at Stagg Field in Chicago with a graphite moderator and cadmium control rods.

Compare requirements for controlled and uncontrolled nuclear chain reactions

To produce an uncontrolled nuclear chain reaction, all that is required is a mass of fissionable material such as Uranium-235 greater than the critical mass specified for that material. The critical mass for a material is the minimum amount of material required so that the neutrons emitted from fission go on to cause further fission reactions in a chain reaction, sustaining the reaction. So for an uncontrolled reaction, all that is required is a source of neutrons, a means of slowing them down, and a super- critical mass of fissionable material. A large lump of fissionable material will generally meet all 3 criteria, as the material itself is a super-critical mass, a source of neutrons, and a means for slowing down neutrons.

To produce a controlled reaction, a reactor is required with a mechanism to capture neutrons so that the overall number of neutrons that cause fission is constant. Normally uranium releases 3 neutrons- the control rods in a reactor capture two of the neutrons so that only one goes on to cause another fission reaction. If all 3 neutrons caused fission, each stage of fission would have triple the number of reactions, resulting in exponential growth of energy release and an uncontrolled reaction. By absorbing excess neutrons, the reaction is controlled and doesnt spiral into an explosion. Also, a moderator is used to slow fast neutrons in the reactor. Fast neutrons travel past nuclei rapidly and have a low chance of being absorbed, causing fission. Slow neutrons spend much longer in the vicinity of atomic nuclei (since they are travelling slower) and so have a much greater chance of being captured by the nucleus. So a controlled nuclear reaction needs a super-critical mass of fissionable material, a source of neutrons, a way to slow those neutrons down, and a mechanism to absorb excess neutrons from the reaction.

Remember- An uncontrolled reaction requires a critical mass of fissionable material, along with a moderator and source of neutrons (which is often the material itself). A controlled reaction also needs a control mechanism to capture excess neutrons.

Matter Waves and the Quantum Atom

Describe the impact of de Broglie’s proposal that any kind of particle has both wave and particle properties

The immediate and most important impact that de Broglie’s proposal had was to provide a model to accompany Bohr’s first assertion that there were stable orbits where electrons did not emit energy. Under the first postulate, Bohr simply claimed they would not emit energy, directly contravening Maxwell’s theories without explanation. This, without a model to explain it, deprived the Bohr atom from receiving scientific credibility, and as such it was rejected by the scientific community. De Broglie’s proposal gave a workable solution to explain stable orbits that don’t emit energy, and this gave the Bohr model the credibility it required to be accepted and developed upon by the scientific community, which proved vital in terms of understanding the structure of the atom. Later, de Broglie’s proposal was used to exploit the wave nature of electrons in electron microscopes which could be used to image objects at far greater resolutions than was possible with light due to the smaller wavelength of electrons. De Broglie’s proposal also reconciled Einstein’s theory of light with classical physics by showing that light could have both wave and particle nature.

Remember- de Broglie’s proposal provided a model to solve one of the biggest problems with the Bohr atom, namely the stability of orbits. Later it was used as the foundation for electron micro- scopes.

Define diffraction and identify that interference occurs between waves that have been diffracted

The diagram for this dotpoint has been deliberately oversimplified to show the obstruction of the original wave, and the generation of point-source waves at the corners of the object. In reality, there will be actual interference between the original and the diffracted waves (not shown in this diagram)

Diffraction is the bending of waves around obstructions. It is solely a wave property, and is observed when the passage of a wave is obstructed by an object. The wave can bend around the object and exist where there should be a shadow from the object- this effect is strongest when the size of the object is of the same order as the wavelength of the wave. The corner of the object acts as a point source for the wave, resulting in a curved wave that radiates outward. There are now two waves- the point source and the main wave, and because they exist in the same location interference occurs between the two waves. This means that the process of diffraction results in an interference pattern. This is because at some points the waves interfere destructively and at others they interfere constructively. This results in lines of light and dark, or light and dark rings if it is a circular obstruction.

Remember- Diffraction is when waves bend around objects, and since the corners of the object act as point sources interference occurs between the original wave and the new point source waves.

Describe the confirmation of de Broglie’s proposal by Davisson and Germer

Davisson and Germer were studying the surface of nickel with an electron beam, expecting that even the smoothest surface would appear rough to the electrons. In their experiment an accident occurred and the nickel oxidised when it was exposed to air. To remove the oxide film, they heated the nickel to near its melting point, resulting in the formation of crystals larger than the width of their electron beam. Then, when they fired the beam at the nickel and reflected it to a detector, they observed an interference pattern very similar to an x-ray diffraction pattern, confirming the wave nature of electrons and confirming their wavelength as being very close to what de Broglie predicted.

Remember- Davisson and Germer observed the electron diffraction de Broglie had predicted.

Explain the stability of the electron orbits in the Bohr atom using de Broglie’s hypothesis

Electrons under the de Broglie hypothesis exist as standing waves around the nucleus. This means that they have a closed orbit with no movement of energy, and therefore the orbit is stable with no energy emission. This was the model to explain the stability of electron orbits. The electron orbits must have a circumference equal to a multiple of the wavelength of the electron, as this allows for a standing wave. Therefore,

which was Bohr’s third postulate. Therefore, using de Broglie’s theory of matter waves not only could there be an explanation for Bohr’s first postulate, but it was also possible to mathematically derive Bohr’s third postulate that had initially been nothing more than an assertion.

Remember- Stable orbits in the Bohr atom exist at radii where the circumference of the orbit is a multiple of the wavelength of an electron.

Gather, process, analyse and present information and use available evidence to assess the contributions made by Heisenberg and Pauli to the development of atomic theory

Heisenberg and Pauli both made very significant contributions to quantum theory, Heisenberg through his uncertainty principle and Pauli through his exclusion principle.

Heisenberg firstly devised matrix mechanics to explain the atom in terms of quantum probabilities, rather than mixing classical and quantum theory as Bohr had done. This led to an entirely quantum theory of the atom, helping to mathematically understand its nature. Secondly, he devised the uncertainty principle which essentially stated that the more was known about the momentum of a particle, the less could be known about its position in space and vice versa. This changed the way science viewed atomic structure, and is perhaps one of the most important central principles of quantum mechanics, that knowledge of one thing can be mutually exclusive to knowledge of another. This isn’t just due to measurements changing quantities- it’s a fundamental property of quantum mechanics. Heisenberg’s work greatly changed the way in which scientists approached quantum physics.

Eventually, it was realised that the position and properties of an electron could be described in terms of 4 quantum numbers. Pauli’s exclusion principle stated that no two electrons could have all 4 numbers exactly the same- this explained the maximum number of electrons in each shell, and provided a quantum explanation for the position of the first 20 elements in the periodic table. Further, Pauli was able to use his work with quantum numbers to explain the Zeeman effect. Pauli also proposed the existence of the neutrino, another significant subatomic particle.

Remember- Heisenberg devised matrix mechanics and the uncertainty principle, and Pauli developed the exclusion principle and the neutrino.

Atomic Structure

Discuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits

The Rutherford atom consisted of a small positive nucleus with negatively charged electrons orbiting the nucleus. The Rutherford atom was devised following Rutherford’s experiments in which he fired positively charged alpha particles at thin gold foil. Rutherford found that while most alpha particles passed straight through the foil, a small proportion of them were reflected back. He hypothesised that they had encountered very dense areas of positive charge. The fact that most alpha particles passed through the gold foil led Rutherford to model the atom with a great deal of empty space. Rutherford modelled the atom with a dense, positively charged nucleus, negatively charged electrons that orbited the nucleus, and free space between the nucleus and the electrons. The model was essentially a simplified version of what we use today- it was groundbreaking at the time as it was a step in the right direction for other scientists to build on, but it lacked a description of where the electrons were and failed to address how atoms had stability without energy emission from accelerating electrons.

figure 14

Remember- The Rutherford atom had a positive nucleus, negative electrons orbiting the nucleus, and empty space between the electrons and the nucleus.

Analyse the significance of the hydrogen spectrum in the development of Bohr’s model of the atom.

Bohr’s model of the atom was quite similar to Rutherford’s, but with two important differences- firstly, it assigned positions to the electrons, but secondly the electron energy levels were quantised. This was radically new, the idea that electrons had energy states and could absorb and emit energy to change states, and had no evidence. Bohr realised that if his model was correct, each atom would have a spectral fingerprint related to the differences between electron energy levels in that atom. The Rydberg equation, otherwise known as the Balmer equation, gave him evidence for the quantised emission of energy from the hydrogen atom, leading to him going on to further his model and define his postulates. So the hydrogen spectrum was very significant to the development of Bohr’s model of the atom, because without an understanding of it Bohr may not have continued to work on his model.

Remember- The hydrogen spectrum was extremely significant because it provided the only evidence at the time for an otherwise purely theoretical model.

Perform a first-hand investigation to observe the visible components of the hydrogen spectrum

In our experiment, we had a discharge tube (vacuum tube with a cathode and anode, powered by a high-voltage induction coil) with low-pressure hydrogen inside it. When high-voltage current was passed through the tube, the hydrogen fluoresced, emitting light that was visible in our darkened room. We observed the visible components of the spectrum with handheld spectrometers that used a diffraction grating to split the light. Using the spectrometer, we could clearly observe the red and blue/violet hydrogen emission lines, although the violet lines were very hard to observe. The red line was very clear and intense compared to the other observed lines.

Remember- Hydrogen discharge tube observed with a spectrometer.

Discuss Planck’s contribution to the concept of quantised energy

The concept of quantised energy is that energy can only occur in small packets of fixed amounts, and distinguished between energy increases due to increased intensity (bigger packets) and energy increases due to greater intensity (more packets). This was developed entirely by Planck in his work on black body radiation, and although Einstein significantly improved upon Planck’s ideas, the underlying idea was Planck’s alone, and so Planck made a huge contribution to the underlying concept of quantised energy. However, his involvement was limited to developing the mere concept- others developed it into a functional model.

Remember- Planck developed the concept of quantised energy, but not a functioning model.

Define Bohr’s postulates

Bohr had 3 postulates. The first was that electrons in an atom exist in stationary states of stability and emit no energy when in these states. The second was that energy is only lost or gained by an electron when it moves from state to state, and when it moves from a high energy state to a low energy state it releases a photon with energy equal to the difference between the states (and therefore a characteristic frequency). His third postulate was that electron angular momentum in a

Describe how Bohr’s postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum (the Rydberg equation)

Balmer originally devised the equation empirically by examining the lines in the hydrogen spectrum and creating a formula to fit them. Rydberg used Bohr’s postulates and manipulated them (especially the third) to create the same formula (derived from calculating differences in energy states). Essentially, there were two paths to the Rydberg equation and one of them used Bohr’s postulates to arrive at the equation, while the other didn’t.

Process and present diagrammatic information to illustrate Bohr’s explanation of the Balmer series

In this diagram, the energy levels described by Bohr are clearly marked. According to Bohr, the Balmer series (shown on the top of the diagram as the hydrogen spectrum) was caused by electrons changing energy levels. The electron makes a transition from a higher energy level to a lower energy level, in the process releasing light. As shown, larger energy changes produce more energetic photons, as seen in the Balmer series, and further, this diagram shows how the Balmer series is formed by successive electron transitions to the 2nd shell (transitions to other shells produce additional lines named after their discoverers).

figure 15

Remember- Bohr explained the Balmer series as being the result of successive electron transitions down to the 2nd shell.

Discuss the limitations of the Bohr model of the hydrogen atom (including “Analyse secondary information to identify the difficulties with the Rutherford- Bohr model, including its inability to completely explain the spectra of larger atoms, the relative intensity of spectral lines, the existence of hyperfine spectral lines, and the Zeeman Effect”)

For all the questions the Bohr model answered, it posed still more. There was still no explanation for there being no energy emission from accelerating electrons as Maxwell predicted- instead it was simply an assumption. Further, there was no evidence for the Bohr model to give it scientific credibility. Finally, in terms of explaining spectral lines there were observed effects that simply could not be explained. These were

  • Relative intensity of spectral lines- When observing spectra, some lines were much brighter than The Bohr model could not explain why some lines were more intense than others (i.e. why some electron transitions were preferred to others)
  • Hyperfine splitting- When the spectral lines were examined closely, it was observed that each line actually consisted of many small lines, the existence of which the Bohr model could not explain as it only predicted one clear line for each transition
  • Larger atoms- The Bohr model could not explain the spectra of larger atoms with more than one electron, a problem that Bohr tried unsuccessfully to
  • Zeeman effect- The Zeeman effect occurs when a magnetic field is passed through the discharge The magnetic field increases the hyperfine splitting of spectral lines, further breaking them up. Again, the Bohr model was unable to explain the experimental evidence

Although the Bohr model lay down the framework for the quantum model of the atom, which ended in a scientific revolution out of which quantum mechanics (a vital part of modern physics) emerged, it was left to future scientists such as Pauli and Heisenberg to fully explain these phenomena.

Remember- Relative intensity, hyperfine splitting, larger atoms, Zeeman effect.


Outline the methods used by the Braggs to determine crystal structure

Diffraction occurs when waves bend around obstructions, and interference patterns result when waves interfere. Diffraction can often result in interference patterns when the bent wave (acting as a point source) interferes with the original wave. A diffraction grating uses small obstructions with separations similar to the wavelength of the wave in question placed side by side to produce a predictable interference pattern that is directly linked to the spacing within the diffraction grating. The Braggs realised that the spacing between layers in a crystal lattice were similar to the wavelength of x-rays, and would therefore act as a diffraction grating. Further, they realised that from the interference pattern they obtained they could calculate the spacing between the lattice layers. The Braggs used an x-ray tube as their x-ray source, and the x-rays travelled through a hole in a shield which acted as a collimator to produce a tightly focussed beam of x-rays. The waves then reflected through a crystal target which acted as a diffraction grating, and then the x-rays travelled to a sensor to analyse the interference pattern. From this they could calculate lattice separation distance, which was of great importance to science and understanding crystal structures.

Remember- The Braggs used diffraction and interference patterns with x-rays to calculate the spacing between crystal lattice layers.

Identify that metals possess a crystal lattice structure

Metals, like many other molecules, have a crystal lattice structure in their solid state. This means that they exist as a 3-dimensional grid of atoms arranged into layers. It is a repeating structure where each atom occupies a well-defined equilibrium distance from its neighbours. In the case of metals, free electrons exist in between lattice layers and conduct electricity.

figure 11

Describe conduction in metals as a free movement of electrons unimpeded by the lattice

A metal has free electrons that exist in the space between metal ions in the lattice. This means that they exist in a more-or-less empty space containing no lattice ions, leaving them free to travel without being impeded by the lattice. However, collisions between the electron and the lattice still occur, as do collisions between electrons and other electrons. When an electric field or potential is applied to a metal, the metal conducts because the electrons move freely between the lattice layers.

Remember- Conduction occurs when electrons travel through the metal lattice, thereby moving charge.

Identify that resistance in metals is increased by the presence of impurities and the scattering of electrons by lattice vibrations

In order to conduct electricity, electrons must travel through the space between lattice layers. Resistance is low when the electrons are free to travel unimpeded, and resistance is high when the passage of electrons is obstructed. Impurities in a metal distort the lattice structure and electrons collide with the impurity, increasing resistance. Similarly, vibrations in the lattice (often caused by heating) destabilize the structure and make it harder for electrons to flow, increasing resistance.

figure 12

Remember- Impurities and lattice vibrations increase resistance.

Describe the occurrence in superconductors below their critical temperature of a population of electrons unaffected by electrical resistance

Phonons are a particular type of quantum particle. They represent quantised vibration states within a crystal structure. It is not necessary to know precisely what they are, only the role that they play in the formation of Cooper pairs.

Superconductors are materials that exhibit no resistance. They only occur at low temperatures because at higher temperatures electron pairs are not capable of forming. In a superconductor, lattice vibrations are eliminated due to the low temperature. As an electron travels through the lattice, it attracts lattice ions causing a lattice distortion- a small region of positive space that attracts another electron. The two electrons then exchange phonons and bind, forming a Cooper pair of electrons which behaves as a single particle. Because the two electrons are interacting with each other they interact less with the lattice, and so travel through it very easily with very little resistance. So below the critical temperature, when a material becomes a superconductor pairs of electrons form that are unaffected by electrical resistance.

Remember- Superconductivity occurs at low temperatures when electrons swap phonons to form Cooper pairs that interact far less with the lattice, resulting in superconductivity.

Process information to identify some of the metals, metal alloys, and com- pounds that have been identified as exhibiting the property of superconductivity and their critical temperatures

You shouldn’t need to remember this list, since in a test they will most likely give you a table of data to use in an answer. However, it will be useful to remember one or two values from this table so you can add them to any answer to show depth of knowledge. Also make sure you remember that 138K is the maximum temperature for superconductivity as of 2009 (though it probably isn’t necessary to remember the exact material).

Material Critical temperature (K)
Zinc 0.85
Aluminium 1.175
Mercury 4.15
Lead 7.196
Tin 7.72
AuBa2Ca3Cu4O11 99
Ba2Ca2Cu3O8.33 138

Discuss the BCS theory

If asked a question like this in an exam, ensure you describe the BCS theory before discussing it. 

The BCS theory of superconductivity is simply the idea that lattice distortions at low temperatures lead to the formation of Cooper pairs. This theory is extremely successful at explaining supercon- ductivity in Type 1 superconductors (substances that have a critical temperature below 30K) as it is almost 50 years old now, and still used. It provided a concrete framework on which to model superconductivity that was vital to understanding how it works. However, it is unable to explain superconductivity in Type 2 superconductors- the ceramic variety that can be superconductors at far higher temperatures. This is because the model predicts 30K as being the maximum temperature at which Cooper pairs are able to form. So while it is extremely important to understanding Type 1 superconductors, it does little to explain Type 2 and so is an incomplete theory.

Remember- The BCS theory explains Type 1 superconductivity but cannot explain Type 2 semicon- ductors.

Discuss the advantages of using superconductors and identify limitations to their use

There are many advantages to using superconductors. These are mainly that they operate with very little loss and so are extremely efficient, and also that they generate no waste heat because they are perfect conductors. They are capable of generating very strong magnetic fields per unit of weight, useful for MRI scanners, and could be used to make very efficient motors, generators and batteries. There are two key limitations to superconductors, however. Firstly, it is very difficult to cool superconductors to below their critical temperatures- they require a constant supply of liquid nitrogen at the moment (given the low temperatures currently needed to achieve superconductivity), and secondly it is very hard to shape ceramic superconductors as they are not ductile, making it difficult to turn superconductors into wires

Remember- Superconductors increase efficiency and can reduce size and weight, but are difficult to manufacture and require cooling.

Analyse information to explain why a magnet is able to hover above a super-
conducting material that has reached the temperature at which it is superconducting

A magnet is able to hover over a superconducting material for two reasons- firstly because magnetic fields are excluded from the superconductor, forcing the magnet to be repelled from the supercon- ductor thus causing it to rise up (this is the Meissner effect), and secondly due to the phenomenon of quantum pinning which stops the magnet from moving horizontally off the superconductor.

The Meissner effect is separate to the induction of eddy currents which would theoretically perfectly oppose the magnetic field of a magnet. This is shown to be true because if a magnet is placed on a superconductor as it is being cooled, it will jump into the air as the superconductor becomes superconducting- this shows it is not an induction phenomenon as change in magnetic flux is required to induce eddy currents. Therefore the levitation occurs due to the exclusion of magnetic fields from the superconductor.

figure 13

Remember- A magnet can float above a superconductor due to the Meissner effect.

Perform an investigation to demonstrate magnetic levitation

In this experiment, we had a ceramic superconducting disk in a Petri dish and a small magnetic cube. We poured liquid nitrogen onto the superconducting disk (and into the dish) to lower it below its critical temperature, making it superconductive. When we used insulated plastic tongs to place the magnet just above the disk, the magnet floated. Nudging it with the tongs caused it to rotate. Eventually, the magnet fell as the disk warmed up and lost its superconductivity. In our second trial, we left the magnet on the disk before pouring liquid nitrogen. As the disk cooled, the magnet suddenly floated upwards off the disk. This showed that the Meissner effect is due to the exclusion of magnetic fields from superconductors, rather than the formation of perfect eddy currents due to changes in flux (because for eddy currents to form there must be an initial change in flux to create them. In the experiment the magnet rose upwards by itself. In fact, the movement of the magnet upwards would have ordinarily induced eddy currents that would drag the magnet down. So this is compelling evidence that the levitation of the magnet is due to the exclusion of the field and not due to eddy currents).

Remember- The exclusion of the magnetic field from the superconductor caused the magnet to levitate.

Gather and process information to describe how superconductors and the effects of magnetic fields have been applied to develop a maglev train

Note that this dotpoint is not only about how superconductors are used in maglev trains, but also how superconductors make maglev trains possible. Often questions will require to you to examine the benefits of using superconductors for maglev trains, in addition to outlining how they are used.

A maglev train relies on superconductors for operation, because superconductors are extremely light, extremely strong magnets, making them well suited to levitate a heavy load such as the train. Superconductors are used in two areas- to levitate the maglev train, and to propel the train. The tracks and the train both have superconductors. Superconductors on the train consist of a looped superconductor on either side of the train. The superconductor is charged with electric current when it is made, and because it is looped (physically, with one end joined to the other), the current flows continuously. This sets up a strong, constant magnetic field. Superconducting electromagnets on the track, positioned above and below the train’s magnetic loops, repel the train from the bottom, and attract the train from the top, causing the train to float. The track magnets are mounted on the vertical sides of the track. Additional superconducting electromagnets on the track serve to propel the train. These electromagnets are situated all along the side of the track. Magnets in front of the train attract the train’s magnets, while magnets on the track behind the train repel the train. By constantly changing the polarity of the track magnets, the train is attracted and repelled in the same direction constantly, causing the maglev train to move rapidly along the track. Superconductors are vital to the development of maglev trains, because permanent magnets would be too heavy to generate the same field strength, and conventional electromagnets would lose too much energy as waste heat due to electrical resistance.

Remember- Superconductors are used in maglev trains because they are light and can produce the incredibly strong magnetic fields required to levitate and propel a train.

Process information to discuss possible applications of superconductivity and the effects of those applications on computers, generators and motors, and trans- mission of electricity through power grids

Superconductors offer great potential in a variety of fields, offering increased performance and ef- ficiency compared to conventional conductors. However, there are still two major obstacles that impede the use of superconductors in virtually all their applications. Firstly, superconductors must be extremely cold, necessitating liquid nitrogen cooling. In some cases this is merely inconvenient, such as in a maglev train, but in applications such as computers, it is extremely difficult and unwieldy to use liquid nitrogen as a coolant, although it has been accomplished by some computer enthusiasts. Secondly, at present type-2 superconductors, (the only realistic option for real-world applications be- cause they only require liquid nitrogen cooling, as opposed to type-1 superconductors with lower critical temperatures), are ceramic compounds that are not ductile. This makes it extremely difficult to use in electrical circuits that rely on ductility to produce a long wire to transfer electricity. Because they are not ductile, they would also be difficult materials to use in computer processors.

However, once these obstacles are overcome, using superconductors in place of standard conductors would bring tremendous benefits. In computers, a great deal of energy is wasted as heat. Further, heating makes it difficult for processors to operate properly, as it changes the properties of the silicon presently used. By using a superconductor, there will be little, if any, waste heat produced, resulting in a processor that can function at far faster speeds. Further, by replacing transistors with superconducting quantum switches (SQUID, or superconducting quantum interference device), the processor can operate faster still. In motors and generators, they can be used to operate at high currents with no losses and no heat production, resulting in extremely efficient motors and generators. According to V = IR, current output will be maximised when there is low resistance, showing that a superconductor will improve current output. Finally, in terms of transmission, a great deal of energy is wasted in the transmission of electricity through conversion to heat in wires. By using superconducting wires, energy loss through the electricity grid will be eliminated, resulting in greater efficiency, with possible impacts such as reduced cost of power, or a reduced need for additional electricity generation capacity. A superconducting electricity grid was successfully trialled in America 4 years ago, and is presently used in parts of the New York grid.

Remember- Superconductors can be used in generators, computer chips and electricity grids, al- though at present there are challenges that need to be resolved first.

Semiconductors and Transistors

Identify that some electrons in solids are shared between atoms and move freely

In solids, electron shells are replaced by electron band structures (because the energy levels of neighbouring atoms shift according to Pauli’s exclusion principle, with the energy levels clustering into broad band structures). These consist of the conduction band and the valence band. The valence band can be thought of as the normal outer shell of an atom where electrons are chained to that particular atom, while the conduction band can be thought of as a level where electrons are free to move between other atoms in the solid structure. Only electrons in the conduction band are shared- those in the valence band are not and remain immobile.

figure 8

Describe the difference between conductors, insulators and semiconductors in terms of band structures and relative electrical resistance

Because electrons can only move between atoms and therefore conduct electricity in the conduction band, the relative positions of the conduction band and valence band play a large part in determining the conducting properties of a material. In conductors, the conduction and valence bands overlap- this means that electrons in their normal valence positions can, without gaining any energy, be in the conduction band and move freely between atoms. Because it is so easy for electrons to move into the conduction band, there is little resistance. With insulators, there is a very large energy gap between the valence and conduction bands- this is known as the forbidden energy gap. In order to conduct electricity, electrons in an insulator must gain enough energy to jump from their normal valence band positions over the forbidden energy gap and into the conduction band- because this process requires a great deal of energy input it is very difficult to cause insulators to be conductive, and so they have high electrical resistance. Intrinsic semiconductors (pure semiconductor crystals consisting of only one element) have band gaps smaller than for insulators but bigger than conductors- they lie in between, so are initially insulators but when heated moderately become conductive. Other moderate energy input will cause conductivity. Extrinsic semiconductors (semiconductor crystals with deliberate impurities consisting of small quantities of a group 3 or group 5 element) also contain an extra energy level inside the forbidden energy gap for electrons to exist, reducing the energy required to get an electron into the conduction band.

Remember- Conductivity depends on the gap between the valence band the conduction band. Insulators have a large gap, semiconductors have a small gap, and conductors have no gap.

Identify absences of electrons in a nearly full band as holes, and recognise that both electrons and holes help to carry current

In a crystal lattice of a pure (intrinsic) semiconductor, all the outer shells are (theoretically) filled and there are no electrons available to conduct electricity (since free electrons in the conduction band are required). When a Group 3 impurity exists, an impurity with one less electron than a Group 4 semiconductor such as silicon or germanium, there is a hole in the crystal lattice structure where there should have been a bond electron. This hole forms a positive region of space, and because it’s charged it is capable of moving charge. To move the hole, bonds within the lattice switch around and change so that the position of the hole in the lattice changes. In this way, holes are able to carry current, helping to make a semiconductor conductive, with holes effectively behaving as if they were positive point charges (although the reality is they are regions of empty space that are positive relative to the lattice). With the application of additional energy to move lattice electrons into the conduction band, electrons can also carry current through the lattice.

figure 9
Remember- Holes are positive points in a crystal lattice that behave as point charges, and both holes and electrons can carry current.

Compare qualitatively the number of free electrons that can drift from atom to atom in conductors, semiconductors and insulators

Under normal conditions, conductors have very many free electrons that can drift from atom to atom (on the order of the number of atoms in the lattice), whereas in semiconductors and insulators very few, if any electrons are free and able to drift from atom to atom. However, with semiconductors if energy is applied to the system in the form of heat or a strong electrical field, the number of free electrons increases greatly causing it to conduct (although not to the same extent as straight conductors).

Remember- Conductors have many free electrons, insulators have very few.

Perform an investigation to model the behaviour of semiconductors, including the creation of a hole or positive charge on the atom that has lost the electron and the movement of electrons and holes in opposite directions when an electric field is applied across the semiconductor

Make sure that you are able to clearly explain this experiment. Don’t forget to talk about how holes and electrons move in opposite directions.

We modelled a semiconductor using marbles in a Petri dish, with each marble representing an electron. Removing a marble from the dish represented the creation of a hole. As the dish is disturbed by moving it, simulating the application of an electric field, the position of the hole changed as marbles moved in to fill it, moving the hole elsewhere in the dish. The gap and the marble moved in opposite directions, as a new gap was created when a marble moved to fill the old gap.

Then we modelled semiconductors using marbles as atoms and a metal ball bearing as an extra free electron that was capable of moving around the dish as the dish moved. When we moved the dish, the ball bearing moved from marble to marble, showing the movement of free electrons.

Remember- Swirling marbles in a dish along with a ball bearing.

Identify that the use of germanium in early transistors is related to lack of ability to produce other materials of sufficient purity

During early research with transistors and semiconductors, germanium was the semiconductor of choice. The main reason germanium was used was because of purification- in order to operate with predictable properties, the semiconductor crystal needed to be very pure. The only two semicon- ductors suitable for transistor use are germanium and silicon, being Group 4 semiconductors and somewhat easily available. Silicon was in fact the superior material, being more abundant and therefore cheaper, easier to dope, and having superior thermal properties (germanium became too conductive with only moderate heating making germanium chip performance highly dependant on temperature).

However, in the 1940’s at the start of semiconductor research, scientists were only able to purify germanium. The techniques that they used to purify germanium crystals could not be applied to silicon crystals. This meant that although silicon was the superior material, it could not be used because silicon crystals could not be manufactured pure enough to make reliable chips. Germanium was therefore used in early transistors until suitable purity silicon was developed.

Remember- Germanium was used in early transistors because scientists couldn’t purify silicon.

Describe how “doping” a semiconductor can change its electrical properties

The process of doping a semiconductor involves adding a Group 3 or Group 5 element as an impurity into the crystal structure of the semiconductor to reduce the energy input required for the semicon- ductor to become conductive. Only tiny amounts of the impurities are added- too much, and the semiconductor’s conductive properties become unpredictable. If a Group 3 element is added, then because it has one less electron in its outer shell the lattice structure will be missing an electron- in this way a hole is produced, and this hole is capable of moving charge. Similarly, if a Group 5 element is added there will be an extra, free electron in the lattice structure which is free to move between atoms and carry charge.

figure 10

Remember- Adding impurities to “dope” a semiconductor makes it more conductive.

Identify differences in p-type and n-type semiconductors in terms of the relative number of negative charge carriers and positive holes

P-type semiconductors have been doped with Group 3 elements whereas n-type semiconductors have been doped with Group 5 elements. This means that although they both are capable of carrying charge, the p-type semiconductor has positive holes to move charge whereas n-type semiconductors have extra electrons- negative charge carriers, to do the same. Holes and electrons flow in opposite directions in the crystal structure to conduct electricity, but they both enable the passage of current through the lattice.

Remember- “P-type” stands for “positive” and so uses Group 3 elements. “N-type” is for “negative” and so uses Group 5.

Describe differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices

This dotpoint is focussed on the differences between solid state and thermionic devices in terms of their performance and usage. See the Extra Content chapter for an overview of the physics behind their operation.

Although thermionic devices and solid state transistors perform exactly the same function (amplification of a signal or electrical switching), solid state devices almost completely replaced thermionic devices because of their vastly superior properties in terms of operation.

Attribute Thermionic device Solid state device
Cost Expensive Cheap
Dimensions Bulky and heavy Small and lightweight
Durability Fragile, easily broken Durable and reliable
Lifespan Short lifespan Long lifespan
Warm-up time Significant None
Energy efficiency Large power requirements Very low power

Although some audio enthusiasts claim valves are still better devices for amplification, it is generally accepted that transistors are superior to valves in almost every way. This led to solid state devices replacing valves.

Gather, process and present secondary information to discuss how shortcom- ings in available communication technology lead to an increased knowledge of the properties of materials with particular reference to the invention of the transistor

The biggest problem with communication technology in the early days of the radio was amplification- the received signal was extremely weak and could not produce a loud sound without being amplified. This meant researchers were always trying to improve amplification technology to address the short- comings with valves such as their high failure rate, high power consumption, their weight and their warm-up time. When they first determined some of the properties of semiconductors this need for better amplifiers fuelled heavy research into the properties of semiconductors and the ways in which they could be used as amplifiers in the form of transistors. So the shortcomings in available com- munications technology led to the rapid development of the transistor which would have otherwise taken many years longer.

Remember- The drive for transistors to replace valves was brought about not only by the limitations of valves but also because of the high demand for communications technology.

Identify data sources, gather, process, analyse information and use available evidence to assess the impact of the invention of transistors on society with particular reference to their use in microchips and microprocessors

The invention of the transistor has dramatically changed society, largely through the use of micro- processors and microchips. They have enabled the building of small, efficient computers that now have widespread applications throughout society as well as in scientific research. It has allowed the automation of repetitive tasks which has led to higher quality of life, at the expense of jobs and a rise in unemployment. However, in terms of communication it has had a tremendous benefit enabling the internet which has drastically changed society for the better. So overall transistors have had an extremely positive impact on society.