IB Physics Syllabus

Physics HL Syllabus (2009) Marc Wierzbitzki Including notes on HL Option E (Astrophysics) and H (Relativity)

IB Session May 201216

Marc W. Hockerill Anglo-‐European College IB Session May 2012

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TABLE OF CONTENT

1.1 THE REALM OF PHYSICS 4 1.2 MEASUREMENT AND UNCERTAINTIES 5 1.3 VECTORS AND SCALARS 7

2.1 KINEMATICS 8 2.2 FORCES AND DYNAMICS 9 2.3 WORK, ENERGY AND POWER 11 2.4 UNIFORM CIRCULAR MOTION 12

3.1 THERMAL CONCEPTS 14 3.2 THERMAL PROPERTIES OF MATTER 15

4.1 KINEMATICS OF SIMPLE HARMONIC MOTION (SHM) 21 4.2 ENERGY CHANGES DURING SIMPLE HARMONIC MOTION (SHM) 22 4.3 FORCED OSCILLATIONS AND RESONANCE 23 4.4 WAVE CHARACTERISTICS 24 4.5 WAVE PROPERTIES 26

5.1 ELECTRIC POTENTIAL DIFFERENCE, CURRENT AND RESISTANCE 29 5.2 ELECTRIC CIRCUITS 31

6.1 GRAVITATIONAL FORCE AND FIELD 34 6.3 MAGNETIC FORCE AND FIELD 36

7.1 THE ATOM 39 7.2 RADIOACTIVE DECAY 41 7.3 NUCLEAR REACTIONS, FISSION AND FUSION 42

8.1 ENERGY DEGRADATION AND POWER GENERATION 46 8.2 WORLD ENERGY SOURCES 46 8.3 FOSSIL FUEL POWER PRODUCTION 47 8.4 NON-‐FOSSIL FUEL POWER PRODUCTION 48 8.5 GREENHOUSE EFFECT 53 8.6 GLOBAL WARMING 55

9.1 PROJECTILE MOTION 58 9.2 GRAVITATIONAL FIELD, POTENTIAL AND ENERGY 58 9.3 ELECTRIC FIELD, POTENTIAL AND ENERGY 60 9.4 ORBITAL MOTION 62

10.1 THERMODYNAMICS 64 10.2 PROCESSES 65 10.3 SECOND LAW OF THERMODYNAMICS AND ENTROPY 66

11.1 STANDING (STATIONARY) WAVES 68


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11.2 DOPPLER EFFECT 69 11.3 DIFFRACTION 70 11.4 RESOLUTION 71 11.5 POLARIZATION 72

12.1 INDUCED ELECTROMOTIVE FORCE (EMF) 75 12.2 ALTERNATING CURRENT 76 12.3 TRANSMISSION OF ELECTRICAL POWER 78

13.1 QUANTUM PHYSICS 80 13.2 NUCLEAR PHYSICS 84

14.1 ANALOGUE AND DIGITAL SIGNALS 88 14.2 DATA CAPTURE; DIGITAL IMAGING USING CHARGE-‐COUPLED DEVICES (CCDS) 90

E1 INTRODUCTION TO THE UNIVERSE 93 E3 STELLAR DISTANCES 99 E4 COSMOLOGY 102 E5 STELLAR PROCESSES AND STELLAR EVOLUTION 107 E6 GALAXIES AND THE EXPANDING UNIVERSE 110

H1 INTRODUCTION TO RELATIVITY 113 H2 CONCEPTS AND POSTULATES OF SPECIAL RELATIVITY 113 H3 RELATIVISTIC KINEMATICS 114 H4 SOME CONSEQUENCES OF SPECIAL RELATIVITY 117 H5 EVIDENCE TO SUPPORT SPECIAL RELATIVITY 119 H6 RELATIVISTIC MOMENTUM AND ENERGY 121 H7 GENERAL RELATIVITY 122 H8 EVIDENCE TO SUPPORT GENERAL RELATIVITY 125


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1.1 The realm of physics Assessment statement Teacher’s notes 1.1.1 State and compare

quantities to the nearest order to magnitude.

Comparison between orders of magnitude can easily be made because working out the ration between two powers of ten is just a matter of adding or subtracting whole numbers. The diameter of an atom, 10-‐10, does not sound that much bigger than the diameter of a proton in its nucleus, 10-‐15, but the ratio between them is 105 or 100,000 times bigger. This is the same ration as between the size of a railway station and the diameter of the Earth.

1.1.2 State the ranges of magnitude of distances, masses and times that occur in the universe, from smallest to greatest.

Distances: from 10–15 m to 10+25 m (sub-‐nuclear particles to extent of the visible universe). Masses: from 10–30 kg to 10+50 kg (electron to mass of the universe). Times: from 10–23 s to 10+18 s (passage of light across a nucleus to the age of the universe).

Distance: 1026 Radius of observable

Universe 1021 Radius of local galaxy 1017 Distance to nearest star 1011 Distance from Earth to Sun 109 Distance from Earth to Moon 107 Radius of Earth 105 Deepest part of the ocean /

highest mountain 102 Tallest building 10-‐2 Length of fingernail 10-‐4 Thickness of piece of paper 10-‐6 Wavelength of light 10-‐10 Diameter of hydrogen atom 10-‐12 Wavelength of gamma ray 10-‐15 Diameter of proton Mass: 1052 Total mass of observable

Universe 1042 Mass of local galaxy 1030 Mass of Sun 1025 Mass of Earth 1021 Total mass of oceans 1018 Total mass of atmosphere 109 Laden oil super tanker 104 Elephant 102 Human 10-‐2 Mouse 10-‐7 Grain of sand 10-‐10 Blood corpuscle 10-‐14 Bacterium 10-‐22 Haemoglobin molecule 10-‐27 Proton 10-‐30 Electron Time: 1018 Age of the universe 1017 Age of the Earth 1014 Age of species – homo

sapiens 109 Typical human lifespan 107 1 year 105 1 day 100 One heartbeat 10-‐3 Period of high-‐frequency

sound


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10-‐8 Passage of light across a room

10-‐13 Vibration of an ion in a solid 10-‐15 Period of visible light 10-‐19 Passage of light across an

atom 10-‐23 Passage of light across a

nucleus

1.1.3 State ratios of quantities as differences of orders of magnitude.

For example, the ratio of the diameter of the hydrogen atom to its nucleus is about 105, or a difference of five orders of magnitude.

See 1.1.1

1.1.4 Estimate approximate values of everyday quantities to one or two significant figures and /or to the nearest order of magnitude.

See 1.1.2

1.2 Measurement and uncertainties Assessment statement Teacher’s notes 1.2.1

State the fundamental units in the SI system.

Students need to know the following: kilogram, metre, second, ampere, mole and kelvin.

Kilogram Metre Second Ampere Mole Kelvin (Candela)

1.2.2 Distinguish between fundamental and derived units and give examples of derived units.

Fundamental units are the SI units stated above. Derived units are combinations of SI units, such as 𝑠𝑝𝑒𝑒𝑑 = !"#$"%

!"#$%&

1.2.3 Convert between different units of quantities.

For example, J and kW h, J and eV, year and second, and between other systems and SI.

1.2.4 State units in the accepted SI format.

Students should use m s–2 not m/s2 and m s–1 not m/s.

1.2.5 State values in scientific notation and in multiples of units with appropriate prefixes.

For example, use nanoseconds or gigajoules

1.2.6 Describe and give examples of random and systematic errors.

Repeating readings does not reduce systematic errors. Sources of random errors include the readability of the instrument, the observer being less than perfect and the effects of a change in the surroundings. Sources of systematic errors include and instrument being wrongly calibrated, the observer being less than perfect in the same way every measurement, …

1.2.7 Distinguish between precision and accuracy.

A measurement may have great precision yet may be inaccurate (for example, if the instrument

An accurate experiment is one that has a small systematic error, whereas a precise experiment is


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has a zero offset error). one that has a small random error.

1.2.8 Explain how the effects of random errors may be reduced.

Students should be aware that systematic errors are not reduced by repeating readings.

Repeat the readings to reduce random human error.

1.2.9 Calculate quantities and results of calculations to the appropriate number of significant figures.

The number of significant figures should reflect the precision of the value or of the input data to a calculation. Only a simple rule is required: for multiplication and division, the number of significant digits in a result should not exceed that of the least precise value upon which it depends. The number of significant figures in any answer should reflect the number of significant figures in the given data.

For multiplication and division, the number of significant digits in a result should not exceed that of the least precise value upon which it depends. The number of significant figures in any answer should reflect the number of significant figures in the given data.

1.2.10 State uncertainties as absolute, fractional and percentage uncertainties.

1.2.11 Determine the uncertainties in results.

A simple approximate method rather than root mean squared calculations is sufficient to determine maximum uncertainties. For functions such as addition and subtraction, absolute uncertainties may be added. For multiplication, division and powers, percentage uncertainties may be added. For other functions (for example, trigonometric functions), the mean, highest and lowest possible answers may be calculated to obtain the uncertainty range. If one uncertainty is much larger than others, the approximate uncertainty in the calculated result may be taken as due to that quantity alone.

For functions such as addition and subtraction, absolute uncertainties may be added. For multiplication, division and powers, percentage uncertainties may be added. For other functions (for example, trigonometric functions), the mean, highest and lowest possible answers may be calculated to obtain the uncertainty range.

1.2.12 Identify uncertainties as error bars in graphs.

1.2.13 State random uncertainty as an uncertainty range (±) and represent it graphically as an “error bar”.

Error bars need be considered only when the uncertainty in one or both of the plotted quantities is significant. Error bars will not be expected for trigonometric or logarithmic functions.

1.2.14 Determine the uncertainties in the gradient and intercepts of a straight-‐line graph.

Only a simple approach is needed. To determine the uncertainty in the gradient and intercept, error bars need only be added to the first and the last data points.


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1.3 Vectors and scalars Assessment statement Teacher’s notes 1.3.1 Distinguish between

vector and scalar quantities, and give examples of each.

A vector is represented in print by a bold italicized symbol, for example, F.

A quantity that has magnitude and direction is called a vector quantity whereas one that has only magnitude is called a scalar quantity.

1.3.2 Determine the sum or difference of two vectors by a graphical method.

Multiplication and division of vectors by scalars is also required.

1.3.3 Resolve vectors into

perpendicular components along chosen axes.

For example, resolving parallel and perpendicular to an inclined plane.


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2.1 Kinematics Assessment statement Teacher’s notes 2.1.1 Define displacement,

velocity, speed and acceleration.

Quantities should be identified as scalar or vector quantities. See sub-‐topic 1.3.

Displacement: the distance moved in a particular direction Velocity: the rate of change of displacement Speed: the rate of change of distance Acceleration: the rate of change of velocity

2.1.2 Explain the difference between instantaneous and average values of speed, velocity and acceleration.

An instantaneous value of speed, velocity or acceleration is one at a particular point in time. An average value of speed, velocity or acceleration is taken over a period of time.

2.1.3 Outline the conditions under which the equations for uniformly accelerated motion may be applied.

. The equations for uniformly accelerated motion can only be applied if the acceleration is constant. They are as follows:

2.1.4 Identify the acceleration

of a body falling in a vacuum near the Earth’s surface with the acceleration g of free fall.

When we ignore the effect of air resistance on an object falling due to gravity we say that the object is under free fall. Free fall is an example of uniformly accelerated motion as the only force acting on the object is gravity. In the absence of air resistance, all falling objects have the same acceleration, independent of their mass.

2.1.5 Solve problems involving the equations of uniformly accelerated motion.

Example: A car accelerated uniformly from rest. After 10s it has travelled 200m. Average acceleration:

𝑠 = 𝑢𝑡 +12𝑎𝑡!

200 = 0 ∗ 10 +12𝑎 ∗ 10!

𝑎 = 4 Instantaneous speed after 10s:

𝑣! = 𝑢! + 2𝑎𝑠 𝑣! = 0 + 2 ∗ 4 ∗ 10

𝑣 = 8.9 2.1.6 Describe the effects of air

resistance on falling objects.

Only qualitative descriptions are expected. Students should understand what is meant by terminal speed.

Air resistance eventually affects all objects that are in motion. Due to air resistance objects can reach terminal velocity. This is a point by which the velocity remains constant and acceleration is zero.


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2.1.7 Draw and analyse distance–time graphs, displacement–time graphs, velocity–time graphs and acceleration–time graphs.

Students should be able to sketch and label these graphs for various situations. They should also be able to write descriptions of the motions represented by such graphs.

2.1.8 Calculate and interpret the

gradients of displacement–time graphs and velocity–time graphs, and the areas under velocity–time graphs and acceleration–time graphs.

2.1.9 Determine relative velocity in one and in two dimensions.

If two things are moving in the same straight line but are travelling at different speeds, then we can work out their relative velocities.

2.2 Forces and dynamics Assessment statement Teacher’s notes 2.2.1 Calculate the weight of a

body using the expression W = mg.

The weight of a body is the gravitational force experienced by that body.

2.2.2 Identify the forces acting on an object and draw free-‐body diagrams representing the forces acting.

Each force should be labelled by name or given a commonly accepted symbol. Vectors should have lengths approximately proportional to their magnitudes. See sub-‐topic 1.3.

2.2.3 Determine the resultant force in different situations.

The resultant force is the overall force acting on an object when all individual forces have been added together.

2.2.4 State Newton’s first law of motion.

‘An object continues in uniform motion in a straight line or at rest unless a resultant external force acts.’ All it says is that a resultant force causes acceleration.

2.2.5 Describe examples of Newton’s first law.

2.2.6 State the condition for translational equilibrium.

If the resultant force on an object is zero then it is said to be in


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translational equilibrium. 2.2.7 Solve problems involving

translational equilibrium.

2.2.8 State Newton’s second law of motion.

A correct statement of Newton’s second law using momentum would be ‘the resultant force is proportional to the rate of change of momentum.’ ‘The resultant force is proportional to the acceleration.’ ‘The resultant force is equal to the product of the mass and the acceleration.’

2.2.9 Solve problems involving Newton’s second law.

If a mass of 3kg is accelerated in a straight line by a resultant force of 12N then the acceleration must be 4m s-‐2 since F=ma.

2.2.10 Define linear momentum and impulse.

Linear momentum is defined as the product of mass and velocity. It is a vector. The change or momentum is called the impulse.

2.2.11 Determine the impulse due to a time-‐varying force by interpreting a force–time graph.

2.2.12 State the law of

conservation of linear momentum.

‘The total linear momentum of a system is of interacting particles remains constant provided there is no resultant external force.’

2.2.13 Solve problems involving momentum and impulse.

2.2.14 State Newton’s third law of motion.

‘When two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.’

2.2.15 Discuss examples of Newton’s third law.

Students should understand that when two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.


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2.3 Work, energy and power Assessment statement Teacher’s notes 2.3.1 Outline what is meant by

work. Students should be familiar with situations where the displacement is not in the same direction as the force.

Work is done when a force moves its point of application in the direction of the force. If the force moves to right angles to the direction of the force, then no work has been done.

𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝐹𝑠𝑐𝑜𝑠𝜃 It is a scalar quantity.

2.3.2 Determine the work done by a non-‐constant force by interpreting a force–displacement graph.

A typical example would be calculating the work done in extending a spring. See 2.3.7.

The total work done is the area under the force-‐displacement graph.

2.3.3 Solve problems involving

the work done by a force.

2.3.4 Outline what is meant by kinetic energy.

The energy that an object has as a result of its motion.

2.3.5 Outline what is meant by change in gravitational potential energy.

2.3.6 State the principle of conservation of energy.

Overall the total energy of any closed system must be constant. Energy is neither created nor destroyed, it just changes form. There is not change in the total energy in the Universe.

2.3.7 List different forms of energy and describe examples of the transformation of energy from one form to another.

-‐ Kinetic energy -‐ Gravitational potential -‐ Elastic potential energy -‐ Electrostatic potential -‐ Thermal energy -‐ Electrical energy -‐ Chemical energy -‐ Nuclear energy -‐ Internal energy -‐ Radiant energy -‐ Solar energy -‐ Light energy

2.3.8 Distinguish between elastic and inelastic collisions.

Students should be familiar with elastic and inelastic collisions and explosions. Knowledge of the coefficient of restitution is not required.

A collision in which no mechanical energy is lost is called an elastic collision. Most collisions are inelastic because kinetic energy is transformed to other forms of energy. If you are asked whether a collision is elastic or inelastic, calculate the kinetic energy before and after the collision

2.3.9 Define power. Power is defined as the rate at which energy is transferred. This


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is the same as the rate at which work is done.

2.3.10 Define and apply the concept of efficiency.

We define efficiency as the ratio of useful energy to the total energy transferred. It is often expressed as a percentage.

2.3.11 Solve problems involving momentum, work, energy and power.

2.4 Uniform circular motion Assessment statement Teacher’s notes 2.4.1 Draw a vector diagram to

illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the centre of the circle.

2.4.2 Apply the expression for centripetal acceleration.

The acceleration of a particle travelling in circular motion is called the centripetal acceleration. The force needed to cause the centripetal acceleration is called the centripetal force. The acceleration of any object moving at constant speed in a circle is given by:

𝑎 =𝑣!

𝑟

The centripetal acceleration is required for an object to move in a circle at constant speed.

2.4.3 Identify the force producing circular motion in various situations.

Examples include gravitational force acting on the Moon and friction acting sideways on the tyres of a car turning a corner.

Centripetal force: 𝐹 = 𝑚𝑎 = !!!

!

(Ball at the end of a string, swung vertically)

2.4.4 Solve problems involving circular motion.

Problems on banked motion (aircraft and vehicles going round banked tracks) will not be included.

Example: A car of mass 1500kg is travelling at constant speed of 20ms-‐1 around a circular track of radius 50m. The resultant force


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that must be acting on it works out to be:

𝐹 =1500 20 !

50= 12000𝑁

The centripetal force does not do any work.


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3.1 Thermal concepts Assessment statement Teacher’s notes 3.1.1 State that temperature

determines the direction of thermal energy transfer between two objects.

Students should be familiar with the concept of thermal equilibrium.

Hot and cold are just labels that identify the direction in which thermal energy will be naturally transferred when two objects are placed in thermal contact. Thermal energy naturally flows from hot to cold. Eventually, two objects would be expected to reach the same temperature – thermal equilibrium.

3.1.2 State the relation between the Kelvin and Celsius scales of temperature.

T/K = t/°C + 273 is sufficient.

3.1.3 State that the internal energy of a substance is the total potential energy and random kinetic energy of the molecules of the substance.

Students should know that the kinetic energy of the molecules arises from their random/translational/ rotational motion and that the potential energy of the molecules arises from the forces between the molecules.

The molecules have kinetic energy because they are moving. To be absolutely precise, a molecule can have either translational kinetic energy (the whole molecule is moving in a certain direction) or rotational kinetic energy (the molecule is rotating about one or more axes). The molecules have potential energy because of the intermolecular forces. If we imagine pulling two molecules further apart, this would require work against intermolecular forces.

3.1.4 Explain and distinguish between the macroscopic concepts of temperature, internal energy and thermal energy (heat).

Students should understand that the term thermal energy refers to the non-‐mechanical transfer of energy between a system and its surroundings. In this respect it is just as incorrect to refer to the “thermal energy in a body” as it would be to refer to the “work in a body”.

The macroscopic point of view considers the system as a whole and sees how it interacts with its surroundings. The microscopic point of view looks inside the system to see how its component parts interact with each other.

3.1.5 Define the mole and molar mass.

Mole: the mole is the basic SI unit for ‘amount of substance’. One mole of any substance is equal to the amount of that substance that contains the same number of atoms as 0.012kg of carbon-‐12. Molar Mass: The mass of one mole of a substance is called the molar mass. A simple rule applies. If an element has a certain mass number, A, then the molar mass will be A grams.

3.1.6 Define the Avogadro constant.

The number of atoms in 0.012kg or carbon-‐12 (6.02x1023)


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3.2 Thermal properties of matter Assessment statement Teacher’s notes 3.2.1 Define specific heat

capacity and thermal capacity.

Thermal capacity: The energy required to raise the temperature of an object by 1K. (C) Specific heat capacity: The energy required to raise a unit mass of a substance by 1K. (c)

3.2.2 Solve problems involving specific heat capacities and thermal capacities.

e.g. When a car brakes, an amount of thermal energy equal to 112500J is generated in the brake drums. If the mass of the brake drums is 28kg and their specific heat capacity is 460.5J kg-‐1 K-‐1, what is the change in their temperature? From 𝑄 = 𝑚𝑐Δ𝑇 we find:

Δ𝑇 =𝑄𝑚𝑐

=11250028 ∗ 460.5

= 8.7°𝐶 e.g. A piece of iron of mass 200g and temperature 300°C is dropped into 1.00kg of water of temperature 20°C. What will be the eventual temperature of the water? (Take c for iron as 470 and for water as 4200) Let T be the final unknown temperature. The iron will also be at this temperature, so Amount of thermal energy lost by the iron

= 𝑚!𝑐!(300 − 𝑇) and the amount of thermal energy gained by the water

= 𝑚!𝑐!(𝑇 − 20) Conservation of energy demands that thermal energy lost = thermal energy gained: 𝑚!𝑐!(𝑇 − 20) = 𝑚!𝑐!(300 − 𝑇)

àT=26°C Note how the large specific heat capacity of water leads to a small increase in the temperature of water compared with the huge drop in the temperature of iron.

3.2.3 Explain the physical differences between the solid, liquid and gaseous phases in terms of molecular structure and particle motion.

Only a simple model is required.

A solid is made up of particles that are arranged in a solid 3D shape. There is a strong force of attraction between the particles. If the solid was to be heated the particles would gain more energy and start to vibrate more vigorously.


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In a liquid the particles are free to move around. A liquid will mould itself to the shape of the container it is in. There is still a force of attraction between the particles.

In a gas the particles are free to move around. The particles have a lot of energy so they move quickly. Collisions between the molecules and the sides of the container are responsible for the pressure a gas exerts. There is almost no force of attraction between the molecules in a gas.

3.2.4 Describe and explain the

process of phase changes in terms of molecular behaviour.

Students should be familiar with the terms melting, freezing, evaporating, boiling and condensing, and should be able to describe each in terms of the changes in molecular potential and random kinetic energies of molecules.

Kinetic theory can be used to explain what happens when a substance is heated. To change from a solid to a liquid the particles must gain sufficient kinetic energy to overcome the forces between them and break away from their fixed positions. While the substance changes state its temperature does not change. Once the phase change has been completed the particles begin to gain more kinetic energy and the temperature of the substance increases again. As the boiling point is reached the particles gain enough kinetic energy to completely overcome the intermolecular forces and escape into the gaseous state.


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3.2.5 Explain in terms of

molecular behaviour why temperature does not change during a phase change.

The input energy is used to break or create bonds and so the energy is not turned into kinetic energy of the particles.

3.2.6 Distinguish between evaporation and boiling.

Evaporation is the change of state from liquid to gas that occurs below the boiling point of that liquid. In a liquid, a small amount of the molecules have enough kinetic energy to leave the surface of the liquid and become a gas. As the high-‐energy molecules leave the liquid, the temperature of the remaining liquid falls. Rate of evaporation depends on: -‐ The surface area of the liquid. As molecules leave from the surface of the liquid only, a bigger surface area will mean a greater rate of evaporation.

-‐ The temperature of the liquid. If the liquid is warmer then more molecules will have sufficient kinetic energy to escape.

-‐ The pressure of the air above the liquid. If the pressure is higher more kinetic energy will be required.

-‐ The movement of air. If there is a drought across the liquid the rate of evaporation will increase. Boiling occurs when the whole liquid is heated to its boiling point. All the molecules have sufficient kinetic energy to turn into a gas.

3.2.7 Define specific latent heat.

The thermal energy required to melt a unit mass of material at its melting point is called the specific latent heat of fusion and the termal energy required to vaporize a unit mass at its boiling point is called the specific latent heat of vaporization. Thus to melt or vaporize a mass m, we require a quantitiy of thermal energy

𝑄 = 𝑚𝐿! 𝑎𝑛𝑑 𝑄 = 𝑚𝐿! respectively. The units are J kg-‐1.

3.2.8 Solve problems involving specific latent heats.

Problems may include specific heat calculations.

e.g. An ice cube of mass 25g and temperature -‐10°C is dropped into a glass of water of mass 300g and temperature 20°C. What is the


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temperature eventually? (c=2200, L=334) Let this final temperature be T. Ignoring thermal energy lost by the glass itself, water will cool down by losing thermal energy. This thermal energy will be taken up by the ice to increase its temperature from -‐10°C to 0°C (thermal energy required 25*10-‐3*2200*10), melt the ice cube into water at 0°C (thermal energy required 25*10-‐3*334*103) and increase the temperature of the former ice cube from 0°C to the final temperature T. Thus, 0.3 ∗ 4200 ∗ 20 − 𝑇 =25 ∗ 10!! ∗ 2200 ∗ 10 + 25 ∗10!! ∗ 334 ∗ 10! + 20 ∗ 10!! ∗4200 ∗ T Solving for T gives T=11.9°C. e.g. Thermal energy is provided at a constant rate of 833J s-‐1 to 1kg of copper at the melting temperature. If it takes 4 minutes to completely melt the copper, find the latent heat of fusion of copper. The thermal energy needed to melt 1kg of copper is L, the specific latent heat of fusion. In 4 minutes the heat supplied is 833*60*4=200kJ, as m=1kg, L=200kJ kg-‐1.

3.2.9 Define pressure. Pressure is the normal force per unit area. The pressure in a gas results from the collision of the gas molecules with the walls of its container (not from collisions between molecules)

3.2.10 State the assumptions of the kinetic model of an ideal gas.

Kinetic theory uses the model of small particles bouncing around to describe the properties of gases and matter. Assumptions: -‐ Matter consists of large numbers of tiny particles

-‐ Particles are in constant motion moving in straight lines and thus have kinetic energy

-‐ All collisions between particles and the sides of the container are elastic

-‐ There are no forces of attraction or repulsion between the particles

-‐ The average kinetic energy per particle is proportional to the Kelvin temperature of the gas

-‐ Molecules move with a range of speeds


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-‐ The volume of the molecules is negligible compared with the volume of the gas itself

-‐ Molecules exert no forces on each other or the container except when in contact

-‐ The duration of collisions is very small compared with the time between collisions

-‐ The molecules obey Newton’s laws of mechanics

3.2.11 State that temperature is a measure of the average random kinetic energy of the molecules of an ideal gas.

Temperature is a measure of the average kinetic energy per particle of an ideal gas.

3.2.12 Explain the macroscopic behaviour of an ideal gas in terms of a molecular model.

Only qualitative explanations are required. Students should, for example, be able to explain how a change in volume results in a change in the frequency of particle collisions with the container and how this relates to a change in pressure and/or temperature.

Pressure law: Macroscopically, at a constant volume the pressure of a gas is proportional to its temperature. Microscopically this can be analysed as follows: -‐ If the temperature of a gas goes up, the molecules have more average kinetic energy – they are moving faster on average

-‐ Fast moving molecules will have a greater change of momentum when they hit the walls of the container

-‐ Thus the microscopic force from each molecule will be greater

-‐ The molecules are moving faster so they hit the walls more often

-‐ For both reasons, the total force on the wall goes up, thus the pressure increases. Charles’s law: Macroscopically, at a constant pressure, the volume of a gas is proportional to its temperature in Kelvin. This can be analysed as follows: -‐ A higher temperature means faster moving molecules

-‐ Faster moving molecules hit the walls with a greater microscopic force

-‐ If the volume of the gas increases, then the rate at which these collisions take place on a unit area of the wall must go down

-‐ The average force on a unit area of the wall can thus be the same

-‐ Thus the pressure remains the same Boyle’s law: Macroscopically, at a constant temperature, the pressure of a gas is inversely proportional to its volume. This can be analysed as follows: -‐ The constant temperature of a gas


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means that the molecules have a constant average speed

-‐ The microscopic fore that each molecule exerts on the wall will remain constant

-‐ Increasing the volume of the container decreases the rate with which the molecules hit the wall – average total force decreases

-‐ If the average total force decreases the pressure decreases


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4.1 Kinematics of simple harmonic motion (SHM) Assessment statement Teacher’s notes 4.1.1 Describe examples of

oscillations. An example of a SHM would be a

mass oscillating between two springs or a pendulum for small angles of osciallation.

4.1.2 Define the terms displacement, amplitude, frequency, period and phase difference.

The connection between frequency and period should be known.

Displacement: The instantaneous distance of the moving object from its mean position. Amplitude: The maximum displacement from the mean position. Frequency: The number of oscillations completed per unit time The SI measurement is the number of cycles per second or Hertz. Period: the time taken for one complete oscillation. Phase difference: This is a measure of how “in step” different particles are. If moving together they are in phase. It is measured in either degrees or radians. A phase difference of 90° is a quarter out of cycle.

4.1.3 Define simple harmonic motion (SHM) and state the defining equation as a = −w2x .

Students are expected to understand the significance of the negative sign in the equation and to recall the connection between w and T.

Simple harmonic motion is defined as the motion that takes place when the acceleration of the object is always directed towards, and is proportional to, its displacement from a fixed point. This acceleration is caused by a restoring force that must always be pointed towards the mean position and also proportional to the displacement from the mean position. The constant of proportionality between acceleration and displacement is often identified as the square of a constant w which is referred o as the angular frequency. a = −w2x

4.1.4 Solve problems using the

defining equation for SHM.

4.1.5 Apply the equations


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𝑣 = 𝑣!𝑠𝑖𝑛𝜔𝑡 , 𝑣 = 𝑣!𝑐𝑜𝑠𝜔𝑡, 𝑣 =±𝜔 (𝑥!! − 𝑥!), 𝑥 =𝑥!𝑠𝑖𝑛𝜔𝑡, 𝑥 = 𝑥!𝑐𝑜𝑠𝜔𝑡 as solutions to defining equations for SHM.

4.1.6 Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM.

4.2 Energy changes during simple harmonic motion (SHM) Assessment statement Teacher’s notes 4.2.1 Describe the interchange

between kinetic energy and potential energy during SHM.

In a SHM the total energy is interchanged between kinetic and potential energy. If no resistive force acts on the motion the total energy is constant and is said to be undamped. Potential energy increases as we move away from the equilibrium position and kinetic energy decreases. EP can be expressed as a sine curve, EK as a cosine curve.

4.2.2 Apply the expressions

for the kinetic energy of a particle undergoing SHM, for the total energy and for the potential energy.

4.2.3 Solve problems, both graphically and by calculation, involving energy changes during SHM.


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4.3 Forced oscillations and resonance Assessment statement Teacher’s notes 4.3.1 State what is meant by

damping. It is sufficient for students to know that damping involves a force that is always in the opposite direction to the direction of motion of the oscillating particle and that the force is a dissipative force.

Damping is a force that is always in the opposite direction of the motion of the oscillating particle and the force is dissipative.

4.3.2 Describe examples of damped oscillations.

Reference should be made to the degree of damping and the importance of critical damping. A detailed account of degrees of damping is not required.

This happens on cars in their suspensions, when it vibrates the damper tries to reduce the number of oscillations, to reduce the possible effects. On a piano the pedals reduce the oscillations of the springs of the piano. One pedal reduces the damping and one cuts the oscillations completely.

4.3.3 State what is meant by natural frequency of vibration and forced oscillations.

The natural frequency is the frequency that an object will oscillate at if it is moved from its equilibrium position and then released. Objects can also be made to oscillate by an external force, which is known as forced oscillation.

4.3.4 Describe graphically the variation with forced frequency of the amplitude of vibration of an object close to its natural frequency of vibration.

Students should be able to describe qualitatively factors that affect the frequency response and sharpness of the curve.

4.3.5 State what is meant by

resonance. Resonance occurs when a system is

subject to an oscillating force at exactly the same frequency as the natural frequency of oscillation of the system.

4.3.6 Describe examples of resonance where the effect is useful and where it should be avoided.

Examples may include quartz oscillators, microwave generators and vibrations in machinery.

Musical instruments: Many musical instruments produce their sounds by arranging for column of air or a string to be driven at its natural frequency which causes the amplitude of the oscillation to increase. Vibrations in machinery: When in operation, the moving parts of machinery provide regular driving forces on the other sections of the machinery. If the driving frequency is equal to the natural frequency, the amplitude of a particular vibration may get dangerously high. E.g. at a particular engine speed a truck’s rear view mirror can be seen to vibrate.


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4.4 Wave characteristics Assessment statement Teacher’s notes 4.4.1 Describe a wave pulse and

a continuous progressive (travelling) wave.

Students should be able to distinguish between oscillations and wave motion, and appreciate that, in many examples, the oscillations of the particles are simple harmonic.

A continuous wave involves a succession of individual oscillations. A wave pulse involves just one oscillation.

4.4.2 State that progressive (travelling) waves transfer energy.

Students should understand that there is no net motion of the medium through which the wave travels.

Light, sound and ripples on the surface of a pond are all examples of wave motion. They all transfer energy from one place to another, they do so without a net motion of the medium through which they travel and they all involve oscillations of one sort or another. The oscillations are SHM.

4.4.3 Describe and give examples of transverse and of longitudinal waves.

Students should describe the waves in terms of the direction of oscillation of particles in the wave relative to the direction of transfer of energy by the wave. Students should know that sound waves are longitudinal, that light waves are transverse and that transverse waves cannot be propagated in gases.

Transverse waves: Oscillations are at right angles to the direction of energy transfer. E.g. water ripples, light wave Longitudinal waves: Oscillations are parallel to the direction of energy transfer. E.g. sound waves, compression waves down a spring.

4.4.4 Describe waves in two dimensions, including the concepts of wavefronts and of rays.

Wave fronts highlight the parts of the wave that are moving together. Rays highlight the direction of energy transfer.

4.4.5 Describe the terms crest, trough, compression and rarefaction.

The top of the wave is known as the crest, whereas the bottom of the wave is known as the trough. A point on the wave where everything is ‘bunched together’ (high pressure) is known as compression. A point where everything is ‘far apart’ (low pressure) is known as a rarefaction.

4.4.6 Define the terms displacement, amplitude, frequency, period, wavelength, wave speed and intensity.

Students should know that intensity is proportional to amplitude2.

Displacement: This measures the change that has taken place that has taken place as a result of a wave passing a particular point. Zero displacement refers to the mean position. Amplitude: This is the maximum displacement from the mean position. If the wave does not lose any of its energy its amplitude is constant. Frequency: This is the number of oscillations that take place in one second. The unit used is Hertz. Period: This is the time taken for


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one complete oscillation. It is the time taken for one complete wave to pass any given point. Wavelength: This is the shortest distance along the wave between two points that are in phase with one another. For example, the distance from one crest to the next crest on a water ripple. Wave Speed: This is the speed at which the wave fronts pass a stationary observer. Intensity: The intensity of a wave is the power per unit area that is received by the observer. The intensity of a wave is proportional to the square of its amplitude: 𝐼 ∝ 𝐴!

4.4.7 Draw and explain displacement–time graphs and displacement–position graphs for transverse and for longitudinal waves.

Displacement-‐time: Represents the oscillation for one point on the wave. All the other points on the wave will oscillate in a similar manner, but they will not start their oscillation at exactly the same time. Displacement-‐position: Represents a ‘snap shot’ of all the points along the wave at one instant of time At a later time, the wave will have moved on but it will retain the same shape. The graphs can be used to represent longitudinal and transverse waves because the y-‐axis records only the value of the displacement. It does not specify the direction of this displacement.

4.4.8 Derive and apply the relationship between wave speed, wavelength and frequency.

There is a simple relationship that links wave speed, wavelength and frequency. It applies to all waves. The time taken for one complete oscillation is the period of the wave, T. In this time, the wave pattern will have moved on by one wavelength. This means that the speed of the wave must be given by:

𝑣 =𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑡𝑖𝑚𝑒

=𝜆𝑇

since !!= 𝑓

𝑣 = 𝑓𝜆


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4.4.9 State that all electromagnetic waves travel with the same speed in free space, and recall the orders of magnitude of the wavelengths of the principal radiations in the electromagnetic spectrum.

4.5 Wave properties Assessment

statement Teacher’s notes

4.5.1 Describe the reflection and transmission of waves at a boundary between two media.

This should include the sketching of incident, reflected and transmitted waves.

In general, when any wave meets the boundary between two different media it is partially reflected and partially transmitted.

Reflection: in this case the law of refraction applies: incident angle = reflected angle when measured from the normal. Refraction: in this case the wave is refracted towards the normal (when slowing down) and away from the normal (when getting faster)

4.5.2 State and apply Snell’s law.

Students should be able to define refractive index in terms of the ratio of the speeds of the wave in the two media and also in terms of the angles of incidence and refraction.

Snell’s law (an experimental law of refraction) states that the ratio

sin 𝑖sin (𝑟)

= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

for a given frequency. The ratio is equal to the ratio of the speeds in the different media

sin 𝑖sin (𝑟)

=𝑣!𝑣!

4.5.3 Explain and discuss qualitatively the diffraction of waves at apertures and obstacles.

The effect of wavelength compared to aperture or obstacle dimensions should be discussed.

When waves pass through apertures they tend to spread out. Waves also spread around obstacles. This is known as diffraction. Diffraction becomes relatively more important when the wavelength is large in comparison to the size of the object. The wavelength needs to be of the same order of magnitude as the aperture for diffraction to be


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noticeable.

4.5.4 Describe examples of

diffraction. E.g. ocean waves diffract through the harbour

opening and spread out, closely spaced tracks on a CD or DVD create the rainbow pattern because light is diffracted. Diffraction provides the reason why we can hear something even if we cannot see it.

4.5.5 State the principle of superposition and explain what is meant by constructive interference and by destructive interference.

Superposition: When two waves of the same type meet, they interfere and we can work out the resulting wave using the principle of superposition. The overall disturbance at any point and at any time where the waves meet is the vector sum of the disturbances that would have been produced by each f the individual waves. Constructive interference: Takes place when the two waves are in phase. There is zero phase difference between them. Destructive interference: Takes place when the two waves are exactly out of phase.

4.5.6 State and apply the conditions for constructive and for destructive interference in terms of path difference and phase difference.

See above. Constructive interference occurs when two waves are exactly in phase, which means that the path difference is zero. Destructive interference occurs when two waves are out of phase.

4.5.7 Apply the principle of superposition to determine the


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resultant of two waves.


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5.1 Electric potential difference, current and resistance Assessment statement Teacher’s notes 5.1.1 Define electric potential

difference. Electric potential difference is

defined as the energy difference per unit charge in an electric field. Electric potential energy: The energy that a charge has as a result of its position in an electric field.

5.1.2 Determine the change in potential energy when a charge moves between two points at different potentials.

To move a charge in an electric field, work must be done. The change in electric potential energy (work done) is the potential difference.

When a charge moves from A to B it gains electrical potential energy. Work must be done to move the charge. Change in p.d. = Force X distance = F x d = Eq x d where E is the electric field strength

5.1.3 Define the electronvolt. The electronvolt is the energy that would be gained by an electron moving through a potential difference of 1 volt.

5.1.4 Solve problems involving electric potential difference.

e.g. Calculate the speed of an electron accelerated in a vacuum by a p.d. of 1000V. KE of electron = V x e = 1000 x 1.6 x 10-‐19 = 1.6 x 10-‐16J 0.5mv2=1.6 x 10-‐16J v=1.87x107 m/s

5.1.5 Define electric current. It is sufficient for students to know that current is defined in terms of the force per unit length between parallel current-‐carrying conductors.

Current is defined as the rate of flow of electrical charge. If a current flows in just one direction it is known as direct current. A current that constantly changes direction is known as an alternating current. Current flows through an object when there is a potential difference across the object.

5.1.6 Define resistance. Students should be aware that R = V/I is a general definition of

Resistance is the mathematical ratio between potential difference


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resistance. It is not a statement of Ohm’s law. Students should understand what is meant by resistor.

and current. If something has a high resistance, it means that you would need a large potential difference across it in order to get a current to flow. We define a new unit, the ohm, to be equal to one volt per amp. A device with constant resistance (in other words an ohmic device) is called a resistor)

5.1.7 Apply the equation for resistance.

𝑅 = 𝑝𝑙𝐴

The resistance of a wire (at constant T) depends upon its length, the cross sectional area and its resistivity. The resistivity of a material tells us how well a material conducts.

5.1.8 State Ohm’s law. Ohm’s law states that the current flowing through a piece of metal is proportional to the potential difference across it providing the temperature remains constant.

5.1.9 Compare ohmic and non-‐ohmic behaviour.

For example, students should be able to draw the I–V characteristics of an ohmic resistor and a filament lamp.

If current and p.d. difference are proportional the device is said to be ohmic. Devices where current and p.d. are not proportional (filament lamp, diode) are said to be non-‐ohmic.

5.1.10 Derive and apply expressions for electrical power dissipation in resistors.

Since potential difference: 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑐ℎ𝑎𝑟𝑔𝑒 𝑓𝑙𝑜𝑤𝑒𝑑

And current: 𝑐ℎ𝑎𝑟𝑔𝑒 𝑓𝑙𝑜𝑤𝑒𝑑𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛

Multiplying current x p.d.:

𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛

This energy difference per time is the power dissipated by the resistor, All this energy is going into heating up the resistor. P = V x I(P = I2 x R)

5.1.11 Solve problems involving potential difference, current and resistance.

e.g. A 1.2 kW electric kettle is plugged into the 250V mains supply. Calculate the current drawn and its resistance. I = 1200 / 250 = 4.8 A R = 250 / 4.8 = 52 Ohm


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5.2 Electric circuits Assessment statement Teacher’s notes 5.2.1 Define electromotive

force (e.m.f.). The total energy difference per unit

charge is called the electromotive force (e.m.f.). However, it is not a force but an energy difference per charge measured in volts. In practical terms, e.m.f. is exactly the same as potential difference if no current flows.

5.2.2 Describe the concept of internal resistance.

When a battery supplies a current to an external circuit it gets warm. This is due to the battery having a small internal resistance.

The e.m.f. of the supply is the sum of the potential dropped across the internal resistor and the external resistor. e.m.f. = Ir + IR When a 6V battery is connected in a circuit some energy will be used up inside the battery itself. In other words, the battery has some internal resistance. The total energy difference per unit charge around the circuit is still 6V, but some of this energy is used up inside the battery. The energy difference per unit charge from one terminal to the other is less than the total made available by the chemical reaction in the battery.

5.2.3 Apply the equations for resistors in series and in parallel.

This includes combinations of resistors and also complete circuits involving internal resistance.

Resistors in series: Rt = R1 + R2 + …


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Resistors in parallel: 1𝑅!=1𝑅!+1𝑅!+⋯

5.2.4 Draw circuit diagrams. Students should be able to recognize and use the accepted circuit symbols.

5.2.5 Describe the use of ideal

ammeters and ideal voltmeters.

A current measuring meter is called an ammeter. It should be connected in series at the point where the current needs to be measured. A perfect ammeter would have zero resistance. This means that no potential difference is dropped across them. A meter that measures potential difference is called a voltmeter. It should be placed in parallel with the component or components being considered. A perfect voltmeter has infinite resistance.

5.2.6 Describe a potential divider.

Two resistors ‘divide up’ the potential difference of the battery. You can calculate the ‘share’ taken by one resistor using from the ratio of the resistances but this approach does not work unless the voltmeter’s resistance also is considered. A variable potential divider is often the best way to produce a variable power supply. When designing the potential divider, the smallest resistor that is going to be connected needs to be taken into account: the potentiometer’s resistance should be significantly smaller.

5.2.7 Explain the use of sensors in potential divider circuits.

Sensors should include light-‐dependent resistors (LDRs), negative temperature coefficient (NTC) thermistors and strain gauges.

A light dependent resistor (LDR) is a device whose resistance depends on the amount of light shining on its surface. An increase in light causes a decrease in resistance. A thermistor is a resistor whose value of resistance depends on its temperature. Most are semi-‐conducting devices that have a negative temperature coefficient (NTC). This means that an increase in temperature causes a decrease in resistance. Both of these devices can be used in potential divider circuits to create sensor circuits. The output potential difference of a sensor circuit depends on an external factor. Another possible sensor is a strain gauge whose output voltage depends


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on a small extension or compression that occurs which results in a change of length.

5.2.8 Solve problems involving electric circuits.

Students should appreciate that many circuit problems may be solved by regarding the circuit as a potential divider. Students should be aware that ammeters and voltmeters have their own resistance.


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6.1 Gravitational force and field Assessment statement Teacher’s notes 6.1.1 State Newton’s universal

law of gravitation. Students should be aware that the masses in the force law are point masses. The force between two spherical masses whose separation is large compared to their radii is the same as if the two spheres were point masses with their masses concentrated at the centres of the spheres.

It is called ‘universal’ gravitation because at the core of this theory is the statemtn that every mass in the Universe attracts all other masses. The value of the attraction between two point masses is given by an equation:

𝐹 =𝐺𝑚!𝑚!

𝑟!

Important: -‐ the law only deals with point masses

-‐ the masses in the equation are gravitational masses, not inertial masses

-‐ there is a force acting on each of the masses, these forces are equal and opposite

-‐ the forces are always attractive -‐ the forces only become significant if one or both objects involved are massive

-‐ the interaction between spherical masses turns out to be the same as if the masses were concentrated at the centres of the spheres

6.1.2 Define gravitational field strength.

The gravitational field is defined as the force per unit mass. Gravitational field strength is the force per unit mass on a particle in a gravitational field.

6.1.3 Determine the gravitational field due to one or more point masses.

Field strengths are vectors and therefore the gravitational field due to one or more point masses can be found by vector addition.

6.1.4 Derive an expression for gravitational field strength at the surface of a planet, assuming that all its mass is concentrated at its centre.

The gravitational field strength at the surface of a planet must be the same as the acceleration due to gravity on the surface. Field strength is defined to be !"#$%!"##

= 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 (𝐹 = 𝑚𝑎)

𝐹 =𝐺𝑚!𝑚!

𝑟!

𝑔 =𝐹𝑚

Combining this gives:

𝑔 =𝐺𝑀𝑟!

For the earth: M = 6.0 x 1024kg r = 6.4 x 106m

𝑔 =6.67 𝑥 10!!!𝑥 6.0 𝑥 10!"

6.4 𝑥 10! !

𝑔 = 9.8 𝑚𝑠!! 6.1.5 Solve problems involving

gravitational forces and


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fields. 6.2 Electric force and field Assessment statement Teacher’s notes 6.2.1 State that there are two

types of electric charge. Two types of charge exist –

positive and negative. Equal amounts of positive and negative charge cancel each other. Matter that contains not charge , or matter that contains equal amounts of positive and negative charge, is said to be electrically neutral.

6.2.2 State and apply the law of conservation of charge.

A very important experimental observation is that charge is always conserved. Charged objects can be created by friction. In this process electrons are physically moved from one object to another -‐ in order for the charge to remain on the object, it normally needs to be an insulator. Charge can be added or removed from an object but it cannot be destroyed. The quantity of electric charge is always conserved.

6.2.3 Describe and explain the difference in the electrical properties of conductors and insulators.

A material that allows the flow of charge through it is called an electrical conductor. If charge cannot flow through a material it is called an electrical insulator. In solid conductors the flow of charge is always as a result of the flow of electrons from atom to atom.

6.2.4 State Coulomb’s law. Students should be aware that the charges in the force law are point charges.

Coulomb’s law is used to calculate the force of attraction or repulsion between two point charges.

𝐹 =𝑘𝑞!𝑞!𝑟!

This can also be stated as:

𝐹 =𝑞!𝑞!4𝜋𝜀!𝑟!

Where 𝜀! is the permittivity of free space. If there are two or more charges near another charge, the overall force can be worked out using vector addition.

6.2.5 Define electric field strength.

Students should understand the concept of a test charge.

Electric field strength is defined as the force experienced per coulomb by a small test charge in an electric field.


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𝐸 =𝐹𝑞!

6.2.6 Determine the electric field strength due to one or more point charges.

Field strengths are vectors and therefore the electric field due to one or more point charges can be found by vector addition.

6.2.7 Draw the electric field patterns for different charge configurations.

These include the fields due to the following charge configurations: a point charge, a charged sphere, two point charges, and oppositely charged parallel plates. The latter includes the edge effect. Students should understand what is meant by radial field.

In a radial field the field lines diverge radially outward from a point source or converge radially inwards towards a point source.


electric charges, forces and fields.

6.3 Magnetic force and field Assessment statement Teacher’s notes 6.3.1 State that moving

charges give rise to magnetic fields.

An electric current can cause a magnetic field.

6.3.2 Draw magnetic field patterns due to currents.

These include the fields due to currents in a straight wire, a flat circular coil and a solenoid.


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6.3.3 Determine the direction

of the force on a current-‐carrying conductor in a magnetic field.

Different rules may be used to determine the force direction. Knowledge of any particular rule is not required.

When a current carrying wire is placed between the poles of a magnet it experiences a force. This causes the wire to move.

The direction in which the wire will move can be predicted using Fleming’s Left Hand rule:

6.3.4 Determine the direction

of the force on a charge moving in a magnetic field.

A single charge moving through a magnetic field also feels a force in exactly the same way that a current feels a force. In this case, the force is proportional to: -‐ the magnitude of the magnetic field B -‐ the magnitude of the charge q -‐ the velocity of the charge v -‐ the sine of the angle between the velocity and the field

𝐹 = 𝐵𝑞𝑣𝑠𝑖𝑛𝜃 Since the force on a moving charge is always at right angles to the velocity of the charge the resultant motion can be circular.


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6.3.5 Define the magnitude and direction of a magnetic field.

The magnetic field strength, B, is defined as follows:

𝐵 =𝐹

𝐼𝐿𝑠𝑖𝑛𝜃

A new unit, the tesla, is introduced. 1T is equal to 1NA-‐1m-‐1. The direction that the North pole of a small test compass would point if placed in the field (N to S)

6.3.6 Solve problems involving magnetic forces, fields and currents.


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7.1 The atom Assessment


7.1.1 Describe a model of the atom that features a small nucleus surrounded by electrons.

Students should be able to describe a simple model involving electrons kept in orbit around the nucleus as a result of the electrostatic attraction between the electrons and the nucleus.

The atomic (nuclear) model describes a very small central nucleus surrounded by electrons arranged in different energy levels. The nucleus itself contains protons and neutrons (collectively called nucleons). All of the positive charge and almost all the mass of the atom is in the nucleus. Overall an atom is neutral. The vast majority of the volume is noting at all – a vacuum.

7.1.2 Outline the evidence that supports a nuclear model of the atom.

A qualitative description of the Geiger–Marsden experiment and an interpretation of the results are all that is required.

One of the most convincing pieces of evidence for the nuclear model comes from the Geiger-‐Marsden experiment. Positive alpha particles were fired at a thin gold leaf. The relative size and velocity of the particles meant that most of them were expected to travel straight through he gold leaf. The idea was to see if there was any detectable structure within the gold atoms. The amazing discovery was that some of the alpha particles were deflected through huge angles.

They were surprised that some of the alpha particles were deflected as they passed through the gold. From this they deduced that the atom was made up of a small positively charged nucleus surrounded by space.


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Evidence for electron energy levels comes from emission and absorption spectra. The existence of isotopes provides evidence for neutrons.

7.1.3

Outline one limitation of the simple model of the nuclear atom.

The problem with this theory was that accelerating charges are known to lose energy. If the orbiting electrons were to lose energy they would spiral into the nucleus. Also, this model does not explain the emission and absorption spectrum. The model does not account for how the protons and neutrons stay together in the nucleus.

7.1.4 Outline evidence for the existence of atomic energy levels.

Students should be familiar with emission and absorption spectra, but the details of atomic models are not required. Students should understand that light is not a continuous wave but is emitted as “packets” or “photons” of energy, each of energy hf.

Evidence for electron energy levels comes from emission and absorption spectra. An energy level of 0 corresponds to the electron escaping from the atom. Electrons attached to an atom have negative energy levels.

7.1.5 Explain the terms nuclide, isotope and nucleon.

Nuclide: The name given to a particular species of atom (one whose nucleus contains a specified number of protons and a specified number of neutrons) Isotope: Elements that contain the same number of protons but a different number of neutrons. Nucleon: Protons and neutrons are collectively called nucleons.

7.1.6 Define nucleon number A, proton number Z and neutron number N.

A: Nucleon number – Number of nucleons (protons + neutrons) in the nucleus Z: Proton number – also called atomic number, equal to number of protons in the nucleus N: Neutron number – Number of neutrons in the nucleus N = A – Z


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7.1.7 Describe the

interactions in a nucleus.

Students need only know about the Coulomb interaction between protons and the strong, short-‐range nuclear interaction between nucleons.

The protons in a nucleus are all positive. Since like charges repel, they must be repelling one another all the time. This means there must be another force keeping the nucleus together. We know a few things about this force: -‐ It must be strong -‐ It must be very short-‐ranged as we do not observe this force anywhere other than inside the nucleus

-‐ It is likely to involve the neutrons as well The name given to this force is the strong nuclear force.

7.2 Radioactive decay Assessment statement Teacher’s notes 7.2.1 Describe the

phenomenon of natural radioactive decay.

The inclusion of the antineutrino in beta−decay is required.

Radioactive decay is a random process and is not affected by external influences. Some nuclei are more stable than others. When an unstable nucleus disintegrates to acquire a more stable state, radiations are emitted.

7.2.2 Describe the properties of alpha and beta particles and gamma radiation.

Property Alpha Beta Gamma Effect on photographic film

Yes Yes Yes

Appropriate number of irons produced in air

104 per mm travelled

102 per mm travelled

1 per mm travelled

Typical material needed to absorb

10-‐2 mm aluminium, piece of paper

A few mm aluminium

10cm lead

Penetration ability

Low Medium High

Typical path length in air

A few cm Less than one m Effectively infinite

Deflection by E and B fields

Behaves like a positive charge

Behaves like a negative charge

Not deflected

speed About 107 m/s About 108 m/s Speed of light

7.2.3 Describe the ionizing properties of alpha and beta particles and gamma radiation.

All three radiations are ionizing, which means that as they go through a substance, collisions occur which cause electrons to be removed from atoms. Atoms that have lost or gained electrons are called ions. When ionisations occur in biologically important molecules, such as DNA, mutations can occur.

7.2.4 Outline the biological effects of ionizing radiation.

Students should be familiar with the direct and indirect effects of radiation on structures within cells. A simple account of short-‐term

At the molecular level, an ionisation could cause damage directly to a biologically important molecule such as DNA. This could cause it to cease functioning. Molecular damage can


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and long-‐term effects of radiation on the body is required.

result in a disruption to the functions that are taking place within the cells that make up the organism. As well as potentially causing the cell to die, this could just prevent cells from dividing and multiplying. If malignant cells continue to grow then this is called cancer.

7.2.5 Explain why some nuclei are stable while others are unstable.

An explanation in terms of relative numbers of protons and neutrons and the forces involved is all that is required.

The stability of a particular nuclide depends greatly on the numbers of neutrons present. -‐ For small nuclei, the number of neutrons tends to equal the number of protons

-‐ For large nuclei there are more neutrons than protons

-‐ Nuclides above the band of stability have too many neutrons and will tend to decay with either alpha or beta decay.

-‐ Nuclides below the band of stability have too few neutrons and will tend to emit positrons.

7.2.6 State that radioactive decay is a random and spontaneous process and that the rate of decay decreases exponentially with time.

Exponential decay need not be treated analytically. It is sufficient to know that any quantity that reduces to half its initial value in a constant time decays exponentially. The nature of the decay is independent of the initial amount.

Radioactive decay is a random process and is not affected by external influences.

7.2.7 Define the term radioactive half-‐life.

The time taken for the number (or mass) of radioactive nuclei present to fall to half its value. This length of time is constant at any point in time -‐ showing that radioactive decay is exponential.

7.2.8 Determine the half-‐life of a nuclide from a decay curve.

7.2.9 Solve radioactive decay

problems involving integral numbers of half-‐lives.

7.3 Nuclear reactions, fission and fusion Assessment statement Teacher’s notes 7.3.1 Describe and give an

example of an artificial Artificial transmutation is the

process whereby a nucleus is


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(induced) transmutation. artificially made from another nucleus. It is different from regular radioactivity in that the reaction is not spontaneous; it is made to happen. When nitrogen gas was bombarded by alpha-‐particles it was found that there were two products: oxygen gas and positively charged particles.

𝑁!!" + 𝐻𝑒!! → 𝑂!!" + 𝑃!!

7.3.2 Construct and complete nuclear equations.

7.3.3 Define the term unified

atomic mass unit. In order to compare atomic

masses we often use unified


44

mass units, u. These are defined in terms of the most common isotope of carbon, cabon-‐12. One unified mass unit is defined as exactly one twelfth the mass of a caron-‐12 atom. Essentially, the mass of a proton and the mass of a neutron are both 1u.

7.3.4 Apply the Einstein mass–energy equivalence relationship.

Students must be familiar with the units MeV c −2 and GeV c −2 for mass.

If an object increases in energy, then its mass also increases. The relationship between mass and energy is described by Einstein’s famous equation:

𝐸 = 𝑚𝑐! When energy is released, there is also a decrease in mass of the products. In Einstein’s equation, 1kg of mass is equivalent to 9x1016J of energy. Since mass and energy are equivalent it is sometimes useful to work in units that avoid having to do repeated multiplications by the speed of light. A new possible unit for mass is thus MeV c −2. If 1 MeV c −2 worth of mass is converted you get 1MeV worth of energy.

7.3.5 Define the concepts of mass defect, binding energy and binding energy per nucleon.

Mass defect: The difference between the mass of a nucleus and the masses of its component nucleons. Binding energy: The amount of energy that is released when a nucleus is assembled from its component nucleons. It comes from a decrease in mass. The binding energy would also be the energy that needs to be added in order to separate a nucleus into its individual nucleons. Binding energy per nucleon: A useful measure of the stability of a nucleus is its binding energy. This is the energy that needs to be supplied to remove a nucleon from the nucleus. Nuclides that have the largest binding energy per nucleon are therefore the most stable. The total binding energy divided by the total number of nucleons.

7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.

Students should be familiar with binding energies plotted as positive quantities.


45

Nucleons in iron have the most binding energy, so they are the most stable. Nuclides therefore become more stable if they change in mass closer to that of the mass of iron. Therefore nuclides heavier than iron tend to break apart (fission) and nuclides lighter than iron tend to join (fuse) with other light nuclides.

7.3.7 Solve problems involving mass defect and binding energy.

7.3.8 Describe the processes of nuclear fission and nuclear fusion.

Fission: Fission is the name given to the nuclear reaction whereby large nuclei are induced to break up into smaller nuclei and release energy in the process. It is the reaction that is used in nuclear reactors and atomic bombs. A typical single reaction might involve bombarding a uranium nucleus with a neutron. This can cause the uranium nucleus to break up into two smaller nuclei. Fusion: Fusion is the name given to the nuclear reaction whereby small nuclei are induced to join together into larger nuclei and release energy in the process. It is the reaction that fuels all stars.

7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion.

7.3.10 State that nuclear fusion is the main source of the Sun’s energy.

Fusion is the name given to the nuclear reaction whereby small nuclei are induced to join together into larger nuclei and release energy in the process. It is the reaction that fuels all stars.

7.3.11 Solve problems involving fission and fusion reactions.


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8.1 Energy degradation and power generation Assessment statement Teacher’s notes 8.1.1 State that thermal energy

may be completely converted to work in a single process, but that continuous conversion of this energy into work requires a cyclical process and the transfer of some energy from the system.

In principle, thermal energy can be completely converted to work in a single process, nut the continuous conversion of this energy into work implies the use of machines that are continuously repeating their actions in a fixed cycle. Any cyclical process must involve the transfer of some energy from the system to the surroundings that is no longer available to perform useful work.

8.1.2 Explain what is meant by degraded energy.

Students should understand that, in any process that involves energy transformations, the energy that is transferred to the surroundings (thermal energy) is no longer available to perform useful work.

Any cyclical process must involve the transfer of some energy from the system to the surroundings that is no longer available to perform useful work. This unavailable energy is known as degraded energy, in accordance with the principle of the second law of thermodynamics.

8.1.3 Construct and analyse energy flow diagrams (Sankey diagrams) and identify where the energy is degraded.

It is expected that students will be able to construct flow diagrams for various systems including those described in sub-‐topics 8.3 and 8.4.

8.1.4 Outline the principal

mechanisms involved in the production of electrical power.

Students should know that electrical energy may be produced by rotating coils in a magnetic field. In sub-‐topics 8.2 and 8.3 students look in more detail at energy sources used to provide the energy to rotate the coils.

In all electrical power stations the process is essentially the same. A fuel is used to release thermal energy. This thermal energy is used to boil water and to make steam, which is used to turn turbines and the motion of the turbines is used to generate electrical energy. Transformers alter the potential difference.

8.2 World energy sources Assessment statement Teacher’s notes 8.2.1 Identify different world

energy sources. Students should be able to recognize those sources

Renewable Non-‐renewable Hydroelectric Coal Photovoltaic Oil


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associated with CO2 emission. Students should also appreciate that, in most instances, the Sun is the prime energy source for world energy.

Solar heaters Natural gas Wind Nuclear Biofuels Most of the energy used by humans can be traced back to energy radiated from the Sun, but not quite all of it. Possible sources are: -‐ The Sun’s radiated energy -‐ Gravitational energy of the Sun and the Moon

-‐ Nuclear energy stored within atoms -‐ The Earth’s internal heat energy

8.2.2 Outline and distinguish between renewable and non-‐renewable energy sources.

Renewable source of energy are those that cannot be used up, whereas non-‐renewable source of energy can be used up, cannot easily be replaced and will eventually run out.

8.2.3 Define the energy density of a fuel.

Energy density is measured in J kg–1.

Energy density provides a useful comparison between fuels and is defined as the energy liberated per unit mass of fuel consumed. 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =

𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑙𝑒𝑎𝑠𝑒 𝑓𝑟𝑜𝑚 𝑓𝑢𝑒𝑙𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑

8.2.4 Discuss how choice of

fuel is influenced by its energy density.

The values of energy density of different fuels will be provided.

Fuel choice can be particularly influenced by energy density when the fuel needs to be transported: the greater the mass of fuel that needs to be transported, the greater the cost.

8.2.5 State the relative proportions of world use of the different energy sources that are available.

Only approximate values are needed.

8.2.6 Discuss the relative

advantages and disadvantages of various energy sources.

The discussion applies to all the sources identified in sub-‐topics 8.2, 8.3 and 8.4.

See sections below + common sense.

8.3 Fossil fuel power production Assessment statement Teacher’s notes 8.3.1 Outline the historical and

geographical reasons for the widespread use of fossil fuels.

Students should appreciate that industrialization led to a higher rate of energy usage, leading to industry being developed near to large deposits of fossil fuels.

As the industrial revolution spread, the rate of energy usage greatly increased and industry tended to develop near to existing deposits of fossil fuels. Infrastructure was created to allow coal and other fossil fuels to be easily transported as the


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higher rates of energy usage demanded the use of fuels with a high energy density. This encouraged the growth of industries near the raw materials.

8.3.2 Discuss the energy density of fossil fuels with respect to the demands of power stations.

Students should be able to estimate the rate of fuel consumption by power stations.

8.3.3 Discuss the relative advantages and disadvantages associated with the transportation and storage of fossil fuels.

Advantages: -‐ Very high energy density -‐ Easy to transport -‐ Still cheap compared to other sources

-‐ Can be built anywhere with good transportation links

-‐ Can be used directly at home to provide heating

Disadvantages: -‐ Combustion products can produce pollution, acid rain, contain greenhouse gases

-‐ Extraction of fossil fuels can damage the environment

-‐ Non-‐renewable 8.3.4 State the overall efficiency

of power stations fuelled by different fossil fuels.

Only approximate values are required.

Fuel Typical E Maximum E Coal 35% 42% Nt. gas 45% 52% Oil 38% 45%

8.3.5 Describe the environmental problems associated with the recovery of fossil fuels and their use in power stations.

-‐ Combustion products can produce pollution, acid rain, contain greenhouse gases

-‐ Extraction of fossil fuels can damage the environment

-‐ Non-‐renewable

8.4 Non-‐fossil fuel power production Assessment


8.4.1 Describe how neutrons produced in a fission reaction may be used to initiate further fission reactions (chain reaction).

Students should know that only low-‐energy neutrons (~1 eV) favour nuclear fission. They should also know about critical mass.

In each individual reaction, an incoming neutron causes a uranium nucleus to split apart. The fragments are moving fast. In other words the temperature is very high. Among the fragments are more neutrons. If these neutrons go on to initiate further reactions then a chain reaction is created. Critical mass: minimum mass for chain reaction to occur.

8.4.2 Distinguish between controlled nuclear fission (power production) and uncontrolled nuclear

Students should be aware of the moral and ethical issues associated with nuclear weapons.

The design of a nuclear reactor needs to ensure that, on average, only one neutron from each reaction goes on to initiate a further reaction. If more reactions took place then the number


49

fission (nuclear weapons).

of reactions would run out of control. If fewer reaction took place, then the number of reactions would be decreasing and the fission process would soon stop.

8.4.3 Describe what is meant by fuel enrichment.

Naturally occurring uranium contains less than 1% of uranium-‐235. Enrichment is the process by which this percentage composition is increased to make nuclear fission more likely (to about 3%).

8.4.4 Describe the main energy transformations that take place in a nuclear power station.

-‐ Water is heated by heat energy created through nuclear fission -‐ Energy is lost to surroundings -‐ Steam turns a turbine (heat to kinetic) -‐ Energy is lost (friction) -‐ Turbine powers a generator -‐ Energy is lost (friction, heat, sound) -‐ Energy is transformed into electrical energy

8.4.5 Discuss the role of the moderator and the control rods in the production of controlled fission in a thermal fission reactor.

Three important components in the design of all nuclear reactors are the moderator, the control rods and the heat exchanger: -‐ Collisions between the neutrons and the nuclei of the moderator slow them down and allow further reactions to take place

-‐ The control rods are movable rods that readily absorb neutrons. They can be introduced or removed from the reaction chamber in order to control the chain reaction.

8.4.6 Discuss the role of the heat exchanger in a fission reactor.

The heat exchanger allows the nuclear reactions to occur in a place that is sealed off from the rest of the environment. The reactions increase the temperature in the core. This thermal energy is transferred to water and the steam that is produced turns the turbine.

8.4.7 Describe how neutron capture by a nucleus of uranium-‐238 (238U) results in the production of a nucleus

Plutonium-‐239 is also capable of sustaining fission reactions. This nuclide is formed as a by-‐product of a conventional nuclear reactor. A uranium-‐238 nucleus can capture fast moving neutrons to form


50

of plutonium-‐239 (239Pu).

uranium-‐239. This undergoes beta-‐decay to neptunium-‐239, which undergoes further beta-‐decay to plutonium-‐239. Reprocessing involves treating used fuel waste from nuclear reactors to recover uranium and plutonium and to deal with other waste products. A fast breeder reactor is one design that utilizes plutonium-‐239.

8.4.8 Describe the importance of plutonium-‐239 (239Pu) as a nuclear fuel.

It is sufficient for students to know that plutonium-‐239 (239Pu) is used as a fuel in other types of reactors.

A fast breeder reactor is one design that utilizes plutonium-‐239. It is capable of sustaining fission reactions.

8.4.9 Discuss safety issues and risks associated with the production of nuclear power.

Such issues involve: • The possibility of thermal meltdown and how it might arise • Problems associated with nuclear waste • Problems associated with the mining of uranium • The possibility that a nuclear power programme may be used as a means to produce nuclear weapons.

-‐ If the control rods were all removed, the reaction would rapidly increase its rate of production. Completely uncontrolled nuclear fission would cause an explosion and thermal meltdown in the core. The radioactive material in the reactor could be distributed around the surrounding area causing many fatalities.

-‐ A significant amount of material produced will remain dangerously radioactive for millions of years. The current solution is to bury this waste in geologically secure sites.

-‐ The uranium is mined from underground and any mining operation involves significant risk.

-‐ The transportation of the uranium from the mine to a power station needs to be secure (same for transportation of waste).

-‐ By-‐products of the civilian use of nuclear power can be used to produce nuclear weapons.

8.4.10 Outline the problems associated with producing nuclear power using nuclear fusion.

It is sufficient that students appreciate the problem of maintaining and confining a high-‐temperature, high-‐density plasma.

Fusion reactors offer the theoretical potential of significant power generation without many of the problems associated with current fission reactors. The fuel used, hydrogen, is in plentiful supply and the reaction (if it could be sustained) would not produce significant amounts of radioactive waste. The reaction requires creating temperatures high enough to ionize atomic hydrogen into a plasma state in which electrons and photons are not bound in atoms nut move independently. Currently the principal design challenges are associated with maintaining and confining the plasma at sufficiently high temperature and density for fusion to take place.

8.4.11 Solve problems on the production of nuclear power.

8.4.12 Distinguish between a photovoltaic cell and a

Students should be able to describe the energy transfers

Photovoltaic cell: Converts a portion of the radiated energy directly into a


51

solar heating panel. involved and outline appropriate uses of these devices.

potential difference. Active solar heater: Designed to capture as much thermal energy as possible. The hot water that it typically produces can be used domestically and would save on the use of electrical energy.

8.4.13 Outline reasons for seasonal and regional variations in the solar power incident per unit area of the Earth’s surface.

Scattering and absorption in the atmosphere means that often less energy arrives at the Earth’s surface. The amount that arrives depends greatly on the weather conditions. Different parts of the Earth’s surface will receive different amounts of solar radiation. It will also vary with the seasons since this will affect how spread out the rays have become.

8.4.14 Solve problems involving specific applications of photovoltaic cells and solar heating panels.

8.4.15 Distinguish between different hydroelectric schemes.

Students should know that the different schemes are based on: • water storage in lakes • tidal water storage • pump storage.

Water storage in lakes: Tidal water storage: Pump storage:

8.4.16 Describe the main

energy transformations that take place in hydroelectric schemes.

The source of energy in a hydroelectric power station is the gravitational potential energy of water. -‐ As part of the water cycle, water can fall as rain. It can be stored in large reservoirs as high up as is feasible

-‐ Tidal power schemes trap water at high tides and release it during a low tide.

-‐ Water can be pumped from a low reservoir to a high reservoir.

Gravitational PE of water à KE of


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water à KE of turbines à electrical energy.

8.4.17 Solve problems involving hydroelectric schemes.

8.4.18 Outline the basic features of a wind generator.

8.4.19 Determine the power

that may be delivered by a wind generator, assuming that the wind kinetic energy is completely converted into mechanical kinetic energy, and explain why this is impossible.

The area swept out by the blades of the turbine = 𝐴 = 𝜋𝑟! In one second the volume of air that passes = vA So mass of air that passes the turbine in one second = vAp (where p is the density of air) Kinetic energy available per second = !

!𝑚𝑣! = !

!𝑣𝐴𝑝 𝑣! = !

!𝐴𝑝𝑣!

à Power available. In practice, the kinetic energy of the incoming wind is easy to calculate, but it cannot all be harnessed – in other words the wind turbine cannot be 100% efficient. A doubling of the wind speed would mean that the available power would increase by a factor of eight.

8.4.20 Solve problems involving wind power.

8.4.21 Describe the principle of operation of an oscillating water column (OWC) ocean-‐wave energy converter.

Students should be aware that energy from a water wave can be extracted in a variety of different ways, but only a description of the OWC is required.

8.4.22 Determine the power We model the waves as squares to simplify the mathematics. Consider


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per unit length of a wavefront, assuming a rectangular profile for the wave.

a wave of Amplitude A, speed v and wavelength 𝜆.

8.4.23 Solve problems

involving wave power.

8.5 Greenhouse effect Assessment statement Teacher’s notes 8.5.1 Calculate the intensity

of the Sun’s radiation incident on a planet.

As the distance of an observer from a point source of light increases, the power received by the observer will decrease as the energy spreads out over a larger area. A doubling of distance will result in the reduction of the power received to a quarter of the original value. 𝐼 = !

!!!!

The intensity of the received radiation is inversely proportional to the square of the distance from the pint source to the observer. This is known as the inverse square law.

8.5.2 Define albedo. The proportion of power (or energy) reflected compared to the total power (energy) received.

=𝑡𝑜𝑡𝑎𝑙 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑒𝑑 𝑝𝑜𝑤𝑒𝑟𝑡𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑝𝑜𝑤𝑒𝑟

8.5.3 State factors that determine a planet’s albedo.

Students should know that the Earth’s albedo varies daily and is dependent on season (cloud formations) and latitude. Oceans have a low value but snow a high value. The global annual mean albedo is 0.3 (30%) on Earth.

The albedo of snowy surfaces is about 0.85 – indicating that this typ of surface reflects 85% of the sun’s radiation back. The global annual mean albedo of the Earth is 30% (so ~70% of the radiation reaching the Earth is absorbed).

8.5.4 Describe the greenhouse effect.

Short wavelength radiation is received from the sun and causes the


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surface of the Earth to warm up. The Earth will emit infrared radiation (longer wavelengths than the radiation coming from the sun because the Earth is cooler). Some of this infrared radiation is absorbed by gases in the atmosphere and re-‐radiated in all directions. The net effect is that the upper atmosphere and the surface of the earth are warmed. The greenhouse effect is a natural process without which the temperature of the Earth would be much lower.

8.5.5 Identify the main greenhouse gases and their sources.

The gases to be considered are CH4, H2O, CO2 and N2O. It is sufficient for students to know that each has natural and man-‐made origins.

Methane CH4: Principal component of natural gas and the product of decay, decomposition or fermentation. Water H2O: The small amounts of water vapour in the upper atmosphere have a significant effect. Carbon dioxide CO2: Combustion releases carbon dioxide into the atmosphere which can significantly increase the greenhouse effect. Nitrous oxide N2O: Livestock and industries are major sources of nitrous oxide. Its effect is significant as it can remain in the upper atmosphere for long periods.

8.5.6 Explain the molecular mechanisms by which greenhouse gases absorb infrared radiation.

Students should be aware of the role played by resonance. The natural frequency of oscillation of the molecules of greenhouse gases is in the infrared region.

These gases absorb infrared radiation as a result of resonance. The natural frequency of oscillation of the bonds within the molecules of the gas is in the infrared region. If the driving frequency (radiation from Earth) is equal to the natural frequency of the molecule, resonance will occur.

8.5.7 Analyse absorption graphs to compare the relative effects of different greenhouse gases.

Students should be familiar with, but will not be expected to remember, specific details of graphs showing infrared transmittance through a gas.

8.5.8 Outline the nature of

black-‐body radiation. Students should know that black-‐body radiation is the radiation emitted by a “perfect” emitter.

The perfect emitter will also be a perfect absorber of radiation. A black object absorbs all of the light falling on it. For this reason the radiation from a theoretical perfect emitter is known as black-‐body radiation.


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8.5.9 Draw and annotate a graph of the emission spectra of black bodies at different temperatures.

8.5.10 State the Stefan–

Boltzmann law and apply it to compare emission rates from different surfaces.

The Stefan-‐Boltzman law links the total power radiated by a black body (per unit are) to the temperature of the black body:

𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ∝ 𝑇! 𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 = 𝜎𝐴𝑇!

8.5.11 Apply the concept of emissivity to compare the emission rates from the different surfaces.

Emissivity is a number (from 0 to 1) measuring how well a surface emits radiation. Good emitters have an emissivity close to 1 (black body emissivity = 1)

8.5.12 Define surface heat capacity Cs.

Surface heat capacity is the energy required to raise the temperature of unit area of a planet’s surface by one degree, and is measured in J m–2 K–1.

Surface heat capacity is the energy required to raise the temperature of unit area of a planet’s surface by one degree, and is measured in J m–

2 K–1. 𝐶!=

𝑒𝑛𝑒𝑟𝑔𝑦∆𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑒𝑡𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑥 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎

8.5.13 Solve problems on the greenhouse effect and the heating of planets using a simple energy balance climate model.

Students should appreciate that the change of a planet’s temperature over a period of time is given by: (incoming radiation intensity – outgoing radiation intensity) x time / surface heat capacity. Students should be aware of limitations of the model and suggest how it may be improved.

See left, important.

8.6 Global warming Assessment statement Teacher’s notes 8.6.1 Describe some possible

models of global warming. Students must be aware that a range of models has been suggested to explain global warming, including changes in the composition of greenhouse gases in the atmosphere, increased solar flare activity, cyclical changes in the Earth’s orbit and volcanic activity.

8.6.2 State what is meant by the It is sufficient for students to be An enhancement of the


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enhanced greenhouse effect.

aware that enhancement of the greenhouse effect is caused by human activities.

greenhouse effect caused by human activities.

8.6.3 Identify the increased combustion of fossil fuels as the likely major cause of the enhanced greenhouse effect.

Students should be aware that, although debatable, the generally accepted view of most scientists is that human activities, mainly related to burning of fossil fuels, have released extra carbon dioxide into the atmosphere.

Although it is still being debated, the generally accepted view is that the increased combustion of fossil fuels has released extra carbon dioxide into the atmosphere, which has enhanced the greenhouse effect.

8.6.4 Describe the evidence that links global warming to increased levels of greenhouse gases.

For example, international ice core research produces evidence of atmospheric composition and mean global temperatures over thousands of years (ice cores up to 420,000 years have been drilled in the Russian Antarctic base, Vostok).

One piece of evidence that links global warming to increased levels of greenhouse gases comes from ice core data. Isotopic analysis allows the temperature to be estimated and air bubbles trapped in the ice cores can be used to measure the atmospheric concentrations of greenhouse gases. The record provides data from over 400,000 years ago to the present. The variations of temperature and carbon dioxide are very closely correlated.

8.6.5 Outline some of the mechanisms that may increase the rate of global warming.

Students should know that: • Global warming reduces ice/snow cover, which in turn changes the albedo, to increase rate of heat absorption • Temperature increase reduces the solubility of CO2 in the sea and increases atmospheric concentrations • Deforestation reduces carbon fixation.

Not only does deforestation result in the release of further CO2into the atmosphere, the reduction in number of trees reduces carbon fixation. Temperature increase reduces the solubility of CO2 in the sea and thus increases atmospheric concentrations.

8.6.6 Define coefficient of volume expansion.

Students should know that the coefficient of volume expansion is the fractional change in volume per degree change in temperature.

The coefficient of volume expansion records the fractional change in volume per degree change in temperature.

𝛾 =∆𝑉𝑉!∆𝑇

8.6 .7 State that one possible effect of the enhanced greenhouse effect is a rise in mean sea-‐level.

Between 0°C and 4°C, the coefficient of volume expansion for water is negative. This means that if the temperature of water is increased within the range of 0°C to 4°C this will cause a decrease in volume. When ice that is floating on seawater melts, the overall water level is predicted to initially decrease. Ice that is on land, however, is not displacing water and when it melts it will increase the sea level.

8.6.8 Outline possible reasons for a predicted rise in mean sea-‐level.

Students should be aware that precise predictions are difficult to make due to factors such as:

See above for explanation.


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• anomalous expansion of water • different effects of ice melting on sea water compared to ice melting on land.

8.6.9 Identify climate change as an outcome of the enhanced greenhouse effect.

“Most of the observed increase in globally averaged temperature since the mid-‐20th century is very likely due to the observed increase in anthropogenic [human caused] greenhouse gas concentrations.” (IPCC)

8.6.10 Solve problems related to the enhanced greenhouse effect.

Problems could involve volume expansion, specific heat capacity and latent heat.

8.6.11 Identify some possible solutions to reduce the enhanced greenhouse effect.

Students should be aware of the following: • greater efficiency of power production • replacing the use of coal and oil with natural gas • use of combined heating and power systems (CHP) • increased use of renewable energy sources and nuclear power • carbon dioxide capture and storage • use of hybrid vehicles.

Look left, should all be common sense.

8.6.12 Discuss international efforts to reduce the enhanced greenhouse effect.

These should include, for example: • Intergovernmental Panel on Climate Change (IPCC) • Kyoto Protocol • Asia-‐Pacific Partnership on Clean Development and Climate (APPCDC).

IPCC: Hundreds of governmental scientific representatives from more than a hundred countries regularly assess the up to date evidence from international research into global warming and human induced climate change. Kyoto Protocol: This is an amendment to UN Framework Convention on Climate Change. By signing the treaty, countries agree to work towards achieving a stipulated reduction in greenhouse gas emission. USA and Australia have not signed. APPCDC: Founding partners are Australia, China, India, Japan, Korea, US. They have agreed to work together to meet goals for energy security, air pollution reduction and climate change in was that promote sustainable economic growth and poverty reduction.


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9.1 Projectile motion Assessment statement Teacher’s notes 9.1.1 State the independence of

the vertical and the horizontal components of velocity for a projectile in a uniform field.

A ball that is thrown up in the air is moving horizontally and vertically at the same time but the horizontal and vertical components of the motion are independent of one another. Assuming the gravitational force is constant, this is always true.

9.1.2 Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance.

Proof of the parabolic nature of the trajectory is not required.

Horizontal: There are no forces in the horizontal direction, so there is no horizontal acceleration. This means that the horizontal velocity must be constant. Vertical: There is a constant vertical force acting down, so there is a constant vertical acceleration (g due to gravity).

9.1.3 Describe qualitatively the effect of air resistance on the trajectory of a projectile.

The path is no longer parabolic. The maximum height and range decrease. The angle at which the projectile impacts the ground steepens.

9.1.4 Solve problems on projectile motion.

Problems may involve projectiles launched horizontally or at any angle above or below the horizontal. Applying conservation of energy may provide a simpler solution to some problems than using projectile motion kinematics equations.

9.2 Gravitational field, potential and energy Assessment statement Teacher’s notes 9.2.1 Define gravitational

potential and gravitational potential energy.

Students should understand the scalar nature of gravitational potential and that the potential at infinity is taken as zero. Students should understand that the work done in moving a mass between two points in a

Gravitational potential: scalar quantity (V), the work done per unit mass in bringing a small point mass from infinity to a point (always negative) Gravitational potential energy: the work done in moving a mass


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gravitational field is independent of the path taken.

form infinity to a point in space (independent of path taken)

9.2.2 State and apply the expression for gravitational potential due to a point mass.

𝑉 =𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒𝑡𝑒𝑠𝑡 𝑚𝑎𝑠𝑠

= −𝐺𝑀𝑟

The units are J kg-‐1.

9.2.3 State and apply the formula relating gravitational field strength to gravitational potential gradient.

The change in potential per metre. Gravitational field strength is the negative of the potential gradient.

9.2.4 Determine the potential

due to one or more point masses.

9.2.5 Describe and sketch the pattern of equipotential surfaces due to one and two point masses.

9.2.6 State the relation between

equipotential surfaces and gravitational field lines.

There is a simple relationship between field lines and lines of equipotential – they are always at right angles to one another.

9.2.7 Explain the concept of escape speed from a planet.

The escape speed of a rocket is the speed needed to be able to escape the gravitational attraction of the planet. This means getting to an infinite distance away.

9.2.8 Derive an expression for the escape speed of an object from the surface of a planet.

Students should appreciate the simplifying assumptions in this derivation.

12𝑚𝑣! =

𝐺𝑀𝑚𝑅

So

𝑣 =2𝐺𝑀𝑅


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9.2.9 Solve problems involving gravitational potential energy and gravitational potential.

9.3 Electric field, potential and energy Assessment statement Teacher’s notes 9.3.1 Define electric potential

and electric potential energy.

Students should understand the scalar nature of electric potential and that the potential at infinity is taken as zero. Students should understand that the work done in moving a point charge between two points in an electric field is independent of the path taken.

Electric potential: The work done per unit charge in bringing a positive test charge from infinity to a point in an electric field. Electric potential energy: Energy that a charge has due to its position in an electric field.

9.3.2 State and apply the expression for electric potential due to a point charge.

If the total work done in bringing a positive test charge q from infinity to a point in an electric field is W, then the electric potential at that point V is defined to be:

𝑉 =𝑊𝑞

9.3.3 State and apply the formula relating electric field strength to electric potential gradient.

𝑉 =𝑄

4𝜋𝜀!𝑟

9.3.4 Determine the potential due to one or more point charges.

If several charges all contribute to the total potential at a point, it can be calculated by adding up the individual potentials due to the individual charges. The electric potential at any point outside the charge conducting sphere is exactly the same as if all the charge had been concentrated at its centre.


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9.3.5 Describe and sketch the pattern of equipotential surfaces due to one and two point charges.

9.3.6 State the relation between

equipotential surfaces and electric field lines.


electric potential energy and electric potential.


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9.4 Orbital motion Assessment statement Teacher’s notes 9.4.1 State that gravitation

provides the centripetal force for circular orbital motion.

Gravitation provides the centripetal force for circular orbital motion.

9.4.2 Derive Kepler’s third law. Expressing the above in formulae 𝐺𝑀𝑚𝑟!

=𝑚𝑣!

𝑟

𝐺𝑀 = 𝑣!𝑟

𝑣 =𝐺𝑀𝑟

Since a satellite does one orbit in time T,

𝑣 =𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒

𝑇=2𝜋𝑟𝑇

Substituting:

𝐺𝑀 =2𝜋𝑟𝑇

!

𝑟 =4𝜋! 𝑟!

𝑇!

As G, M and 4𝜋! are all constants, !!

!!= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

9.4.3 Derive expressions for the

kinetic energy, potential energy and total energy of an orbiting satellite.

Kinetic energy: 12𝑚𝑣!

From above:

𝑣 =𝐺𝑀𝑟

=>1 2𝐺𝑀𝑚𝑟

Potential energy:

−𝐺𝑀𝑚𝑟

Total energy: Total energy = Kinetic energy + Potential energy 1 2𝐺𝑀𝑚𝑟

−𝐺𝑀𝑚𝑟

= −1 2𝐺𝑀𝑚𝑟


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9.4.4 Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite.

9.4.5 Discuss the concept of

“weightlessness” in orbital motion, in free fall and in deep space.

If the lift cables beak and the lift (and passenger) accelerate down at 10m s-‐2, the person would appear to be weightless for the duration of the fall. Given the ambiguity of the term weight, it is better to call this situation the apparent weightlessness of objects in free fall together. In a space station, the gravitational pull on the astronaut provides the centripetal force needed to stay in the orbit. This resultant force causes the centripetal acceleration. The same is true for the gravitational pull on the satellite and the satellite’s acceleration. There is no contact force between the satellite and the astronaut so, once again, we have apparent weightlessness.

9.4.6 Solve problems involving orbital motion.


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10.1 Thermodynamics Assessment statement Teacher’s notes 10.1.1 State the equation of state

for an ideal gas. Students should be aware that an ideal gas is one that has the equation of state PV = nRT and that this equation also defines the universal gas constant R.

The three ideal gas laws can be combined together to produce one mathematical relationship:

𝑝𝑉𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

This constant will depend on the mass and type of gas. If we compare the value of this constant for different masses of different gases, it turns out to depend on the number of molecules that are in the gas -‐not their type. In this case we use the definition of the mole to state that for n moles of ideal gas: 𝑝𝑉𝑛𝑇

= 𝑎 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 This constant is called the molar gas constant R. à pV=nRT

10.1.2 Describe the difference between an ideal gas and a real gas.

Students should be aware of the circumstances in which real gas behaviour approximates to ideal gas behaviour. Students should also appreciate that ideal gases cannot be liquefied.

An ideal gas is one that follows the gas laws for all values of p, V and T and this cannot be liquefied. Real gases, however, can approximate to ideal behaviour providing that the intermolecular forces are small enough to be ignored. For this to apply, the pressure/density of the gas must be low and the temperature must be reasonably high. For an ideal gas, there are no intermolecular forces, collisions between particles are elastic and particles are considered to be points (small size).

10.1.3 Describe the concept of the absolute zero of temperature and the Kelvin scale of temperature.

The linear relationship can be extrapolated back to -‐273K, known as absolute zero. At absolute zero, a body would not have any volume.

10.1.4 Solve problems using the equation of state of an ideal gas.

e.g. What volume will be occupied by 8g of helium (mass number 4) at room temperature (20°C) and atmospheric pressure (1x105Pa)


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n=8/4=2 moles T=20+273=293K

𝑉 =𝑛𝑅𝑇𝑝

=2𝑥8.314𝑥293

1𝑥10!= 0.049𝑚!

10.2 Processes Assessment statement Teacher’s notes 10.2.1 Deduce an expression for

the work involved in a volume change of a gas at constant pressure.

For a gas in the cylinder expanding by a small distance dx

𝑑𝑊 = 𝐹𝑑𝑥 and

𝐹 = 𝑃 𝐴 𝑑𝑊 = 𝑃 𝐴 𝑑𝑥

and 𝐴 𝑑𝑥 = 𝑑𝑉 𝑑𝑊 = 𝑃 𝑑𝑉

𝐴𝑑𝑉!!

!!

10.2.2 State the first law of thermodynamics.

Students should be familiar with the terms system and surroundings. They should also appreciate that if a system and its surroundings are at different temperatures and the system undergoes a process, the energy transferred by non-‐mechanical means to or from the system is referred to as thermal energy (heat).

Thermodynamic system: Most of the time when studying the behaviour of an ideal gas in particular situations, we focus on the macroscopic behaviour of the gas as a whole. In terms of work and energy, the gas can gain or lose thermal energy and it can do work or work can be done on it. In this context, the gas can be seen as a thermodynamic system. Surroundings: If we are focusing our study on the behaviour of an ideal gas, then everything else can be called its surroundings. First law: The first law is simply a statement of the principle of energy conservation as applied to a system. If an amount of thermal energy ∆𝑄 is given to a system,


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then one of two things must happen (or a combination of both). The system can increase its internal energy ∆𝑈 or it can do work ∆𝑊.

10.2.3 Identify the first law of

thermodynamics as a statement of the principle of energy conservation.

Stated above.

10.2.4 Describe the isochoric (isovolumetric), isobaric, isothermal and adiabatic changes of state of an ideal gas.

In each process, the energy transferred, the work done and the internal energy change should be addressed. Students should realize that a rapid compression or expansion of a gas is approximately adiabatic.

Isochoric: constant volume Isobaric: constant pressure Isothermal: constant temperature Adiabatic: no thermal energy transfer between gas and surrounds

10.2.5 Draw and annotate

thermodynamic processes and cycles on P–V diagrams.

See above.

10.2.6 Calculate from a P–V diagram the work done in a thermodynamic cycle.

The area under the graph (between the lines of change in a cycle) represents the work done.

10.2.7 Solve problems involving state changes of a gas.

10.3 Second law of thermodynamics and entropy Assessment statement Teacher’s notes 10.3.1 State that the second law

of thermodynamics implies that thermal energy cannot spontaneously transfer from a region of low temperature to a region of high temperature.

No heat engine, operating in a cycle, can take in heat from its surroundings and totally convert it into work. No heat pump can transfer thermal energy from a low-‐temperature reservoir to a high-‐temperature reservoir without work being done on it. Heat flows from hot to cold objects. The entropy of the universe can


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never decrease. 10.3.2 State that entropy is a

system property that expresses the degree of disorder in the system.

Entropy is a property that expresses the disorder in the system.

10.3.3 State the second law of thermodynamics in terms of entropy changes.

A statement that the overall entropy of the universe is increasing will suffice or that all natural processes increase the entropy of the universe.

When thermal energy flows from a hot object to a colder object, the overall entropy has increased.

10.3.4 Discuss examples of natural processes in terms of entropy changes.

Students should understand that, although local entropy may decrease, any process will increase the total entropy of the system and surroundings, that is, the universe.

Water freezes at 0°C because this is the temperature at which the entropy increase of the surroundings (when receiving the latent heat) equals the entropy decrease of the water molecules becoming more ordered. It would not freeze at a higher temperature because this would mean that the overall entropy of the system would decrease.


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11.1 Standing (stationary) waves Assessment statement Teacher’s notes 11.1.1 Describe the nature of

standing (stationary) waves.

Students should consider energy transfer, amplitude and phase.

A standing wave will be formed if waves of the same amplitude and frequency travelling in different directions interfere.

11.1.2 Explain the formation of one-‐dimensional standing waves.

Students should understand what is meant by nodes and antinodes.

Node: Points along the wave that are always at rest. Antinode: Points along the wave where maximum movement takes place.

11.1.3 Discuss the modes of vibration of strings and air in open and in closed pipes.

The lowest-‐frequency mode is known either as the fundamental or as the first harmonic. The term overtone will not be used.

The lowest-‐frequency mode is known either as the fundamental or as the first harmonic.

11.1.4 Compare standing waves

and travelling waves. Stationary wave Travelling wave All points on the wave have different amplitudes. The maximum amplitude is 2A at the antinodes. It is zero at the nodes.

All points on the wave have the same amplitude.

All points oscillate with the same frequency.

All points oscillate with the same frequency.

Wavelength is twice the distance from one node to the next node.

Wavelength is the shortest distance along the wave between two points


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that are in phase. All points between one node and the next node are moving in phase.

All points along a wavelength have different phases.

Energy is not transmitted by the wave, but it does have energy associated with it.

Energy is transmitted by the wave.

11.1.5 Solve problems involving standing waves.

An organ pipe (open at one end) is 1.2m long. Calculate its fundamental frequency. The speed of sound is 330ms-‐1. L=1.2m à !

!= 1.2𝑚 𝜆 = 4.8𝑚 𝑣 = 𝑓𝜆

𝑓 =3304.8

= 69𝐻𝑍

11.2 Doppler effect Assessment statement Teacher’s notes 11.2.1 Describe what is meant by

the Doppler effect. A change in frequency of a wave

due to a moving source or observer. The Doppler effect is the name given to the change of frequency of a wave as a result of the movement of the source or the movement of the observer.

11.2.2 Explain the Doppler effect by reference to wave front diagrams for moving-‐detector and moving-‐source situations.

11.2.3 Apply the Doppler effect

equations for sound. The frequency of a car’s horn is

measured by a stationary observer as 200Hz when the car is at rest. What frequency will be heard if the car is approaching the observer at 30ms-‐1 (speed of sound is 330ms-‐1).

𝑓! = 200𝐻𝑧 𝑓! =?

𝑣! = 30𝑚𝑠!! 𝑐 = 330𝑚𝑠!!


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𝑓! = 2001

1 − 30330

𝑓! = 220𝐻𝑧 11.2.4 Solve problems on the

Doppler effect for sound. Problems will not include situations where both source and detector are moving.

See above for an example problem.

11.2.5 Solve problems on the Doppler effect for electromagnetic waves using the approximation

Δf =𝑣𝑐𝑓

Students should appreciate that the approximation may be used only when v << c.

Unfortunately, the above equations do not apply to light – the velocities cannot be worked out relative to the medium. It is, however, possible to derive an equation for light that turns out to be in exactly the same form as the equation for sound as long as two conditions are met: 1. The relative velocity of source and detector is used in the equation

2. This relative velocity is a lot less than the speed of light.

11.2.6 Outline an example in which the Doppler effect is used to measure speed.

Suitable examples include blood-‐flow measurements and the measurement of vehicle speeds.

Radar detectors can be used to measure the speed of a moving object. They do this by measuring the change in the frequency of the reflected wave.

11.3 Diffraction Assessment statement Teacher’s notes 11.3.1 Sketch the variation with

angle of diffraction of the relative intensity of light diffracted at a single slit.

Diffraction is a wave effect. The objects involved have a size that is of the same order of magnitude as the wavelength of visible light.

11.3.2 Derive the formula 𝜃 = !

!

for the position of the first minimum of the diffraction pattern produced at a single slit.

We can treat the slit as a series of secondary wave sources. In the forward direction (𝜃=0) these are all in phase so they add up to give a maximum intensity. At any other angle, there is a path difference between the rays that depends on the angle. The overall result is the


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addition of all the sources. The condition for the first minimum is that the angle must make all of the sources across the slit cancel out.


single-‐slit diffraction.

11.4 Resolution Assessment statement Teacher’s notes 11.4.1 Sketch the variation with

angle of diffraction of the relative intensity of light emitted by two point sources that has been diffracted at a single slit.

Students should sketch the variation where the diffraction patterns are well resolved, just resolved and not resolved.

11.4.2 State the Rayleigh

criterion for images of two sources to be just resolved.

Students should know that the criterion for a circular aperture is 𝜃 = 1.22 !

!

For a slit, the first minimum is at the angle: 𝜃 = !

!; for a circular

aperture, the first minimum is at the angle: 𝜃 = 1.22 !

!.

If two sources are just resolved, then the first minimum of one diffraction pattern is located on top of the maximum of the other diffraction pattern. This is known as the Rayleigh criterion.

11.4.3 Describe the significance of resolution in the development of devices such as CDs and DVDs, the electron microscope and radio telescopes.

CDs and DVDs: The maximum amount of information that can be stored depends on the size and the method used for recording information. Electron microscope: Resolves items that cannot be resolved using a light microscope. The electrons have an effective wavelength that is much smaller than the wavelength of visible light. Radio telescope: The size of the dish limits the maximum resolution possible. Several radio telescopes can be linked together in an array to create a virtual radio telescope with a greater diameter


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and with a greater ability to resolve astronomical objects.

11.4.4 Solve problems involving resolution.

Problems could involve the human eye and optical instruments.

Late one night, a student was observing a car approaching from a long distance away. She noticed that when she first observed the headlights of the car, they appeared to be one point of light. Later, when the car was closer, she became able to see two separate points of light. If the wavelength of the light can be taken as 500nm and the diameter of her pupil is approximately 4mm, calculate how far away the car was when she could first distinguish two points of light. Take the distance between the headlights to be 1.8m. When just resolved:

𝜃 = 1.22𝜆𝑏

𝜃 = 1.225𝑥10!!

0.004

𝜃 = 1.525𝑥10!! Since the angle is very small:

𝜃 =1.8

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑥 =1.8

1.525𝑥10!!= 12𝑘𝑚

11.5 Polarization Assessment statement Teacher’s notes 11.5.1 Describe what is meant

by polarized light. Any EM wave is said to be

unpolarised if the plane of vibration varies randomly whereas place-‐polarised light has a fixed plane of vibration. A mixture of polarized light and unpolarised light is partially plane-‐polarised. If the plane of polarisation rotates uniformly the light is said to be circularly polarised.

11.5.2 Describe polarization by

reflection. This may be illustrated using light or microwaves. The use of polarized sunglasses should be included.

A ray of light incident on the boundary between two media will, in general, be reflected and refracted. The reflected ray is always partially plane-‐polarized. If the reflected ray and the


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refracted ray are at right angles to one another, then the reflected ray is totally plane-‐polarised. The angle of incidence for this condition is known as the polarising angle.

11.5.3 State and apply Brewster’s law.

Brewster’s law relates the refractive index of medium 2, n, to the incident angle:

𝑛 =𝑠𝑖𝑛𝜃!𝑠𝑖𝑛𝜃!

=𝑠𝑖𝑛𝜃!𝑐𝑜𝑠𝜃!

= 𝑡𝑎𝑛𝜃!

11.5.4 Explain the terms polarizer and analyser.

A polariser is any device that produces plane-‐polarized light from an unpolarised beam. An analyser is a polariser used to detect polarised light.

11.5.5 Calculate the intensity of a transmitted beam of polarized light using Malus’ law.

The intensity of light is proportional to the amplitude squared. 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝐼 ∝ 𝐸!

𝐼 ∝ 𝐸!!𝑐𝑜𝑠!𝜃 Malus’s law:

𝐼 = 𝐼!𝑐𝑜𝑠!𝜃 I is the transmitted intensity 𝐼! is the incident intensity 𝜃 is the angle between the plane of vibration and the analyser’s preferred direction

11.5.6 Describe what is meant by an optically active substance.

Students should be aware that such substances rotate the plane of polarization.

An optically active substance is one that rotates the plane of polarisation of light that passes through it. Many solutions are optically active.

11.5.7 Describe the use of

polarization in the determination of the concentration of certain solutions.

A polarimeter is a device that measures 𝜃 for a given solution. It consists of two polarisers (1 polariser and 1 analyser) that are initially aligned. The optically active solution is introduced between the two and the analyser is rotated to find the maximum transmitted light.

11.5.8 Outline qualitatively how polarization may be used in stress analysis.

Glass and some plastics become birefringent when placed under stress. When polarised white light is passed through stressed plastics and then analysed, bright coloured lines are observed in the regions of maximum stress.


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11.5.9 Outline qualitatively the action of liquid-‐crystal displays (LCDs).

Aim 8: The use of LCD screens in a wide variety of different applications/devices can be mentioned.

1. Polarising filter film with a

vertical axis to polarise light as it enters

2. Glass substrate with ITO electrodes. The shapes of these electrodes will determine the shapes that will appear when the LCD is turned on.

3. Twisted liquid crystal 4. Glass substrate with common

electrode film with horizontal ridges to line up with the horizontal filter

5. Polarising filter film with a horizontal axis to block/pass light

6. Reflective surface to send light back to viewer. (in a backlit LCD, this layer is replaced with a light source)

11.5.10 Solve problems involving the polarization of light.


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12.1 Induced electromotive force (emf) Assessment statement Teacher’s notes 12.1.1 Describe the inducing of

an emf by relative motion between a conductor and a magnetic field.

When a conductor moves through a magnetic field, an emf is induced. It depends on: -‐ The speed of the wire -‐ The strength of the magnetic field

-‐ The length of the wire in the magnetic field

12.1.2 Derive the formula for the emf induced in a straight conductor moving in a magnetic field.

Students should be able to derive the expression induced emf = Blv without using Faraday’s law.

Electrical force due to emf:

𝐹! = 𝐵 ∗ 𝑞 =𝑉𝐼∗ 𝑞

Magnetic force due to movement 𝐹! = 𝐵 ∗ 𝑞 ∗ 𝑣

So

𝐵 ∗ 𝑞 ∗ 𝑣 =𝑉𝐼∗ 𝑞

𝑉 = 𝐵𝑙𝑣 As no current is flowing, the emf=potential difference

𝑒𝑚𝑓 = 𝐵𝑙𝑣 12.1.3 Define magnetic flux and

magnetic flux linkage. Magnetic flux: A measurement of

the amount of field lines passing through an area, at right angles to that area.

𝜙 = 𝐵𝐴𝑐𝑜𝑠𝜃 B = magnetic field strength A = area (square metres) 𝜃 = angle between flux and line normal to area 1Wb (weber) = 1Tm2 Magnetic flux linkage: A measure of the number of turns of wire ‘linked’ to (passing through) magnetic flux. (flux linkage = flux x number of turns)

12.1.4 Describe the production of an induced emf by a time-‐changing magnetic flux.

An emf is induced in a conductor whenever flux is cut. If the magnetic flux ∆𝜙 is perpendicular to the surface, the magnetic flux passing through the area ∆𝐴 is defined in terms of the magnetic flied strength B as follows:

∆𝜙 = 𝐵∆𝐴 → 𝐵 =∆𝜙∆𝐴

The alternative name for ‘magnetic field strength’ is ‘flux density’. If the area is not perpendicular, but at an angle to the field lines:

𝜙 = 𝐵𝐴𝑐𝑜𝑠𝜃 Since 𝜀 = 𝐵𝑙𝑣 and 𝑣 = ∆!

∆! à 𝜀 = !"∆!

∆!

Also, 𝑙∆𝑥 = ∆𝐴 à 𝜀 = !∆!∆!

Then, 𝐵∆𝐴 = ∆𝜙 à 𝜺 = ∆𝝓∆𝒕

In words, the emf induced is equal to the rate of cutting of flux. If the conductor is kept stationary and the magnets are moved, the same effect is produced.

12.1.5 State Faraday’s law and Lenz’s law.

Faraday’s law: The induced emf in a circuit is equal to the rate of change of flux linkage through


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the circuit.

𝜀 = −𝑁∆𝜙∆𝑡

The minus sign indicates that the emf is always induced so as to oppose the change causing it (Lenz’s law) Lenz’s law: ‘The direction of the induced emf is such that if an induced current were able to flow, it would oppose the change which caused it.’

12.1.6 Solve electromagnetic induction problems.

12.2 Alternating current Assessment statement Teacher’s notes 12.2.1 Describe the emf

induced in a coil rotating within a uniform magnetic field.

Students should understand, without any derivation, that the induced emf is sinusoidal if the rotation is at constant speed.

12.2.2 Explain the operation of

a basic alternating current (ac) generator.

The coil of wire rotates in the magnetic field due to an external force. As it rotates the flux linkage of the coil changes with time and induces an emf causing a current to flow. A coil rotating at constant speed will produce a sinusoidal induced emf. Increasing the speed of rotation will reduce the time period of the oscillation and increase the amplitude of induced emf.

12.2.3 Describe the effect on

the induced emf of Students will be expected to compare the output from

Changing the frequency will affect the time between peaks and the emf


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changing the generator frequency.

generators operating at different frequencies by sketching appropriate graphs.

(amplitude peaks). For example, it the frequency is doubled, the period will be halved and the emf (amplitude) will be doubled:

12.2.4 Discuss what is meant by

the root mean squared (rms) value of an alternating current or voltage.

Students should know that the rms value of an alternating current (or voltage) is that value of the direct current (or voltage) that dissipates power in a resistor at the same rate. The rms value is also known as the rating.

If the output of an a.c. generator is connected to a resistor an alternating current will flow. A sinusoidal potential difference means a sinusoidal current.

The graph shows that the average power dissipation is half the peak power dissipation for a sinusoidal current.

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 =𝐼!!𝑅2

Thus the effective current through the resistor is called the root mean square or rms current.

𝐼!.!.!. =𝐼!2

The rms value is also known as the rating.

12.2.5 State the relation between peak and rms values for sinusoidal currents and voltages.

12.2.6 Solve problems using

peak and rms values.

12.2.7 Solve ac circuit problems


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for ohmic resistors. 12.2.8 Describe the operation of

an ideal transformer. The value of output potential difference can be changed by changing the runs ratio. A step-‐up transformer increases the voltage, whereas a stet-‐down transformer decreases the voltage.

12.2.9 Solve problems on the

operation of ideal transformers.

12.3 Transmission of electrical power Assessment statement Teacher’s notes 12.3.1 Outline the reasons for

power losses in transmission lines and real transformers.

• Resistance of the windings of a transformer result in the transformer warming up.

• Eddy currents are unwanted currents induced in the iron core. The currents can be reduced by laminating he core into individually electrically insulates thin strips.

• Hysteresis losses cause the iron core to warm up as a result of the continues cycle of changes to its magnetism

• Flux losses are caused by magnetic ‘leakage’. A transformer is only 100% efficient if all of the magnetic flux that is produced by the primary links with the secondary.

• The wires cannot have zero resistance. This means they must dissipate some power.

• Over large distance, the power wasted would be very significant.

• A very high potential difference is much more efficient but very dangerous to the user.

12.3.2 Explain the use of high-‐

voltage step-‐up and step-‐Students should be aware that, for economic reasons, there is

Use step-‐up transformers to increase the voltage for the


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down transformers in the transmission of electrical power.

no ideal value of voltage for electrical transmission.

transmission stage and then use step-‐down transformers for the end user. (power dissipated is 𝑃 = 𝐼!𝑅 à if the current is large then the power dissipated will be large)

12.3.3 Solve problems on the operation of real transformers and power transmission.

12.3.4 Suggest how extra-‐low-‐frequency electromagnetic fields, such as those created by electrical appliances and power lines, induce currents within a human body.

Electrical power lines carry alternating current, which means they produce changing extra-‐low-‐frequency electromagnetic fields. These changing fields are theoretically able to induce currents within any conductor, including human bodies.

12.3.5 Discuss some of the possible risks involved in living and working near high-‐voltage power lines.

Students should be aware that current experimental evidence suggests that low-‐frequency fields do not harm genetic material. Students should appreciate that the risks attached to the inducing of current in the body are not fully understood. These risks are likely to be dependent on current (density), frequency and length of exposure.

Electrical power lines on pylons are not insulated along their length and are thus extremely dangerous if they become unattached from the pylon. In addition, some statistical evidence exists which suggests that there are regions (near power lines) where more children are diagnosed with leukaemia, a cancer of the blood, than usual. Students should appreciate that the risks attached to the inducing of current in the body are not fully understood. These risks are likely to be dependent on current (density), frequency and length of exposure.


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13.1 Quantum physics Assessment statement Teacher’s notes 13.1.1 Describe the

photoelectric effect. Under certain conditions, when light

(ultra-‐violet) is shone onto a metal surface, electrons are emitted from the surface. Below a certain threshold frequency, not photoelectrons are emitted. Above the threshold frequency, the maximum kinetic energy of the electrons depends on the frequency of the incident light. The number of electrons emitted depends on the intensity of the light and does not depend on the frequency. There is no noticeable delay between the arrival of the light and the emission of electrons. These observations cannot be reconciled with the view that light is a wave. A wave of any frequency should eventually bring enough energy to the metal plate.

13.1.2 Describe the concept of the photon, and use it to explain the photoelectric effect.

Students should be able to explain why the wave model of light is unable to account for the photoelectric effect, and be able to describe and explain the Einstein model.

Einstein introduced the idea of thinking of light as being made up of particles. His explanation was: • Electrons at the surface need a certain minimum energy in order to escape from the surface. This minimum energy is called the work function of the metal (𝜙)

• The UV light energy arrives in lots of little packets of energy – photons

• The energy in each packet is fixed by the frequency of UV light that is being used, whereas the number of packets arriving per second is fixed by the intensity of the source

• The energy carried by a photon is given by E=hf

13.1.3 Describe and explain an experiment to test the Einstein model.

Millikan’s experiment involving the application of a stopping potential would be suitable.

13.1.4 Solve problems

involving the photoelectric effect.

Example: What is the maximum velocity of electrons emitted from a zinc surface (𝜙 = 4.2eV) when illuminated by EM radiation of wavelength 200nm? 𝜙 = 4.2eV = 4.2 x 1.6 x 10-‐19 J = 6.72 x 10-‐19 J Energy of photon: ℎ !

!= !.!"∗!"!!"∗!∗!"!

!∗!"!!= 9.945 ∗ 10!!"𝐽

K.E. of electron = 9.945 − 6.72 ∗ 10!!"𝐽 = 3.225 ∗ 10!!"𝐽

Therefore: 𝑣 = !!"!= 8.4 ∗ 10!𝑚 𝑠!!


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13.1.5 Describe the de Broglie hypothesis and the concept of matter waves.

Students should also be aware of wave–particle duality (the dual nature of both radiation and matter).

Every particle has a wavelength associated to its momentum p. 𝑝 = !

!

The hypothesis assigns wave-‐like properties to something that is normally thought to be a particle. This state of affairs is called the duality of matter.

13.1.6 Outline an experiment to verify the de Broglie hypothesis.

A brief outline of the Davisson–Germer experiment will suffice.

In the Davisson-‐Germer experiment, electrons of kinetic energy 54eV were directed at a surface of nickel where a single crystal had been grown and were scattered b it. Using the Bragg formula and the known separation of the crystal atoms allowed the determination of the wavelength, which was then seen to agree with the de Broglie formula.

13.1.7 Solve problems involving matter waves.

For example, students should be able to calculate the wavelength of electrons after acceleration through a given potential difference.

Find the de Broglie wavelength of a proton that has been accelerated from rest by a potential difference of 500V. The kinetic energy of the proton is given by:

𝐾𝐸 =𝑝!

2𝑚

The work done in accelerating the proton through a potential difference V is qV and this work goes into kinetic energy. Thus

𝑝!

2𝑚= 𝑞𝑉

𝑝 = 2𝑚𝑞𝑉 Hence

𝜆 =ℎ2𝑚𝑞𝑉

𝜆 = !.!"∗!"!!"

!∗!.!"∗!"!!"∗!.!"∗!"!!"∗!""

𝜆 = 1.3 ∗ 10!!" 𝑚 13.1.8 Outline a laboratory

procedure for producing and observing atomic spectra.

Students should be able to outline procedures for both emission and absorption spectra. Details of the spectrometer are not required.

When hydrogen gas is heated to a high temperature or exposed to a high electric field, it will glow, emitting light. In the laboratory, this can be seen with a tube of hydrogen whose ends are at a high potential difference. The emitted light may be analysed by letting it go through a spectrometer. In the case of hydrogen, the emitted light consists


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of a series of bright lines. This is the emission spectrum of hydrogen.

A similar phenomenon takes pace when white light is allowed to pass through hydrogen gas. When the light that has been transmitted through the gas is analysed, a series of dark lines superimposed on the continuous band of colours is seen. This is the absorption spectrum of hydrogen. absorption:

emission The dark lines in the absorption spectrum are at precisely the same wavelengths as the coloured bright lines in the emission spectrum.

13.1.9 Explain how atomic spectra provide evidence for the quantization of energy in atoms.

An explanation in terms of energy differences between allowed electron energy states is sufficient.

13.1.10 Calculate wavelengths

of spectral lines from energy level differences and vice versa.

e.g. Calculate the wavelength of the photon emitted in the transition from n=3 to n=2. The energy of the level n=3 is:

−13.63!

𝑒𝑉 = −1.51𝑒𝑉


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The energy of the level n=2 is:

−13.62!

𝑒𝑉 = −3.40𝑒𝑉 The energy difference is the 1.89eV and that is the energy of the emitted photon.

1.89𝑒𝑉 = ℎ𝑓

𝑓 =1.89 ∗ 1.6 ∗ 10!!"𝐽6.63 ∗ 10!!" 𝐽𝑠

= 4.56 ∗ 10!"𝐻𝑧 13.1.11 Explain the origin of

atomic energy levels in terms of the “electron in a box” model.

The model assumes that, if an electron is confined to move in one dimension by a box, the de Broglie waves associated with the electron will be standing waves of wavelength !!

! where L is the

length of the box and n is a positive integer. Students should be able to show that the kinetic energy EK of the electron in the box is !

!!!

!!!!!.

Imagine that an electron is confined within a box of linear size L. The electron, treated as a wave, according to de Broglie, has a wavelength associated with it given by 𝑝 = !

!.

Since the electron cannot escape from the box, it is reasonable to assume that the electron wave is zero at the edges of the box. In addition, since the electron cannot lose energy, it is also reasonable to assume that the wave associated with the electron in this case is a standing wave. So we want a standing wave that will have nodes at x=0 and x=L. This implies that 𝜆 = !!

! where n is an integer.

Therefore the momentum of the electron is

𝑝 =ℎ2𝐿𝑛=𝑛ℎ2𝐿

The kinetic energy is then

𝐾𝐸 =𝑛ℎ2𝐿

!

2𝑚=𝑛!ℎ!

8𝑚𝐿!

13.1.12 Outline the Schrödinger model of the hydrogen atom.

The model assumes that electrons in the atom may be described by wavefunctions. The electron has an undefined position, but the square of the amplitude of the wavefunction gives the probability of finding the electron at a particular point.

The theory gives probabilities for finding an electron somewhere – it does not pinpoint an electron at a particular point in space. The probability of finding a particle at any point in space within the atom is given by the square of the amplitude of the wave function at that point. The wave function provides a way of working out the probability of finding an electron at that particular radius. The (amplitude)2 of the wave at any given point is a measure of the probability of finding the electron at that distance away from the nucleus in any direction. The exact position of the electron is not known but we know where it is more likely to be.


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13.1.13 Outline the Heisenberg uncertainty principle with regard to position–momentum and time–energy.

Students should be aware that the conjugate quantities, position–momentum and time–energy, cannot be known precisely at the same time. They should know of the link between the uncertainty principle and the de Broglie hypothesis. For example, students should know that, if a particle has a uniquely defined de Broglie wavelength, then its momentum is known precisely but all knowledge of its position is lost.

The basic idea behind the principle is the wave-‐particle duality. Particles sometimes behave like waves and waves sometimes behave like particles, so that we cannot cleanly divide physical objects as either particles or waves. The Heisenberg uncertainty principle applied to position and momentum states that it is not possible to measure simultaneously the position and momentum of something with indefinite precision. The uncertainty ∆𝑥 in position and in ∆𝑝 momentum are related by:

∆𝑥∆𝑝 ≥ℎ4𝜋

This says that making momentum as accurate as possible makes position inaccurate, whereas accuracy in position results in inaccuracy in momentum. If one is made zero, the other has to be infinite. If a particle has a uniquely defined de Broglie wavelength, then its momentum is known precisely but all knowledge of its position is lost.

13.2 Nuclear physics Assessment statement Teacher’s notes 13.2.1 Explain how the radii of

nuclei may be estimated from charged particle scattering experiments.

Use of energy conservation for determining closest-‐approach distances for Coulomb scattering experiments is sufficient.

Consider an alpha particle (charge 2e) that is shot head-‐on toward a stationary nucleus of charge Q=Ze. Initially the system has a total energy consisting of the alpha particle’s kinetic energy 𝐸 = 𝐸! . We take the separation of the alpha particle and the nucleus to be large so no potential energy exists. At the point of closest approach, a distance d from the centre of the nucleus, the alpha particle stops and is about to turn


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back. Thus, the total energy now it the electric potential energy of the alpha particle and the nucleus, given by:

𝐸 = 𝑘𝑄𝑞𝑑= 𝑘

(2𝑒)(𝑍𝑒)𝑑

= 𝑘2𝑍𝑒!

𝑑

Then, by conservation of energy:

𝐸! = 𝑘2𝑍𝑒!

𝑑

𝑑 = 𝑘2𝑍𝑒!

𝐸!

As the energy of the incoming particle is increased, the distance of closest approach decreases, The smallest it can get is, however, of the same order as the radius of the nucleus.

13.2.2 Describe how the masses of nuclei may be determined using a Bainbridge mass spectrometer.

Students should be able to draw a schematic diagram of the Bainbridge mass spectrometer, but the experimental details are not required. Students should appreciate that nuclear mass values provide evidence for the existence of isotopes.

Ions enter through the collimating slits S1. They then enter a region of magnetic and electric fields and approach a second slit, which only allows ions of a given velocity to pass. A second magnetic field bends these ions into circular paths according to their mass. If the beam contains atoms of equal mass, all atoms will hit the plate at the same point. If, however, isotopes are present, the heavier atoms will follow a longer radius and will hit the plate further to the right. Measurement of the radius of each isotope’s paths thus allows for the determination of its mass.

13.2.3 Describe one piece of evidence for the existence of nuclear energy levels.

For example, alpha particles produced by the decay of a nucleus have discrete energies; gamma-‐ray spectra are discrete. Students should appreciate that the nucleus, like the atom, is a quantum system and, as such, has discrete energy levels.

The main evidence for the existence of nuclear energy levels comes form the fact that the energies of the alpha particles and gamma ray photons are discrete (in contrast to beta decays, where the electron has a continuous range of energies).


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13.2.4 Describe beta+ decay,

including the existence of the neutrino.

Students should know that beta energy spectra are continuous, and that the neutrino was postulated to account for these spectra.

The beta decay originates from a decay of a neutron inside an atomic nucleus:

𝑛!! → 𝑝!! + 𝑒!!! + 𝑣!! !

The neutron decays into a proton, an electron and an antineutrino. Sine the energy of a beta decay has a range of possible values, it means that a third very light particle must also be produced so that it carries the remainder of the available energy. Neutrino stands for the ‘little neutral one’.

13.2.5 State the radioactive decay law as an exponential function and define the decay constant.

Students should know that the decay constant is defined as the probability of decay of a nucleus per unit time.

The law of radioactive decay states that the number of nuclei that will decay per second is proportional to the number of atoms present that have not yet decayed:

𝑑𝑁𝑑𝑡

= −𝜆𝑁 Here lambda is the decay constant. Its physical meaning is that it represents the probability of decay per unit time. If the number of nuclei originally present (at t=0) is 𝑁!, by integrating the above equation it can be sen that the number of nuclei of the decaying element present at time t is:

𝑁 = 𝑁!𝑒!!" 13.2.6 Derive the relationship

between decay constant and half-‐life.

After one half-‐life, 𝑇!!, half of the

nuclei present have decayed, so: 𝑁!2= 𝑁!𝑒

!!!!!

Taking logarithms we find: 𝜆𝑇!

!= 𝑙𝑛2

This is the relationship between the decay constant and the half-‐life.

13.2.7 Outline methods for measuring the half-‐life of an isotope.

Students should know the principles of measurement for both long and short half-‐lives.

When measuring the activity of a source, the background rate should be subtracted. -‐ If the half-‐life is short, then readings can be taken of


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activity against time. à A simple graph of activity against time would produce the normal exponential shape. Several values could be read from the graph and then averaged. This method is simple and quick but not the most accurate. à a graph of ln(activity) against time could be produced. This should give a straight lines and the decay constant can be calculated from the gradient.

-‐ If the half-‐life is long, then the activity will effectively be constant over a period of time. In this case one needs to find a way to calculate the number of nuclei present and then use !"!"= −𝜆𝑁

13.2.8 Solve problems involving radioactive half-‐life.

e.g. Carbon-‐14 has a half-‐life of 5730yr and in living organisms it has a decay rate of 0.25Bq g-‐1. A quantity of 20g of carbon-‐14 was extracted from an ancient bone and its activity was found to be 1.81 Bq. What is the age of the bone? The decay constant is

𝜆 =𝑙𝑛25730

= 1.21 ∗ 10!!𝑦𝑟!! When the bone was part of the living body the 20g would have had an activity of 5Bq. If the activity now is 1.81Bq, then

𝐴 = 𝐴!𝑒!!" 1.81 = 5𝑒!!.!"∗!"!!!

𝑡 = 8400𝑦𝑟


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14.1 Analogue and digital signals Assessment statement Teacher’s notes 14.1.1 Solve problems involving

the conversion between binary numbers and decimal numbers.

Students should be aware of the term bit. An awareness of the least-‐significant bit (LSB) and most-‐significant bit (MSB) is required. Problems will be limited to a maximum of five bits in digital numbers.

e.g. 13=1101 (1+4+8) where 1 is the most significant bit (highest power) and 1 is the least significant bit (smallest power)

14.1.2 Describe different means of storage of information in both analogue and digital forms.

Students may consider LPs, cassette tapes, floppy disks, hard disks, CDs, DVDs, and so on.

14.1.3 Explain how interference of light is used to recover information stored on a CD.

Students must know that destructive interference occurs when light is reflected from the edge of a pit.

A phase difference between successive laser beams means that a land has changed to a pit or vice versa. 1. The speed of rotation of the

disc is controlled so that a constant length of track is scanned in a given time.

2. The CD has a higher speed of revolution when the laser is reading near the centre compared with the outer edge.

3. The laser beam is focused on the track.

4. When the beam reflects from a land or a pit, a strong signal is received.

5. When the beam reflects from the edge between a land and a pit, destructive interference takes place and a weak signal is received.

6. A strong signal = 0, a weak signal = 1.

Bump height is always equal to !!

for destructive interference to occur.

14.1.4 Calculate an appropriate depth for a pit from the wavelength of the laser light.

e.g. Laser light of frequency 6x1014 Hz is used in a laser. Calculate the appropriate depth of a pit on a CD.

𝜆 =𝐶𝑓=3 ∗ 10!

6 ∗ 10!"= 500𝑛𝑚

Depth of pit: 𝜆4=5004

= 125𝑛𝑚


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14.1.5 Solve problems on CDs and DVDs related to data storage capacity.

e.g. Information is stored on a CD at a rate of 44100 words per second. The information consists of 32-‐bit words. A CD lasts for 74 minutes. Calculate the storage capacity of the CD. The number of bits imprinted on the CD is 44100x32x74x60 = 6.27 ∗ 10! bits. Since 1 byte = 8 bits this corresponds to

6.27 ∗ 10!

8= 780𝑀𝑏𝑦𝑡𝑒𝑠

14.1.6 Discuss the advantage of the storage of information in digital rather than analogue form.

Students should consider quality, reproducibility, retrieval speed, portability of stored data and manipulation of data.

Quality: There must be a complex set of rules for the conversion of input into digital signal and from digital to output. Reproducibility: Optical techniques can ensure that each subsequent retrieval is virtually identical. Retrieval speed: Text and simple data can be retrieved at great speed. More complex data takes longer but selecting different sections of information often does not add significant time. Portability: Modern miniaturization techniques have ensured that large quantities of data can be stored in a very small device. Manipulation: Manipulation of data can be easily achieved without significant corruption.

14.1.7 Discuss the implications for society of ever-‐increasing capability of data storage.

Teachers should consider moral, ethical, social, economic and environmental implications.

Common sense + see revision guide for table.


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14.2 Data capture; digital imaging using charge-‐coupled devices (CCDs) Assessment statement Teacher’s notes 14.2.1 Define capacitance. The amount of charge that can be

stored on a body per unit electrical potential.

14.2.2 Describe the structure of a charge-‐coupled device (CCD).

Students should know that a CCD is a silicon chip divided into small areas called pixels. Each pixel can be considered to behave as a capacitor.

A CCD is a silicon microchip that can be used to electronically record an image focused onto its surface. The surface is divided into a large number of small areas called pixels.

14.2.3 Explain how incident light causes charge to build up within a pixel.

Students are required to use the photoelectric effect.

1. During a photo exposure, each element within the CCD generates a charge proportional to the incident light as a result of the photoelectric effect

2. The charge is collected in different pixels. The pixel behaves as a capacitor and a charge builds up.

3. The charge collected from each pixel is transferred in turn by ‘coupling’ charges from one pixel to the next in turn.

4. Individual charge packets are converted to an output voltage and then digitally encoded. The value of the p.d. is converted into a digital signal in binary code. The light intensity information from each pixel can be stored along with other digital signal representing the position of the pixel on the surface.

14.2.4 Outline how the image on a CCD is digitized.

Students are only required to know that an electrode measures the potential difference developed across

See above.


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each pixel and this is then converted into a digital signal. The pixel position is also stored.

14.2.5 Define quantum efficiency of a pixel.

Quantum efficiency is the ratio of the number of photoelectrons emitted to the number of photons incident on the pixel.

The ratio of the number of electrons emitted to the number of incident photons on a pixel.

14.2.6 Define magnification. Students are required to know that magnification is the ratio of the length of the image on the CCD to the length of the object.

Magnification is the ratio of the length of the image on the CCD to the length of the object.

14.2.7 State that two points on an object may be just resolved on a CCD if the images of the points are at least two pixels apart.

If we think about an image looking like this: | | | then if the white squares are on adjacent pixels on the CCD, they will look like one object and not like two separate. Therefore, there has to be at least on pixel in a different colour between them.

14.2.8 Discuss the effects of quantum efficiency, magnification and resolution on the quality of the processed image.

The greater the quantum efficiency, the greater the sensitivity of the device. A greater magnification means that more pixels are used for a given section of the image. The image will be more detailed. The greater the resolution, the greater the amount of detail recorded. An improvement in resolution will mean a given image will occupy more memory.

14.2.9 Describe a range of practical uses of a CCD, and list some advantages compared with the use of film.

Students should appreciate that CCDs are used for image capturing in a large range of the electromagnetic spectrum. They should consider items such as digital cameras, video cameras, telescopes, including the Hubble Telescope, and medical X-‐ray imaging.

See left + common sense.

14.2.10 Outline how the image stored in a CCD is retrieved.

See 14.2.3

14.2.11 Solve problems involving the use of CCDs.

A digital camera is used to photograph an object. Two points on the object are separated by 0.002cm. The CCD in the camera has a collecting area of 16cm2 and contains 4 megapixels. The magnification of the camera is 1.5. Can the image of the points be resolved? Area corresponding to each pixel: !"∗!"

!!

!∗!"!= 4.0 ∗ 10!!"𝑚!

Separation of pixels:


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4.0 ∗ 10!!" = 2.0 ∗ 10!! Equivalent separation on the object:

2.0 ∗ 10!!

1.5= 0.0013

Distance between two pixels < 0.0020cm so the image can be resolved.


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E1 Introduction to the universe Assessment statement Teacher’s notes E.1.1 Outline the general

structure of the solar system.

Students should know that the planets orbit the Sun in ellipses and moons orbit planets. (Details of Kepler’s laws are not required.) Students should also know the names of the planets, their approximate comparative sizes and comparative distances from the Sun, the nature of comets, and the nature and position of the asteroid belt.

Asteroid belt: consists of thousands of small objects (small planets) in orbit around the sun. One theory about the asteroid belt involves the disruption of one planet into many pieces. Another invokes the effect of nearby Jupiter, whose large mass did not allow the material that was there at the time of the formation of the solar system to assemble into a planet. Comets: A small body (mainly ice and dust) orbiting the sun in an elliptical orbit.

Planet Distance from Sun

Mass (Earth =1)

Mercury 60m km 0.055 Venus 110m km 0.814 Earth 150m km 1 Mars 230m km 0.107 Jupiter 780m km 320 Saturn 1400m

km 95

Uranus 2900m km

15

Neptune 4500m km

17

Pluto 6000m km

0.0026

E.1.2 Distinguish between a stellar cluster and a constellation.

Stellar cluster: A group of stars that are physically near each other in space, created by the collapse of the


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same gas cloud. Constellation: A group of stars in a recognizable pattern that appear to be near each other in space when observed from Earth. Stars in a constellation are not necessarily close to one another.

E.1.3 Define the light year. We define the light year (ly) as the distance travelled by light in one year. Thus: 1ly = 9.46 x 1015m

E.1.4 Compare the relative distances between stars within a galaxy and between galaxies, in terms of order of magnitude.

The average distance between stars in a galaxy is about 1pc. The average distance between galaxies varies from about 100kpc for galaxies within the same cluster to a few Mpc for galaxies belonging to different clusters.

E.1.5 Describe the apparent motion of the stars/constellations over a period of a night and over a period of a year, and explain these observations in terms of the rotation and revolution of the Earth.

This is the basic background for stellar parallax. Other observations, for example, seasons and the motion of planets, are not expected.

The constellations appear to move over the period of one night. They appear to rotate around one direction. In the northern hemisphere everything seems to rotate about the pole star. The same movement is continued during the day. The Sun rises in the East and sets in the West, reaching its maximum height at midday. At this time in the northern hemisphere the sun is in a southerly direction. Every night, the constellations have the same relative positions to each other, but the location of the pole star 8and thus the portion of the night sky that is visible above the horizon) changes slightly from night to night. Over the period of a year this slow change returns back to the exact same position. The sun continues to rise in the East and set in the West, but as the year goes from winter into summer, the arc gets bigger and the sun climbs higher in the sky.

E2 Stellar radiation and stellar types Assessment


E.2.1 State that fusion is the main energy source of stars.

Students should know that the basic process is one in which hydrogen is converted into helium. They do not need to know about the fusion of elements with higher proton numbers.

The source of the energy of the sun is nuclear fusion in the interior, where nuclei of hydrogen fuse to produce helium and release energy in the process. Because of the high temperatures in the interior of the star, the electrostatic repulsion between protons can be overcome and hydrogen nuclei can fuse. Because of the high pressure in stellar interiors, the nuclei are sufficiently close to each other to give a high probability of


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collision and hence fusion. The sequence of nuclear fusion reactions that take place is called the proton-‐proton cycle.

E.2.2 Explain that, in a stable star (for example, our Sun), there is an equilibrium between radiation pressure and gravitational pressure.

Nuclear fusion provides the energy that is needed to keep the star hot, so that the radiation pressure is high enough to oppose further gravitational contraction.

E.2.3 Define the

luminosity of a star. Luminosity is the amount of energy radiated by

a star per second; that is the power radiated by the star. Luminosity depends on the surface temperature and surface area of the star.

E.2.4 Define apparent brightness and state how it is measured.

The received energy per second per unit area of detector is called the apparent brightness and is given by 𝑏 = !

!!!!. The units of apparent

brightness are Wm-‐2. Apparent brightness is measured using a CCD. A CCD has a photosensitive silicon surface that releases and electron when it is hit by a photon. The number of electrons released is proportional to the number of photons that hit the surface. Thus, the amount of charge is a direct measure of the brightness of the object being observed. CCDs are more than 50 times more efficient in recording the photons arriving at the device than conventional photographic film.

E.2.5 Apply the Stefan–Boltzmann law to compare the luminosities of different stars.

The amount of energy per second radiated by a star of surface area A and absolute surface temperature T is given by:

𝐿 = 𝜎𝐴𝑇! e.g. A star has half the sun’s surface temperature and 400 times its luminosity. How many times bigger is it? We have that

400 =𝐿

𝐿!"#

=𝜎4𝜋𝑅!𝑇!

𝜎4𝜋𝑅!"#!𝑇!!"!

=𝑅! 𝑇!"#

2!

𝑅!"# !𝑇!"#!

=𝑅!

16 𝑅!"# !

𝑅!

16 𝑅!"# ! = 400

𝑅𝑅!"#

= 80


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E.2.6 State Wien’s (displacement) law and apply it to explain the connection between the colour and temperature of stars.

The Wien displacement law relates the wavelength 𝜆! to surface temperature T:

𝜆!𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 2.9 ∗ 10!!𝐾 𝑚 which implies that the higher the temperature, the lower the wavelength at which most of the energy is radiated. Most energy is emitted around the peak wavelength 𝜆!. We see that the colour of the star is mainly determined by the colour corresponding to 𝜆!. The area under the black-‐body curve is the total power radiated from a unit area, irrespective of wavelength, and is thus given by 𝜎𝑇!.

E.2.7 Explain how atomic

spectra may be used to deduce chemical and physical data for stars.

Students must have a qualitative appreciation of the Doppler effect as applied to light, including the terms red-‐shift and blue-‐shift.

Temperature: The surface temperature of the star is determined by measuring the wavelength at which most of the radiation is emitted. Chemical composition: In the absorption spectrum each dark line represents the absorption of light of a specific frequency by a specific chemical element in the star’s atmosphere. It has been found, that most stars have essentially the same chemical composition, yet show different absorption spectra. The reason for this difference is that different stars have different temperatures. Radial velocity: If a star moves away from or toward us, its spectral lines will show a Doppler shift. The shift will be toward red if the star moves away, and toward blue if it comes toward us. Measurement of the shift allows the determination of the radial velocity of the star. Rotation: If a star rotates, then part of the star is moving toward the observer and part away from the observer. Thus, the light from the different parts of the star will again show Doppler shifts. Magnetic fields: In a magnetic field a spectral line may split into two or more lines. Measurement of the amount of splitting yields information on the magnetic field of the star.

E.2.8 Describe the overall classification system of spectral classes.

Students need to refer only to the principal spectral classes (OBAFGKM).

Stars are divided into seven spectral classes according to their colour (therefore surface temperature). Class Colour Temperature O Blue 25000-‐50000 B Blue-‐

white 12000-‐25000


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A White 7500-‐12000 F Yellow-‐

White 6000-‐7500

G Yellow 4500-‐6000 K Yellow-‐

red 3000-‐4500

M Red 2000-‐3000 Oh Be A Fine Girl Kiss Me.

E.2.9 Describe the different types of star.

Students need to refer only to single and binary stars, Cepheids, red giants, red supergiants and white dwarfs. Knowledge of different types of Cepheids is not required.

Single star: Our sun is a single star. Binary star: Stars that are in orbit around each other (their common centre of mass). A visual binary star is one that can be distinguished as two separate stars using a telescope. (also see below) Cepheid: Stars that are a little unstable. They are observed to have a regular variation in bright ness and hence luminosity. This is thought to be due to an oscillation in the size of the star. Red giant: Very large, cool stars with a reddish appearance. The luminosity of red giants is considerably greater than the luminosity of main sequence stars of the same temperature. The mass of a red giant can be as much as 1000 times the mass of our sun, but their huge size also implies small densities. A red giant will have a central hot core surrounded by an enormous envelope of extremely tenuous gas. Red supergiant: A bigger version of a red giant. The production of energy does not stop with at helium or carbon. White dwarf: Very common, but their faintness makes them hard to detect. Very small in size and white in colour. Since they are white, they are comparatively hot. They turn out to be one of the final stages for some stars, Fusion is no longer taking place, and a white dwarf is just a hot remnant that is cooling down. Eventually it will cease to give out light when it becomes sufficiently cold. It is then known as a brown dwarf.

E.2.10 Discuss the characteristics of spectroscopic and eclipsing binary stars.

Spectroscopic: Identified from the analysis of the spectrum of light from the ‘star’. Over time the wavelengths show a periodic shift or splitting in frequency.

Eclipsing: Identified from the analysis of the brightness of the light from


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the ‘star’. Over time the brightness shows a periodic variation. The explanation for the ‘dip’ in brightness is that as a result of its orbit, one star gets in front of the other. If the stars are of equal brightness, then this would cause the total brightness to drop to 50%.

E.2.11 Identify the general

regions of star types on a Hertzsprung–Russell (HR) diagram.

Main sequence, red giant, red supergiant, white dwarf and Cepheid stars should be shown, with scales of luminosity and/or absolute magnitude, spectral class and/or surface temperature indicated. Students should be aware that the scale is not linear. Students should know that the mass of main sequence stars is dependent on position on the HR diagram.

Once we know the temperature of a star (for example, through its spectrum), the HR diagram can tell us the luminosity of the star with an acceptable degree of accuracy, provided it’s a main sequence star.


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E3 Stellar distances Assessment statement Teacher’s notes E.3.1 Define the parsec. When a star with a parallax angle

of exactly one second of arc, its distance must be 3.26ly. This distance is defined as on parsec (pc). [One parsec is the distance to a star whose parallax is 1 arc second.]

E.3.2 Describe the stellar parallax method of determining the distance to a star.

The parallax angle can be measured by observing the changes in a star’s position over the period of a year. Then: tan 𝜃 = ! !"#$! !" !"#

! !"# !" !"#$

Sine the angle is small, tan 𝜃 = 𝜃. Then: 𝑑 𝑝𝑐 = !!"#"$$"% !"#$% (!)

E.3.3 Explain why the method

of stellar parallax is limited to measuring stellar distances less than several hundred parsecs.

The parallax method can be sued to measure stellar distances that are less than several hundred parsecs. The parallax angle for stars that are at greater distances becomes too small to measure accurately.

E.3.4 Solve problems involving stellar parallax.

e.g. A star is 1.32 ∗ 10!"𝑚 away. Calculate its parallax angle.

𝑑 = 1.32 ∗ 10!"𝑚 1.32 ∗ 10!"

3.08 ∗ 10!"𝑝𝑐 = 42.9𝑝𝑐

Then the parallax angle is: 1

42.9= 0.023′′

E.3.5 Describe the apparent magnitude scale.

Students should know that apparent magnitude depends on luminosity and the distance to a star. They should also know that a magnitude 1 star is 100 times brighter than a magnitude 6 star.

The scale was introduced over 2000 years ago as a way of classifying stars. They were all assigned to one of six classifications according to their brightness as seen by the naked eye. Very bright stars were called magnitude 1 stars, whereas the faintest stars were called


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magnitude 6. With the aid of telescopes, we can now see stars that are fainter than the magnitude 6 stars. A magnitude 1 star is 100 times brighter than a magnitude 6 star and the scale is logarithmic. As the magnitude numbers get bigger, the stars are getting dimmer. Magnitudes are negative for very bright stars. Each step on the scale equates to a brightness increase/decrease of 2.512.

E.3.6 Define absolute

magnitude. The absolute magnitude M of a

star is the apparent magnitude it would have if place d at a distance of 10pc from earth.

E.3.7 Solve problems involving apparent magnitude, absolute magnitude and distance.

e.g. Calculate the absolute magnitude of a star whose distance is 25ly and whose apparent magnitude is 3.45. We must first change light years into parsecs:

25𝑙𝑦 =253.26

𝑝𝑐 = 7.67𝑝𝑐

𝑚 −𝑀 = 5 log𝑑10

𝑀 = 𝑚 − 5 log𝑑10

= 4.03

E.3.8 Solve problems involving apparent brightness and apparent magnitude.

E.3.9 State that the luminosity of a star may be estimated from its spectrum.

Temperature can be deduced from examining the spectrum of a star. Knowing the temperature and using the HR diagram (assuming the star is a main sequence star) allow us to determine the luminosity.

E.3.10 Explain how stellar distance may be determined using apparent brightness and luminosity.

Assuming that we know the luminosity and apparent brightness of a star, we can find the distance, since:

𝑏 =𝐿

4𝜋𝑑!→ 𝑑 =

𝐿4𝜋𝑏

E.3.11 State that the method of spectroscopic parallax is limited to measuring stellar distances less than about 10 Mpc.

The term spectroscopic parallax refers to a method of finding the distance to a star given the star’s luminosity and apparent brightness.The method of spectroscopic parallax is limited to measuring stellar distances less than about 10 Mpc.

E.3.12 Solve problems involving stellar distances, apparent

e.g. A main sequence star emits most of its energy at a


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brightness and luminosity. wavelength of 2.4x10-‐7m. Its apparent brightness is measured to be 4.3x10-‐9. How far is the star? From Wien’s law we find the temperature of the star to be

𝜆!𝑇 = 2.9 ∗ 10!!𝐾 𝑚

𝑇 =2.9 ∗ 10!!

2.4 ∗ 10!!= 12000𝐾

From the HR diagram we see that such a temperature corresponds to a luminosity of about 100 times that of the sun, that is

𝐿 = 3.9 ∗ 10!"𝑊 Thus

𝑑 =𝐿4𝜋𝑏

𝑑 =3.9 ∗ 10!"

4𝜋 ∗ 4.3 ∗ 10!!

= 8.5 ∗ 10!"𝑚 E.3.13 Outline the nature of a

Cepheid variable. Students should know that a Cepheid variable is a star in which the outer layers undergo a periodic expansion and contraction, which produces a periodic variation in its luminosity.

Cepheid variable stars are stars whose luminosity is not constant in time but varies from a minimum to a maximum periodically. The brightness increases sharply and then fades off more gradually. The reason for this has to do with the interaction of radiation with matter in the atmosphere of the star. This interaction causes the outer layers of the star to undergo periodic expansions and contractions.

E.3.14 State the relationship between period and absolute magnitude for Cepheid variables.

The longer the period, the larger the luminosity of a Cepheid variable.

E.3.15 Explain how Cepheid variables may be used as “standard candles”.

It is sufficient for students to know that, if a Cepheid variable is located in a particular galaxy, then the distance to the galaxy may be determined.

If the luminosity of the Cepheid is found, the distance between earth and the galaxy that contains the Cepheid can be found.

E.3.16 Determine the distance to a Cepheid variable using the luminosity–period relationship.

e.g. A Cepheid has a period of about 22 days, this corresponds to a luminosity of about 7000 solar luminosities, or about L=2.73x1030W.


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The peak apparent magnitude is about m=3.7. The peak apparent brightness can be found from

𝑏2.52 ∗ 10!!

= 2.512!!

𝑏 = 2.52 ∗ 10!! ∗ 2.512!!.! = 8.34 ∗ 10!!"𝑊𝑚!! Then,

𝑑 =𝐿4𝜋𝑏

= 1.6 ∗ 10!"𝑚 = 1700𝑙𝑦 = 520𝑝𝑐

E4 Cosmology Assessment statement Teacher’s notes E.4.1 Describe Newton’s model

of the universe. Students should know that Newton assumed an infinite (in space and time), uniform and static universe.

Newton used an extreme version of the cosmological principle when he suggested that the universe is infinite in extent, has no beginning an is static, meaning it has been uniform and isotropic at all times. He assumed and infinite, uniform and static universe.

E.4.2 Explain Olbers’ paradox. Students should be able to show quantitatively, using the inverse square law of luminosity, that Newton’s model of the universe leads to a sky that should never be dark.

If the universe is really like Newton imagined then the night sky should be bright. Imagine a universe that is infinite and contains an infinite number of stars more or less uniformly distributed in space. The very distant stars contribute very little light to an observer on earth but there are very many of them. Mathematically, let n stand for the number of stars per unit volume of space. At a distance d from a star of luminosity L, the received energy per area per second is


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𝑏 =𝐿

4𝜋𝑑!

The number of stars in a thin shell of thickness t at a distance d from the observer is number density * volume = 4𝜋𝑑!𝑛𝑡.

Hence the received energy per second per area from all the stars in the thin shell is

𝐿4𝜋𝑑!

∗ 4𝜋𝑑!𝑛𝑡 = 𝐿𝑛𝑡 This number does not depend on the distance d to the shell. Since there is an infinite number of such shells surrounding the observer, and since each contributes a constant amount of energy, the total energy received must be infinite, making the night sky infinitely bright, which it is not. This is Olbers’ paradox.

E.4.3 Suggest that the red-‐shift of light from galaxies indicates that the universe is expanding.

Hubble interpreted the redshift of the spectral lines as evidence of a velocity of the galaxy away from us, as in the Doppler shift. The faster the galaxy, the larger the redshift.

Hubble’s observations thus suggest an expanding universe with galaxies moving away from us and from each other. It also suggests that in the past the universe was much smaller. The universe appears to have started from a kind of explosion that set matter moving outward. This is the idea of the Big Bang model of cosmology.

E.4.4 Describe both space and time as originating with the Big Bang.

Students should appreciate that the universe is not expanding into a void.

See above. It is important to realize that the universe is not expanding into


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empty space. The galaxies that are moving away from us are not moving into another, previously unoccupied, part of the universe. Space is being created in between the galaxies and so the distance between them increases, creating the illusion of motion of one galaxy relative to another.

E.4.5 Describe the discovery of cosmic microwave background (CMB) radiation by Penzias and Wilson.

Two radio astronomers working at Bell Laboratories accidentally discovered CMB when an antenna they designed was picking up a signal the persisted no matter what part of the sky the antenna was pointing at. The spectrum of this signal turned out to be a black-‐body spectrum corresponding to a temperature of 2.7K.

E.4.6 Explain how cosmic radiation in the microwave region is consistent with the Big Bang model.

A simple explanation in terms of the universe “cooling down” is all that is required.

Today we observe the background radiation at 2.7K. This is consistent with a small, hot universe in the distant past, which began to cool down as it expanded. Penzias and Wilson realised that the radiation detected was the remnant of the hot explosion at the beginning of time. It was the afterglow of the enormous temperature that existed in the early universe. As the universe has expanded, the temperature has kept falling to reach its present value of 2.7K.

E.4.7 Suggest how the Big Bang model provides a resolution to Olbers’ paradox.

Big Bang theory says that the universe is not infinite, therefore a solution to the paradox.

E.4.8 Distinguish between the terms open, flat and closed when used to describe the development of the universe.

If the distance between two galaxies was 𝑥! at some arbitrary time, then the separation of these two galaxies at some time t later is given by the expression

𝑥 𝑡 = 𝑅(𝑡)𝑥! Where R(t) is the scale factor of the universe , that can be interpreted as follows: 1. R(t) starts from zero, increases to a maximum value and then decreases back to zero again. The universe collapses after an initial period of expansion. This is called the closed universe.

2. The scale factor R(t) increases without limit, the universe continues to expand forever. This is called the open universe.

3. The universe does expand forever, but the rate of expansion decreases, this is called a flat universe.


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Depending on which solution is taken, the age of the universe is different.

E.4.9 Define the term critical density by reference to a flat model of the development of the universe.

The theoretical value of density that would create a flat universe is called the critical density (5x10-‐26 kg m-‐3). If the density is equal to the critical density, the universe expands forever at a rate that approaches zero (flat universe).

E.4.10 Discuss how the density of the universe determines the development of the universe.

Let p be the actual density and pc the critical density. Then: • p < pc the universe expands forever at a slowing rate. (open)

• p = pc the universe expands forever at a slowing rate that approaches zero. (flat)

• p > pc the universe collapses after a period of expansion (closed)

E.4.11 Discuss problems associated with determining the density of the universe.

This statement is included to give the students a flavour for the ongoing and complex current nature of research. They should be able to discuss relevant observations and possible explanations. They should recognize that, in common with many other aspects of our universe, much about the phenomena is currently not well understood. Teachers should include dark matter, MACHOs and WIMPs.

The density of the universe is not an easy quantity to measure, It is reasonably easy to estimate the mass in a galaxy by estimating the number of stars and their average mass. This calculation results in a galaxy mass that is to small. We know this because we can use the mathematics or orbital motion to work out how much mass there must be keeping the outer stars in orbit around the galactic centre. We think we can see a maximum of 10% of the matter that must exist in the galaxy. This means that much of the mass of a galaxy and indeed the universe itself must be dark matter – in other words we cannot observe it because it is not radiating


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sufficiently for us to detect it. The matter could be found in MACHOs (massive astronomical compact halo objects). There is some evidence that lots of ordinary matter does exist in these groupings. These can be thought of as low-‐mass ‘failed’ stars of high-‐mass planets. They could even be black holes, which would produce little or no light. There also could be few particles that we do not know about. These are WIMPs (weakly interacting massive particles) Furthermore, it might be that our current theories of gravit are not completely correct.

E.4.12 State that current scientific evidence suggests that the universe is open.

Current scientific evidence suggests that the universe is open. There is also evidence that the rate of expansion may have increased.

E.4.13 Discuss an example of the international nature of recent astrophysics research.

It is sufficient for students to outline any astrophysics project that is funded by more than one country.

e.g. The Cassini spacecraft that has been in orbit around the Saturn for several years sending information about the planet back to Earth and it is designed to continue doing so for many more years. The Cassini-‐Huygens spacecraft was funded by ESA, NASA and ASI. As well as general information about Saturn, an important focus of the mission was a moon of Saturn called Titan. The Huygens probe was released and sent back information as it descended towards the surface. The information discovered is shared among the entire scientific community.

E.4.14 Evaluate arguments related to investing significant resources into researching the nature of the universe.

Students should be able to demonstrate their ability to understand the issues involved in deciding priorities for scientific research as well as being able to express their own opinions coherently. Advantages Disadvantages Understanding the nature of the universe (why are we here?, is there life elsewhere in the universe?)

The money could be more usefully spent providing food, shelter and medical care

All fundamental research will give rise to technology that may eventually improve the quality of life

If money is to be allocated on research, it is much more worthwhile to invest limited resources into medical research.

Life on earth will, at some time in the distant future, become and impossibility. Therefore we must be able to travel to

It is better to fund a great deal of small diverse research rather than concentrating all funding into one expensive


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distant stars and colonise new planets

area. Is the information gained really worth the cost?

E5 Stellar processes and stellar evolution Assessment statement Teacher’s notes E.5.1 Describe the conditions

that initiate fusion in a star.

In order for the proton-‐proton cycle to take place, two positively charges particles need to come close to each other for interactions to take place. Obviously, they will repel one another. This means that they must be at a high temperature. If a large cloud of hydrogen is hot enough, then these nuclear reactions can take place spontaneously. As the cloud comes together, the loss of gravitational potential energy must mean an increase in kinetic energy and hence temperature. In simple terms the gas molecules speed up as they fall in towards the centre to form a proto-‐star.

E.5.2 State the effect of a star’s mass on the end product of nuclear fusion.

The star cannot continue in its main sequence state forever. It is fusing hydrogen into helium and at some point hydrogen in the core will become rare. The route that is followed after the red giant phase depends on the initial mass of the star. An important critical mass is called the Chandrasekhar limit and is equal to approximately 1.4 times the mass of our sun. • If a star has a mass less than 4 solar masses, its remnant will be less than 1.4 solar masses and so it is below the Chandrasekhar limit. In this case the red giant forms a planetary nebula and becomes a white dwarf.

• If a star is greater than 4 solar masses, its remnant will have a mass greater than 1.4 solar masses. In this case the red supergiant experiences a supernova, it then becomes a neutron star or collapses to a black hole


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E.5.3 Outline the changes that take place in nucleosynthesis when a star leaves the main sequence and becomes a red giant.

Students need to know an outline only of the processes of helium fusion and silicon fusion to form iron.

If it has sufficient mass, a red giant can continue to fuse higher and higher elements and the process of nucleosynthesis can continue. This process of fusion as a source of energy must come to an end with the nucleosynthesis of iron. The iron nucleus has the greatest binging energy per nucleon of all nuclei. On other words the fusion of iron to form a higher mass nucleus would need to take in energy rather than release energy.

E.5.4 Apply the mass–

luminosity relation. For stars in the main sequence,

there is a mass-‐luminosity relation: 𝐿 ∝ 𝑀! where n is between 3 and 4. The uncertainty in the value of a comes from the fact that the composition of stars is not precisely known. One application of the mass-‐luminosity relation is to estimate the lifetime of a star on the main sequence. Sine luminosity is the power radiated by the star:

𝐸𝑇 ∝ 𝑀!

For the purpose of an estimate, we may assume that the total energy the star can radiate will come from converting all its mass into energy according to 𝐸 = 𝑚𝑐!

𝐸𝑇 ∝ 𝑀! →

𝑀𝑐!

𝑇∝ 𝑀!

𝑇 ∝ 𝑀!!! E.5.5 Explain how the

Chandrasekhar and Oppenheimer–Volkoff limits are used to predict the fate of stars of different masses.

If the mass of the core of a star is less than the Chandrasekhar limit of about 1.4 solar masses, the star will become a stable white dwarf in which electron pressure keeps the star from collapsing further. If the core is more massive than the Chandrasekhar limit, the core will collapse further until electrons are driven into protons, turning them into neutrons. Neutron pressure now keeps the star from collapsing further and the star has become a neutron star.


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If the core is substantially more massive than the Oppenheimer-‐Volkhoff limit of about 2-‐3 solar masses, neutron pressure will not be enough to oppose the gravitational collapse and the star will become a black hole.

E.5.6 Compare the fate of a red giant and a red supergiant.

Students should know that: • a red giant forms a planetary nebula and then becomes a white dwarf • a white dwarf is stable due to electron degeneracy pressure • a red supergiant experiences a supernova and becomes a neutron star or collapses to a black hole • a neutron star is stable due to neutron degeneracy pressure.

See E.5.2

E.5.7 Draw evolutionary paths of stars on an HR diagram.

E.5.8 Outline the characteristics

of pulsars. A neutron star may have a

magnetic field of quite large magnitude (108T) and may rotate as well, with a period ranging from 30ms to 0.3s. Rotating neutron stars emit


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electromagnetic waves in the radio part of the spectrum and so neutron stars can be detected by radio telescopes. Rotating neutron stars that radiate n this way are called pulsars. The radiation emitted by the pulsar is in a narrow cone around the magnetic field. If the magnetic field is not aligned with the axis of rotation, then, as the star rotates, the cone containing the radiation precesses around the rotation axis. An observer who can receive some of this radiation will then do so every time the cone sweeps past.

E6 Galaxies and the expanding universe Assessment


E.6.1 Describe the distribution of galaxies in the universe.

Students should understand the terms galactic cluster and galactic supercluster.

Galaxies are not distributed randomly throughout space. They tend to be found clustered together. For example, in the region of the Milky Way there are twenty or so galaxies in less then 2.5m light years. On a larger scale, galactic clusters are grouped into huge superclusters of galaxies. In general, these superclusters often involve galaxies arranged together in joined bands that are arranged as though randomly throughout empty space.

E.6.2 Explain the red-‐shift of light from distant galaxies.

Students should realize that the red-‐shift is due to the expansion of the universe.

The velocity of recession is found by an application of the Doppler effect to light. Light from galaxies arrives on earth redshifted. This means that the wavelength of the light measured upon arrival is longer than the wavelength at emission. According to the Doppler effect, this implies that the source of light (galaxy) is moving away from the observer on earth.

E.6.3 Solve problems involving red-‐shift and the

e.g. A hydrogen line has a wavelength of 434nm. When received from a distant galaxy, this line is measured on earth at


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recession speed of galaxies.

486nm. What is the speed of recession of this galaxy? From formula booklet:

∆𝜆𝜆≅𝑣𝑐

(486 − 434)

434∗ 3 ∗ 10! ≅ 𝑣

𝑣 ≅ 3.6 ∗ 10!𝑚𝑠!! E.6.4 State Hubble’s

law. Hubble studied a large number of

galaxies and found that, the more distant the galaxy, the faster it moves away from us. This is Hubble’s law, which states that the velocity of recession is directly proportional to the distance: v = Hd, where d is the distance between the earth and the galaxy, and v its velocity of recession. The constant of proportionality, H, is the slope of the graph and is known as the Hubble constant.

E.6.5 Discuss the limitations of Hubble’s law.

The uncertainties in H come mainly from the enormous difficulties in measuring distances to remote galaxies accurately. It is also not clear whether the expansion of the universe happened at a constant rate. Some theories suggest that the expansion is currently accelerating.

E.6.6 Explain how the Hubble constant may be determined.

E.6.7 Explain how the

Hubble constant may be used to estimate the age of the universe.

Students need only consider a constant rate of expansion.

Imagine a galaxy, which is now a distance r from us. Its velocity is thus v = Hr. In the beginning the galaxy and the earth were at zero separation from each other. If the present separation of r is thus covered at the same constant velocity Hr, the time, T, taken to achieve this separation must be given by

𝑣 =𝑟𝑇= 𝐻𝑟 → 𝑇 =

1𝐻

Where T is the age of the universe. E.6.8 Solve problems

involving Find the age of the universe (𝐻 = 72 ∗

10!𝑚𝑠!!𝑀𝑝𝑐!!)


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Hubble’s law. 𝑇 =1𝐻=

172 ∗ 10!𝑚𝑠!!𝑀𝑝𝑐!!

172 ∗ 10!𝑚𝑠!!

∗ 10!𝑝𝑐 1

72 ∗ 10!𝑚𝑠!!∗ 10! ∗ 3.09 ∗ 10!"𝑚

4.29 ∗ 10!"𝑠 = 13.6 ∗ 10!𝑦𝑟𝑠 E.6.9 Explain how the

expansion of the universe made possible the formation of light nuclei and atoms.

Students should appreciate that, at the very high temperatures of the early universe, only elementary (fundamental) particles could exist and that expansion gave rise to cooling to temperatures at which light nuclei could be stable.

At t=10-‐2s, the temperature had fallen sufficiently to 1011K for quarks to bind together and to form protons and neutrons and their antiparticles. The universe had a size of 10-‐10 of its present size. At t=1s after the Big Bang, T=1010K, and protons, neutrons, electrons and their antiparticles were in thermal equilibrium with each other. [At the very high temperatures of the early universe, only elementary (fundamental) particles could exist and that expansion gave rise to cooling to temperatures at which light nuclei could be stable.]


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H1 Introduction to relativity Assessment statement Teacher’s notes H.1.1 Describe what is meant by

a frame of reference. A frame of reference may refer to

a coordinate system or set of axes, within which to measure the position, orientation and other properties of objects in it. [The observer along with the rulers and clocks that he or she uses to measure distances and times constitute what is called a frame of reference.]

H.1.2 Describe what is meant by a Galilean transformation.

It is possible to formalise the relationship between two different frames of reference. The idea is to use the measurement in one frame of reference to work out the measurements that would be recorded in another frame of reference. The equations that do this without taking the theory of relativity into consideration are called Galilean transformations. [The relation between coordinates of events when one frame moves past the other with uniform velocity on a straight line.]

H.1.3 Solve problems involving relative velocities using the Galilean transformation equations.

Simple e.g. A ball rolls on the floor of a train at 2ms-‐1 (with respect to the floor). The train moves with respect to the ground to the right at 12ms-‐1 (a); to the left at 12ms-‐1 (b) What is the velocity of the ball relative to he ground?

a) v = 14ms-‐1 b) v = -‐10ms-‐1

H2 Concepts and postulates of special relativity Assessment statement Teacher’s notes H.2.1 Describe what is meant by

an inertial frame of reference.

Frames moving with uniform velocity past each other on straight lines are called inertial frames of reference. These are non-‐accelerating frames.

H.2.2 State the two postulates of the special theory of relativity.

1) The laws of physics are the same in all inertial frames

2) The speed of light in a vacuum is the same for all inertial observers

H.2.3 Discuss the concept of simultaneity.

Students should know that two events occurring at different

Events that are simultaneous for one observer and which take


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points in space and which are simultaneous for one observer cannot be simultaneous for another observer in a different frame of reference.

place at different points in space, are not simultaneous for another observer in motion relative to the first. On the other hand, if two events are simultaneous for one observer and take place at the same point in space, they are simultaneous for all other observers as well. [Simultaneity, like motion, is a relative concept. Our notion of absolute simultaneity is based on the idea of absolute time: events happen at specific times that all observers agree on. Einstein has taught us that the idea of absolute time, just like the idea of absolute motion, must be abandoned.]

H3 Relativistic kinematics Assessment statement Teacher’s notes H.3.1 Describe the concept of a

light clock. Only a very simple description is required here. For example, a beam of light reflected between two parallel mirrors may be used to measure time.

A light clock is an imaginary device. A beam of light bounces between two mirrors – the time taken by the light between bounces is one ‘tick’ of the light clock. The path taken by light in a light clock that is moving at constant velocity is longer. We know that the speed of light is fixed so the time between the ‘ticks’ on a moving clock must also be longer. This effect – that moving clocks run slow – is called time dilation.

H.3.2 Define proper time interval.

A proper time interval is the same separating two events that take place at the same point in space. It turns out to be the shortest possible time that any observer could correctly record for the event.

H.3.3 Derive the time dilation formula.

Students should be able to construct a simple derivation of the time dilation formula based on the concept of the light clock and the postulates of relativity.

If we imagine a stationary observer with one light clock then t is the time between ‘ticks’ on the stationary clock. In this stationary frame, a moving clock runs slowly and t’ is the time between ‘ticks’ on the moving clock: t’ is greater than t.


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In the time t’, the clock has moved on a distance = vt’. Distance travelled by the light:

𝑙! = 𝑣𝑡! ! + 𝑙!

𝑡! =𝑙!

𝑐=

𝑣𝑡! ! + 𝑙!

𝑐

𝑡!" =𝑣𝑡! ! + 𝑙!

𝑐!

𝑡!" 1 −𝑣!

𝑐!=𝑙!

𝑐!

Since 𝑙!

𝑐!= 𝑡!

𝑡!" 1 −𝑣!

𝑐!= 𝑡!

𝑡! =1

1 − 𝑣!

𝑐!

∗ 𝑡

This equation is true for all measurements of time, whether they have been made using a light clock or not.

H.3.24 Sketch and annotate a graph showing the variation with relative velocity of the Lorentz factor.

H.3.5 Solve problems involving

time dilation. e.g. The time interval between

the ticks of a clock carried on a fast rocket is half of what observers on earth record. How fast is the rocket moving with respect to earth? From the time dilation formula:

2 =1

1 − 𝑣!

𝑐!

1 −𝑣!

𝑐!=12

𝑣!

𝑐!=34

(Lorentz factor)


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𝑣 = 8.866𝑐 H.3.6 Define proper length. The proper length of an object is

the length recorded in a frame where the object is at rest.

H.3.7 Describe the phenomenon of length contraction.

The derivation of the length contraction formula is not required.

One of the peculiar aspects of Einstein's theory of special relativity is that the length of objects moving at relativistic speeds undergoes a contraction along the dimension of motion. An observer at rest (relative to the moving object) would observe the moving object to be shorter in length. That is to say, that an object at rest might be measured to be 200 feet long; yet the same object when moving at relativistic speeds relative to the observer/measurer would have a measured length which is less than 200 ft. This phenomenon is not due to actual errors in measurement or faulty observations. The object is actually contracted in length as seen from the stationary reference frame. The amount of contraction of the object is dependent upon the object's speed relative to the observer. [Note that it is only lengths in the direction of motion that are contracted.]

H.3.8 Solve problems involving length contraction.

e.g. An unstable particle has a life time of 4.0 ∗ 10!!𝑠 in its own rest frame. If the frame is moving at 98% of the speed of light, calculate (a) its life time in the lab frame and (b) the length travelled in both frames.

a) 𝛾 = !!!!.!"!

= 5.025

∆𝑡 = 𝛾∆𝑡! = 5.025 ∗ 4.0 ∗ 10!!= 2.01 ∗ 10!!𝑠

b) in the lab frame, the particle moves: length = speed x time.

0.98 ∗ 3 ∗ 10! ∗ 2.01 ∗ 10!!= 59.1𝑚

in the particle’s frame, the laboratory moves

∆𝑙 =59.1𝛾

= 11.8𝑚


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H4 Some consequences of special relativity Assessment statement Teacher’s notes H.4.1 Describe how the concept

of time dilation leads to the “twin paradox”.

Different observers’ versions of the time taken for a journey at speeds close to the speed of light may be compared. Students should be aware that, since one of the twins makes an outward and return journey, this is no longer a symmetrical situation for the twins.

A rocket follows a long circular path. It sets off from space station P and will eventually come back. The passenger in the rocket sets his clock by looking at the station’s clock. The time is 0. When he returns, he looks at his watch and finds that it is slow compared with the station clock. Thus, if the trip lasted, say 6 years, by the passenger’s watch, the passenger is 6 years older. However, the passenger’s twin brother, who is the stationmaster, is older by 10 years. (assuming that v=0.8c). The stationmaster may claim that it was he who moved away. So when the stationmaster again meets the rocket passenger, he will claim that his clock is slower than the passenger’s. So the stationmaster is only 6y older while the passenger is 10y older. Which of the twins is older when they meet again? This is often referred to as the twin paradox. At all times the stationmaster was in an inertial frame. However, the rocket had been moving in a circle (thus experiencing centripetal acceleration) and so the rocket’s frame had not been inertial. Careful application of the laws of relativity to this asymmetric situation leads to the conclusion that the stationmaster has aged by 10y and the passenger by 6y. Even if the roket moves in a straight line and then reverses direction to return to the space station, this does not help because in this case the rocket must decelerate and then accelerate.

H.4.2 Discuss the Hafele–Keating experiment.

Atomic clocks were put into aircraft and flown, both eastwards and westwards, around the world. Before and after the flights, the times on the clocks were compared with clocks that remained fixed in the same location on the surface of the Earth. An observer in the centre of the Earth would describe the clock


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flying eastwards as moving the fastest, the clock that is on the same location on the Earth’s surface as also moving eastwards (due to the rotation of the earth) but not as fast as the clock in the airplane, and the clock flying westwards as moving the slowest. The results of their experiment agreed with the predictions within the uncertainties of the experiment.

H.4.3 Solve one-‐dimensional problems involving the relativistic addition of velocities.

The derivation of the velocity addition formula is not required.

𝒖 =𝒖! + 𝒗

𝟏 + 𝒖!𝒗𝒄𝟐

e.g. An electron has a speed of 2.00x108ms-‐1 relative to a rocket, which itself moves at a speed of 1.00x108ms-‐1 with respect to the to the ground. Applying the formula above with 𝑢! = 2.00 ∗ 10!𝑚𝑠!! and 𝑣 = 1.00 ∗ 10!𝑚𝑠!! we find 𝑢 = 2.45 ∗ 10!𝑚𝑠!!

H.4.4 State the formula representing the equivalence of mass and energy.

Mass and energy are equivalent. This means that energy can be converted into mass and vice versa. The energy required to create a particle at rest is called the rest energy and can be calculated from the rest mass: 𝐸! = 𝑚!𝑐!

H.4.5 Define rest mass. Students should be aware that rest mass is an invariant quantity. Students should be familiar with the unit MeV c−2 for mass.

The rest mass of an object is its mass as measured in a frame where the object is at rest. A frame that is moving with respect to the object would record a higher mass.

H.4.6 Distinguish between the energy of a body at rest and its total energy when moving.

See H.4.8

H.4.7 Explain why no object can ever attain the speed of light in a vacuum.

𝐸 = 𝛾𝑚!𝑐!

𝐸 =𝑚!𝑐!

1 − 𝑣!

𝑐!

It is very important to notice that, as the speed of a particle approaches the speed of light, the total energy approaches infinity. Therefore, a particle with mass cannot reach the speed of light. Only particles without mass, such as photons, can move at the speed of light.

H.4.8 Determine the total energy of an accelerated particle.

Students should be able, for example, to calculate the total energy of an electron after acceleration through a known potential difference.

If a particle is accelerated by a potential difference of V volts, its total energy will increase by an amount qV, where q is its charge. Thus, if a particle is initially at


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rest, its total energy is the rest energy 𝐸! = 𝑚!𝑐!. After going through the potential difference, the total energy will be 𝐸 = 𝑚!𝑐! + 𝑞𝑉. e.g. An electron f rest energy 0.511 MeV is accelerated through a potential difference of 5.0MV in a lab. (a) What is its total energy with respect to the lab? (b) What is its speed with respect to the lab? a) The total energy will increase by 𝑞𝑉 = 1𝑒 ∗ 5.0 ∗ 10!𝑣𝑜𝑙𝑡 =5𝑀𝑒𝑉 And so the total energy is:

𝐸 = 𝑚!𝑐! + 𝑞𝑉 𝐸 = 0.511𝑀𝑒𝑉 + 5.0𝑀𝑒𝑉

= 5.511𝑀𝑒𝑉 b) We know that

𝐸 = 𝛾𝑚!𝑐! 5.511 = 𝛾 ∗ 0.511

𝛾 =5.5110.511

= 10.785 Since

𝛾 =1

1 − 𝑣!

𝑐!

10.785 =1

1 − 𝑣!

𝑐!

𝑣 = 0.966𝑐

H5 Evidence to support special relativity Assessment statement Teacher’s notes H.5.1 Discuss muon decay as

experimental evidence to support special relativity.

Muons are particles with properties similar to those of the electrons except that they are more massive, unstable and they decay into electrons. Muons are created high up in the atmosphere (10km). Cosmic rays from the sun can cause them to be created with huge velocities: 0.99c. As they travel towards the earth some of them decay but there is still a detectable number of arriving at the surface of the Earth. Without relativity, no muons would be expected to reach the surface at all. A particle with a lifetime of 2.2 ∗ 10!!𝑠 which is travelling near the speed of light would be expected to travel less than a kilometre


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before decaying. The muons’ speed means that the Lorentz factor is high. 𝛾 = !

!!!.!!!= 7.1.

Therefore an average lifetime of 2.2 ∗ 10!!𝑠 in the muons’ frame of reference will be time dilated to a longer time as far as a stationary observer on earth is concerned. Many muons will still decay but some will make it through to the surface – this is exactly what is observed. In the muons’ frame of reference they exist for 2.2 ∗ 10!!𝑠 on average. They make it down to the surface because they atmosphere is moving with respect to the muons. This means that the atmosphere will be length-‐contracted. The 10km distance will only be !"

!.!= 1.4𝑘𝑚. A

significant number of muons will exist long enough for the Earth to travel this distance.

H.5.2 Solve problems involving the muon decay experiment.

H.5.3 Outline the Michelson–Morley experiment.

Students should be able to outline the principles of the Michelson interferometer using a simple sketch of the apparatus.

The aim of the experiment was to measure the speed of the earth through space (the ether). It involved two beams of light travelling down two paths at right angles to one another. Having travelled different paths, the light was brought together where it interfered and produced fringes of constructive and destructive interference. If the apparatus were rotated around, the speed down the paths would change. This would move the interference pattern. The idea was to measure the change and thus work out the speed of the Earth through the ether. The experiment was tried but the rotation of the apparatus did not produce any observable change in the interference pattern.

H.5.4 Discuss the result of the Michelson–Morley experiment and its

The implication that the ether does not exist and that the result is consistent with the

The above null result can be easily understood from the first postulate of relativity – the


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implication. constancy of the speed of light is the accepted explanation.

constancy of the speed of light. The interference pattern does not change because the speed of light along the paths is always the same. It is unaffected by the motion of the Earth. Also, it can be concluded that the ether does not exist.

H.5.5 Outline an experiment that indicates that the speed of light in vacuum is independent of its source.

Students should be familiar with pion decay experiments involving the decay of a fast-‐moving pion into two gamma-‐ray photons.

The first conclusive experiment that demonstrated the constancy of the speed of light with great accuracy was performed at CERN in 1964. In this experiment, neutral pions moving at 0.99975c decayed into a pair of photons moving in different directions. The speed of the photons in both directions was measured to be c with extraordinary accuracy. The speed of light does not depend on the speed of its source.

H6 Relativistic momentum and energy Assessment statement Teacher’s notes H.6.1 Apply the relation for the

relativistic momentum 𝑝 = 𝛾𝑚!𝑣 of particles.

Students should be familiar with momentum expressed in the unit MeV c−1.

In classical mechanics, the momentum is given by the product of mass times velocity, but in relativity this is modified to 𝑝 = 𝛾𝑚!𝑣. We still have the usual law of momentum conservation, which states that, when no external forces act on a system, the total momentum stays the same.

H.6.2 Apply the formula 𝐸! = 𝛾 − 1 𝑚!𝑐! for the kinetic energy of a particle.

The kinetic energy 𝐸! is defined as the total energy minus the rest energy:

𝐸! = 𝐸 −𝑚!𝑐! This can be rewritten as

𝐸! = 𝛾 − 1 𝑚!𝑐! This definition ensures that the kinetic energy is zero when v = 0. The familiar result from mechanics that the work done by the new force equals the change in kinetic energy holds in relativity as well.

H.6.3 Solve problems involving relativistic momentum and energy.

Students should be able to calculate, for example, the kinetic energy, total energy, speed and momentum of an accelerated particle and for particles produced in reactions.

e.g. Find the kinetic energy of an electron whose momentum is 1.5 𝑀𝑒𝑉𝑐!! The total energy of the electron is given by

𝐸! = 𝑚!!𝑐! + 𝑝!𝑐!

to give 𝐸 = 1.58𝑀𝑒𝑉


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And 𝐸! = 𝐸 −𝑚!𝑐!

So 𝐸! = 1.07𝑀𝑒𝑉

H7 General relativity Assessment statement Teacher’s notes H.7.1 Explain the difference

between the terms gravitational mass and inertial mass.

Inertial mass: The property of an object that determines how it responds to a given force (different masses have different accelerations when a force acts on them)

𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑚𝑎𝑠𝑠 𝑚! =𝐹𝑎

Gravitational mass: The property of an object that determines how much gravitational force it feels when close to another object.

𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 ∝ 𝑚! H.7.2 Describe and discuss

Einstein’s principle of equivalence.

Students should be familiar with Einstein’s closed elevator “thought experiment”.

Gravitational and inertial effects are indistinguishable.

H.7.3 Deduce that the

principle of equivalence predicts bending of light rays in a gravitational

Consider an elevator motionless in space (so that there is no gravity inside and any occupants are in “freefall”). This elevator has a pin-‐sized hole in the wall, through which a tiny beam of light enters, creating a speck of light on the opposite wall, directly across from the


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field. hole (if one measured the distance from the floor of the elevator to the hole and to the speck of light, it would be equal). Now, if this elevator began to be pulled forward through space, the inertial mass would pull the occupants to the floor of the elevator (mimicking the pull of gravity), and something peculiar would happen to the beam of light: As the elevator’s acceleration increases, the prick of light will appear to move downward, for in the time it takes for the light to travel from the hole to the opposite wall, the elevator would already have moved forward slightly (though it would have to be moving rather quickly for this effect to be at all noticeable). In other words, because of the motion of the elevator, the beam of light would “bend” as it enters the elevator. Now, carrying this thought through to its conclusion – remember that the occupants of this elevator would have no way of knowing if the sensation they are feeling is caused by the elevator’s inertia or by some gravitational force (it could feel to them that they are on the surface of the Earth), so to these people, the bending of the beam of light appears to be caused by gravity.

H.7.4 Deduce that the principle of equivalence predicts that time slows down near a massive body.

Consider two waves on a wavefront AB, which are bend as they pass near a massive object:

AC is longer than BD, however the speed of light is constant. As:

𝑐 =𝑑𝑡

time must slow down near a massive object. H.7.5 Describe the concept of

spacetime. Space-‐time is a four-‐dimensional

world with three space and one time coordinates. [The mass and energy content of space determine the geometry of that space and time. The geometry of space-‐time determines the motion of mass and energy in the space-‐time.]

H.7.6 State that moving objects follow the shortest path between two points in spacetime.

In the absence of any forces, a body moves in this four-‐dimensional world along paths of shortest length, called geodesics.

H.7.7 Explain gravitational attraction in terms of the warping of spacetime by matter.

The motion of a planet around the sun is, according to Einstein, not the result of a gravitational force acting on the planet (as Newton would have it) but rather due to the curved geometry in the space and time around the sun created by the large


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mass of the sun. The planet follows a geodesic in the curved space-‐time around the earth. This geodesic appears as a circular path if we view space-‐time as flat.

H.7.8 Describe black holes. Students should know that

black holes are a region of spacetime with extreme curvatures due to the presence of a mass.

Some objects contract under the influence of their own gravitation, becoming ever smaller objects. The object is expected to become a hole in space-‐time around this point. This creates an immense bending of space-‐time around this point and it is known as a black hole – since noting can escape from it.

H.7.9 Define the term Schwarzschild radius.

The Schwarzschild radius is not the actual radius of a black hole (the black hole is a point) – it is the distance from the hole’s centre that separates space into a region from which an object can escape and a region from which no object can escape. Any object closer to the centre of the black hole than this radius will fall into the hole; no amount of energy supplied to this body will allow it to escape from the black hole.

H.7.10 Calculate the Schwarzschild radius.

𝑅! =2𝐺𝑀𝑐!

For the sun:

𝑅! =2 ∗ 6.67 ∗ 10!!! ∗ 2 ∗ 10!"

3 ∗ 10! !

𝑅! ≈ 3 ∗ 10!𝑚 H.7.11 Solve problems

involving time dilation close to a black hole.

Two observers who are at different points in a gravitational field measure the time interval between the same two events differently. This is an example of how masses curve not just space but also time.

∆𝑡!"# =∆𝑡!"#$

1 − 𝑅!𝑟

e.g. Consider a theoretical observer approaching a black hole. This observer sends signals to a far-‐away observer in a spacecraft of his position. When his distance from the centre of the black hole is 𝑟 = 1.5𝑅!, the observer stops and sends two signals one second apart (as


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measured by his clocks). The spacecraft observers will receive the signals a time apart given by

∆𝑡!"# =∆𝑡!"#$

1 − 𝑅!𝑟

∆𝑡!"# =1

1 − 11.5

= 1.73𝑠

H.7.12 Describe the concept of gravitational red-‐shift.

Students should be aware that gravitational red-‐shift is a prediction of the general theory of relativity.

Gravitational redshift is an effect that the general theory of relativity predicts – clocks slow down in a gravitational field. In other words a clock on the ground floor of a building will run slowly when compared with a clock in the attic – the attic is further away from the centre of the earth.

H.7.13 Solve problems involving frequency shifts between different points in a uniform gravitational field.

A UFO travels at such a speed to remain above one point on the Earth at a height of 200km above the Earth’s surface. A radio signal of frequency of 110MHz is sent to he UFO. What is the frequency received by the UFP?

𝑓 = 1.1 ∗ 10!𝐻𝑧 𝑔 = 10𝑚𝑠!!

∆ℎ = 2.0 ∗ 10!𝑚 ∆𝑓𝑓=𝑔∆ℎ𝑐!

∆𝑓 =10 ∗ 2.0 ∗ 10!

3 ∗ 10! ! ∗ 1.1 ∗ 10!

∆𝑓 = 2.4 ∗ 10!!𝐻𝑧 Therefore, the received frequency: 1.1 ∗ 10! − 2.4 ∗ 10!!= 109999999.998𝐻𝑧 ≈ 1.1 ∗ 10!𝐻𝑧

H.7.14 Solve problems using the gravitational time dilation formula.

H8 Evidence to support general relativity Assessment statement Teacher’s notes H.8.1 Outline an experiment for

the bending of EM waves by a massive object.

An outline of the principles used in, for example, Eddington’s measurements during the 1919 eclipse of the Sun is sufficient.

When the sun is between the earth and the star, the sun’s light would completely wipe out the light from the star. This is why such an observation is possible only during a total solar eclipse. The bending of light that Eddington measured in 1919 was in agreement with the Einstein prediction, within experimental error, but the accuracy was not enough for this to constitute a test of the theory.


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The measurements have since been refined to include radio signals from distant galaxies, and these agree with general relativity predictions.

H.8.2 Describe gravitational lensing.

The bending of light is important to astronomers in the following was. Light from a distant star will be bent on its way to earth if it goes past a massive star or galaxy. This means that the star will not be observed to be at its true position. In some cases this leads to the formation of multiple images of the star. In this way massive objects act as a king of gravitational lens.

H.8.3 Outline an experiment that provides evidence for gravitational red-‐shift.

The Pound–Rebka experiment (or a suitable alternative, such as the shift in frequency of an atomic clock) and the Shapiro time delay experiments are sufficient.

Pound-‐Rebka: The decrease in frequency of a photon as it climbs out of a gravitational field can be measured in the lab. The measurements need to be very sensitive, but they have been successfully achieved on many occasions. The frequencies of gamma-‐ray photons were measured after they ascended or descended a tower at Harvard university. Atomic clock: Because they are so sensitive, comparing the difference in tie recorded by two identical atomic clocks can provide a direct measurement of gravitational redshift. One of the clocks is taken to high altitude by a rocket, whereas a second one remains on the ground. The clock that is at the higher altitude will run faster. Shapiro time delay: The time taken for a radar pulse to travel to another nearby planet and back can be accurately recorded. The gravitational field of the sun can affect the time taken. The extent of the effect depends on the orientation of the planets and the sun. The experiment was first performed in the 1960s and the result confirmed the predictions of general relativity.

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