Individual Study Plan: Unit 2 Waves and Light

Subject:PhysicsTeacher:N.Lad

Student:UnitUnit 2

Study Task outline:Waves ISPThis ISP covers the Waves and Light part of Physics Unit 2: Physics at Work. While much of the topic contains mathematical concepts which you will be asked quantitatively (using numbers) you may be asked to explain these ideas qualitatively (using words). This ISP will help you build the skills required to answer both styles of questions and aid you to build your knowledge for your assessments. It is of imperative importance that you can answer both quantitative and qualitative Physics questions; without this skill you will struggle to do any better than a D grade. To achieve a grade A-C you will need to be able to bring together different areas of Physics you have learnt (e.g. electrical energy anf the photoelectric effect). You must also be able to use an equation in a qualitative manner, explaining the implications of changing a variable in the equation. The specification points in this ISP are all of the specification points within the Waves module (see your specification or the Revision checklist in your revision guide).

The A* grade is awarded if the candidate meets two requirements: Grade A for the overall Advanced GCE 90% of the total available uniform marks for the A2 unitsThis means that you can get a B grade in AS Physics and still achieve an A*in A2 PhysicsA* students should be able to: apply principles and concepts in familiar and new contexts involving only a few steps in the argument describe significant trends and patterns shown by data presented in tabular or graphical form and interpret phenomena with few errors and present arguments and evaluations clearly explain and interpret phenomena with few errors and present arguments and evaluations clearly carry out structured calculations with few errors and demonstrate good understanding of the underlying relationships between physical quantities.

Exam questions where you are asked to describe experiments are often worth 6 marks, which can help you reach the next grade up of an exam paper. In these questions you need to:Clearly state the equipment required, explain and justify the equipment you have chosen, the measurement you make, and calculations you will do and also how you will make the experiment reliable to satisfy the criteria for full marks.Past papers and other examboard information is available on the following website: http://www.edexcel.com/quals/gce/gce08/physics/Pages/default.aspx

TaskTask detailResources neededCompleted/comments

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Spec points covered: 63,64,65,66,67, 69,70,71,72You should read through the notes from lessons and p146-155. The areas to focus on are:1. Being able to explain why we believe light behaves like a particle and a wave2. Explaining the importance of the photoelectric effect3. Define and use radiation flux as power per unit areaComplete Explaining Everything 1Complete questions for Task 1 from your Waves ISP booklet and mark using mark schemeA*/A link the photoelectric effect to electrical energy

Edexcel AS Physics textbook

Waves Booklet and mark scheme

Explaining everything 1Q E MWWW

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Specification points covered : 68You should read through the notes from lessons and p156-163. The areas to focus on are:1. Explaining atomic line spectra in terms of transitions between discrete energy levelsComplete Explaining Everything 2Complete questions for Task 2 from your Waves ISP booklet and mark using mark schemeA/A* - link the photoelectric effect to atomic line spectraEdexcel AS Physics textbook

Waves Booklet and mark scheme

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Specification points covered : 28, 30, 32, 31, 32, 35You should read through the notes from lessons and p76-91. The areas to focus on are:1. Stating the types of waves2. Describing the key features of a wave3. Describe standing and progressive wavesComplete Explaining Everything 2Complete questions for Task 2 from your Waves ISP booklet and mark using mark schemeA*/A research how electron behave as standing wavesEdexcel AS Physics textbook

Waves Booklet and mark scheme

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Spec points covered : 33,34, 36, 37, 38, 41,42, 43, 44, 45You should read through the notes from lessons and 92-99. The areas to focus on are:1. Describing the process of superposition2. Describing the formation of stationary waves and comparing them to progressive wave3. Describing reflection , refraction and how to find the refractive index4. Describing diffractionComplete Explaining Everything 4Complete questions for Task 4 from your Waves ISP booklet and mark using mark schemeA*/A research de Broglie diffractionEdexcel AS Physics textbook

Waves Booklet and mark scheme

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Specification points covered : 29, 39, 40, 46, 47, 48, 49You should read through the notes from lessons and p100-113. The areas to focus on are:1. Describing the process of polarisation2. Describing the EM spectrum3. Describe the process of the Doppler shift, and pulse echo techniques such as ultrasoundComplete Explaining Everything 5Complete questions for Task 5 from your Waves ISP booklet and mark using mark schemeA*/A link the EM spectrum to the photoelectric effectEdexcel AS Physics textbook

Waves Booklet and mark scheme

Explaining everything 5Q E MWWW

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Q - Questions completedE - Explaining everything completedM - Questions markedWWW want went wellEBI even better ifMRI My response is (student response)Explaining Everything 1The Photoelectric effectExplain why the wave model of light cannot explain emission of photoelectrons, but how the particle model of light does. Describe the experiment performed to prove the particle model of light, and include a labelled diagram.

Keywords to include are:Photon Wave Energy Work function Threshold frequency Photoelectron Emission Interactions Kinetic energyInclude any equations related to the photoelectric effect, and any graphs to help you determine constants/variable involved n the photoelectric effect equations.

Draw V vs I (or I vs V) graphs for a diode, a filament lamp and a resistor. Explain the shape of the graphs.Terms/Words to include:Resistancetemperaturecurrentpotential differenceconstantone direction

Explaining everything 2Energy LevelsUsing a diagram, show how absorption or emission of photons can be shown using an energy level diagram. Explain how we can calculate the wavelength or frequency of the photon. Include any equations.

Keywords to include are:Electron Volts (eV)AbsorptionEmission

Explaining Everything 3In the space below complete the following tasks:Draw a wave labelling the key features (wavelength, frequency, amplitude)Include the wave equation and how to calculate the periodCompare and contrast longitudinal waves and transverse wavesCompare and contrast standing and progressive waves

Explaining Everything 4: Light as a WaveExplain the evidence that supports the theory that light acts as a wave.

Keywords to include are:Interference constructive destructive path difference superposition

fdfds

Explaining everything 5Describe how light can become polarised:

Describe an experiment that you could use to test how the much sugar solution rotates the plane of polarisation of light:

Describe the Doppler Effect:

Describe pulse echo techniques:

List of data, formulae and relationshipsDataAcceleration of free fallg = 9.81 m s2(close to Earths surface)Boltzmann constant k = 1.38 1023 J K1Coulombs law constantk = 1/4 = 8.99 109 N m2 C2Electron chargee = 1.60 1019- CElectron massme = 9.11 1031 kgElectronvolt1 eV = 1.60 1019 JGravitational constantG = 6.67 1011 N m2 kg2Gravitational field strengthg = 9.81 N kg1(close to Earths surface)Permittivity of free space= 8.85 1012 F m1Planck constant h = 6.63 1034 J sProton massmp = 1.67 1027 kgSpeed of light in a vacuumc = 3.00 108 m s1Stefan Boltzmann constant5.67 108 W m2 K4Unified atomic mass unitu = 1.66 1027 kg

2.MechanicsKinematic equations of motionv = u + ats = ut + at2v2 = u2 + 2asForcesF = mag = F/mW = mgWork and energyW = FsEk = mv2Egrav = mgh

MaterialsStokes lawF = 6rvHookes lawF = kxDensity = m/VPressurep = F/AYoungs modulusE = / whereStress = F/AStrain = x/xElastic strain energyEel = FxAppendix 8 Formulae

3.WavesWave speed v = fRefractive index12 = sin i/sin r = v1/v2 ElectricityPotential differenceV = W/QResistanceR = V/IElectrical power, energy andP = VIefficiencyP = I2RP = V2/RW = VIt% efficiency = [useful energy (or power) output/total energy (or power) input] 100%ResistivityR = l/ACurrentI = Q/tI = nqvA

Quantum physicsPhoton modelE = hfEinsteins photoelectrichf = + mv2maxequation

Waves ISP BookletTask 1

1.(a)The maximum wavelength of electromagnetic radiation which can release photoelectrons from the surface of caesium is 6.5 107 m.(i)State the part of the electromagnetic spectrum to which this radiation belongs............................................................................................................................(1)(ii)Show that caesium has a work function of about 3 1019 J........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2) (b)The caesium cathode of a photocell is illuminated by radiation of frequency f. The circuit shown is used to measure the stopping potential Vs for a range of frequencies.

(i)Explain what is meant by the term stopping potential.................................................................................................................................................................................................................................................................................................................................................................................. (2)

(ii)The experiment is repeated, using different photocells, to measure the stopping potentials of calcium and beryllium. The graph shows how the stopping potentials Vs for all three metals vary with frequency f.

Use the relationshiphf = eVs + to explain why all three graphs are parallel.................................................................................................................................................................................................................................................................................................................................................................................. (2)(Total 7 marks)2.(a)Magnesium has a work function of 5.89 1019 J. Explain the meaning of this statement................................................................................................................................................................................................................................................................................................................................................................................................................(2) (b)Ultraviolet radiation from an extremely faint source is incident normally on a magnesium plate. The intensity of the radiation is 0.035 W m2. A single magnesium atom occupies an area of about 8 1020 m2 on the surface of the plate.(i)Show that, if the radiation is regarded as a wave motion, it should take at least 200 s for a magnesium atom to absorb 5.89 1019 J of energy.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)

(ii)In practice, it is found that photoemission from the plate begins as soon as the radiation source is switched on. Explain how the photon model of electromagnetic radiation accounts for this. You may be awarded a mark for the clarity of your answer................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (4)(Total 9 marks)3.A leaf of a plant tilts towards the Sun to receive solar radiation of intensity 1.1 kW m 2, which is incident at 50 to the surface of the leaf.

(a)The leaf is almost circular with an average radius of 29 mm. Show that the power of the radiation perpendicular to the leaf is approximately 2 W.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3) (b)Calculate an approximate value for the amount of solar energy received by the leaf during 2.5 hours of sunlight...........................................................................................................................................................................................................................................................................Energy = ..........................................(2)(Total 5 marks)

4.The photoelectric effect supports a particle theory of light but not a wave theory of light. Below are two features of the photoelectric effect. For each feature explain why it supports the particle theory and not the wave theory.(a)Feature 1: The emission of photoelectrons from a metal surface can take place instantaneously.Explanation ....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (2)(b)Feature 2: Incident light with a frequency below a certain threshold frequency cannot release electrons from a metal surface.Explanation ....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (2)(Total 4 marks)5.The table gives information about two beams of monochromatic light.Intensity /Wm2Colour

Beam A6.0red

Beam B0.2blue

Each beam is shone in turn onto a barium plate. Beam B causes photoemission but beam A does not. A student says that this is because the blue beam is more energetic than the red beam.(a)In one sense the students statement is correct. In another sense the statement is incorrect. Explain how it is:correct...............................................................................................................................................................................................................................................................................................................................................................................................................incorrect...........................................................................................................................................................................................................................................................................(3)

(b)Barium has a work function of 3.98 1019 J.(i)Explain the meaning of the term work function..................................................................................................................................................................................................................................................................................................................................................................................(2)(ii)Calculate the photoelectric threshold frequency for the barium plate..................................................................................................................................................................................................................................................................................................................................................................................Threshold frequency = ......................................(2)(Total 7 marks)

Task 26.The diagram shows some of the energy levels of a tungsten atom.

(a)An excited electron falls from the 1.8 keV level to the 69.6 keV level. Show that the wavelength of the emitted radiation is approximately 0.02 nm............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (4) (b)To which part of the electromagnetic spectrum does this radiation belong?.....................................................................................................................................(1)(Total 5 marks)7.Astronomers can identify different gases present in the outer parts of stars by analysing the line spectra of the starlight.Explain the meaning of line spectra.(2)Explain how line spectra provide evidence for the existence of energy levels in atoms.(3)(Total 5 marks)8.The following is a simplified energy level diagram for atomic hydrogen.

State the ionisation energy of atomic hydrogen.............................................................................................................................................................................................................................................................................................

Account for the labelling of the energy levels with negative numbers.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)Calculate the wavelength of the photon emitted when an electron moves from the1.51 eV energy level to the 3.40 eV energy level.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Wavelength = ........................................(3)Describe how you would produce a line spectrum of atomic hydrogen in a laboratory.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Sketch what you would expect to see.

(1)(Total 9 marks)

9.Below is a simplified energy level diagram for atomic hydrogen.__________________0 eVfirst excited state__________________3.4 eVground state__________________13.6 eV (a)A free electron with 12 eV of kinetic energy collides with an atom of hydrogen. As a result the atom is raised to its first excited state. Calculate the kinetic energy of the free electron, in eV, after the collision.

Kinetic energy = ...................... eV(2)(b)Calculate the wavelength of the photon emitted when the atom returns to its ground state.

Wavelength =................................(3)(Total 5 marks)10.The diagram shows the lowest three energy levels of atomic hydrogen.(a)Excited hydrogen atoms can emit light of wavelength 656 nm. By means of a suitable calculation, determine which transition between energy levels is responsible for this emission..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Transition: from level ............ to level .............(4)

(b)The spectrum of light from the Sun contains a dark line at a wavelength of 656 nm. With reference to the energy level diagram, explain how this line is produced. You may be awarded a mark for the clarity of your answer.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)(c)In the spectrum of light received from another galaxy, the same line appears at a wavelength of 695 nm.How can we deduce from this that the galaxy is receding from the Earth?..........................................................................................................................................................................................................................................................................(1)(Total 9 marks)

Task 311.A source of light emits a train of waves lasting 0.04 s. The light has a wavelength of 600 nm and the speed of light is 3 108 m s1. How many complete waves are sent out?A2.0 107B4.5 107C2.0 1010D4.5 1013(Total 1 mark)12.Electromagnetic waves are produced by oscillating charges. Sound waves are produced by oscillating tuning forks. How are these waves similar?Athey are both longitudinal waves.Bthey are both transverse waves.Cthey both have the same frequency as their respective sources.Dthey both require a medium to travel through.(Total 1 mark)13.It has been suggested that tigers use infrasound low frequency sounds inaudible to humans to keep rivals away from their territory and to attract mates.Sound is a longitudinal wave. Describe how sound travels through the air........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (3)State what is meant by frequency..............................................................................................................................................................................................................................................................................................................................................................................................................................................(1)The frequency range of the sound produced by the tigers extends down to 18 Hz. Calculate the wavelength in air for sounds of this frequency. Speed of sound in air = 330 m s1..............................................................................................................................................................................................................................................................................................................................................................................................................................................Wavelength = ....................................(2)(Total 6 marks)

14.The photograph shows a laboratory machine for illustrating a sinusoidal transverse wave.Beside the machine is a rule. (You may make marks on the photograph if you wish.)

Find a value for the wavelength of the waves...............................................................................................................................................................................................................................................................................................Wavelength = .......................................(2)Find a value for the amplitude of the waves...............................................................................................................................................................................................................................................................................................Amplitude = ..........................................(2) The machine is operated so that every rod along the wave moves up and down through a complete cycle every 2 s. Calculate the frequency of the waves...............................................................................................................................................................................................................................................................................................Frequency = ............................................(1) What is meant by a transverse wave?.............................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Discuss whether this machine would be helpful in illustrating how a sound wave travels............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total 10 marks)15.A loudspeaker emits a sound wave of wavelength 0.66 m. The diagram shows how displacement varies with distance from the loudspeaker at one instant of time.

(a)Which letter indicates the wavelength of the sound wave?.....................................................................................................................................(1) (b)Sound travels at 330 m s1 in air. Calculate the period of the wave...........................................................................................................................................................................................................................................................................Period = ..........................(3)(Total 4 marks)16.Two points on a progressive wave differ in phase by radian. The distance between them is 0.50 m. The frequency of the oscillations is 10 Hz. The maximum speed of the wave isA2.50 m s1B5.00 m s1C12.5 m s1D40.0 m s1(Total 1 mark)17.Which of the following statements about standing waves is true?Aparticles between adjacent nodes all have the same amplitude.Bparticles undergo no disturbance at an antinode.Cparticles immediately either side of a node are moving in opposite directions.Dparticles between adjacent nodes are out of phase with each other.(Total 1 mark)

18.The diagram shows a wave on a rope. The wave is travelling from left to right.

At the instant shown, point L is at a maximum displacement and point M has zero displacement. Which row in the table correctly describes the motion of points L and M during the next half cycle of the wave?Point LPoint M

ARisesfalls

BRisesfalls then rises

Crises then fallsrises

Drises then fallsfalls then rises

(Total 1 mark)19.The following apparatus is set up. When the frequency of the vibrator is 60 Hz, the standing wave shown in the diagram is produced.

(a)What is the wavelength of this standing wave?Wavelength = ...................(1)(b)The frequency of the vibrator is altered until the standing wave has two more nodes.Calculate the new frequency.

Frequency = ................(2)(Total 3 marks)

Task 420.(a)The diagram shows three possible stationary waves on a string of length 1.20 m stretched between fixed points X and Y. (i)Wave A has a frequency of 110 Hz.Complete the table below to show the wavelengths and frequencies of the three waves.WaveWavelength / mFrequency / Hz

A110

B

C

(3) (ii)Each of the waves has nodes at X and Y. Explain why these points must be nodes.......................................................................................................................................................................................................................................................(1)(b)There is a similarity between the behaviour of the string in part (a) and that of the electron in a hydrogen atom. Electron states can be represented by stationary waves which have to fit inside the atom. Stationary waves with greater numbers of nodes represent electrons in higher energy levels. Explain why this is the case................................................................................................................................................................................................................................................................................................................................................................................................................ (2)(Total 6 marks)21.(a)Explain what is meant by the term transverse wave. You may wish to illustrate your answer with the help of a simple diagram.

....................................................................................................................................................................................................................................................................................................................................................................................................................(3)(b)State two differences between a stationary wave and a progressive wave.Difference 1 ........................................................................................................................................................................................................................................................Difference 2 .......................................................................................................................................................................................................................................................(2)(c)Spiders are almost completely dependent on vibrations transmitted through their webs for receiving information about the location of their prey. The threads of the web are under tension. When the threads are disturbed by trapped prey, progressive transverse waves are transmitted along the sections of thread and stationary waves are formed.Early in the morning droplets of moisture are seen evenly spaced along the thread when prey has been trapped.

(i)Explain why droplets form only at these points............................................................................................................................. (1) (ii)The speed of a progressive transverse wave sent by trapped prey along a thread is 9.8 cm s1. Use the diagram to help you determine the frequency of the stationary wave.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Frequency = ................................................(4)(Total 10 marks)

22.State two conditions necessary for total internal reflection to occur at an interface between air and water.Condition 1 ...........................................................................................................................................................................................................................................................................Condition 2 ...........................................................................................................................................................................................................................................................................(Total 2 marks)23.(a)Light changes direction when it passes from air to water.(i)Name the process of light changing direction in this way............................................................................................................................(1)(ii)Explain why this process takes place.......................................................................................................................................................................................................................................................(1) (b)The diagram represents some fish under water and a butterfly above the water. (i)Draw a ray to show the path of light travelling from the butterfly to the eye of fish B.(2)(ii)Explain what is meant by critical angle........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(iii)Explain how rays from fish A could reach the eye of fish B along two different paths. Add rays to the diagram to illustrate your answer..................................................................................................................................................................................................................................................................................................................................................................................(4)(Total 10 marks)24.(a)Explain what is meant by superposition of waves.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(b)The electron micrograph shows a small area of the surface of a copper sample.

Electrons move over the surface and behave like waves. When a wave reaches an edge or an atom, it reflects, and a standing wave is formed. (i)Explain how a standing wave is formed by the reflection of a wave............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. (3) (ii)A standing wave is visible on the micrograph between atoms A and B. There are nodes at X and Y. The dark lines between X and Y show antinodes. The distance between points X and Y on the micrograph is 4.2 109 m. Use this information to calculate the wavelength of the electron waves.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Wavelength = .................................................................(2)(iii)Explain the meaning of the terms amplitude and antinode.Amplitude.................................................................................................................................................................................................................................................................................................................................................................................(1)Antinode.................................................................................................................................................................................................................................................................................................................................................................................(1)(Total 9 marks)25.A ray of light travelling in air, strikes the middle of one face of an equilateral glass prism as shown.State what happens to the following properties as the light goes from the air into the glass.Frequency ..............................................................................................................................Wavelength ............................................................................................................................Speed .....................................................................................................................................(Total 3 marks)

Task 526.(a)Ultrasound images of the body are a useful diagnostic tool for doctors. A single transducer can be used both to send and receive pulses of ultrasound.The diagram shows a lateral cross-section through part of the abdomen. The diagram is not to scale.(i)Calculate the time interval between sending out a single pulse and receiving its echo from interface B. The speed of ultrasound in the abdominal wall is1500 m s1.

Time interval = .......................(3) (ii)The time between pulses being emitted by the transducer is 200 s. At what frequency are the pulses emitted?

Frequency = ................(2) (iii)The time interval before the echo returns from interface D is 250 s. Suggest why this time interval will make reflections from D difficult to interpret and what could be done to overcome this problem.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(iv)State one reason why ultrasound rather than X-rays is now used to scan expectant mothers..................................................................................................................................................................................................................................................................................................................................................................................(1)(b)Ultrasound is also used to measure blood flow in the body. It uses the Doppler shift of the reflected pulse to measure the speed of blood through the arteries of the body.Describe the principle of this method and how it can be used to determine the speed of blood....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)(Total 13 marks)27.(a)Explain with the aid of diagrams why transverse waves can be polarised but longitudinal ones cannot be polarised.

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(b)Describe with the aid of a diagram how you could demonstrate that light can be polarised.

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total 6 marks)28.When the Moon is full, bright moonlight makes it difficult for astronomers to study the stars. Moonlight is scattered by atoms in the atmosphere causing it to become plane polarised.Draw labelled diagrams to show how the polarised light differs from unpolarised light.

Polarised lightUnpolarised light(2)Explain how an astronomers telescope could be adapted to overcome the problem of the bright moonlight..............................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 4 marks)

29.The Raman effect can be used to identify the chemical composition of solid materials in a non-destructive manner, e.g. to identify fake diamonds. When laser light is shone on a material some of the light is scattered. The scattered light will be plane polarised and some of it will have been frequency shifted (that is, it will have undergone a change in frequency). This happens when the light interacts with the vibrating atoms in the materials. Analysis of these Raman frequency shifts reveals the chemical composition of the material.What is meant by plane polarised light?.............................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Some of the light will have been frequency shifted by the moving atoms. Name the phenomenon which caused the frequency shifting and briefly explain how it arises..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)

The Raman frequency shifts are often measured in wavenumbers. Wavenumber is defined as .When laser light of wavelength 1064 nm is shone onto diamond, the scattered light has a Raman wavenumber shift of +1300 cm1. Show that the frequency of this scattered light is about 3 1014 Hz.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(5) A student correctly remarks that if the frequency of the laser light has changed then its energy must have changed.State which model of light is being used by this student, and explain what must have happened to the vibrating atoms in the diamond during this interaction...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total 14 marks)

30.Explain what is meant by the term polarisation when referring to light.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Sugar is produced from plants such as sugar cane. The stems are crushed and the juice extracted. The concentration of sugar in the juice is used to value the crop.The concentration can be determined using polarised light.Explain how to measure the angle of rotation of polarised light when it passes through a sugar solution.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (4)A student has carried out this experiment and obtains three results. He has plotted them on the graph below. He takes three more results and tabulates them.Angle of rotation/degreesConcentration of solution/kg per litre

170.25

330.50

500.75

Add these results to the graph.(3)

Use your graph to determine the concentration of an unknown sample which gives a rotation of38.Concentration: .................................................................................................... kg per litre(1)(Total 10 marks)31.A photographer uses a polarising filter over the camera lens. She notices that the intensity of the light received from the blue sky changes as she rotates the filter.What does this suggest about light from the sky?............................................................................................................................................... (1)Explain the change in intensity as the filter is rotated............................................................................................................................................................................................................................................................................................... (2) The use of a polarising filter makes a blue sky appear darker, but the clouds remain bright.Suggest why there is little change in the intensity of the light from the clouds..............................................................................................................................................................................................................................................................................................................................................................................................................................................(1) Astronomers notice the same effect with the radio waves emitted by some galaxies.What does this suggest about these radio waves?..............................................................................................................................................................................................................................................................................................(1)State why radio waves should behave in the same way as light...............................................................................................................................................................................................................................................................................................(1)(Total 6 marks)32.A food packaging factory is moving soup through a 0.075 m diameter pipe when an obstruction occurs in the pipe. An ultrasound probe, connected to an oscilloscope, is moved along the pipe to find the obstruction (figure 1). The oscilloscope trace is shown below(figure 2).Figure 1 Figure 2

Oscilloscope time base = 20 106 s cm1.

On figure 2, pulse A is the outgoing signal from the probe and pulse B is the reflected signal from the other side of the pipeCalculate the speed of the ultrasound in the liquid in the pipe................................................................................................................................................Speed = .............................................................(2) State one way in which the oscilloscope trace will change when the ultrasound probe is above the obstruction..............................................................................................................................................................................................................................................................................................................................................................................................................................................(1)After the obstruction has been cleared, a Doppler ultrasound probe is used to measure the speed of the soup in the pipe. Describe the principle of this method.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)What must be measured to determine the speed of the soup?..............................................................................................................................................................................................................................................................................................(1) Someone says that this would be easier if the soup contained lumps like vegetables. Comment on this suggestion..............................................................................................................................................................................................................................................................................................................................................................................................................................................(1)(Total 8 marks)

1.(a)Part of spectrumLight / Visible / red (1)1Calculation of work functionUse of = hc/ (1)3.06 1019 [2 sig fig minimum] (1)2(6.63 1034 J s)(3.00 108 m s1)/(6.5 107 m)= 3.06 1019 J(b)(i)Meaning of stopping potentialMinimum potential difference between C and A / across thephotocell (1)Which reduces current to zero OR stops electrons reaching A /crossing the gap / crossing photocell (1)2(ii)Why the graphs are parallelCorrect rearrangement giving Vs = hf/e /e (1)Gradient is h/e which is constant / same for each metal (1)[Second mark can be awarded without the first if norearrangement is given, or if rearranged formula is wrong butdoes represent a linear graph with gradient h/e]2[7]2.(a)Meaning of statement(5.89 1019 J / work function) is the energy needed to remove anelectron [allow electrons] from the (magnesium) surface/plateConsequent markMinimum energy stated or indicated in some way [e.g. at least /or more] (1)2 (b)(i)Calculation of timeUse of P = IA (1)Use of E = Pt (1)[use of E = IAt scores both marks]Correct answer [210 (s), 2 sig fig minimum, no u.e.] (1)[Reverse argument for calculation leading to either intensity,energy or area gets maximum 2 marks]Example calculation:t = (5.89 1019 J)/(0.035 W m2 8 1020 m2)3(ii)How wave-particle duality explains immediate photoemissionQOWC (1)Photon energy is hf / depends on frequency / depends on wavelength (1)One/Each photon ejects one/an electron (1)The (photo)electron is ejected at once/immediately (1)[not just photoemission is immediate]4[9]3.(a)Solar PowerUse of P = Ir2 [no component needed for this mark] (1)Use of cos 40 or sin 50 (with I or A) (1)2.2 [2 sf minimum. No ue] (1)3e.g. P=1.1 103 W m2 cos 40 (29 103 m)2= 2.2 W(b)EnergyUse of E = Pt (1)1.8 104 J/2.0 104J (1)2e.g. E = 2.2W (2.5 3600 s)= 2.0 104 J[5]

4.Photoelectric effect(a)Explanation:Particle theory: one photon (interacts with) one electron (1)Wave theory allows energy to build up, i.e. time delay (1)2(b)Explanation:Particle theory: f too low then not enough energy (is released byphoton to knock out an electron) (1)Wave theory: Any frequency beam will produce enough energy (to release an electron, i.e. should emit whatever the frequency) (1)2[4]5.(a)Why statement correctBlue photon has more energy than red photon (1)Why statement incorrectBlue beam carries less energy per unit area per second / Blue beamcarries less energy per second / Blue beam carries less energy perunit area / Blue beam has lower intensity and intensity = energy per unitarea per secondAdditional explanation[Under correct] Blue has a higher frequency (OR shorter wavelength) /[Under incorrect] Blue beam has fewer photons (1)[Allow reverse statements about Red throughout part a]3(b)(i)Meaning of work functionEnergy to remove an electron from the surface (ORmetal OR substance) (1)[Dont accept from the atom. Dont accept electrons.]Minimum energy / Least energy / Energy to just/ without giving the electron any kinetic energy (1)2(ii)Calculation of threshold frequencyUse of = hf0 (1)Correct answer [6.00 1014 Hz] (1)e.g.(3.98 1019 J)/(6.63 1034 J s) = 6.00 1014 Hz2[7]6.(a)WavelengtheV to J (1)Use of E = hf (1)Use of c = f (1)1.8 1011 [2 sf minimum. No ue] (1)4e.g. f =(1.8 keV ( 69.6 keV)) (103 1.6 1019 J keV1) / 6.6 1034 J s= 1.64 1019 Hz = 3.00 108 m s1/1.64 1019Hz= 1.8 1011 m(b)TypeX rays [Accept gamma rays] (1)1[5]7.Explanation of line spectra:Specific frequencies or wavelengths (1)Detail, e.g. absorption/emission (1)OR within narrow band of wavelengths2Explanation how line spectra provide evidence for existence or energy levels in atoms:Photons (1)Associated with particular energies (1)Electron transitions (1)Discrete levels (to provide line spectra) (1)3[5]

8.Ionisation energy of atomic hydrogen:13.6 eV OR 2.18 1018 J [ sign, X](1)1Why energy levels are labelled with negative numbers:Work/energy is needed to raise the electrons/atoms to an energy of 0 eV, somust start negative(1)(1)ORWork/energy is given out when the electrons/atoms move to the groundstate, so energy now less than 0, i.e. negative(1)(1)ORthe ground state is the most stable/lowest energy level of theelectrons/atoms and must be less than 0, i.e. negative(1)(1)2[1st mark essential: e highest/maximum/surface/ionised/free has energy = 0eV2nd mark: raising levels means energy in OR falling levels means energy out negative levels]Wavelength of photon:E = 1.89 (eV)(1)Convert E to joules, i.e. (1.6 1019)OR = [Their E](1)= 6.6 107 (m) [6.5 6.7](1)3Production of line spectrum of atomic hydrogen in a laboratory:Source hydrogen discharge tube/hydrogen lamp/low p hydrogen with high V across(1)(view through) diffraction grating/prism/spectrometer/spectroscope(1)2 Sketch:

A few vertical lines on a blank background OR sharp bandsDark on light/light on dark NOT equally spaced(1)1Absorption spectrum:White light through gas in container (1)Diffraction grating/prism/spectrometer (1)Must be dark lines on bright background (1)[9]9.(a)Calculation of energy required by atom (1)Answer [1.8 (eV)] (1)Example of answer:Energy gained by atom = 13.6 eV 3.4 eV = 10.2 eVKE of electron after collision = 12 eV 10.2 eV = 1.8 eV2(b)Use of E = hf and c = f (1)Conversion of eV to Joules (1)Answer = [1.22 107 m] (1)Example of answerE = hf and c = f E = hc/ = (6.63 1034 J s 3 108 m s2) (10.2 eV 1.6 1019 C) = 1.21 107 m3[5]

10.(a)Which transitionUse of ()E = hc/ OR ()E = hf and f = c/ (1)Use of 1.6 1019 (1)Correct answer [1.9 eV] (1)C to B / 1.5 to 3.4 (1)[Accept reverse calculations to find wavelengths]e.g.(6.63 1034 J s)(3.00 108 m s1)/(656 109 m)(1.6 1019 J eV1)= 1.9 eV4(b)Explanation of absorption lineQOWC (1)Light of this wavelength is absorbed by hydrogen (1)In the outer part of the Sun (OR Suns atmosphere) (1)Absorbed radiation is reemitted in all directions (1)Transition from B to C (OR 3.4 to 1.5) (1)Max 4(c)Why galaxy recedingWavelength increased (OR stretched) / red shift /frequency decreased1[9]11.A[1]12.C[1]13.Description of soundParticles/molecules/atoms oscillate/vibrate (1)(Oscillations) parallel to/in direction of wave propagation / wavetravel / wave movement [Accept sound for wave] (1)Rarefactions and compressions formed [Accept areas of high and low pressure] (1)3 Meaning of frequencyNumber of oscillations/cycles/waves per second / per unit time (1)1 Calculation of wavelengthRecall v = f (1)Correct answer [18 m] (1)2Example of calculationv = f = 330 m s1 18 Hz= 18.3 m[6]14.Value of wavelength = 13.9 cm 0.5 cm (using interpolated sine curve) (1)= 13.4 cm [accept 13.2 to 13.6 cm] (1)2[12.3 to 12.5 cm for distance using rods (1) ]Value of amplitudePeak to peak = 4.5 cm [Accept 4.3 cm to 4.7 cm] (1)Amplitude = peak to peak= 2.25 cm [Accept 2.15 cm to 2.35 cm] [Allow ecf for 2nd mark if (1)2first part shown]Calculation of frequencyf = 1/T= 1 2 s = 0.5 Hz (1)1Explanation of why waves are transverseOscillation/vibration/displacement/disturbance at right angle (1)to direction of propagation/travel of wave (1)2[Oscillation not in direction of wave (1)] Description of use of machine to illustrate sound waveSound is longitudinal/not transverse (1)with oscillation along the direction of propagation / compressions and rarefactions (1)so model not helpful (1)3[10]15.(a)D1 (b)WavelengthUse of v = f (1)Use of f = 1/T (1)Answer T = [0.002 s] (1)[give full credit for candidates who do this in 1 stage T = /v]Example of answerv = ff = 330 / 0.66T = 1/f = 0.66 / 330T = 0.002 s3[4]16.D[1]17.C[1]18.B[1]19.(a)[1.0 m] (1)1 (b)Ratio of (5 or 6 / 3 ) 60 (1)Answer [f = 100 Hz] (1)2[3]20.(a)(i)Tablef2.4(110)1.22200.8330All wavelengths correct (2)[One or two wavelengths correct gets 1]Both frequencies correct (1)[Accept extra zero following wavelength figure, e.g. 2.40.Accept units written into table, e.g. 2.4 m, 220 Hz]3(ii)Why nodesString cannot move / no displacement / zero amplitude /no oscillation / phase change of on reflection / two wavescancel out / two waves are exactly out of phase (1)(OR have phase difference of OR half a cycle) /destructive interference1 (b)Why waves with more nodes represent higher energiesMore nodes means shorter wavelength (1)Momentum will be larger (1)[OR Allow 1 mark for More nodes means higher frequency and E = hf]2[6]

21.(a)Transverse wave(Line along which) particles/em field vectors oscillate/vibrate (1)Perpendicular to (1)Direction of travel or of propagation or of energy flow or velocity (1)3(b)DifferencesAny two:Standing wavesProgressive waves1. store energy1. transfer energy (1)2. only AN points have max2. all have the max ampl/displ (1) ampl/displ3. constant (relative) phase3. variable (relative) phaserelationshiprelationship (1)Max 2(c)(i)DropletsFormed at nodes / no net displacement at these points (1)1(ii)SpeedUse of = f (1)Evidence that wavelength is twice nodenode distance (1)Wavelength = 1.2 (cm) (1)Frequency = 8.0 [8.2 / 8.16] Hz or s1 only (1)4[10]22.Direction of travel of light is water air (1)Angle of incidence is greater than the critical angle (1)2[2]23.(a)(i)Name processRefraction (1)1(ii)Explanation of refraction taking placechange in speed / density / wavelength (1)1(b)(i)Draw ray from butterfly to fishrefraction shown (1)refraction correct (1)2(ii)Explain what is meant by critical angleIdentify the angle as that in the denser medium (1)Indicate that this is max angle for refraction OR total internalreflection occurs beyond this (1)[angles may be described in terms of relevant media]2(iii)Explain two paths for rays from fish A to fish Bdirect path because no change of medium/refractive index/density (1)(total internal) reflection along other path /angle of incidence > critical angle (1)direct ray correctly drawn with arrow (1)total internal reflection path correctly drawn with arrow (1)[lack of ruler not penalised directly] [arrow penalised once only]4[10]24.(a)Meaning of superpositionWhen vibrations/disturbances/waves from 2 or more sources coincideat same position (1)resultant displacement = sum of displacements due to individual waves (1)2(b)(i)Explanation of formation of standing wavedescription of combination of incident and reflected waves/waves in opposite directions (1)described as superposition or interference (1)where in phase, constructive interference / antinodesOR where antiphase, destructive interference / nodesOR causes points of constructive and destructive interferenceOR causes nodes and antinodes (1)3(ii)Calculate wavelengthIdentify 2 wavelengths (1)Correct answer [2.1 109 m] (1)2Example of calculation:(NANANANAN) X to Y is 2 = 4.2 109 m 2= 2.1 109 m(iii)Explain termsamplitude maximum displacement (from mean position)(can use diagram with labelled displacement axis) (1)antinode position of maximum amplitudeOR position where waves (always) in phase (1)2[9]25.Frequency unaltered (1)Wavelength decreases (1)Speed decreases (1)3[3]26.(a)(i)Use of speed = distance over time (1)Distance = 4 cm (1)Answer = [2.7 105 s] (1)Example of answert = 4 cm 1500 m s1t = 2.7 105 s3(ii)Use of f = 1/T (1)Answer = [5000 Hz] (1)2(iii)Time for pulse to return greater than pulse interval (1)All reflections need to reach transducer before next pulse sent. (1)Will result in an inaccurate image. (1) (Max 2)Need to decrease the frequency of the ultrasound. (1) (Max 3)Max 3(iv)X-rays damage cells/tissue/foetus/baby but ultrasound doesnot (need reference to both X-rays and ultrasound) (1)1 (b)The answer must be clear, use an appropriate style and be organisedin a logical sequence (QWC)Doppler shift is the change in frequency of a wave when the source or thereceiver is moving (1)Requirement for a continuous set of waves (1)Two transducers required (one to transmit and one to receive) (1)Change in frequency is directly related to the speed of the blood (1)4[13]27.(a)Transverse waves oscillate in any direction perpendicular to wave direction (1)Longitudinal waves oscillate in one direction only OR parallel to wavedirection. (1)Polarisation reduces wave intensity by limiting oscillations and wavedirection to only one plane OR limiting oscillations to one direction only. (1)(accept vibrations and answers in terms of an example such as a ropepassing through slits)3(b)Light source, 2 pieces of polaroid and detector e.g. eye, screen, LED ORlaser, 1 polaroid and detector (1)Rotate one polaroid (1)Intensity of light varies (1)3[6]28.Unpolarised and plane polarised light:Correct diagrams showing vibrations in one plane only and in all planes (1)

Vibrations/oscillations labelled on diagrams (1)2 Telescope adaptation:Fit polarising filter / lens [must be lens not lenses] (1)At 90 to polarisation direction to block the moonlight / rotate until2cuts out moonlight (1)[4]29.Meaning of plane polarisedOscillations/vibrations/field variations (1)Parallel to one direction, in one plane [allow line with arrow at both ends] (1)2Doppler effectDoppler (1)If source/observer have (relative) movement [reflections off vibrating/moving atoms] (1)Waves would be bunched/compressed/stretched or formula quoted [accept diagram] (1)Thus frequency / wavelength changes [accept red /blue shift] (1)4Frequency about 3 1014 HzEvidence of use of 1/wavelength = wavenumber (1)laser wavenumber = 9400 or wavelength change =7.69104 (1)New wavenumber = 10700 [or 8100] or conversion of wavelength change to m [7.69 106] (1)New wavelength = 935 nm [or 1240 nm] Use of frequency = c / wavelength [in any calculation] (1)f = 3.2 1014 Hz [note answer of 2.8 1014 = 3 , 3.4 1014 = 4](1)5Model of lightParticle/photon/quantum model (1)Photon energy must have changed / quote E = hf (1)Energy of atoms must have changed [credit vibrating less/more/faster/slower] (1)3[14]30.PolarisationThe (wave) oscillations (1)occur only in one plane (1)2[OR shown with a suitable diagram]How to measure angle of rotationAny four points from:Polaroid filter at one/both endswith no sugar solution, crossed Polaroids (top and bottom oftube) block out lightsugar solution introduced between Polaroidsone Polaroid rotated to give new dark viewdifference in angle between two positions read from scale (1) (1) (1) (1)Max 4 GraphPoints plotted correctly [1 for each incorrect; minimum mark 0] (1) (1)Good best fit line to enable concentration at 38 to be found (1)3Concentration0.57 ( 0.01) kg l11[10]31.Light from sky:Light is polarised (1)1Change in intensity:Filter allows through polarised light in one direction only (1)When polarised light from the sky is aligned with filter, light is let through (1)When polarised light is at right angles with polarising filter, less light passes (1)Turn filter so that polarised light from blue sky isnot allowed through, so sky is darker (1)Max 2 Clouds:Light from clouds must be unpolarised (1)1Radio waves:Radio waves can be polarised OR transverse (1)1 Why radio waves should behave in same way as light:Both are electromagnetic waves/transverse (1)[Transverse only, credited for 1 answer]1[6]32.Speed of ultrasoundUse of = s/t (1) = 150 103 (m) 132 106 (s)= 1140 m s1 (1)2 Change of traceExtra pulse(s)ORReflected pulse moves closer1Principle of Doppler probe3 points from: Arrange probe so that soup is approaching OR Soup reflects ultrasound /with changed frequency/wavelength OR change in frequency/wavelength depends on speed OR Probe detects frequency of reflected ultrasound OR Use of diagrams showing waves3 Determination of speed1 point from:Frequency/wavelength change /Angle between ultrasound direction and direction of flow of soup1CommentLumps give larger reflections/ Lumps travel slower1 [8]