1. D[1]

2. C[1]

3. B[1]

4. A[1]

5. B[1]

6. C[1]

7. B[1]

8. C[1]

9. D[1]

10. (a) D 1

(b) Wavelength

Use of v = f λ (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 = fλf = 330 / 0.66T = 1/f = 0.66 / 330T = 0.002 s 3

[4]

11. Direction of travel of light is water → air (1)Angle of incidence is greater than the critical angle (1) 2

[2]

12. (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]

13. Frequency unaltered (1)Wavelength decreases (1)Speed decreases (1) 3

[3]

14. (a) Use of sensorEvent happens very quickly OR cannot take readings fast enough (1)Sampling rate: 50+ samples per second (1) 2

(b) Initially the temperature is low so current is highResistance of filament increases as temperature increasesCurrent falls to steady value when temperature is constantMaximum heating is when lamp is switched on / when current is highestFilament breaks due to melting caused by temperature rise Max 4

[6]

15. The answer must be clear and the answer must be organised in a logicalsequence (QWC

• It was known that X penetrated (1)

• It was not known that X rays were harmful (1)

• Doctors died because of too much exposure (1)

• Lack of shielding (1)

• New treatments may have unknown side effects (1)

• Treatments need to be tested / time allowed for side effects to appear (1) Max 4[4]

16. (a) [1.0 m] (1) 1

(b) Ratio of (5 or 6 / 3 ) × 60 (1)Answer [f = 100 Hz] (1) 2

[3]

• New treatments may have unknown side effects (1)

• Treatments need to be tested / time allowed for side effects to appear (1) Max 4[4]

17. Use of sin i / sin r = µ (1)Use of either 80° or 1.33 (1)[r = 48°] (1) 3

Example of answersin 80 / sin r = 1.33[r = 48°]

Both rays refracted towards the normalViolet refracted more than red 2

[5]

18. (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 eV 2

(b) Use of E = hf and c = fλ (1)Conversion of eV to Joules (1)

Answer = [1.22 × 10–7 m] (1)

Example of answerE = hf and c = fλ E = hc/λλ = (6.63 × 10–34 J s × 3 × 108 m s–2) ÷ (10.2 eV × 1.6× 10–19 C)

λ = 1.21 × 10–7 m 3[5]

19. The answer must be clear, use an appropriate style and be organised in alogical sequence. (QWC)Reference to I = nqvA (1)

For the lampIncreased atomic vibrations reduce the movement of electrons (1)Resistance of lamp increases with temperature (1)

For the thermistorIncreased atomic vibrations again reduce movement of electrons (1)But increase in temperature leads to a large increase in n (1)Overall the resistance of the thermistor decreased with increase intemperature. (1) Max 5

[5]

20. (a) Diffraction is the change in direction of wave or shape orwavefront (1)when the wave passes an obstacle or gap (1) 2

(b) The energy of the wave is concentrated into a photon (1)One photon gives all its energy to one electron (1) 2

(c) Energy of photon increases as frequency increases OR reference toE = hf (1)Electrons require a certain amount of energy to break free and thiscorresponds to a minimum frequency (1) 2

[6]

21. (a) (i) Use of speed = distance over time (1)Distance = 4 cm (1)

Answer = [2.7 × 10–5 s] (1)

Example of answer

t = 4 cm ÷ 1500 m s–1

t = 2.7 × 10–5 s 3

(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]

22. (a) Voltmeter is across resistor should be across cell (1) 1

(b) (i) Plot of graphCheck any three points (award mark if these are correct) (3)Line of best fit 3

(ii) e.m.f. = [1.36 – 1.44 V] (1) 1

(iii) Attempt to find gradient (1)Answer [0.38 – 0.42 Ω ] (1) 2

(c) Intercept would twice value above (1) (accept numerical value 2× value(b)(ii))Gradient would be twice value above (1) (accept numerical value 2× value(b)(iii)) 2

[9]

23. (a) Diode or LED (1) 1

(b) (i) Use of R = V / I current between 75 and 90 ignoring powers of 10 (1)answer 6.7 – 8.0 Ω (1)

Example of answer

R = 0.60 V ÷ (85 × 10–3) AR = 7.06 Ω 2

(ii) Infinite OR very high OR ∞ 1

(c) ANY ONERectification / AC to DC / DC supply [not DC appliances]Preventing earth leakageStabilising power outputTo protect componentsA named use of LED if linked to LED as component in (a)(egcalculator display / torch)A voltage controlled switch(Allow current in only one direction) 1

[5]

24. (a) Resistivity definition

Resistivity = resistance × (1)× cross sectional area / length (1)

ρ= RA/l with symbols defined scores 2/2equation as above without symbols defined scores ½equation given as R = ρl/A with symbols defined scores 1/2

(1st mark is for linking resistivity to resistance with some other terms) 2

(b) (i) Resistance calculationConverts kW to W (1)Use of P = V2/R OR P = VI and V = IR (1)Resistance = 53 Ω (1)

Example of answer

R = (230 V)2 ÷ 1000 WR = 53 Ω 3

(ii) Length calculationRecall R = ρl/A (1)Correct substitution of values (1)Length = 6.3 m (accept 6.2 m) (1)ecf value of R

Example of answer

l = (52.9 Ω × 1.3 × 10–7 m2) ÷ (1.1× 10–6 Ω m)l = 6.3 m 3

(iii) Proportion methodIdentifies a smaller diameter is needed (1)Diameter = 0.29 mm (1)ORCalculation methodUse of formula with l = half their value in (b)(ii) (1)Diameter = 0.29 mm (1)(Ecf a wrong formula from part ii for full credit)

Example of answerdnew = 0.41 mm ÷ √2dnew = 0.29 mm 2

[10]

B ˆ25.(a) Definition of E.M.F.Energy (conversion) or work done (1)Per unit charge (1)[work done/coulomb 1/2, energy given to a charge 1/2, energygiven to a charge of a coulomb 2/2]OR ORE = W/Q (1) E = P/ISymbols defined (1) Symbolsdefined(E = 1 J/C scores 1) (E = 1 W/A scores 1)

((Terminal) potential difference when no current is drawn 1/2) 2

(b) (i) Internal resistance calculationAttempt to find current (1)Pd across r = 0.2 V (1)r = 0.36 (Ω) (1)[You must follow through the working, I have seen incorrectmethods getting 0.36 Ω]

Example of answerI = 2.8 V ÷ 5.0 Ωr = (3.0 – 2.8) V ÷ 0.56 A = 0.36 Ω 3

(ii) Combined resistanceUse of parallel resistor formula (1)Resistance = 3.3 Ω [accept 3 1/3 but not 10/3] (1) 2

(iii) Voltmeter reading(ecf bii)Current calculation using 3 V with either 3.3 Ω or 3.7 Ω (1)Total resistance = 3.7 Ω [accept 3.66 to 3.73 Ω]OR use of V = E – Ir (1)Voltmeter reading = 2.7 V (1)

ORPotential divider method, ratio of resistors with 3.7 Ω on bottom (1)Multiplied by 3.0 V (1)2.7 V (1)

Example of answerRtotal = 3.7 ΩI = 3 V ÷ 3.7 Ω = 0.81 AVvoltmeter = 3.3 Ω × 0.81 A = 2.7 V 3

(c) Ideal voltmeterIdeal voltmeter has infinite resistance OR extremely high resistanceOR highest possible R OR much larger resistance than that ofcomponent it is connected across OR quotes value > 1 M Ω (1)

Current through voltmeter is zero (negligible) OR doesn’t reduce theresistance of the circuit OR doesn’t reduce the p.d. it is meant tobe measuring. (1) 2

[12]

26. (a) Circuit diagramPotentiometer correctly connected i.e potential divider circuit (1)Ammeter in series and voltmeter in parallel with bulb (1)

(light bulb in series with resistance can score second mark only) 2

(b) (i) Graph+I, +V quadrant; curve through origin with decreasing gradient (1)

[do not give this mark if curve becomes flat and then starts goingdown i.e. it has a hook]

–I, –V quadrant reasonably accurate rotation of +I,+V quadrant (1) 2

(ii) Shape of graphAs current/voltage increases, temperature of the lamp increases /lamp heats up (1)Leading to increase in resistance of lamp (1)Rate of increase in current decreases OR equal increases in Vlead to smaller increases in I (1)Qowc (1)

Ecf if a straight line graph is drawn max 3R constant (1)V α I (1)Qowc (1) 4

[8]

27. (a) (i) Demonstrating the stationary wave

Move microphone between speaker and wall OR perpendicular to wallOR left to right OR towards the wall [could be shown by labelledarrow added to diagram] (1)

Oscilloscope/trace shows sequence of maxima and minima (1) 2

(ii) How nodes and antinodes are produced

Superposition/combination/interference/overlapping/crossingof emitted/incident/initial and reflected waves (1)

Antinodes: waves (always) in phase OR reference to coincidenceof two compressions/rarefactions/peaks/troughs /maxima/minima,hence constructive interference/reinforcement (1)

Nodes: waves (always) in antiphase/exactly out of phase ORcompressions coincide with rarefactions etc, hence destructiveinterference / cancellation (1) 3

(iii) Measuring the speed of sound

Measure separation between (adjacent) nodes / antinodes anddouble to get λ/this is ½λ [not between peaks and troughs] (1)

Frequency known from/produced by signal generator ORmeasured on CRO / by digital frequency meter (1)

Detail on measurement of wavelength OR frequencye.g. measure several [if a number is specified then ≥3] nodespacings and divide by the number [not one several times]OR measure several (≥3) periods on CRO and divide by the numberOR adjust cro so only one full wave on screen (1)

Use v (allow c) = fλ 4

(b) (i) Application to concert hall

Little or no sound /amplitudeOR you may be sat at a node (1)

(ii) Sensible reason

Examples:Reflected wave not as strong as incident waveOR walls are covered to reduce reflectionsOR waves arrive from elsewhere [reflections/different speakers]OR such positions depend on wavelength / frequency (1) 2

[11]

28. (a) Meaning of statement

(5.89 × 10–19 J / work function) is the energy needed to remove anelectron [allow electrons] from the (magnesium) surface/plate

Consequent markMinimum energy stated or indicated in some way [e.g. at least /or more] (1) 2

(b) (i) Calculation of time

Use 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 × 10–19 J)/(0.035 W m–2 × 8 × 10–20 m2) 3

(ii) How wave-particle duality explains immediate photoemission

QOWC (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]

29. (a) (i) Condition for reflection

Angle of incidence greater than critical angle [accept i > c] (1) 1

(ii) Description of path of ray

Any two from:• Ray refracted at A and C• Description of direction changes at A and C• Total internal reflection at B (1)(1) 2

(b) (i) Things wrong with the diagram

Angle of refraction can–t be 0 / refracted too much (1)

No refraction on emergence from prism (1)[Allow 1 mark for correct reference to partial reflection] 2

(ii) Corrected diagram• emergent ray roughly parallel to the rest of the emergent rays (1)• direction of refraction first surface correct (1)• direction of refraction second surface correct (1) 3

[8]

30. (a) Calculation of adaptor–s input

Recall of: power = IV (1)

Correct answer [0.01 A] (1)

Example of calculation:

power = IVI = P/V = 25 W / 230 V= 0.01 A 2

(b) (i) Explain why VA is a unit of power

Power = voltage × current so unit = volt × amp (1) 1

(ii) Calculation of efficiency of adaptor

Use of efficiency equation (1)

Correct answer [24%] (1)

Example of calculation:

efficiency = (0.6 VA / 2.5 W) × 100%= 24 % [0.24] 2

(iii) Reason for efficiency less than 100%

Resistance (accept explanations beyond spec, e.g. eddy currents) (1)

hence heat loss to surroundings (1) 2

(c) (i) Calculation of charge

Recall of: Q = It (1)

Correct answer [4000 C] (1)

Example of calculation:

Q = It= 0.2 A × 6 h= 0.2 A × (6 × 60 × 60) s= 4000 C (4320 C) 2

(ii) Calculation of work done

Recall of: W = QV OR Recall of W = Pt (1)

Correct substitution (1)

Correct answer [13 000 J] (1)

Example of calculation:

W = QVW = 4320 C × 3 V [ecf]= 13 000 J (12 960 J)ORW = PtW = 0.6 W × 6 hW = 0.6 W × (6 × 60 × 60) s= 13 000 J 3

[12]

31. (a) (i) Add standing waves to diagrams

Mark for each correct diagram (1)(1) 2

(ii) Mark place with largest amplitude of oscillation

antinode marked [allow clear indication near centre of wave otherthan an X, allow correct antinode shown on diagrams B or C] (1) 1

(iii) Name of place marked

(Displacement) Antinode [allow ecf from (a) (ii)] (1) 1

(b) (i) Calculation of wavelength

Correct answer [5.6 m]

Example of calculation:= 2 × 2.8 m= 5.6 m (1) 1

(ii) Calculation of frequency

Recall of v = f λ (1)

Correct answer [59 Hz] [ecf] (1)

Example of calculation:v = f λ

f = 330 m s–1 / 5.6 m= 58.9 Hz 2

(c) (i) Explanation of difference in sound

as the room has a standing wave for this frequency / wavelength /it is the fundamental frequency(allow relevant references to resonance) (1) 1

(ii) Suggest another frequency with explanation

Appropriate frequency [a multiple of 59 Hz] [ecf] (1)

Wavelength 1/2, 1/3 etc (stated or used) (1) 2

(d) Explain change in frequencies

wavelengths (of standing waves) bigger / f = v/2l (1)

hence frequencies smaller/lower (1) 2[12]

32. (a) Blue light:Wavelength / frequency / (photon) energy 1

(b) (i) Frequency:Conversion of either value of eV to JoulesUse of f = E / h

Correct frequency range [4.8 × 1014 – 8.2 × 1014 Hz or range =

3.4 × 1014 Hz][no penalty for rounding errors]

eg.2 eV = 2 × 1.6x 10–19 = 3.2 × 10–19 J

= 6.63 × 10–34 × f

f = 4.8 × 1014 Hz

3.4 eV = 3.4 × 1.6 × 10–19 = 5.4 × 10–19 J

f = 8.2 × 1014 Hz 3

(ii) Diagrams:Downward arrow from top to bottom levelOn larger energy gap diagram 2

(c) (i) Resistivity drop:Less heating / less energy lost / greater efficiency / lowervoltage needed / less power lost 1

(ii) Resistance:Recall of R = ρL/AUse of R = ρL/ACorrect answer [80(Ω)] [allow 80–84 (Ω) for rounding errors]

Eg.

R = (2 × 10–2 × 5.0 × 10–3) / (3.0 × 10–3 × 4.0 × 10–4)= 83 Ω 3

[10]

33. (a) Angles:Normal correctly added to raindrop (by eye)

An angle of incidence correctly labelled between normal andincident ray and an angle of refraction correctly labelledbetween normal and refracted ray 2

(b) Angle of refraction:Use of µ = sin i / sin rCorrect answer [20°][allow 20°–21° to allow for rounding errors]

eg.sin r = sin 27°/ 1.3r = 20° 2

(c) (i) Critical angle:

The angle beyond which total internal reflection (of the light)occurs [allow T.I.R] / r = 90° 1

(ii) Critical angle calculation:

Use of µ = 1 / sin CCorrect answer [50.3°] [allow 50° – 51°]

Eg.Sin C = 1/1.3C = 50.3° 2

(d) Diagram:i = 35° [allow 33° –37°]

Ray of light shown refracting away from normal on leavingraindropSome internal reflection of ray also shown with i = r [by eye]

Reflected ray shown refracting away from the normal as it leavesthe front of the raindrop / angle of refraction correctlycalculated at back surface 4

(e) Refractive index:(Red light has) lower refractive index (than violet light) 1

[12]

34. (a) n is (number of) charge carriers per unit volume or

number density or (number of) charge carriers m–3 orcharge carrier density(1)

[allow electrons]

v is drift speed or average velocity or drift velocity(of the charge carriers) (1)

[just speed or velocity scores zero] 2

(b) / A and Q A s or / Cs–1 and Q C (1)

n m–3 (1)

A m2 and v m s–1 (1)[If no equation written assume order is that of equation] 3

(c) (i) n 1 and Q Need all three 1

(i) Ratio vA/ vB less than 1 following sensible calculation (1)Ratio = ¼ // 0.25 // 1:4 (1)(ratio 4:1 scores 1)[4vA:1vB scores 1] 2

[8]

35. (a) Use of P = IV (1)Current in lamp A – 2 A (1)

[0.5 A scores zero unless 24 = I × 12 seen for 1st mark] 2

Example of answerI = P ÷ V = 24 W ÷ 12VI = 2A

(b) (i) Voltmeter reading = 12 V (1) 1

(ii) p.d. across R2 = 6 V or their (b)(i) minus 6V (1)Use of R = V/I (1) conditional on first markR2

Answer to this part must be consistent withvoltmeter reading and if voltmeter reading is wrongthis part has a max 2. If (b)(i) = 15 V then need to see

If (b)(i) = 6V or less they are going to score zero for this section. 3

(iii) current through R1 = 5 A (1) ecf answers from (a) 1

Example of answerCurrent through R1 = 2 A + 3 A = 5 A

(iv) p.d. across R1 = 3 V (1) ecf (15V minus their (b)(i)) 1

Example of answerp.d. across R1 = 15 V – 12 V = 3 V

(v) R1 1

Example of answerR1 = 3 V ÷ 5A = 0.6[accept fraction 3/5]

[9]

36. (a) (i) EI (1) 1

(ii) I2R (1) 1

(iii) I2r (1) 1

(b) EI = I2R + I2r or E = IR + Irecf Must use values (a)(i)-(iii) 1

(c) I for circuit given by Imax = E / r or substitution of5000V into the equation (1)(for safety) need I to be as small as possible (1) 3

[7]

37. (a) (i) How we know the speed is constant

Crest spacing constant / circular crestsOr wavelength constant / equal wavelength (1)

[Accept wavefront for crests][Don’t accept wave] 1

(ii) Calculation of speedλ is 10 mm (1)[Allow 9 to 11]Use of v = fλ (1)

0.40 m s–1 (1)

[Allow 0.36 to 0.44Allow last two marks for correct calculation from wrong wavelength] 3

(40Hz)(10 × 10–3 m)

= 0.40 m s–1

(b) Line X

1st constructive interference line below PQ, labelled X (1)

[Accept straight lineIgnore other lines provided correct one is clearly labelled X] 1

(c) (i) Superposition along PQConstructive interference / reinforcement / waves of largeramplitude / larger crests and troughs (1)Crests from S1 and S2 coincide / waves are in phase / zero phasedifference / zero path difference (1)Amplitude is the sum of the individual amplitudes (OR twice theamplitude of the separate waves) (1) 3

(ii) Table

A constructive (1)B destructive (1) 2

[10]

38. (a) Part of spectrumLight / Visible / red (1) 1

Calculation of work functionUse of ϕ = hc/λ (1)

3.06 × 10–19 [2 sig fig minimum] (1) 2

(6.63 × 10–34 J s)(3.00 × 108 m s–1)/(6.5 × 10–7 m)

= 3.06 × 10–19 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 – ϕ /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]

39. (a) (i) Calculate maximum current

Recall of P = IV (1)

Correct answer [0.49 A] (1)

Example of calculation:P = IVI = 5.9 W / 12.0 V= 0.49 A 2

(ii) Show that resistance is about 24 Ω

Recall of V = IR (1)

Correct answer to 3 s.f. [24.5 Ω] [no u.e.] (1)

Example of calculation:R = 12 V / 0.49 A= 24.5 Ω 2

(b) (i) Calculate current

Use of correct circuit resistance (1)

Correct answer [0.45 A] (1)

Example of calculation:I =V / R= 12 V ÷ (24.5 Ω + 2 Ω)= 0.45 A 2

(ii) Calculate power

Recall of P = IV and V = IR (accept P = I2R) (1)

or P = R

V 2

Correct answer [5.0 W] (1)

Example of calculation:

P = I2R

= (0.45 A)2 × 24.5 Ω= 5.0 W 2

(c) Increase in power available to pump

e.g. lower resistance in wire thicker wire, panel nearer to motor (1)(accept relevant answers relating to panels, e.g. more panels) 1

[9]

40. (a) (i) Name process

Refraction (1) 1

(ii) Explanation of refraction taking place

change in speed / density / wavelength (1) 1

(b) (i) Draw ray from butterfly to fish

refraction shown (1)

refraction correct (1) 2

(ii) Explain what is meant by critical angle

Identify 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 B

direct 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]

41. (a) Ultrasound:High frequency sound / sound above human hearing range / soundabove 20 kHz / sound too high for humans to hear (1) 1

(b) (i) Pulses used:to prevent interference between transmitted and reflected signals /allow time for reflection before next pulse transmitted / to allow forwave to travel to be determined (1)

(ii) High pulse rate:

Greater accuracy in detection of prey–s motion / position / continuousmonitoring / more frequent monitoring (1) 2

(c) Size of object:Use of λ = v/f (1)Correct answer (0.0049 m or 4.9 mm) (1)[accept 0.0048 m or 0.005 m]

example:

λ = 340 m s–1/ 70000 Hz= 0.0049 m = 4.9 mm (accept 5 mm) 2

(d) Time interval:Use of time = distance / speed (1)

Correct answer (2.9 × 10–3 s) [allow 3 × 10–3 s][allow 1 mark if answer is half the correct value ie. Distance = 0.5mused] (1)

example:

time = 1 m / 340 m s–1

= 2.9 × 10–3 s 2

(e) Effect on frequency:Frequency decreases (1)Greater effect the faster the moth moves / the faster themoth moves the smaller the frequency (1) 2

[9]

42. (a) Diffraction diagram:Waves spread out when passing through a gap / past an obstacle(1)λ stays constant (1) 2

(b) Diagrams:Diagram showing 2 waves in phase (1)Adding to give larger amplitude (1) 2

(c) Information from diffraction pattern:Atomic spacing (similar to λ)Regular / ordered structureSymmetrical structureDNA is a double helix structure (2) Max 2

(d) Electron behaviour:(Behave) as waves (1) 1

[7]

43. (a) (i) Diagram:i and r correctly labelled on diagram (1)i = 25 +/– 2° (1)r = 38 +/– 2° (1)[allow 1 mark if angles measured correctly from interfaceie. i = 65 +/– 2°, r = 52 +/– 2°] (1) 3

(ii) Refractive index:Use of gµa = sin i / sin r [allow ecf] (1)Use of aµg = 1/gµa (1)

example:gµa = sin 25 / sin 38 = 0.686

aµg = 1/ gµa = 1.46 2

(b) Ray diagram:Ray added to diagram showing light reflecting at interface withangles equal (by eye) (1) 1

(c) Observation:Incident angle > critical angle (1)T.I.R occurs (1) 2

(d) largest angle:sin C = 1/1.46 (allow ecf) (1)

C = sin–1 (1/1.46) = 43° (1) 2[10]

44. (i) J C –1 Potential difference (1)

(ii) Product of two quantitiesPotential difference (1)

(iii) Rate of changecurrent (1)

(iv) Base quantitycurrent (1)

(for any part if two answers are given score is zero) 4[4]

45. (a) (As temperature of thermistor increases) its resistancedecreases [Do not credit the converse] (1)any TWO(slight) decrease in v (symbol, velocity or drift velocity)Large increase in n increases [accept electrons/charge carriers for n]A, Q and (pd) remain constant (1)(1)[ignore any reference to v staying constant] 3(n constant, can’t score mark for 3,4)

(b) (i) ammeter reading decreases (1)voltmeter reading unaltered (1)

(ii) ammeter is used to indicate temperature (1)

(iii) Assumption: ammeter; ideal/ has zero/negligible resistance (1)(Reference to meters is zero mark) 4

[7]

46. (a) Tungsten filamentQowc (1)I is not (directly) proportional to VTemperature of filament increases/ filament heats up/gets hotter as current/pd increases[accept bulb or lamp but not wire]Links temperature increase to resistance increasestungsten filament does not obey Ohm’s law/not anOhmic conductor or resistor. (1)(1)(1)Any THREE 4

(b) (i) Reading current from graph 1.5 A (1)answer 5.3 Ω (1)(misread current → 0/2)

Example of answerV = IRR = 8.0 ÷ 1.5 = 5.3 Ω 2

(ii) Addition of two currents (1)OR use of R = V/I and resistors in parallel formula1.5 + 1.2 = 2.7 A (1)ecf candidates’ current from above[If you see 2.7 A give 2marks] 2

[8]

47. (a) (i) Use of P = V2 / R OR P = IV and V = IR (1)Total R = 4.5 Ω (1) 2

Example of answer

R = V2 ÷ P = 12 V × 12 V ÷ 32 WR = 4.5 Ω

(ii) Use of 1/R = 1/R1 + 1/ R2……………… OR ΣR = 1/5R (1)[OR find total current, divide that by 5 and use V = IR]Resistance of strip = 22.5 Ω (1)ecf candidates’ R. 2

[common error is to divide by 5 → 0.9 Ω scores 0/2 butecf to next part gives l = 0.033 m which will then score 3/3]

(b) R = ρl / A or correct rearrangement (1)Correct substitution (1)Length = 0.82 m (1)ecf candidates’ R 3

Example of answer

l = RA/ρ = (22.5 Ω × 4.0 × 10–8 m2) ÷ 1.1 × 10–6 Ω ml = 0.82 m

(c) See P = V2 / R OR P = IV leading to increase in currentor decrease in resistance (1)more strips in parallel / material of lower resistivity (1)[not greater conductivity] 2

[9]

48. (a) E.M.F. = work done / charge OR energy transferred / charge (1)

OR power / current

[There is only one mark here and this is consistent withspecification but it must not be Joules or coulombs] 1

(b) (i) Use of V = IR (1)I = 2.0 A (1) 2

Example of answerI = V / R = 8.0 V / 4.0 ΩI = 2.0 A

(ii) Uses p.d. = 4.0 V (1)r = 2.0 Ω ecf their I (1) 2

Example of answerr = V / I = 4.0 V / 2.0 Ar = 2.0 Ω

(iii) Use of P = VI // I2R // V2/R (1)P = 16 W ecf their I (1) 2

Example of answerP = VI = 8 V × 2 AP = 16 W

(iv) Uses 4V or 2A × 2Ω or their I × r (1)see 5 × 60 s in an energy equation (1)energy = 2400 J (1) 3

Example of answerE = VIt = 4 V × 2 A × 5 × 60 sE = 2400 J

[10]

49. (a) Experiment

[Marks may be earned on diagram or in text]

Named light source plus polaroid (OR polariser ORpolarising filter) / Laser / Named light source andsuitable reflector (e.g. bench) (1)

2nd Polaroid plus means to detect the transmitted light (1)(i.e. eye OR screen OR LDR OR light detector ORinstruction to e.g. look through polaroids)Rotate one Polaroid [Only award if expt would work] (1)Detected intensity varies / No light when polaroids are at 90° (1)Maxima and minima 90° apart / changes from dark to light every 90° (1)[Use of microwaves, slits or “blockers”: 0/5Use of filters or diffraction gratings: lose first two marksUse of “sunglasses” to observe: lose mark 2] 5

(b) Why sound can’t be polarised

They are longitudinal / They are not transverse / Only transversewaves can be polarised / Longitudinal waves cannot be polarised /Because the (*) is parallel to the (**) (1)

(*) = vibration OR displacement OR oscillation OR motion of particles

(**) = direction of travel OR direction of propagation OR motion ofthe wave OR direction of energy transfer 1

[6]

50. (a) (i) Table

λ f

2.4 (110)

1.2 220

0.8 330

All 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 nodes

String cannot move / no displacement / zero amplitude /no oscillation / phase change of π on reflection / two wavescancel out / two waves are exactly out of phase (1)(OR have phase difference of π OR half a cycle) /destructive interference 1

(b) Why waves with more nodes represent higher energies

More 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]

51. (a) Why statement correct

Blue photon has more energy than red photon (1)

Why statement incorrect

Blue 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 second

Additional 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 function

Energy to remove an electron from the surface (ORmetal OR substance) (1)[Don’t accept “from the atom”. Don’t accept “electrons”.]Minimum energy… / Least energy… / Energy to just…/ …without giving the electron any kinetic energy (1) 2

(ii) Calculation of threshold frequency

Use of φ = hf0 (1)

Correct answer [6.00 × 1014 Hz] (1)

e.g.

(3.98 × 10–19 J)/(6.63 × 10–34 J s) = 6.00 × 1014 Hz 2[7]

52. (a) Which transition

Use of (∆)E = hc/λ OR (∆)E = hf and f = c/λ (1)

Use of 1.6 × 10–19 (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 × 10–34 J s)(3.00 × 108 m s–1)/

(656 × 10–9 m)(1.6 × 10–19 J eV–1)= 1.9 eV 4

(b) Explanation of absorption line

QOWC (1)Light of this wavelength is absorbed by hydrogen (1)In the outer part of the Sun (OR Sun’s 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 receding

Wavelength increased (OR stretched) / red shift /frequency decreased 1

[9]

53. (a) Describe propagation of longitudinal waves

Particles oscillate / compressions/rarefactions produced (1)

oscillation/vibration/displacement parallel to direction of propagation (1) 2

(b) Calculation of wave speed

Recall of v = f λ (1)

Correct answer [7.2 km s–1] (1)

Example of calculation:

v = f λ

v = 9 Hz × 0.8 km

= 7.2 km s–1 [7200 m s–1] 2

(c) Determine if elephants can detect waves more quickly

Recall of v = s / t (1)

Correct answer for t in minutes or hours [about 6 minutes] orrelevant comment with 347 s or calculation of tidal wave speed

[0.35 km s–1] with comment [allow ecf] (1) 2

Example of calculation:

v = s / t

t = 2500 km ÷ 7.2 km s–1 OR v = 2500 km ÷ (2 × 60 × 60 s)

t = 347 s OR v = 0.35 km s–1

t = about 6 minutes (stated) / much less than hours / 2 h is 7200 s

OR 7.2 km s–1 >> 0.35 km s–1

[6]

54. (a) Meaning of superposition

When 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 wave

description 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 wavelength

Identify 2 wavelengths (1)

Correct answer [2.1 × 10–9 m] (1) 2

Example of calculation:

(NANANANAN) X to Y is 2 × λ

λ = 4.2 × 10–9 m ÷ 2

= 2.1 × 10–9 m

(iii) Explain terms

amplitude – 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]

55. (a) (i) Calculate resistance

Recall of R = V/I (1)

Correct answer [8.65 Ω] (1) 2

Example of calculation:

R = V/I

R = 2.68 V ÷ 0.31 A

= 8.65 Ω

(ii) Show that internal resistance is about 0.4 Ω

Recall of relevant formula [V = ε – Ir OR lost volts = (ε – V) (1) ORε = I(R + r)] including emf

Correct answer [0.39 Ω] [no ue] [allow ecf if ε = I(R + r)] (1) 2

Example of calculation:

V = ε – Ir

r = (ε – V)/I

= (2.8 V – 2.68 V)/0.31 A

= 0.39 Ω

(iii) Comment on match to maximum power

Not matched [ecf for R in (a) (i) and r in (a)(ii)] (1)

Max power when internal resistance = load resistance (1) 2

(b) (i) Show that charge is about 14 000 C

Recall of Q = It (1)

Correct answer [14 400 C] [no ue] (1) 2

Example of calculation:

Q = It

= 2 × 2 A × 60 × 60 s

= 14 400 C

(ii) Calculate time for which battery maintains current

Use of Q = It OR use of W = Pt (1)

Correct answer [46 450 s or 12.9 h] (1) 2

Example of calculation:

t = Q/I

= 14 400 C / 0.31 A

= 46 450 s

(c) Explain effect on efficiency

Efficiency = I2R / I2(r + R) / Efficiency depends on R /(r + R) /more heat dissipated in cells / Efficiency is V/ and V decreases (1)

so efficiency is less (1)

[Must attempt explanation to get 2nd mark] 2[12]

56. (a) Work function:Energy needed for an electron to escape the surface /to be released (from the metal) (1) 1

(b) How current produced:Any 3 from:Photon of light passes energy to an electronIf energy above the work function/frequency above threshold (1)(1)Electron released as a photoelectron / photoelectron released /surface electron released (1)Moving electrons produce a current 3

(c) (i) Intensity of light increased:More electrons released (1)

(ii) Frequency of light increased:Electrons gain more (kinetic) energy (1) 2

(d) Photon energy:Use of f = v/λ or E = hc/λ (1)

Correct answer for E (4.7 × 10–19 J or 2.96 eV) (1)[allow 3.0 eV] 2

Example:

f = v/λ = 3 × 108 / 4.2 × 10–7 = 7.1 × 1014 Hz

E = hf = 4.7 × 10–19 J or 2.96 eVOR

E = hc/λ = 3 × 108 × 6.63 × 10–34/ 4.2 × 10–7

= 4.7 × 10–19 J or 2.96 eV

(e) Max kinetic energy:Knowledge that kemax = energy calculated in (d) – φ (1)

Correct answer for kemax (0.26 eV or 4.2 × 10–20 J)

[allow 0.25–0.26 eV or 4.1 – 4.2 × 10–20 J and allow ecf from (d)] (1) 2

Example:kemax = 2.96 eV – 2.7 eV= 0.26 eV

(f) (i) Why current reduced:Many / some electrons will not have enough (kinetic) energyto reach the anode / only electrons with large (kinetic) energywill reach the anode (1) 1

(ii) Stopping potential:eV = (–) keV = ke / e = 0.26V (1) 1

[12]

57. (a) Plane polarised:Vibrations / oscillations (1)in one plane (1)ORdouble-headed arrow diagram (1)with vibrations / oscillations labelled (1) 2

(b) Polarising filter:• Intensity goes from maximum to minimum (1)• Twice per rotation / after 90° (1)• As filter only lets through vibrations in a particular plane (1) 3[marks may be gained from a clearly labelled diagram]

(c) Response of beetle:Changed direction by 90° / turned through a right-angle (1) 1

(d) No moon:Beetle moves in a random direction / in circles / appears disorientated (1) 1

[7]

58. (a) Circuit:Potential divider (1) 1

(b) Relay potential difference:4 V (1) 1

Example:5/15 × 12 = 4V

(c) (i) Resistance:Recall of R = ρL/A (1)Correct substitution of values into formula (1)Correct answer [98(Ω)] (1)[allow 97 – 98Ω to allow for rounding errors] [no u.e.] 3

Example:

R = (3.4 × 102 × 1.44) / (100 × 0.05)= 98 Ω

(ii) Combined resistance:Use of 1/RTot = 1/R1 + 1/R2 (1)Correct answer for R [4.8Ω] (1)[allow 4.7Ω – 4.8Ω to allow for rounding errors] 2

1/R = 1/98 + 1/5 (or = 1/100 + 1/5)R = 4.8 Ω

(iii) Relay voltage:P.d. across relay with ballast very similar to p.d acrossthe relay alone / p.d. = 3.9 V / p.d. lower (slightly) (1) 1

(iv) Train on track:Relay voltage becomes very small / zero (1) 1

(v) Wet ballast:

Any two–• Combined resistance now small / RT = 0.45 Ω• Relay voltage now small / V = 0.52 V• Relay voltage too small to trigger green light /

signal remains red (1)(1) 2[11]

59. Tungsten filament bulb

(a) Resistance

Use of P = V2/R or P = VI with V = IR (1)answer 960 Ω (1) 2

Example of answerR = (240 V × 240 V) ÷ 60 WR = 960 Ω

(b) Drift speedrearrangement of I = nAvQ (1)

Use of Q = 1.6 × 10–19 (C) (1)

answer 0.15/0.148 m s–1 (1) 3

Example of answer

v = 0.25 A ÷ (3.4 × 1028 m–3 × 1.6 × 10– 19 C × 3.1 × 10–10 m2)

(c) ExplanationQowc (1)

Any THREE• Resistance due to collisions between electrons & ions/atoms/particles• (as T increases) ions/atoms/particles have more energy• (as T increases) ions/atoms/particles vibrate through larger

amplitude /vibrate faster OR amplitude if lattice vibrationincreases.

• more chance/increased frequency of collision/interactionOR impedes the flow of electrons (1)(1)(1) 4

[9]

60. Emf and Internal resistance

(a) DerivationE = I (R + r) OR E = IR + Ir (1) 1

(b) (i) Correct working (allow even if evidence of working backwards) (1)

Example of answerE/I = R + rRearranging R = E/I – r

(ii) EmfAttempt to use gradient (1)answer 1.5 V (bald answer 1.5 V scores 0/2) (1) 2

(iii) PowerFrom graph find value of 1/I when R = 5 Ω (1)

Use of P = I2R (1)answer 0.31 (W) (1) 3

Example of answer

1/I = 4 A–1 → I = 0.25 AP = 0.25 A × 0.25 A × 5 Ω = 0.3125 W

(c) GraphIntercept at –2 (ohms) (1)Graph steeper than original (1)Gradient is 3.0 V i.e. line passes through [10, 27-29] [no ecf] (1) 3

[10]

61. Potential divider

(a) First circuitMiddle terminal MOuter terminals L and K (any order) (1) 1

(b) (i) P.d across lamp.External resistance in circuit is 25 or (20+5) ohms (1)See ratio of resistances (denominator larger) × 6.0V (1)OR current = 6/25 Aanswer 4.8 V (1) 3

(ii) AssumptionThe resistance of the ammeter is zero/negligible. (1) 1

(c) Second circuitSee 2 resistors in parallel with supply (1)Supply across ends of variable resistor (10 Ω) (1)Fixed resistor across one end and slider (consequent mark) (1) 3

[8]

62. (a) Why transverse waves can be polarised but not longitudinal waves[Marks can be earned in diagram or text]Transverse waves have * perpendicular to direction of ** (1)

* = vibration/displacement/oscillation/motion of particles

** = travel/propagation/motion of wave/energy transfer/wave

In a transverse wave, * can be in different planes but polarisationrestricts it to one plane (1)Longitudinal waves have * parallel to ** (1) 3

[Don’t accept “motion” for **Diagrams to earn marks must be clearly labelled, but don’t insiston a label “looking along direction of travel” in the usual diagramsto illustrate polarised and unpolarised waves]

(b) (i) Effect of Polaroid on intensityIntensity is reduced (OR halved) [not zero] (1)[Accept slightly reduced and greatly reduced]Polaroid stops (OR absorbs) vibrations (OR waves OR light) inone plane/ (1)Polaroid only lets through vibrations (OR waves OR light)in oneplane/Light has been polarised 2

(ii) Effect of rotating PolaroidNo effect (1)[ignore incorrect reasons accompanying statements of effect] 1

[6]

63. (a) (i) How the bow causes the wave patternEITHERBow alternately pulls and releases string (or sticks and slips) (1)Creates travelling wave (OR travelling vibration ) (on string) (1)Wave reflects at the end (OR bounces back) (1)Incident and reflected waves (OR waves travelling in opposite (1)directions) superpose (OR interfere OR combine)[Don’t accept collide] max 3

ORBow alternately pulls and releases string (or sticks and slips) (1)Produces forced oscillation/acts as a driver/exerts periodic force (1)[Don’t accept makes it vibrate] At a natural frequency of the string (1)Causing resonance (OR large amplitude oscillation) (1) max 3

(ii) Determination of wavelengthUse of node to node distance = λ/2 / recognise diagram shows 2λ] (1)Correct answer [0.4 m] (1) 2

e.g. λ = 2 × 0.2 m = 0.4 m

(iii) Differences between string wave and sound wave

Any TWO points from:– String wave is transverse, sound wave is longitudinal /…can be polarised, … can’t– String wave is stationary (OR standing), sound wave is travelling(OR progressive) / … has nodes and antinodes, …doesn’t /…doesn’t transmit energy, …does…– The waves have different wavelengths– Sound wave is a vibration of the air, not the string (1)(1) 2

[Don’t accept travel in different directions / can be seen, can’t beseen / can’t be heard, can be heard / travel at different speedsThe first two marking points require statements about both waves,e.g. not just “sound waves are longitudinal”]

(b) Sketch of the waveformSinusoidal wave with T = 1 ms (1)[Zero crossings correct to within half a small squareAccept a single cycle]Amplitude 1.6 cm (1)[Correct to within half a small square] 2

[9]

64. (a) Conditions for observable interferenceAny THREE of:• Same type of wave / must overlap (OR superpose) / amplitude

large enough to detect / fringes sufficiently far apart todistinguish [Only one of these points should be credited]

• (Approximately) same amplitude (OR intensity)• Same frequency (OR wavelength)• Constant phase difference (OR coherent OR must come from

the same source) (1)(1)(1) 3

[Accept two or more points appearing on the same line

Don’t accept– must be in phase– must be monochromatic– must have same speed– no other waves present– must have similar frequencies– answers specific to a particular experimental situation, e.g.

comments on slit width or separation]

(b) (i) Experiment description[Marks may be scored on diagram or in text](Microwave) transmitter, 2 slit barrier and receiver (1)[Inclusion of a screen loses this mark, but ignore a single slit in frontof the transmitter]Barrier, metal sheets (1)[Labels indicating confusion with the light experiment, e.g. slitseparations or widths marked as less than 1 mm, lose this mark]Appropriate movement of receiver relevant to diagram [i.e. move inplane perpendicular to slits along a line parallel to the plane of theslits, or round an arc centred between them] (1) 3

(ii) Finding the wavelengthLocate position P of identified maximum/minimum 1st/2nd/3rd etc. (1)away from centreMeasure distance from each slit to P (1)Difference = λ OR λ/2 (consistent with point 1) (1) 3

[Accept use of other maxima and corresponding multiple of λ][9]

65. (a) [Treat parts (i) and (ii) together. Look for any FIVE of thefollowing points. Each point may appear and be credited in eitherpart (i) or part (ii)]

(i) • Light (OR radiation OR photons) releases electrons from cathode

• Photon energy is greater than work function / frequency oflight > threshold frequency / flight > fo / wavelength of light isshorter than threshold wavelength / λ < λ0

• PD slows down the electrons (OR opposes their motion ORcreates a potential barrier OR means they need energy to crossthe gap)

• Electrons have a range of energies / With the PD, fewer (ORnot all) have enough (kinetic) energy (OR are fast enough) tocross gap

• Fewer electrons reach anode / cross the gap

(ii) • (At or above Vs) no electrons reach the anode / cross the gap

• Electrons have a maximum kinetic energy / no electrons haveenough energy (OR are fast enough) to cross

ANY FIVE (1)(1)(1)(1)(1)

[Don’t worry about whether the candidate is describing the effect ofincreasing the reverse p.d. (as the question actually asks), or simplythe effect of having a reverse p.d.] 5

(b) Effects on the stopping potential

(i) No change (1)

(ii) Increases (1) 2

[Ignore incorrect reasons accompanying correct statements of the effect][7]

66. (a) Circuit diagram and explanation

ammeter and voltmeter shown in series and parallel respectively (1)

current measured with ammeter and voltage / p.d. with voltmeter (1) 2

(b) Calculation of resistance

Recall of R = V/I (1)

Correct answer [25.0 Ω] (1) 2

Example of calculation:

R = V/I

R = 3.00 V ÷ 0.12 A

= 25.0 Ω

(c) Calculation of resistance

Recall of P = V2/R (1)

Correct answer [29.4 Ω] (1) 2

Example of calculation:

P = V2/R

R = (230 V)2 ÷ 1800 W

R = 29.4 Ω

[Accept calculation of I = 7.8 A (1), calculation of R = 29.4 Ω (1)]

(d) Explanation of difference in values of resistance

At higher voltage value element is at a higher temperature (1)

(resistance higher because) increased lattice ion vibrations impedecharge flow (more) (1) 2

[8]

67. (a) Explain how vapour emits light

electrons excited to higher energy levels (1)

as they fall they emit photons/electromagnetic radiation/waves/energy (1) 2

(b) (i) Meaning of spectral line

(when the light is split up) each frequency/wavelength/photon energy isseen as a separate/discrete line (of a different colour) (1) 1

(ii) Calculation of frequency

Recall of v = f λ (1)

Correct answer [f = 5.1 × 1014 Hz] (1) 2

Example of calculation:

v = f λ

3.0 × 108 m s–1 = f × 589 × 10–9 m

f = 5.1 × 1014 Hz

(c) Explanation of different colours

different colours = different freq/wavelengths / photons of differentenergies (1)

photon energy/frequency/wavelength depends on difference betweenenergy levels (1)

diff atoms have diff energy levels/diff differences in levels (1) 3

(d) Explanation of transverse waves

variation in E or B-field /oscillations/vibrations/displacementat right angles/perpendicular to direction of travel/propagation

[not just motion or movement for both 1st and 3rd part] (1) 1[9]

68. (a) Explanation of maximum or minimum

path difference = 2 × 125 × 10–9 m = 250 × 10–9 m (1)

= half wavelength /antiphase (1)

→ destructive interference / superposition (1) 3

(→ minimum intensity)

(b) Meaning of coherent

remains in phase / constant phase relationship (1) 1[4]

69. (a) (i) Energy level diagram:• Arrow showing electron moving from lower level to a

higher level (1)• Arrow downwards from higher to lower level [must

show smaller energy change than upward arrow] (1) 2

(ii) Missing energy:Causes a rise in temperature of a named item (1) 1

(iii) Range of energies:

Minimum energy when λ = 400 × 10–9 m (1)Use of f = c/λ (1)Use of E = hf (1)Correct answer [3.1 eV] (1)[allow 3.0 – 3.3 eV for rounding errors] [no u.e] 4

eg. f = 3 × 108 / 400 ×10–9

= 7.5 × 1014 Hz

E = hf = 5.0 × 10–19 JE = 3.1 eV

(b) Detecting forgeries:Forgery would glow / old painting would not glow (1) 1

[8]

70. (a) (i) Critical angle calculation:Use of sin C = 1/dµa (1)Correct answer [24.4° – only acceptable answer] [no u.e] (1) 2

eg. Sin C = 1/dµa = 1/2.42C = 24.4°

(ii) Ray diagram:Small angle ray shown passing into air, away from the normal (1)Large angle ray showing T.I.R. with angles equal [by eye] (1) 2

(iii) Labelling of angles:An incident angle correctly labelled between normal and rayin diamond (1)An angle of refraction correctly labelled between normaland ray in air (1) 2

(iv) Amount of trapped light:Any 3 of the following:• The higher the refractive index the greater the amount of

trapped light• The higher the refractive index the lower the critical angle• T.I.R occurs at angles greater than the critical angle• So, if critical angle is smaller, more light is reflected (1)(1)(1) Max 3

(b) Comment on angle:Lower critical angle so more sparkle (1) 1

[10]

71. (a) Graph scale:Log scale (1) 1

(b) (i) Choice of material:Any 2 of the following:• (almost) all of the voltage is dropped across the carbon rod• gives the greatest speed variation• others need to be longer (to have same resistance as carbon)• others need to be thinner (to have same resistance as

carbon) (1)(1) Max 2

(ii) Resistance calculation:Use of R = ρL/A (1)Correct units used for all terms [all in cm or all in m] (1)Correct answer [1.9 Ω] (1) 3[allow 1.8 Ω for rounding errors – no u.e]

eg. R = 1.4 × 10–5 × 0.4 / 3.0 × 10–6

= 1.9 Ω

(iii) Available voltage:X – 12 V Y – 0 V (1) 1

(iv) Effect of connecting wires:Less voltage available for train set as some wasted across wires (1)0.5 Ω is (relatively) large % of total resistance, so effect ishigh / not negligible (1)

or

Calculation of potential difference available now (1)[9.5 V] [allow 9.5 – 9.6 V]Significant drop from 12 V (1)Vxy = (Rxy / RTotal) × Vsupply = (1.9/ (1.9 + 0.5)) × 12 = 9.5 V 2

[9]

72. (a) (i) Potential difference = work (done)/(unit) chargeOR Potential difference = Power/current (1) 1

(ii) J = kg m 2 s –2 (1)

C = A s or W = J s1 (1)

V = kg m2 A–1 s–3 (1) 3

(b) Converts 2 minutes to 120 seconds (1)Multiplication of VI∆ t or V∆ Q (1)Energy = 1440 J (1) 3

Example of answer:Energy = 6.0 V × 2.0 A × 120 s

= 1440 J[7]

73. (a) n = number of charge carriers per unit volume OR

n = number of charge carriers m–3 ORn = charge carrier density (1)

v = drift speed/average velocity/drift velocity (of the charge carriers) (1) 2

(b) n is greater in conductors / n less in insulators. (1)[There must be some comparison]larger current flows in a conductor. Dependant on havingreferred to n (1) 2(statement that n large in conductor and so current large max1)

(c) (In series), so same current and same n and Q (1)vB greater vA (1)vA/vB = ¼ // 0.25 (1) 3

[7]

74. (a) pd = 3.6 V (1) 1

Example of answer;p.d. = 0.24 A × 15 Ω = 3.6 V

(b) Calculation of pd across the resistor (1)[6.0 – 3.6 = 2.4 V]Recall V = IR (1)I1 calculated from their pd / 4Ω (1)

[correct answer is 0.60 A. Common ecf is 6V/4Ω gives 1.5 A] 3

Example of answer:I1 = 2.4 V / 4.0 Ω = 0.6 A

(c) Calculation of I2 from I1 – 0.24 [0.36 A] (1)[allow ecf of their I1. common value = 1.26 A]Substitution V = 3.6 V (1)R = 10 Ω (1) 3

[7]

75. (a) (i) (– gradient =) r = 1.95 – 2 Ω (1)E = 8.9 – 9 V (1) 2

(ii) I = 2.15 – 2.17 A (1) 1

(iii) Use of V = IR (1)R = 2.1 – 2.2 Ω (1) 2

(b) (i) Battery or cell with one or more resistive component (1)Correct placement of voltmeter and ammeter (1) 2

(ii) Vary R e.g. variable resistor, lamps in parallel (1)Record valid readings of current and pd (consequent mark) (1) 2

[Do not give these marks if the candidate varies the voltage as well][9]

76. (a) Solar Power

Use of P = Iπr2 [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) 3

e.g. P=1.1 × 103 W m–2 × cos 40 × π(29 × 10–3 m)2

= 2.2 W

(b) EnergyUse of E = Pt (1)

1.8 × 104 J/2.0 × 104J (1) 2

e.g. E = 2.2W × (2.5 × 3600 s)

= 2.0 × 104 J[5]

77. (a) GraphStraight line with positive gradient (1)Starting the straight line on a labelled positive fo (1)[Curved graphs get 0/2. Straight line below axis loses mark 2unless that bit is clearly a construction line.] 2

(b) Work functionFrom the y intercept (1)[Accept if shown on graph]OR Given by gradient × f0 (or h × f0) [Provided that f0 is markedon their graph, or they say how to get it from the graph]OR Read f and Ek off graph and substitute into Ek = hf – φ[Curved graph can get this mark only by use of hf0 or equation methods.] 1

(c) GradientGradient equals Planck constant (1) 1

[Curved graph can’t get this mark][4]

78. (a) WavelengtheV to J (1)Use of ∆ E = hf (1)Use of c = fλ (1)

1.8 × 10–11 [2 sf minimum. No ue] (1) 4

e.g. f =

(–1.8 keV – (– 69.6 keV)) × (103 × 1.6 × 10–19 J keV–1) / 6.6 × 10–34 J s

= 1.64 × 1019 Hz

λ = 3.00 × 108 m s–1/1.64 × 1019Hz

= 1.8 × 10–11 m

(b) TypeX rays [Accept gamma rays] (1) 1

[5]

79. Meaning of energy level

Specific allowed energy/energies (of electron in an atom)(1) 1

Meaning of photon

Quantum/packet/particle of energy/radiation/light/electromagnetic wave (1) 1

Formula for photon energy

E2 – E1 (1) 1[Allow E1 + Ephoton = E2]

Explanation of photon wavelengths

Same energy change / same energy difference / energy the same (1) 1

Meaning of coherent

Remains in phase / constant phase relationship(1) 1

80. Description of sound

Particles/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 frequency

Number of oscillations/cycles/waves per second / per unit time (1) 1

Calculation of wavelength

Recall v = fλ (1)

Correct answer [18 m] (1) 2

Example of calculation

v = fλ

λ = 330 m s–1 ÷ 18 Hz

= 18.3 m[6]

81. Explanation of standing waves

Waves reflected (at the end) (1)

Superposition/interference of waves travelling in opposite directions (1)

Where in phase, constructive interference/superpositionOR where antiphase, destructive interference/superpositionOR causes points of constructive and destructiveinterference/superposition [Do not penalise here if node/antinode mixed up] (1) 3

Mark node and antinode

Both marked correctly on diagram (1) 1

Label wavelength

Wavelength shown and labelled correctly on diagram (1) 1

Explain appearance of string

Any two from:

• light flashes twice during each oscillation / strobefrequency twice that of string [accept light or strobe]

• string seen twice during a cycle

• idea of persistence of vision (2) max 2

Calculate speed of waves

Use of v = √T/µ (1)

Correct answer [57 m s–1] (1) 2

Example of calculation:

v = √T/µ

= √(1.96 N / 6.0 × 10–4 kg m–1)

= 57.2 m s–1

[9]

82. Distance to aircraft:

Use of distance = speed × time(1)

Correct answer [7.2(km) / 7200(m) is the only acceptable answer. No u.e.] (1) 2

e.g.

Distance = speed × time = 3 × 108 × 24 × 10–6

= 7.2 km

Why pulses are used:

Any two of the following:

• Allow time for pulse to return before next pulse sent

• To prevent interference/superposition

• A continuous signal cannot be used for timing

• Can’t transmit / receive at the same time (2) max 2

Doppler shift:

Any three of the following

• Change in frequency/wavelength of the signal [allow specified change,either increase or decrease]

• Caused by (relative) movement between source and observer[accept movement of aircraft/observer]

• Size of change relates to the (relative) speed of the aircraft[Allow frequency increasing; do not allow frequency decreasingunless linked to aircraft moving away]

• Quote v/c = ∆f/f (3) max 3[7]

83. Area of wire:

Use of A = πr2 (1)

Correct answer [1.9 × 10–7 (m2). Allow 1.9 × 10–7 and 2.0 × 10–7 (m2)](1) 2[No u.e.]

e.g.

A = πr2 = π ×(2.5 × 10–4)2

= 1.96 × 10–7 m2

Table + graph:

Length / Area / × 106 m–1

0.0

0.5

1.0

1.5

2.0

2.5

3.1

3.6 (1)

4.0 – 4.1

First 2 points plotted correctly to within 1 mm (1)Rest of points in straight line with origin by eye (1) 3

Resistivity calculation:

Drawn through the origin, ignoring first 2 points (1)Recall ρ= R /(L/A) [in any form] (1)Large triangle drawn on graph OR accept the use of a pair of values (1)read from the line

[ x> 3 × 10-6 m–1)is required in both cases][x-axis allowed as bottom of triangle]

Correct answer [1.2 × 10-7 Ω m)] (1)

[allow 1.1 – 1.3 × 10-7(Ω m)] [no u.e.] 4

e.g.

0.4 / 3.4 × 106 = 1.2 × 10–7 Ω m

Anomalous results:Any two of the following:

• Higher current/lower resistance for shorter lengths/at these points

• At shorter lengths/at these points wire gets hotter

• Non-uniform area/diameter

• Cable / contact resistance

• Sensitivity of meters

• Effect on resistance of any of the above (2) max 2[11]

84. 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 until 2cuts out moonlight (1)

[4]

85. Meaning of plane polarised

Oscillations/vibrations/field variations (1)

Parallel to one direction, in one plane [allow line with arrow at both ends] (1) 2

Doppler effect

Doppler (1)

If source/observer have (relative) movement [reflections offvibrating/moving atoms] (1)

Waves would be bunched/compressed/stretched or formula quoted[accept diagram] (1)

Thus frequency / wavelength changes [accept red /blue shift] (1) 4

Frequency about 3 × 10 14 Hz

Evidence of use of 1/wavelength = wavenumber (1)

laser wavenumber = 9400 or wavelength change =7.69×10–4 (1)

New wavenumber = 10700 [or 8100] or conversion of wavelength

change to m [7.69 × 10–6] (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) 5

Model of light

Particle/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]

86. (a) (i) Lamp brightness

Lamp A (1)

Larger current through it (at 9.0 V)/greater power (1) 2(at 9.0 V)/smaller resistance (at 9.0 V)

(ii) Battery current

Addition of currents (1)

Current = 1.88 – 1.92 A (1) 2

(iii) Total resistance

R = 9 V/1.9 A or use of parallel formula (1)

R = 4.6 – 4.9 Ω (1) 2[full ecf for their current]

(b) Lamps in seriesCurrent same in both lamps/current in A reduced from original value (1)Pd across A less than pd across B (1)Lamp A has a lower resistance than lamp B (1)

P = VI or P = RI2 (1) Any 2

Lamp A will be dimmer than B [conditional on scoring ONE of (1) 1the above marks]

[9]

87. (a) (i) Resistance

Use of V/I [ignore 10x] (1)

3800 Ω (3784 Ω ) (1) 2

(ii) Resistance of thermistor

Use THTH

R

RR

VV

= OR 9V/.74mA – R OR (1)

6.2 V = 0.74 mA × RTH

8400 Ω [8378 Ω ] [substituting 4000 Ω gives 8857 Ω ie 8900 Ω ] (1)[method 2 substituting 3800 Ω gives 8362 Ω : substituting 4000 Ωgives8162 Ω ] 2

(b) Suggestion and Explanation

The milliammeter reading increases (1)

Thermistor resistance ‘becomes zero’ /Short circuit (1)

Since supply voltage is constant / I = 9.0 V/R (1)ORCircuit resistance reduced 3

[7]

88. (a) Definition of E.M.F.

Energy (conversion) or work done (1)

Per unit charge (1)

ORE = W/Q (1)Symbols defined (1)[E = 1J/C scores 1]

ORE = P/I (1)Symbols defined (1)

[terminal pd when no current drawn or open circuit scores max 1] 2

(b) Voltmeter calculation

Any attempt to find any current (1)

Attempt to calculate pd across 10 Ω resistor (1)

5.77 V 2

ORPotential divider method; ratio of resistors with 10.4 Ω on the bottom (1)

Multiplied by 6.0 V (1)

5.77 V (1) 3[For either method, an answer of 0.23 V scores max 1]

(c) Second battery added

Voltmeter reading increased (1)

Any two of:

EMF unchanged

Total resistance reduced

current increases or “lost volts” decreases (2) 3[8]

89. Frequency

(a) (i) 1.0(3) × 1010 Hz (1) 1

Electromagnetic Spectrum

(ii) IR, microwave & radio in correct order above visible (1)UV with either X rays / Gamma rays / both in correct order belowvisible (1)

(iii) Wavelength at boundary 1 × 10–8 m / 1 × 10–9 m (1) 3

Plane polarised

(b) (i) Vibrations/oscillations (of electric field/vector) (1)In one direction/plane (of oscillation) (1) 2

Description

(ii) Diagram showing generator labelled transmitter/generator/source/emitter (1)And suitable detector eg shows how signal is observed by using (1)(micro)ammeter/cro/loudspeaker/computer with interface[Ignore anything drawn between generator and detector but for eachmark do not give credit if a grille etc is attached]

To detect max and min (1)(Rotate through) 90° between max and min (1) 4

[10]

90. (a) Explanation

QOWC (1)

UV/red photon (1) 2

EUV > ER / fuv (1)EUV > Φ / fuv > fTH (so electron can break free) (1)One photon absorbed by one electron (1)Both metal plate and electron are negative or repel (each other) (1) max 2

(b) (i) Intensity red light increased

nothing / no discharge (1)

(ii) Intensity of UV increased

(Coulombmeter) discharges quicker (1) 2

(c) Max KE

Use of E = hc/λ (1)

conversion of eV to J or vice versa i.e. appropriate use of 1.6 × 10–19 (1)Subtraction hc/λ – Φ [must use same units] or use of full equation (1)

max KE = 2.2 × 10–19J (1) 4

[Candidates may convert photon energy to eV leading to max KE = 1.4 eV][10]

91. Explanation increase of resistance with temperature

Temperature increase leads to increased lattice vibrations (1)

scattering flowing electrons / increased collisions of electrons. (1) 2

Calculation of resistance at 200 °C

R = V/I [stated or implied] (1)

= 7.4 V ÷ 0.19 A

= 39 Ω (1) 2

Discuss whether results support hypothesis

No. Resistance is not increasing with temperature. (1) 1

Calculation of mains voltage

P = V2 ÷ R (1)

V2 = PR

= 1200 W × 41 Ω [Mark for rearrangement OR substitution] (1)

[Accept 39 – 41 Ω ] [ecf]

V = 220 V (1)

[Allow P = I2R (1), 3

calculate I = 5.4 A and use in 1200 W = 5.4 A × V (1),V= 220 V (1)][8]

92. Explanation of ‘excited’

Electrons/atoms gain energy (1)

and electrons move to higher (energy) levels (1) 2

[Credit may be gained for diagrams in this and the next 3 parts]

Explanation of how radiation emitted by mercury atoms

Electrons (lose energy as they) drop to lower levels (1)

Emit photons / electromagnetic radiation (1) 2

Explanation of why only certain wavelengths are emitted

Wavelength (of photon) depends one energy (1)

Photon energy depends on difference in energy levels (1)

Levels discrete / only certain differences / photon energies possible (1) 3(and therefore certain wavelengths)

Why phosphor emits different wavelengths to mercury

Different energy levels / different differences in energy levels (1) 1

Calculation of charge

Q = It (1)

= 0.15 A × 20 × 60s

= 180 C (1) 2[10]

93. Why microwaves are reflected

Wave is reflected when passing from one medium to another / when density changes / when speed changes (1) 1

Varying amplitude

Any two of the following:

Varying differences in density of the two mediums produce different intensities of signal (1)

Different distances travelled give different amplitudes (1)

Following a reflection there is less energy available (1) Max 2

Varying time

Different thicknesses of medium (1) 1

What is meant by Doppler shift

Change in frequency/wavelength (1)

Caused by movement of a source (1) 2

Changes due to Doppler shift

Wavelength increases (1)

Frequency decreases (1)

[Allow e.c.f. from incorrect wavelength]

Any one of the following:

• Each wave has further to travel than the one before to reach the heart

• The waves are reflected from the heart at a slower rate (1) 3[9]

94. Resistance calculation

Use of R = ρL/A (1)

Substitution R = 1.6 × 10–4 × 0.02/(5 × (10–3) × 0.02 × (10–3)) (1)

= 32 Ω (1) 3

Total resistance

Either Section 2 = ½ × R1 (16 Ω ) OR Section 3 = 31 × R1 (10.7 Ω ) (1)

Use of RTotal =R1 + R2 + R3 (1)

RTotal = 58.7 Ω [55 Ω if 30 Ω used as starting point] (1) 3

[ecf if section 3 calculated as ¼ × R1 = 56 Ω OR 52.5 Ω if 30 Ω usedas starting point]

Why thermochromic ink becomes warm

Current produces heat / reference to I2RORThermal conduction from conductive ink (1) 1

[Mark for identifying that the heating effect originates in the conductive ink]

Why only thin section transparent

Thinner / section 1 has more resistance (1)

So even a small current will heat it/Power (heating effect) given by

I2R / current will heat it more (1) 2

[OR opposite argument explaining why thicker section is harder to heat][9]

95. Example of light behaving as a wave

Any one of:

• diffraction

• refraction

• interference

• polarisation (1) 1

What is meant by monochromatic

Single colour / wavelength / frequency (1) 1

Completion of graph

Points plotted correctly [–1 for each incorrect point] (1) (1)

Line of best fit added across graph grid (1) 3

What eV s tells us

Maximum (1)

Kinetic energy of the electrons / ½ mv2 of electrons (1) 2

Threshold frequency for sodium

Correct reading from graph: 4.3 × 1014 Hz (1) 1

[Accept 4.1 × 1014 – 4.7 × 1014Hz]

Work function

f = hf0 = 6.63 × 10–34 J s × 4.3 × 1014 Hz (1)

= 2.9 × 10–19J [Allow ecf] (1) 2

Why threshold frequency is needed

• Electron requires certain amount of energy to escape from surface (1)

• This energy comes from one photon of light (1)

• E = hf (1) Max 2[12]

96. Adding angles to diagram

Critical angle C correctly labelled (1) 1

Calculation of critical angle

Use of µ = 1/sin C (1)

Sin C = 1/1.09

C = 66.6° (1) 2

Why black mark not always seen

At (incident) angles greater than the critical angle (1)

t.i.r. takes place (so black mark not visible) (1)

light does not reach X / X only seen at angles less than C (1) 3

[OR opposite argument for why it is seen at angles less than C]

Comparison of sugar concentration

Lower µ means greater density (1)

Greater density means more sugar (1) 2[8]

97. Circuits

Base unit: ampere OR amperes OR amp OR amps (1)Derived quantity: charge OR resistance (1)Derived unit: volt OR volts OR ohm OR ohms (1)Base quantity: current (1) 4

[If two answers are given to any of the above, both must be correct to gain the mark][4]

98. (a) Io and Jupiter: Time taken for electrons to reach Jupiter

t = s/υ = (4.2 × 108 m)/(2.9 × 107 m s–1) = 14.48 s

Correct substitution in υ= s/t (ignore powers of ten) (1)

Answer: 14.48 s, 14.5 s [no ue] (1) 2

(b) Estimate of number of electrons

Q = ne = It

n = It/e

n = (3.0 × 106 A) (1s)/(1.6 × 10–19 C)

Use of ne = It (1)

(1.8 – 2.0) × 1025 (1) 2

(c) Current direction

From Jupiter (to Io) / to Io / to the moon (1) 1[5]

99. (a) p.d. across 4 Ω resistor

1.5 (A) × 4 (Ω)

= 6 V (1) 1

(b) Resistance R2

Current through R2 = 0.5 A (1)

R2 = 0.5(A)

(V) 6

R2 = 12 Ω (1) 2

[allow ecf their pd across 4 Ω]

(c) Resistance R1

p.d. across R1 = 12 − 6 − 4

= 2 V (1)

Current through R1 = 2 A (1)

R1 = 2(A)

V)(2 = 1Ω (1)

[allow ecf of pd from (a) if less than 12 V]

Alternative method

Parallel combination = 3Ω (1)

Circuit resistance = 12(V)/2 (A) = 6Ω (1)

R1 = 6 – (3 + 2) = 1 Ω (1) 3

[allow ecf of pd from (a) and R from (b)][6]

100. (a) Current in filament lamp

P = VI or correct rearrangement (1)

2 A (1) 2

(b) (i) Sketch graph

Correct shape for their axes (1)

−I−V quadrant showing fair rotational symmetry (1) 2

I

V

(ii) Explanation of shape

(As the voltage/p.d. increases), current also increases (1)

(As the current increases), temperature of lamp increases (1)

(This leads to) an increase in resistance of lamp (1)

so equal increases in V lead to smaller increases in I OR rate ofincrease in current decreases OR correct reference to their correct (1) 4gradient

[8]

[If a straight line graph was drawn though the origin then (1)(0)(0)(1) forthe following:

V is proportional to Rtherefore the graph has a constant gradient]

101. (a) (i) Replacement

V1 (1) 1

(ii) Explanation

[ONE pair of marks]Resistance: resistance of V1 [not just the voltmeter] is much largerthan 100 Ω OR combined resistance of parallel combination is (1)approximately 100 Ω

Voltage: p.d. across V1 is much greater than p.d. across 100 Ω OR (1)all 9 V is across V1

OR

Current: no current is flowing in the circuit / very small current (1)Resistance: because V1 has infinite/very large resistance (1)

OR

(Correct current calculation 0.9 x 10 –6 A and) correct pd calculation

90 x 10 –6 A (1)This is a very small/negligible pd (1) 2

(b) Circuit diagram

(i) or equivalent resistor symbol labelled 10 MΩ (1)

or equivalent resistor symbol labelled 10 MΩ (1) 2

[They must be shown in a correct arrangement with R]

(ii) Value of R

6 (V): 3 (V) = 10 (MΩ): 5 (MΩ) / Rtotal of parallel combination is 5 (1)MΩ

1/5 (MΩ) = 1/10 (MΩ) + 1/R OR some equivalent correct (1)substitution to show working

R = 10 MΩ (1) 3[8]

102. Table 6

Wavelength of light in range 390 nm – 700 nm (1)

Wavelength of gamma ≤ 10–11 m (1)

Source (unstable) nuclei (1)

Type of radiation radio (waves) (1)

Type of radiation infra red (1)

Source Warm objects / hot objects /above 0 K

(1)

[6]

103. (a) AmplitudeMaximum distance/displacementFrom the mean position / mid point / zero displacement line / (1) 1equilibrium point[If shown on a diagram, at least one full wavelength must be shown,the displacement must be labelled “a” or “amplitude” and the zerodisplacement line must be labelled with one of the terms above.]

(b) Progressive wave

Displacement at A: 2.0 (cm) [accept 2] (1)Displacement at B: 2.5 (cm) to 2.7 (cm) (1)Displacement at C: 1.5 to 1.7 (cm) (1) 3

Diagram

[Minimum] one complete sinusoidal wavelength drawn (1)

Peak between A and B [accept on B but not on A] (1)

y = 0 (cm) at x = +2.6 cm with EITHER x = +6.2 cm OR x = − 1.0 (1)cm 3

[7]

104. (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) Differences

Any two:

Standing waves Progressive waves

1. store energy 1. transfer energy (1)2. only AN points have max 2. all have the max ampl/displ (1) ampl/displ3. constant (relative) phase 3. variable (relative) phaserelationship relationship (1) Max 2

(c) (i) DropletsFormed at nodes / no net displacement at these points (1) 1

(ii) Speed

Use of υ= fλ (1)Evidence that wavelength is twice node–node distance (1)Wavelength = 1.2 (cm) (1)

Frequency = 8.0 [8.2 / 8.16] Hz or s–1 only (1) 4[10]

105. 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]

106. Description of photon

Packet/quantum/particle of energy [accept E = hf for energy] (1) (1)

[allow packet/quantum/particle of light/e-m radiation/e-m wave etc for (1) X] 2[zero marks if error of physics such as particle of light with negative charge]

Show that energy to move electron is about 8 × 10 –20 J

W = QV (1)

= 1.6 × 10–19 C × 0.48 V

= 7.7 × 10–20 J [no ue] (1) 2

Calculate efficiency of photon energy conversion

Efficiency = (7.7 × 10–20 J ÷ 4.0 × 10–19 J) [ecf] (1)

= 0.19 or 19 % (1) 2[6]

107. Explanation of pressure nodes or antinodes

Pressure constant (1)

Node as a result (1) 2

Relationship between length and wavelength

l = λ/2 or λ = 2l (1) 1

Calculation of fundamental frequency

λ = 2 × 0.28 m = 0.56 m [ecf for relationship above] (1)

v = f λ (1)

f = v/λ = 330 m s–1 ÷ 0.56 m

= 590 Hz (1) 3

Calculation of time period

T = 1/f (1)

T = 1 ÷ 590 Hz [ecf]

= 0.0017 s (1) 2

State another frequency and explain choice

e.g. 590 Hz × 2 = 1180 Hz (or other multiple) (1)

multiple of f0 or correct reference to changed wavelength (1)

diagram or description, e.g. N A N A N, of new pattern [ecf for A & N] (1) 3[11]

108. Explain zeroing of meter

No resistance when leads touched together/short circuit/calibration forzero error (1) 1

Show that resistance is about 70 Ω

R = V ÷ I (1)

= 0.54 V ÷ 0.0081 A

= 67 Ω [no ue] (1) 2

Explain section from passage

Other currents/voltages/resistances present (1)

change in current changes reading for resistance (1) 2

Explain changes in meter reading with temperature increase

Increased lattice vibrations/vibration of atoms/molecules (1)

scattering flowing electrons/more collisions (1)

increased resistance/increase meter reading (1) 3[8]

109. Name process of deviation

Refraction (1) 1

Completion of ray diagram

B – no deviation of ray (1)

A and C – refraction of ray away from normal on entering hot air region (1)

A and C – refraction of ray towards normal on leaving hot air region/ (1) 3

Show positions of tree trunks

B the same (1)

[consistent with ray diagram]

A and C closer to B (1) 2

Explanation of wobbly appearance

Hot air layers rise/density varies/layers uneven (1)

Change in the amount of refraction [accept refractive index]/changein direction light comes from (1) 2

[8]

110. Circuit diagram

Ammeter and power source in series (1)

Voltmeter in parallel with electrodes (1) 2

[Allow both marks if diagram shows an ohmmeter without a powerpack –1 if power pack]

Calculation of resistance

Use of area = πr2 (1)

R = 2.7 × 10–3 Ω m × 5.0 × 10–4 m/A (1)

= 172 Ω (171.9 Ω) (1) 3

Plotting graph

Axis drawn with R on y-axis and labelled with units (1)

Points plotted correctly [−1 for each incorrect] (1)

Sensible scale (1)

Curve added passing through a minimum of 4 points (1) 4

Diameter of hole

Correct reading from graph = 0.23 mm [Allow 0.22 – 0.26 mm] (1) 1[10]

111. Unpolarised and plane polarised light

Minimum of 2, double-headed arrows indicating more than 1 planeand 1 double-headed arrow indicating 1 plane labelled unpolarisedand polarised (1)

Vibrations/oscillations labelled (1) 2

Appearance of screen

Screen would look white/bright/no dark bits/light [not dark = 0] (1)

Explanation

As no planes of light prevented from leaving screen/all light getsthrough/all polarised light gets through (1) 2

Observations when head is tilted

Screen goes between being bright/no image to image/dark bits (1)

Every 90°/as the polarising film on the glasses becomes parallel/perpendicular to the plane of polarisation of the light (1) 2

Comment on suggestion

Image is clear in one eye and not the other (1)

If plane of polarisation is horizontal/vertical (1)

ORImage is readable in both eyes (1)

As the plane of polarisation is not horizontal or vertical (1) 2[8]

112. How sound from speakers can reduce intensity of sound heard by driver

Any 6 from:

• graphs of 2 waveforms, one the inverse of the other

• graph of sum showing reduced signal

• noise detected by microphone

• waveform inverted (electronically)

• and fed through speaker

• with (approximately) same amplitude as original noise

• causing cancellation/destructive superposition

• error microphone adjusts amplification 6[6]

113. Temperature calculation

Current = 4.5 × 10–3 A (1)

p.d. across thermistor is 4.2 V (1)

Rthermistor = 930 Ω ecf their current and pd subtraction error (1)

Temperature = 32 °C − 34 °C [Allow ecf for accurate reading] (1) 4

Supply doubled

Any two from:

• Current would increase / thermistor warms up / temperature increases

• Resistance of thermistor would decrease (1) (1)

• Ratio of p.d.s would change

No OR voltmeter reading / pd across R more than doubles (1) 3

[This mark only awarded if one of the previous two is also given][7]

114. Diagram

Labelled wire and a supply (1)

Ammeter in series and voltmeter in parallel (1)

OR

Labelled wire with no supply (1)

Ohmmeter across wire (1) 2

Readings

Current and potential difference OR resistance ( consistent with diagram) (1)

Length of wire (1)

Diameter of wire (1) 3

Use of readings

R = V/I OR ρ = RA/l (1)

Awareness that A is cross–sectional area (may be seen above and credited here) (1)

Repetition of calculation OR graphical method (1) 3

Precaution

Any two from:

• Readings of diameter at various places /different orientations

• Contact errors

• Zeroing instruments

• Wire straight when measuring length

• Wire not heating up / temperature kept constant (1) (1) 2[10]

115. Conductor resistance

R = ρ l/A (1)

Correct substitution of data (1)

R = 4.3 × 10–2 Ω (1) 3

Manufacturer’s recommendation

Larger A has a lower R (1)

Energy loss depends on I2R / reduces overheating in wires (1) 2[5]

116. Car battery

Voltmeter reading: 12.2 (V) (1) 1

Equation

Terminal p.d. = 12 V + (5.0 A × 0.04 Ω )

See 12V (1)

See 5.0 A × 0.04 Ω (1)

Addition of terms (1) 3

Wasted power

See 0.04 Ω + 0.56 Ω OR 2.8 V + 0.2 V OR 5 x (15 – 12) W (1)

Power = 15 W (1) 2

Efficiency

(same current) 12 V / 15 V OR POUT/PIN = 60 W/75 W (1)

Efficiency = 0.8/80% Efficiency = 0.8/80% (1) 2

Explanation

Any two from:

• Starter motor / to start car needs (very) large current

• I = rR

E

+

• (E and R fixed) rmin ⇒ Imax (1) (1) (1) 2[10]

117. Wavelength

0.30 m (1) 1

Letter A on graph

A at an antinode (1) 1

Wavespeed

Use of υ = fλ (1)

11(10.8) m s–1 (1) 2

[allow ecf λ = 0.15 m ie υ = 5.4 m s–1]

Phase relationship

In phase (1) 1

Amplitude

2.5 mm (1) 1[6]

118. Diagram

One arrow straight down (from −3.84 to − 5.02) (1)Two arrows down (from −3.84 to −4.53, then − 4.53 to − 5.02) (1) 2

Transition T

T from − 5.02 to − 1.85 upwards (1) 1

Kinetic energy values and explanation of what has happened to lithium atomin each case

0.92 eV (1)Atom stays in −5.02 (eV) level/nothing happens to it (1)

0.43 eV (1)

Atom excited to − 4.53 (eV) level (1) 4

Full credit is given to candidates who take the k.e. of the electron to be 0.92 Jafter collision. Any TWO correct energies with correct statement.

[7]

119. Value of wavelength

λ = 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 amplitude

Peak to peak = 4.5 cm [Accept 4.3 cm to 4.7 cm] (1)

Amplitude = ½ × peak to peak

= 2.25 cm [Accept 2.15 cm to 2.35 cm] [Allow ecf for 2nd mark if (1) 2

first part shown]

Calculation of frequency

f = 1/T

= 1 ÷ 2 s

= 0.5 Hz (1) 1

Explanation of why waves are transverse

Oscillation/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 wave

Sound is longitudinal/not transverse (1)

with oscillation along the direction of propagation / compressions and rarefactions (1)

so model not helpful (1) 3[10]

120. Circuit diagram

Variable voltage (1)

Includes ammeter and voltmeter (1)

…. in series and parallel respectively (1) 3

[No penalty for LED bias]

Description of current variation in LEDs

Initially, increasing voltage still gives zero current

OR

Current doesn’t flow until a specific minimum voltage (1)

Current then increases… (1)

….with an increasing rate of increase (1) 3

Discussion of whether LEDs obey Ohm’s law

No (1)

I not proportional to V

OR

R not constant / V/I not constant / R decreases (1) 2

Calculation of resistance of green LED at 1.9 V

R = V/I [Stated or implied] (1)

= 1.9 V ÷ 1.46 × 10–3 A

= 1300 Ω (1) 2

Calculation of power dissipated by red LED at 1.7 V

P = IV [Stated or implied] (1)

= 3.89 × 10–3 A × 1.7 V [do not penalise mA twice]

= 6.6 × 10–3 W (1) 2[12]

121. Process at A

Refraction [Accept dispersion] (1) 1

Ray diagram

Diagram shows refraction away from normal (1) 1

Explanation of condition to stop emergence of red light at B

Angle greater than critical angle (1)

Correctly identified as angle of incidence [in water] (1) 2

Calculation of wavelength of red light in water

c = fλ [stated or implied] (1)

λ = 2.2 × 108 m s–1 ÷ 4.2 × 1014 Hz

= 5.24 × 10–7 m (1) 2[6]

122. Difference between polarised and unpolarised light

Polarised: vibrations in one plane (at right angles to direction of travel) (1)

Unpolarised: vibrations in all planes [NOT 2 planes] (1) 2

OR

Correct drawing (1)

Vibrations labelled (1)

Meaning of advertisement

(Light vibrations are) in one plane (1) 1

Evidence that glare comprises polarised light

Glare is eliminated, so must be polarised light (1) 1

Sunglasses turned through 90°

Glare would be seen through glasses (1)

since they now transmit the reflected polarised light (1) 2[6]

123. Charge

Charge is the current × time (1) 1

Potential difference

Work done per unit charge [flowing] (1) 1

Energy

9 V × 20 C (1)

= 180 J (1) 2[4]

124. Number of electrons

(–64 × 10–9 C) / (–1.6 × 10–19 C) = 4.0 × 1011 electronsUse of n = Q/e (1)

Seeing 1.6 × 10–19 C (1)

Answer of 4.0 × 1011 (electrons) (1) 3

[Use of a unit is a ue]

[–ve answer: 2/3]

Rate of flow

(6.4 × 10–8 C)/3.8 s = 16.8/17 [nC s–1] OR 16.8/17 × 10–9 [C s–1]

(6.4) / 3.8 s i.e. use of I = Q/t [Ignore powers of 10] (1)

Correct answer [No e.c.f.] [1.7 or 1.68 x 10–8 or 1.6 × 10–8] (1) 2

Unit

Amp(ere)/A (1) 1[6]

125. Explanation of observation

Any two from:

• LED on reverse bias/R in LED infinite/ LED wrong way round

• so current is zero /LED does not conduct / very small reversebias current

• since V = IR

• V = 0 × 1K = 0 V (1) (1) 2

Explanation of dimness

RV very large / RV much greater than RLED (1)Current very low / pd across LED very small (not zero) (1) 2

Circuit diagramLED in forward bias (1)Variation of pd across LED (1)Voltmeter in parallel with LED alone (1) 3[LED in series with voltmeter 0/3]

[7]

126. Circuit diagramAmmeter in series with cell and variable resistor (correct symbol) (1)Voltmeter in parallel with cell OR variable resistor (1) 2

Power output at point XPower = voltage × current (1)= 0.45 V × 0.6 A= 0.27 W (1) 2

Description of power output

Any three from:

• Current zero; power output zero/small/low

• As current increases power output also increases

• Then (after X ) power decreases

• Maximum current; power output zero (1) (1) (1) 3

[Accept reverse order]

e.m.f. of cell

0.58 V (1) 1

Internal resistance

Attempt to use current

"lost volts" OR ε = V+ IR (1)

= A6.0

0.45V–V58.0

= 0.217 / 0.2 Ω (1) 2

[ecf an emf greater than 0.45 V][10]

127. Statement 1

Statement is false (1)

Wires in series have same current (1)

Use of I = nAeυ with n and e constant (1) 3

[The latter two marks are independent]

Statement 2

Statement is true (1)

Resistors in parallel have same p.d. (1)

Use of Power = V2/R leading to R ↑, power ↓ (1) 3

OR as R ↑, I ↓ leading to a lower value of VI 3rd mark consequenton second

[6]

128. Description + diagram

Diagram to show:Microwave source/transmitter and detector (not microphone) (1)Transmitter pointing at metal plate/second transmitter from same source (1)Written work to include:Move detector perpendicular to plate/to and fro between /accept ruler on diagram (1)Maxima and minima detected/nodes and antinodes detected (1) 4[Experiments with sound or light or double slit 0/4]

Observation

In phase/constructive interference → maximum/antinode (1)Cancel out/out of phase/Antiphase/destructive interference → minimum /node (1) 2

How to measure wavelength of microwavesDistance between adjacent maxima/antinodes = λ /2 (1)Measure over a large number of antinodes or nodes (1) 2

[8]

129. Incident photon energies

Use of E = hf (1)

Use of c = f λ [ignore × 10X errors] (1)÷ e (1)For 320 nm E = 3.9 (eV) and 640 nm E = 1.9 (eV) (1) 4

Photocurrent readings

Work function of Al > 3.9 / energies of the incident photonsOR threshold frequency is greater than incident frequencies (1)

For Li (ϕ = 2.3 eV / f = 5.6 × 1014 Hz / λ = 540 nm hence) a photocurrentat 320 nm but not 640 nm (1)If intensity × 5 then photocurrent × 5 (1) 3

Stopping PotentialKEmax = 4.00/3.88 –2.30 = 1.7/1.58 [ignore anything with only e] (1)Vs = 1.7/1.58 V (1) 2

[9]

130. Wavelength and wave speed calculation

λ = 0.96 m (1)

seeing f = 2 their λ (f = 2.1 Hz) (1) 2

Qualitative description

(Coil) oscillates / vibrates (1)

With SHM / same frequency as wave (their value) (1)

Parallel to spring / direction of wave (1) 3[5]

131. Explanation of emission of radiation by hydrogen atoms

Electrons excited to higher energy levels (1)

as they fall they emit photons / radiation (1) 2

[Accept 21 cm line arises from ground state electron changing spinorientation (1) / relative to proton (1)]

Why radiation is at specific frequencies

Photon frequency related to energy / E = hf (1)

Energy of photon = energy difference between levels / hf = E1 – E2 (1)

Energy levels discrete/quantised / only certain energy differences possible (1) 3

Show that hydrogen frequency corresponds to λ = 21 cm

f = 4.4623 × 109 ÷ π= 1.42 × 109 Hz (1)

c = f λ

λ = 3 × 108 ÷ (1.42 × 109 Hz) (1)

λ = 0.211 m or 21.1 cm [no up] (1) 3[8]

132. Explanation of assumption that voltmeter does not affect values

Voltmeter has very high resistance/takes very small current (1) 1

Current through X

4.8 A ÷ 6 = 0.8 A

OR 48 V ÷ 60 Ω = 0.8A (1) 1

Value missing from E7

P =IV

P = 4.4 A × 53 V = 233 W (1) 1

Description of appearance of lamp X as lamps switched on

Gets dimmer

from table, voltage decreasing / current in X decreasing / power per lamp decreasing (1)

So P decreases (1) 3

Formula for cell C6

I = ε / Rtot (1)

I = 120 / (15 + B6) (1) 2

Effect of internal resistance on power

Power has a maximum value (1)

when external resistance = internal resistance (1) 2[10]

133. Fundamental frequency of note

440 Hz (1) 1

Frequencies of first three overtones

880 Hz

1320 Hz

1760 Hz

Two correct frequencies (1)

Third correct frequency (1) 2

Comment on the pattern

Any 2 from the following:

[Allow ecf]

880 Hz = 2 × 440 Hz

1320 Hz = 3 × 440 Hz

1760 Hz = 4 × 440 Hz

1760 Hz = 2 × 880 Hz (1) (1) 2

[OR They are multiples (1) of the fundamental (or similar qualification) (1)]

[Allow 1 mark for amplitude decreasing with frequency]

Measurement of period

Example: 7 cycles takes (0.841 – 0.825) s [at least 5 cycles] (1)

Period = 0.016 s ÷ 7

= 2.3 × 10–3 s [in range 2.2 × 10–3 s to 2.4 × 10–3 s] (1) 2

Calculation of frequency

f = 1/T (1)

= 1 ÷ 2.2 × 10–3 s [Allow ecf]= 454 Hz (1) 2

[9]

134. Mark on diagram

Correctly drawn normal (1)

Correctly labelled angles to candidate’s normal (1) 2

Show that refractive index of water is about 1.3

Angles correctly measured:

i = 53 (± 2)°

r = 39 (± 2)° (1)

µ = sin i / sin r = sin 53° / 39°

= 1.27 [Allow ecf] [Should be to 2 d.p. min] (1) 2

Critical angle

µ = 1/sinC (1)

so sin C = 1/1.27 so C = 52° [ecf] (1) 2[use of 1.3 gives 50°]

Explanation of reflection of ray

Internal angle of incidence = 39° ± 1° (1)

Compare i with critical angle (1)

Valid conclusion as to internal reflection being total/partial (1) 3

Refractive index

It varies with colour (1) 1[10]

135. Explanations

(i) Refraction:e.g. bending of wave when travelling from one medium to another [OR change of speed] (1)

(ii) Diffraction:e.g. spreading of wave when it goes through a gap (1) 2

Diagram of wavefronts near beach

Gradual bend in wavefronts (1)

Smaller wavelengths (1)

Waves bending upwards as they approach shore (1) 3

Diagram of wavefronts in bay

Constant wavelength (1)

Waves curve (1) 2

Explanation

Refraction/diffraction causes waves to bend towards the beach (1) 1[8]

136. Measurement needed

Any three from:

• Resistance

• Distance between probes

• Effective area/cross sectional area

• R = ρ A

L (1) (1) (1) 3

Equation of line A

Intercept = –3.5 (Ω m) (+/– 0.3) (1)

Gradient = 1.5 (Ω mm–1) (+/– 0.05) (1)

So equation is ρ = 1.5 d – 3.5 [Or equivalent, e.c.f. allowed] (1) 3

Addition of line

Points correctly plotted (–1 for each error, allow ½ square tolerance) (1) (1)

Line of best fit drawn (1) 3

Best distance

Between 1.90 and 1.99 km (1) 1[10]

137. Ultrasound

Ultrasound is very high frequency sound (1)

How ultrasound can be used

Any three from:

• gel between probe and body

• ultrasound reflects

• from boundaries between different density materials

• time taken to reflect gives depth of boundary

• probe moved around to give extended picture

• size of reflection gives information on density different (1) (1) (1) 3

How reflected ultrasound provides information about heart

Any two from:

• Doppler effect

• frequency changes

• when reflected from a moving surface

• gives speed of heart wall

• gives heart rate (1) (1) 2[6]

138. Physics principles

Requires 9 V battery:

Battery required for electronic circuitry / microphone / speaker (1)

Rubberized foam ear cups:

Air filled material / material has large surface area (1)

Air molecules collide frequently with material (1)

Foam deforms plastically/collisions are inelastic (1)

Sound converted to heat in material (1)

Active noise attenuation:

Noise picked up by microphone (1)

Feedback signal inverted / 180° out of phase with noise / antiphase (1)

Amplified [OR amplitude adjusted] and fed to earphones / speaker (1)

Sound generated cancels/superimposes/minimum noise (1)

Diagrams of superposing waves showing (approx.) cancellation (1)

Amplifier gain automatically adjusted if noise remains (1)

Device only works over frequency range 20 – 800 Hz (1) Max 6

Where does the energy go?

Some places will have constructive interference (1)

More intense noise (1)

Some noise dissipated as heat in air / foam (1)

increased kinetic energy of air [OR foam] molecules (1) Max 2[8]

139. Resistance calculationsEvidence of 20 Ω for one arm (1)

20

1

20

11 +=R

(1)

R = 10 Ω (1) 3

Comment

This combination used instead of a single 10 Ω resistor [or samevalue as before] (1)

because a smaller current flows through each resistor/reduce heatingin any one resistor/average out errors in individual resistors (1) 2

[5]

140. Graphs

Diode:

RH quadrant: any curve through origin (1)Graph correct relative to labelled axes (1)LH side: any horizontal line close to axes (1) 3

I

V

L i n e o n o r c l o s e t ov o l t a g e a x i s

Filament lamp

I

V

RH quadrant:Any curve through origin (1)Curve correct relative to axes (1)LH quadrant:Curve correct relative to RH quadrant (1) 3[Ohmic conductor scores 0/3]

[6]

141. CircuitAmmeters and two resistors in series (1) 1[1 mark circuit penalty for line through cell or resistor]Cell e.m.f

E= 150 x 10–6 (A) x 40 x 103 (Ω ) total R (1)Powers of 10 (1) 2E = 6.0 (V)

New circuit

Voltmeter in parallel with 25 (kΩ ) resistor (1) 1Resistance of voltmeter

(Total resistance) = A)(10 170

V)(66–×

= (35.3 kΩ )

(Resistance of ll combination) = 35 – 15 kΩ= (20 Ω ) [e.c.f. their total resistance]

VR

1

25

1

20

1 +=

100

4–51 =VR

RV = 100 kΩ [108 kΩ if RT calculated correctly]

(1)

(1)

(1)

Alternative route 1: 3

p.d. across 15 kΩ = 2.55 V(∴ p.d. across ll combination = 3.45 V)resistance combination = 20 kΩ→ RV = 100 kΩ

(1)

(1)

(1)

[7]

Alternative route 2: 3

p.d. across parallel combination = 3.45 VI through 25 kΩ = 138 µ A→ RV = 100 kΩ

(1)

(1)

(1)

142. Resistance of strain gauge

State R = A

lρ (1)

Use of formula (1)x 6 (1)R = 0.13 Ω [ecf their l] (1) 4

Ω=Ω×=

××××Ω×==

13.0

106.129

m101.1

6m104.2m109.9ρ

3–

27–

2–8–

R

A

lR

Change in resistance

∆ R = 0.13 Ω × 0.001

∆ R = 1.3 × 10–4 (Ω ) [no e.c.f.]OR∆ R = 0.02 × 0.001

∆ R = 2.0 × 10–5 Ω

0.1% → 0.001 (1)Correct number for ∆ R (1) 2

Drift velocity

Stretching causes R to increase (1)Any two from:• Current will decrease• I = nAυ Q• Drift velocity υ decreases• nAe constant (1) (1) 3

[9]

[For R decreasing, max 1:Any one from:• I will increase• I = nAυ Q• υ will increase• nAe constant]

143. ExplanationClarity of written communication (1)Wave reflects off bench (1)(Incident and reflected) waves superpose/stationary wave is formed (1)Maxima or antinodes where waves in phase or constructiveinterference occurs (1)Minima or nodes where waves exactly out of phase or destructiveinterference occurs (1) 5

Speed of soundSee a value between 5.0 and 5.6 (cm) (1)Use of υ = fλ (1)λ = 2 × spacing (1)

320 m s–1 to 360 m s–1 (1) 4

Explanation of contrastAs height increases, incident wave gets stronger, reflected wave weaker (1)So cancellation is less effective [consequent mark] (1) 2

[11]

144. Ionisation energy

(10.4 eV) × (1.6 × 10 –19 J eV–1) (1)

(–) 1.66 × 10–18(J) (1) 2

Kinetic energy

0.4 (eV) (1) 1

Transition

Use of E = hc/λ (1)3.9 (eV) (1)Transition is from (–)1.6 eV to (–)5.5 eV 3

[6]

145. Deductions about incident radiations

(i) Radiations have same frequency/same wavelength/ same photon energy (1)

(ii) Intensity is greater in (a) than in (b) (1) 2

Sketch graph (c)

Line of similar shape, starting nearer the origin on negative V axis (1) 1

Maximum speed

Use of E = hf (1)

Subtract 7.2 × 10–19 (J) (1)

Equate to ½ mυ 2 (1)

3.1 × 106 ms–1 (1) 4[7]

146. Wavelength

Distance between two points in phase (1)

Distance between successive points in phase (1) 2

[May get both marks from suitable diagram]

Sunburn more likely from UV

UV (photons) have more energy than visible light (photons) (1)

Since shorter wavelength / higher frequency (1) 2

What happens to atoms

Move up energy levels/excitation/ionization (1)

Correctly related to electron energy levels (1) 2[6]

147. Emitted pulse

Greater amplitude/pulse is larger/taller (1) 1

Depth of rail

2d = vt = 5100 m s–1 × 4.8 × 10–5 s

= 0.24 m

Hence d = 0.12 m

Reading from graph [4.8 or 48 only] (1)

Calculation of 2d [their reading × timebase × 5 100] (1)

Halving their distance (1) 3

Description of trace

A reflected peak closer to emitted/now 3 pulses (1)

Exact position e.g. 1.6 cm from emitted (1) 2

Diagram

Shadow region (1)

Waves curving round crack (1) 2[8]

148. Resistance in darkness

In the dark R = 4 kΩ (1)

so resistance per mm = 4000 Ω /40 mm = 100 Ω (mm–1) (1) 2

Resistance of 8 mm length

In the light R = 200 Ω (1)so resistance of 8 mm strip = (8 mm/40 mm) × 200 Ω [= 40 Ω ] (1) 2

Calculations

Resistance of remainder = 32 mm × 100 Ω mm–1 = 3200 Ω (1) 1

(i) Total resistance = 3240 Ω (1)

Current = V/R = 1.2 V/3240 Ω = 3.7 × 10–4 A (1)

(ii) Across 8 mm, p.d. = IR = 3.7 × 10–4 A × 40 Ω (1)= 0.015 V (1) 4

Explanation of why current decreases

Any two points from:

• more of strip is now in the dark

• greater total resistance

• I = V/R where V is constant (1) (1) Max 2[11]

149. Error in circuit diagram

Cell needs to be reversed (1)Any one point from:

• electrons released from the magnesium

• copper wire needs to be positive to attract electrons (1) 2

Completion of sentence

UV is made up of particles called photons (1) 1

UV and visible light

(i) UV has shorter wavelength/higher frequency/higher photonenergy (1)

(ii) Both electromagnetic radiation/both transverse waves/samespeed (in vacuum) (1) 2

Explanation of why low intensity UV light produces a current

Any three points from:

• reference to photons or E = hf

• frequency > threshold frequency

• electron must have sufficient energy to be released

• UV photons have more energy

• electron is released by ONE photon

• brighter light just means more photons (1) (1) (1) Max 3

Why current stopped

Glass prevents UV reaching magnesium (1) 1[9]

150. Total internal reflection

Any two points from:

• from a more dense medium to a less dense medium/high to low refractive index

• incident angle greater than the critical angle

• light is reflected not refracted/no light emerges (1) (1) Max 2

Critical angle

Sin i / sin r = µ ; gives sin 90°/sin C = µ (1)

C = 42° (1) 2

Diagram

Reflection (TIR) at top surface (air gap) (1)

Reflection (TIR) at bottom surface and all angles equal by eye (1) 2

Path of ray A

Passing approximately straight through plastic into glass (1)

Emerging at glass–air surface (1)

Refraction away from normal (1) 3

Why there are bright and dark patches on image

Bright where refracted/reference to a correct ray A in lower diagram (1)

Dark where air gap (produces TIR)/reference to correct top diagram (1) 2[11]

151. Polarisation

The (wave) oscillations (1)occur only in one plane (1) 2[OR shown with a suitable diagram]

How to measure angle of rotation

Any four points from:

• Polaroid filter at one/both ends

• with no sugar solution, crossed Polaroids (top and bottom oftube) block out light

• sugar solution introduced between Polaroids

• one Polaroid rotated to give new dark view

• difference in angle between two positions read from scale (1) (1) (1) (1) Max 4

Graph

Points plotted correctly [–1 for each incorrect; minimum mark 0] (1) (1)

Good best fit line to enable concentration at 38° to be found (1) 3

Concentration

0.57 (± 0.01) kg l–1 1[10]

152. Explanations of observations

Speed of light is much greater than speed of sound (1)

Speed of sound in soil is greater than speed of sound in air (1) 2[2]

153. Resistance of lamps

P = R

V 2

OR I= 60/12 = (5 A) 1

R = W60

V12V12 × R = V/I 1

R = 2.4 Ω 1

Resistance variation

Lamp A: resistance of A decreases with current increase 1

Lamp B: resistance of B increases with current increase 1

Dim filament

Lamps are dim because p.d. across each bulb is less than 12 V 1

Why filament of lamp A is brighter

Bulbs have the same current 1

p.d. across A > p.d. across B/resistance A> Resistance B 1

OR

power in A > power in B 2[8]

154. Current in heating element

p = VI

I=V230

W500

I = 2.2 A

p = R

V 2

R = )(8.105/500

2302

Ω

I = 2.2 A

1

1

1

Drift velocity

Drift velocity greater in the thinner wire / toaster filament 1

Explanation

Quality of written communication 1

See I = nAQυ 1

I is the same (at all points ) 1

(probably) n (and Q) is the same in both wires 1[8]

155. Resistance of films

R = A

lρ1

R = t

l

ωρ

or A = ω t [consequent on first mark] 1

[i.e. product = ω t]

Resistance calculation

R = m)10(0.001m)10(3

)10(8Ωm)10(6.03–3–

–3–5

××××××

OR

R = )m100.1)(mm0.3(

)mm8()m100.6(6–

5–

××Ω×

R = 160 Ω

Correct substitution except powers of 10 1

Correct powers of 10 1

Answer 1

Resistance of square film

l = ω 1

R = t

ρ1

[7]

156. Definition of e.m.f. of a cell

Work/energy (conversion) per unit charge 1

for the whole circuit / refer to total (energy) 1

OR

Work/energy per unit charge 1converted from chemical to electrical (energy) 1

OR

E = Q

W for whole circuit 1

All symbols defined 1

OR

E = I

P for whole circuit 1

All symbols defined 1

[Terminal p.d. when no current drawn scores 1 mark only]

Circuit diagram

A AR R

V

R 1 R (can be variable) 1 2A in series 1 A and V correct 1

V as shownOr across R + AOr across battery

[2nd mark is consequent on R(fixed, variable) or lamp]

Sketch graph

R V

1 / I I

Graph correctly drawn with axes appropriately labelled andconsistent with circuit drawn 1

Intercept on R axes Gradient ≡ (–)r [Gradient mark consequent 1≡ (–)r on graph mark]

[Gradient may be indicated on graph][6]

157. Wavefront

Line/surface joining points in phase 1

Addition to diagrams

Wavefront spacing ≈ as for incident waves (min. 3 for each) 1

1st diagram: wavefronts nearly semicircular 1

2nd diagram: much less diffraction 1

Reception

L W has longer wavelength 1

so is more diffracted around mountains [consequent] 1[6]

158. Total e.m.f of cells in series

e.m.f. in series add up / 6000 × 40 × 10-3 V (1)

= 240 V (1) 2

Internal resistance of cells in series

6000 × 0.70 Ω = 4200 Ω (1) 1

Calculation of current

I= V ÷ R (1)

= 240 V ÷ 4200 Ω = 0.057 A (1) 2

Calculation of total current

20 × 0.057 A = 1.1 A (1) 1

Explanation of voltmeter reading

Since V = IR (1)

and R = 0 Ω (1)

V = 0 V ≠ e.m.f.

OR

V= E – Ir (1)

r ≠ 0 Ω (1)

so V < E

OR

Lost volts (1)

across internal resistance (1) 2

Voltmeter reading

0 V [No u.e.] (1) 1[9]

159. Path difference

2 × 1.11 × 10-7 m = 2.22 × 10-7 m (1) 1

Explanation of why light appears dim

Path difference = ½ × wavelength (1)

so waves in antiphase/destructive interference/superposition (1) 2

Reason for increase in film thickness

Because of gravity/soap runs down (1) 1

Explanation of whether film further down appears bright or dark

Path difference = wavelength (1)

Waves in phase/constructive interference (so appears bright) (1) 2

Explain bright and dark stripes

Different positions have different thicknesses/path differences (1)

So some points in phase, some in antiphase/some points have constructive interference, some destructive (1) 2

Movement of bright and dark stripes

Soap flows down/thickness profile changes (1)

so positions of destructive/constructive interference changing (1) 2

Alternative path added to diagram

One or more extra reflections at each internal soap surface (1) 1[11]

160. Diagram

(i) Any angle of incidence marked and labelled I (1)

(ii) Any angle of refraction marked and labelled R (1)

(iii) Angle of incidence/reflection at lower surface marked and labelled G (1) 3

Refraction of light

Velocity of light is lower in glass (1) 1

Velocity of light in hot air

Velocity is greater (1) 1

Property of air

(Optical) density / refractive index (1) 1[6]

161. Resistor

(i) A = π r2 = π × (4.0 × 10-4)2 (1)

= 5.03 × 10-7 m2 (no u.e) (1) 2

(ii) Recall of R = ρ l/A (1)

Length l = RA/ρ

= 0.12 × 5.0 × 10-7 / 1.8 × 10-8 [substitutions]

= 3.3 m (1) 3

Advantage of using iron wire of same diameter

Shorter piece of wire needed (if iron chosen) (1) 1[6]

162. Description

Electron (near surface of cathode) absorbs photon and gains energy (1)

Work function is energy needed for electron to escape from surface (1)

Electrons released in this way are called photoelectrons (1) 3

Lowest frequency of radiation

f0 = E/h (1)

= 2.90 × 10-19 J/6.63 × 10-34 J s (1)

= 4.37 × 1014 Hz (1) 3

Suitability of potassium

λ = 3 × 108 m s-1 / 4.37 × 1014 Hz [use of lowest frequency] (1)6.86 × 10-7 m [with suitable comment] (1)

OR

f = 3 × 108 m s-1 / 4.0 ×10-7 and f = 3 × 108 m s-1 / 7.0 × 10-7 [usesrange of λ ] (1)f = 7.5 × 1014 Hz to 4.3 × 1014 Hz [with suitable comment] (1) 2

[Suitable comment – e.g. this is within range of visible light/almostall of the visible light photons will emit photoelectrons]

Maximum kinetic energy

Use of E = hc/λ AND minimum wavelength (1)

Max photon energy = hc/λ = 6.63 × 10-34 J s × 3 × 108 m s-1/(400 ×10-9m)

= 4.97 × 10-19 J [no u.e]

Max k.e. = max photon energy – work function [or use equation]

= 4.97 × 10-19 J – 2.90 × 10-19 J

= 2.07 × 10-19 J [allow ecf if wrong wavelength used] [no u.e] (1) 3

Why some photoelectrons will have less than this k.e.

One point from:

• photon energy might be transferred to electron below surface• so some energy transferred to atoms on the way to surface• hence electron leaves surface with less energy than max• max is for electron from the surface• lower energy photon responsible for emission (1) 1

[12]

163. Unit of current

Amps/ampere (1) 1

Base units of p.d.

For V = IR method

Any three from:

• V = J C–l

• C = A s• J = N m

• N = kg m s–2

[kg m2 s–3 A–1]

[See J = kg, m2 s–2 (1) (1)]

OR

For P = VI method

• Watt is J s-1 / J/s

• V = J s–1 A–1

• J = Nm

• N = kg m2 s–2 (1) (1)] (1) (1) (1)

[See kg m2 s–2 (1) (1)] 3[4]

164. Show that resistance is approximately 45 Ω

A

lR

ρ=

7

5

100.8

m65.0m105.5R

−

−

××Ω×=

= 44.7 Ω [No u.e.] (1)

[Must see this value and not 45]

Table

Switch X Switch Y Resistance of heater/ΩOpen Closed 22.5/22.35 (1)

Closed Open 45/44.7 (1)Closed Closed 15/14.9 (1)

[No u.e.] 3

Calculation of maximum power

R

VP

2

= Use of equation with 15 Ω OR their minimum value (1)

= 3526 W,3500 W [full ecf] (1) 2

Explanation of power output fall

increases)metalsof(cetanresishot/hottergetsitasOR

increasesheatertheofetemperaturtheAs

Since V is constant P = R

V 2

OR P = VI and V = IR

[Then P ↓ as R ↑] (1) 2

OR P ∝ R

1 [so P↓ as R↑]

[10]

165. Explanation of greater drift velocity

(Electrons have greater drift velocity) in the thinner wire (1)

Any two from:

• Same current in both wires• Reference to I = nAQυ• nQ same in both wires (1) (1) 3

Explanation of higher dissipation of power

(Higher power is dissipated) by the smaller(er)/ low resistor (1)

Any two from:

• Resistors have same p.d. across them• The small resistor has the largest current [or reverse]

• Power = voltage × current, OR voltage2 ÷ resistance [NOT I2R] (1) (1) 3[6]

166. Circuit diagram

Resistor with another variable resistor/potential divider/variable powerpack (1)

Ammeter reading current through resistor (1)

Voltmeter in parallel with resistor (1) 3

Graph labels

labelledBothlamp–Curve

resistor–lineStraight

(1) 1

Potential difference

At 0.5 A p.d.= 3.5 V / 3.4 V + 7.8 V / idea of adding p.d. [for same current] (1)

= 11.2 V/11.3 V (1) 2

[Accept 11.0 –11.5 V]

Resistance of lamp

A5.0

V5.3 [OR their value of p.d. across lamp ÷ 0.5 A] (1)

= 7.0 Ω (1) 2

[e.c.f. their value][8]

167. Table

Radio waves Sound wavesTransverse LongitudinalTravel much faster than sound Travel more slowly(Can) travel in a vacuum Cannot travel in a vacuumCan be polarised Not polarisedElectromagnetic Pressure/Mechanical wave

Any three of the above Max 3

Assumption

Attempt to calculate area (1)

Intensity = 0.02 kW m–2 OR 20 W m–2 (1)

Efficiency at collector is 100%/beam perpendicular to collector

Power

Use of I P/4π r2 (1)

Power = 3.3 × 1017 W [ecf their I]

No energy “lost” due to atmosphere (not surroundings) OR Inversesquare applies to this situation (1)

More efficient method

Use a laser (maser) / reference to beaming/ray (1) 1[10]

168. Ionisation energy

Use of × 1.6 × 10–19

2.2 × 10–18[No u.e.] (1) 2

Addition to diagram

(i) From 4 to 3 labelled R / (i) (1)

(ii) From 1 to 4 labelled A / (ii) (1) 2

Emission spectrum

Hydrogen ‘excited’ in a discharge/thin tube/lamp [not bulb] (1)

Viewed through a diffraction grating/prism/spectrometer (1)

Appearance of emission spectrum

A series of lines / colours on a dark background [accept bands] (1) 3

Region of spectrum

Radio/microwave (1) 1

Speed of galaxy and deduction

∆ λ = 8 (mm) / 211 - 203 (mm) (1)

Use of 3 × 108 (1)

υ = 1. 1(4) × 107 ms–1 [No e.c.f.] (1)

Moving towards Earth (us) (1) 4[12]

169. Photoelectric effect

Any two features and explanation from the following:

Feature: Experiments show k.e(max) ∝ f, OR not intensity[Accept depends upon] (1)

Explanation: Photon energy ∝ f (1)[Consequent]

k.e(max) ∝ intensity is a wave theory (1)

Feature: Emission of photoelectrons immediate (1)

Explanation: One photon releases one electron particle theory (1)[Consequent] Wave theory allows energy to “build up” (1)

Feature: (Light) below a threshold frequency cannot releaseelectrons (1)

Explanation: Particle theory-f too low as not enough energy is released[Consequent] by photon to knock out an electron (1)

Wave theory- if leave a low frequency beam on longenough, it will produce enough energy to release anelectron (1)

[Max 5]

170. How stationary waves could be produced on a string

Diagram showing:

String and arrangement to apply tension (1)

Vibration generator and signal generator (1) 3

Vary f / tension / length until wave appears (1)

Determination of speed of travelling waves

QOWC (1)

Determine node-node spacing; double to obtain λ (1)

Read f off signal generator / cro / use a calibrated strobe (1)

Use υ = f λ for υ (1) 4[7]

171. Explanation of superposition

When 2 (or more) waves meet / cross / coincide /interfere...(1)

Reference to combined effect of waves, e.g. add displacement / amplitude - maybe a diagram [constructive/destructive interference not sufficient withoutimplication of addition] (1) 2

Calculation of thickness of fat layer

Thickness = half of path difference

= 0.5 × 3.8 × 10–7 m

= 1.9 × 10–7 m (1) 1

Explanation of constructive superposition

Path difference of 3.8 × 10–7 m same as a wavelength of green light (1)

Waves are in phase / phase difference 2 π or 360° (1) 2

Explanation of what happens to other wavelengths

Path difference greater than/less than/not one wavelength/ waves not in phase / out of phase (1)

Constructive interference will not take placeOR (1)These colours will not appear bright (1) 2

Explanation of why colours are seen at other places

Thickness of fat variesORLight seen at a different angle to the meat surface (1)

Other wavelengths may undergo constructive interference/be in phaseOR (1)Path difference will vary 2

[9]

172. Meaning of m

× 10–3 (1) 1

Calculation of resistance for reading 3

R = V/I OR R = 74 × 103 V ÷ 150 × 10–9 A [ecf for milli] (1)

R = 4.9 × 105 Ω (1) 2

Calculation of power for reading 4

P = I × V OR P = R

V 2

OR P = I 2R (1)

= 210 × 10–9 A × 57 × 10–3 (1) 2

= 1.2 × 10–8 W

Plotting points on graph

Two correct points (1)Third correct point (1)Best fit straight line for points as they appear on student’s graph (1) 3

Predicting short-circuit current

Correct from graph, e.g 450 nA (1) 1

Suggested e.m.f

Correct from graph, or table, 110 mV (1) 1

Explanation of why voltage falls

Cell has internal resistance/ “lost volts” (1)

“Lost volts” = Ir, so lost volts increase as current increasesORV = E – Ir, so V decreases as I increases (1) 2

[12]

173. Diameters of dark ring

Diameter in frame 1 = 9 mm (± 1 mm)Diameter in frame 2 = 19 mm (± 1 mm) [No ue] (1) 1

Show that ripple travels about 25 Mm

Difference between diameters = 19 mm – 9 mm = 10 mmDistance travelled by one part = 10 mm ÷ 2 = 5 mm (1)

Scale: 200 Mm = 40 mm (39 mm to 41 mm)Distance = 5 mm × 200 Mm ÷ 40 mm = 25.0 Mm [No ue] (1) 2

Calculation of speed of ripple

Speed = distance ÷ time (1)

= 25.0 ×106 m ÷ (10 × 60) s (1)

= 41 600 m s–1 [no ue] (1) 3

How to check speed constant

Use third frame to calculate speed in this time intervalOR plot diameter (or radius) against time to get a straight lineOR compare distance travelled between frames 3 and 2 with distance travelledbetween frames 2 and 1 (1) 1

Cross-section of wave

Wavelength (1)Amplitude (1) 2

Calculation of frequency of waves

Wavespeed = frequency × wavelength (1)

Frequency = wavespeed ÷ wavelength = 41 700 m s–1 ÷ 1.4 × 107m (1) = 3.0 × 10–3 Hz (1) 2

[11]

174. Movement of water molecules

Molecules oscillate/vibrate (1)

Movement parallel to energy flow (1) 2

Pulses

To prevent interference between transmitted and reflected signals (1) 1

OR allow time for reflection before next pulse transmitted

Calculation

Time for pulse to travel to fish and back again = distance ÷ speed

υ∆=∆ x

t

= 1ms1500

m3002−

× (1)

= 0.4 s (1) 2

[0.2 s = 1 mark]

Effect used in method

Doppler effect (1)

Any two from:

• a change in frequency of the signal• caused by relative movement between the source and the observer• size and sign of change relate to the relative speed and direction of the

movement between shoal and transmitter• frequency increase - moving towards• frequency decrease - moving away (1) (1) 3

[8]

175. Equation to define resistivity

l

RA=ρ (1)

All symbols defined (resistivity, resistance, length, cross-sectional area) (1) (1)

[3 symbols only defined (1)] 3

Resistance meter

Any two from:

• the resistance between the two probes is measured, not the resistivity• because you cannot measure the cross-sectional area of skin between the probes• A and 1 both vary; cannot calculate resistivity (1) (1) 2

Whether results support claims

Yes (1)

Any two from:

• resistance chances with programme content• least resistance with political programme• sweat reduces resistance / is a better conductor (1) (1) 3

[8]

176.

Word Equation Quantity Defined

Voltage ÷ Current Resistance (1)

Voltage × Current Power (1)

Charge ÷ Time Current (1)

Work done ÷ Charge Voltage/p.d./e.m.f (1)

[4]

177. Charge calculation

Q = 20 000 × 4.0 × 10−4 s [substitution]

Q = 8.0 C/A s 2

Resistance calculation

R = Alρ

= )m100.1(

)m50)(107.1(23–

8–

×Ω×

R = 8.5 × 10−4 Ω

Formula (1)

Correct substitution (1)Answer (1) 3

Potential difference calculationV = IR

= (20 000 A) × (85 × 10−5 Ω ) [or their value] (1)

= 17 V [Allow full e.c.f] (1) 2

ExplanationFor the tree: R or p is larger (1) 1

[8]

178. NetworksFirst network: 2.5(Ω ) (1)Second network: 25 (Ω ) (1)Third network: 10 (Ω ) (1) 3

Meter readingsAmmeter: 25 (mA) (1)Voltmeter V1: 25 × 10 OR 50 × 5 [ignore powers of 10] (1)= 0.25 V (1)Voltmeter V2: 50 × 25 [ignore powers of 10] (1)= 1.25 V (1) 5

[Allow full e.c.f. for their resistance for 2nd network OR their V1 answer][8]

179. Potential difference across resistors

2.0 MΩ : 6.0 V 5.99998 V (1) OR

4.0 Ω : 0V 1.2 × 10−5 V (1) 2

Second potential divider circuit

p.d. across 45 Ω :

( 5045

× 6.0 V) = 5.4 V (1)

p.d. across diode:

(6.0 V – 5.4 V) = 0.6 V (1) 2

[Allow e.c.f. for 2nd mark if candidate uses

455 × 6.0 V = 0.7 V for diode

then

6.0 V – 0.7 V5.3 V for 45 Ω ]

Graph

I / A

0 V / Vm u s t h a v e t h i si n i t i a l s h a p e

B e l e n i e n t e h e r e p r o v i d e dg e n e r a l l y r i g h t s h a p e

(1) 1[5]

180. Wavelength0.80 m (1)

Out of phaseEither X as in diagram below (1)

At restY at crest or trough as in diagram below (1)

Y

Y

YA

X X

Direction of movementArrow at C up the page (1) 4

Time calculation (1)Use of t = λ /υ (1)0.25 s [ecf λ ] 2

[6]

181. Electromagnetic waves experiment

EITHER

‘Lamp’, 1 polaroid // LASER (1)

2nd polaroid, suitable detector [e.g. eye, screen, LDR] (1)Rotate one polaroid [ consequent on 2 polaroids] [one if LASER] (1)Varies [consequent] (1)

OR

Microwave transmitter (and grille) [not polaroid or grating] (1)Receiver (or and grille) (1)Rotate ANY [if 2 grilles; must rotate a grille] (1)Varies [consequent] (1) 4

Nature of wavesransverse (1) 1

[5]

182. Planck constantRealise that h is the gradientCorrect attempt to find gradient [but ignore unit errors here]

h = (6.3 to 6.9) × 10–34 J s [No bald answers] 3

Work functionUse of hf0 / use intercept on T axis/use of φ = hf – T (1)

φ = (3.4 to 3.9) × 10–19 J [ -1 if –ve] [2.1 to 2.4 eV] (1) 2[5]

183. Calculation of current

P = IV

I = P/V (stated or implied) (1)

= 0.78 W ÷ 6 V

= 0.13 A (1) 2

Calculation of resistance

P = V2/R

OR R = V/I

OR R = (6 V)2 ÷ 0.78 W

OR R = 6 V ÷0.13 A [ecf] (1)

= 46 Ω (1) 2

Explanation of operation from mains

In series (1)

240 V ÷40 lamps = 6 V per lamp (1) 2

Explanation of constant brightness of lamps

Current equal – justified, e.g. in series or same V/R or same P/V 1

Statement and explanation of different resistance with ohmmeter

Lower resistance with ohmmeter (1)

Identify lower temperature with ohmmeter [may be implied] (1)

(Lattice) ions’/atoms’ vibrations impede electrons/current (1)

[Require interaction]

[Allow converse argument] 3[10]

184. Explanation of amp hours

Charge = current × time or Q = It (1)

Amp →current and hour → time (so amp × hour → charge) (1) 2

Show that charge about 5000 C

Charge = 1.5 A × 1 × 60 × 60 s

= 5400 C 1

Calculation of energy stored

W = QV OR W = I v t (1)

= 5400 C × 3 V [may use 5000 C]

= 16 200 J [up] (1) 2

Show that energy is about 20 000 J

Energy = Ivt (1)

= 0.3 A × 3.1 V × 6 × 60 × 60 s

= 20 100 J (20 088 J) (1) 2

Calculation of efficiency

Efficiency = (stored energy/input energy) × 100%

= 16 200 J [allow ecf from 3rd part] (1)

÷ 20 088 J [allow 20 000 J from 4th part] (1)

× 100%

= 80.6% [Accept fractional answers. Allow ecf, but check nos.] (1) 3[10]

185. Reason for non – destructive testing

Sensible reason e.g.

• destroyed rails would require replacement

• trains continuously using tracks, so removing them would cause greater disruption

• saves money 1

Description of sound wave

Particles oscillate / vibrate (not move)

… in direction of wave propagation/longitudinal

causes rarefactions and compressions

[Marks may be gained from suitable diagram] 3

Show that wavelength about 1.5 × 10 –3 m

Wavespeed = frequency × wavelength, v = fλ , any correct arrangt (1)

Wavelength = wavespeed ÷ frequency

= 5900 m s–1 ÷ 4 000 000 Hz

= 1.48 × 10–3 m (1) 2

Meanings

Frequency:

Number of oscillations/waves per second/unit time (may be 4 000 000 oscillations per second) (1)

Wavelength: [may be from diagram]

Distance between 2 points in phase/2 compressions/2 rarefactions (1)

Distance between successive points in phase etc. (1) 3

Calculation of length of track

Length of track = area under graph (or sign of finding area, e.g. shading) or 3 calculated distances using const acceleration formulae (1)

Use of 18 m s–1 as a speed x a time in a calculation (1)

E.g., distance = 0.5 × (116 s + 96 s) × 18 m s–1

= 1908 m (1) 3[12]

186. Explanation of superposition

When 2 (or more) waves meet / cross / coincide … (1)

Reference to combined effect of waves, e.g. add displacement /amplitude – may be a diagram [constructive/destructive interferencenot sufficient without implication of addition] (1) 2

Explanation of cancellation effect

Any 3 from the following:

• path/phase difference between direct and reflected waves

• destructive interference/superposition

• path difference is (n + ½)λ / phase diff π /180o / waves in antiphase / out of phase

• “crest” from one wave cancels “trough” from other 3

Reason for changes

Any 3 from the following:

• movement changes path of reflected waves

• so changes path difference

• A movement of 75 cm is about ¼ wavelength

• waves reflected so path difference changed to ½ wavelength

• enough to change from antiphase to in phase / change in phase difference

• causes constructive interference/superposition 3[8]

187. Device

Potential divider or potentiometer 1

Voltmeter reading

A 9.0 V (1)

B 0 V (1) 2

Diagram

Label X two thirds of the way down from A [Allow e.c.f.] 1

Explanation

Any 3 points from the following:

• lamp in parallel with lowest 1/3 of AB

• when resistors in parallel, resistance decreases

• p.d. across lamp reduced to below 3 V

• current divides

• no longer enough current to light lamp 3[7]

188. Speed of ultrasound

Use of υ = s/t (1)

= 150 × 10–3 (m) ÷132 × 10–6 (s)

= 1140 m s–1 (1) 2

Change of trace

Extra pulse(s)

OR

Reflected pulse moves closer 1

Principle of Doppler probe

3 points from:

• Arrange probe so that soup is approaching

• Soup reflects ultrasound

• with changed frequency/wavelength

• change in frequency/wavelength depends on speed

• Probe detects frequency of reflected ultrasound

• Use of diagrams showing waves 3

Determination of speed

1 point from:

• Frequency/wavelength change

• Angle between ultrasound direction and direction of flow of soup 1

Comment

Lumps give larger reflections

Lumps travel slower 1[8]

189. Wavelength range

465 – 720 nm (± ½ square) 1

Sketch graph

Scale (No more than 90 – 100%)

AND all graph between 96% and 99% (1)

Inversion – in shape with 2 peaks (at 510 and 680 nm) (1) 2

Wavelength

(µ = υ 1 / υ 2 = fλ 1 /fλ 2) λ 1 = 360 nm × 1.38 (1)

(= 497 nm) 1

Explanation

Thickness = λ /4 OR path difference = 180 nm (1)

Path difference = λ /2 (1)

Minimum reflection needs destructive interference betweenreflected rays from front and back of coating (1) 3

Difference between unpolarised and plane polarised light

Unpolarised light consists of waves vibrating in all planes(perpendicular to direction of propagation) (1)

Polarised light consists of waves vibrating in one plane only (1)

OR

Diagrams showing:

Waves / rays in 1 plane (1)

Waves / rays in many planes (1) Max 2[9]

190. Energy of photon of light

E = hf = 6.63 × 10–34 J s × 6.0 × 1014 Hz = 3.98 × 10–19 (J) 1

Graph

Points correct (± ½ square) (2)

Single straight line of best fit (NOT giving intercept below 4.5 × 1014) (1)

Line drawn as far as f axis (1) 4

Value for h

Large triangle [at least 7 cm on K.E. axis] (1)

Gradient = e.g. (6.05 – 4.55) × 10 14 / 1.0 × 10–19 = 1.5 × 1033 (1)

h = 1/gradient = 6.67 × 10–34 J s (1) 3

Value of φ

Reading co-ordinates of a fixed point on graph (e.g. 0, 4.55 × 1014) (1)

φ from equation, e.g.

so φ = frequency intercept × h

= e.g. 4.55 × 1014 × 6.67 × 10–34

= 3.03 × 10–19 J (1) 2

Explanation

Not enough energy [OR frequency too low]

For 2nd mark, numerical/added detail required,

e.g calculation: E = 6.63 × 10–34 × 4.5 × 1014 Hz = 2.98 × 10–19 < φ

OR threshold frequency read from graph 2[12]

191. Circuit diagram

(Variable) resistor symbol (1)

Voltmeter in parallel with cell/resistor (1)

Ammeter in series [even if R missing] (1) 3

A V

Maximum power available

Use of P = IV (1)

Any pair of values which round to 1.4 W (1) 2

Analysis of data

1000 W m–2 P = 1.4 W

100 W m–2 P = 0.11 W OR at least one further value of P (1)

Inspection of ratio

[e.g. 100 ÷ 1000, 0.11 ÷ 1.4, 1000 ÷ 1.4] (1)

Comment based on candidate’s result

[e.g. Yes as ratio is similar] (1) 3

Graph

E/V + scale : 2 large squares = 0.05 V (1)

Points (1)

Straight line good fit (1)

t/°C + scale: 1 large square = 10° (1)

[No penalty if t vs E] 4

Determination of mathematical relationship

Intercept = 0.640 →0.655 (1)

A gradient evaluated (1)

= 2.1 → 2.3 × 10–3 (1)

E = –2.2 × 10–3t + 0.65 [e.c.f.] (1) 4

Other axes:

Intercept 290 → 310 (1)

Gradient (1)

440 → 460 (1)

t = –450E + 300 (1)

Determination of light power from the sun

Attempted evaluation of an area (1)

= 0.13 → 0.17 [–1 if 10–6] (2)

[0.10 → 0.20 (1)]

Their answer × 4.0 = ……….(W) [no u.e.] (1) 4[20]

192. Explanation of “coherent”

In / constant phase (difference) (1)

symbol 51 \f "Monotype Sorts" \s 123 (1) 1

Power delivered by laser

P = 1510400

40−× (1)

= 1 × 1014 W (1) 2

Energy level change

υ = fλ / f = 9

8

101050

103−×

× [–1 if omit 10–9] (1)

Use of E = hf / 6.6 × 10–34 × 9

8

101050

103−×

× (1)

[If f = 1/T used, give this mark]

= 1.9 × 10–19 J (1) 3[6]

193.

Base unit Derivedunit

Basequantity

Derivedquantity

Mass

Charge

Joule

Ampere

Volt

(1)

(1)

(1)

(1)

(1)

5[5]

194. Explanation:As the temperature rises, the resistance decreases (1)As the resistance decreases, so the ammeter reading/current increases (1)[No mention of resistance 0/2][Current controls temperature → controls R is wrong physics – 0/2][If T changes so R changes OR vice versa so I changes 1 mark only][Correct static relationship (extremes) 1 mark only]

Reading on milliammeter:At 20 °C R = 1.4 (kΩ ) (1)Substitute correctly in V = IR i.e. 6 V = I × 1400 Ω (1)

[Allow their incorrect R; ignore 10x] (1)Milliammeter reading = 0.0043 A OR 4.3 mA [no e.c.f.] (1)[Accept 4 mA/4.2 mA] 5

[5]

195. Current:

Conversion, i.e. 0.94 × 10–3 m s–1 (1)

Use of 1.6 × 10–19 C (1)Answer 3.0 A

1.0 × 1029 m–3 × 0.20 × 10–6 m2 × 1.6 × 10–19 C × 0.94 × 10–3 mm s–1 (1)

Current = 3.0 A [Accept 2.8 A if 0.9 × 10–3 used.] 3

Resistance:

Recall R = A

lρ (1)

Substitution:

R = 26-

8

m10 0.20

m4.0mΩ107.1

××× −

(1)

Resistance = 0.34 Ω (1) 3

Potential difference:Potential difference = 3.0 A × 0.34 Ω (1)= 1.0 V (1.02 V)[Mark for correct substitution of their values or for the answer of 1.0 V] 1

Explanation:(Increasing resistivity) increases resistance (1)Leads to a smaller current (1) 2

Comparison:Drift velocity decreases (in second wire) (1) 1[Allow V1/V2 = I1/I2][Allow e.c.f. answer consistent with their current answer][Resistivity up, current down

ρ up, I down / 2 (2nd mark)][10]

196. E.m.f.Use of intercept mentioned/indicated on graph/when I = 0 (1)e.m.f. = 1.5 V (1) 2

Use of graph:Internal resistance: mention use of gradient/use of numbers/triangle on graph (1)Internal resistance = 0.5 Ω (1) 2

[Finds r and/or V by substitution, can score answer mark, but NOT method mark]

[Gradient = Ω5.00.1

0.15.1 =−

They might write gradient = 5.10.1

5.1 = Ω OR gradient = 2.1

5.1 - ignore signs]

Graph:Negative gradient of a straight line starting anywhere (1)from (0.0, 3.0) [No e.c.f.] (1)heading for (1.0, 2.0[1.9 → 2.1])/gradient of −1 [Consequent mark] 1 3

Filament lamp: any two ofif the variable resistor is set to zero [NOT, as RVR down] (1)the lamp prevents I from becoming too large (1)and overloading/damaging the ammeter (1)bulb acting like a fuse OR prevents short circuit (1)bulb means there is still resistance in circuit (1) Max 2

[9]

197. Table:

Description Type of wave

A wave capable of causing photo-electric emission of electrons

Ultraviolet (1)

A wave whose vibrations are parallel to the direction of propagation of the wave

Sound (1)

A transverse wave of wavelength

5 × 10–6 m

Infrared (1)

The wave of highest frequency Ultraviolet (1)

4[4]

198. Explanation:

• waves diffracted from each slit/each slit acts as a source

• these superpose/interfere (1)

• maxima/reinforcement – waves in phase/pd = nλ (1)[or on a diagram][crest & crest] (1)

• minima/cancellation – waves in antiphase/pd = (n+1/2)λ (1)[or on a diagram][crest and trough] [not just ‘out of phase’] (1)

• phase or path difference changes as move around AB (1) Max 4

Determination of wavelength:Use of wavelength = p.d. [incorrect use of xs/D 1/3 max] (1)3 × (path difference. e.g.78 − 66 mm) (1)= 36 mm[Range 30 – 42 mm] (1) 3

Explanation:Less/No diffraction/spreading (1)∴ waves will not superimpose/overlap as much (1) 2

Explanation:Fixed phase relationship/constant phase difference (1)Both waves derived from single source [transmitter ⇒] (1) 2

[11]

199. Calculation:

E = hc/λ [seen or implied] (1)physically correct substitutions (1)

÷ 1.6 × 10–19 eV J–1 (1)5.78 eV (1) 4

Maximum kinetic energy:3.52 eV [ecf but not if –ve.] (1)

Stopping potential:3.52 V [Allow e.c.f., but not signs] (1) 2

Annotated graph:Position of S (1)Cuts V axis between origin and existing graph (1)Similar shape [I levels off up/below existing line] (1) 3

[9]

200. Completion of circuit:

Ammeter and voltmeter used [correct symbols required] (1)

Ammeter in series, voltmeter in parallel (1)

[Do not penalise variable resistor in series] 2

Explanation of difference voltages:

Any two from:

• Internal resistance of cell/battery

• When current flows, energy transferred to / lost by internal resistance/heating in cell

• Hence voltage across internal resistance/ “lost volts”

• Reduced terminal p.d. / V = E – Ir / E = V + Ir 2

Show that internal resistance is about 0.6 Ω :

R = V/I

= (1.5 – 1.25) (1)

¸ 0.4 A

= 0.63 Ω [No u.e.] (1) 2

Calculation of resistance of bulb:

R = V/I (1)

= 1.25 V ÷ 0.4 A

= 3.1 Ω (1) 2

Explanation of lower resistance with ohmmeter:

Identify lower temperature with ohmmeter (1)

Lattice ions’/atoms’ vibrations impede electrons (1)

[Allow converse argument] 2[10]

201. Energy of photon of green light:

f = m105.58

sm1037–

–18

××

=λc

= 5.38 × 1014 Hz (1)

E = hf = 6.63 × 10–34 J s × 5.38 × 1014 Hz (1)

= 3.56 × 10–19 J 2

Diagram:

Larger gap identified (1)

Downwards arrow between levels of same element (1)

2[4]

202. Diffraction:

The spreading out of waves when they pass through a narrow slit or around an object (1)

Superposition:

Two or more waves adding (1)

to give a resultant wave [credit annotated diagrams] (1)

Quantum:

A discrete/indivisible quantity (1) 4

Particles:

Photon/electron (1) 1

What the passage tells us:

Any 2 points from:

• large objects can show wave-particle duality

• quantum explanations now used in “classical” solutions

• quantum used to deal with sub-atomic particles andclassical with things we can see Max 2

[7]

203. Explanation of why it is a good approximation:

Resistance of connecting lead is (very) small (1)

So I × R(very) small = (very) small p.d./e–1s do little work so p.d. small/r small (1)

compared with rest of the circuit so p.d. small

2

Circumstances where approximation might break down:

If current is large OR resistance of rest of circuit is small (1)

[Not high voltage/long lead/thin lead/high resistivity lead/hot lead]

1

Calculation:

Use of R = A

lρ with A attempted × sectional area (1)

Correct use of 16 (1)

Use of V = IR (1)

0.036 V (1)

4[10]

204. Wavelength of the microwaves:

λ = 442 mm – 420 mm (1)

= 22 mm [2.2 cm, 0.22 m] (1)

Frequency of microwaves:

Use of c = fλ with λ from above substituted OR if no attempt, then (1)

C = 3. × 108 substituted

1.4 × 1010 Hz [e.c.f. λ above] (1)

4

Maximum Q and minimum D marked on diagram:

Either Q (1)

Any D (1)

2

I n c o m i n g

m i c r o w a v e s

M e t a l s h e e tw i t h t w o s l i t s

4 2 0 m m

4 4 2 m m

P

R

D

D

D

D

Q

Q

Q

X

X

Why a maximum would not be detected at P:

Wavelength of sound wave = 0.3 m (1)

Path difference at P is not whole wavelength (1)

2

[OR valid reference to phase difference OR λ sound greater so no

diffraction with this slit width OR valid reference to λ = xs/D][8]

205. Ionisation energy of atomic hydrogen:

13.6 eV OR 2.18 × 10–18 J [– sign, X] (1)

1

Why 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)

OR

Work/energy is given out when the electrons/atoms move to the groundstate, so energy now less than 0, i.e. negative (1)(1)

OR

the 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 = 0eV

2nd 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 × 10–19)

OR

λ = )106.1(89.1

1031063.619

834

−

−

×××××

[Their E] (1)

= 6.6 × 10–7 (m) [6.5 – 6.7] (1)

3

Production 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 bands

Dark on light/light on dark NOT equally spaced (1)

1

Absorption spectrum:

White light through gas in container (1)Diffraction grating/prism/spectrometer (1)Must be dark lines on bright background (1)

[9]

206. Number of carriers or electrons per unit volume / per m3 /carrier density/electron density (1)

[Not charge density / concentration]

Drift velocity OR drift speed OR average/mean/net/overall velocity (1) 2

[Not just velocity; not speed unless drift]

m–3 (1)

m2 As m s–1 (1)

Multiply and reduce to A (1) 3

[Base units not needed][Mixed units and symbols could get the third mark]

[mA = m–1 loses 1 mark]

Metal:

M : l a r g e s o t h e r e i s a c u r r e n tn

I n s u l a t o r

I : z e r o ( n e g l i g i b l e ) / v e r y s m a l l s o l e s sc u r r e n t ( o r z e r o c u r r e n t )

n

n : i n m e t a l l a r g e rn m u c h

C u r r e n t i n m e t a l i s l a r g e r

( 1 )

( 1 )

2

[Ignore anything about v. Allow e.g. electron density for n][7]

207. Use R = ρ l/A OR correct rearrangement OR plot R → l gradient = ρ /A (1)[Symbols or words]

With A = tw (1) 2

l = RA/ρ [Rearrangement mark symbols or numbers] (1)

Use of A = tw (1)

[Correct physical quantities substituted but ignoring unit errors, powers of 10]

= 110 m

[111 m] (1) 3

Reduce width/w of strip OR use thinner/t foil [Not reduce A; not increase T, V, I] (1)

Smaller w/t/A will be less accurate OR have larger error OR larger R will be more accurate (1) 2

[Increase w or t, could give e.c.f. to increased accuracy][7]

208. I 2 R / (ε I – I 2 r) / R

Ir 2)( −ε (1)

I 2 r / (ε I – I 2 r) R

Ir 2)( −ε(1)

ε I OR I 2 R + I 2 r / ε 2 / (R + r) (1)

ε I = I 2 R + I 2 r OR (It = I 2 RT + I 2 rt / their (iii) = their (i) + their (ii) (1)

Cancel I (OR I and t) and arrange [only if energy equation is correct] (1) 5

Maximum current occurs when R = 0 (1)

Imax = ε /r (1) 2

OR larger r means smaller I (1 mark)

1 MΩ [Could be underlined OR circled] (1)

It gives the smallest current (1)

[If 100 kΩ this reason: 1 only] 2[9]

209. No, because V is not proportional to I OR not straight line through origin / (1)only conducts above 0.5 V / resistance changes 1

Use of R = 0.74 / current from graph (1)

= 9.25 Ω [9.0 – 9.5 Ω ] [Minimum 2 significant figures] (1) 2

Calculation of p.d. across R [8.26]

Calculation of total resistance[109 – 115]

Ratio R: ratio V E = Σ IR (1)

÷ I – diode resistance [9] Correct substitutions

Correct substitutions (1)

103 Ω [100 – 106] (1)

3

[If not vertical line, 0/2]

0 . 7 0 . 7≠ 0 . 7

A n y t h i n g ( g a p , c u r v e , b e l o w a x i s )

( 1 )( 1 )

( 1 )( 0 )

( 1 )( 0 ) (1)(1) 2

[Otherwise 0 0 ][8]

210. Potential difference = charge

ywork/energ OR

current

power

OR in words: work done in moving 1 coulomb of charge between two points. (1) 1

Unit: volt OR J C–1 OR V (1) 1

Base units: kg m2 A–1 s–3 (1)(1) 2

[2/2 possible even if final answers wrong for recognising that As = C J = Nm][4]

211. Current in motor:

V

PI = = 300 000 W / 420 V

= 714A [allow710][no u.e.] (1) 1

Problem:

Overheating in wires OR circuit/motor becomes hot

OR Need thick/large/heavy cables

OR other sensible comment (1) 1

Why e.m.f. of battery must be more than 420 V:

Mention of internal resistance (1)

Detail e.g. loss of p.d. inside battery when current delivered/ lost volts (1)

OR equations used correctly 2 marks 2

Overall efficiency of motor:

K.E. gained = ½ mυ 2

= ½ × 1160 ×107 2 J

= 6.64 M J (1)

Energy input = P × t

= 300 000 × 100 J

= 30 M J (1)

⇒ efficiency = out/in × 100

= 6.64/30 × 100 =22% (1) 3

Reasons for energy losses: (1)

Work (done) against air resistance (1)

Work (done) against friction (1)

Heating in wires of circuit (1)

Heating, in battery (1)

Heating in motor coils (1) Max 2

OR other sensible comments e.g. sound

[Friction or heat loss scores zero unless detailed e.g. heat due to friction/air

resistance √ but heat to surroundings ×]

[9]

212. Explanation of formula:

(For fundamental) λ = 2 l (1)

⇒ υ = λ × f [stated or used]H3 2 × B3 × D3 (1) 2

How value is calculated:

Volume = π r2 × l

3

23–

m12

105.2π ×

××= (1)(1)

OR 23–

2

10mmindiameterπ

×

OR P1 * (0.001 * C5/2) Λ 2

OR similar valid route

[ for 2

)diam( 2

× π , for factor 10–3] 2

Value in G4:

Mass/metre = ρ × volume/metre

OR

= 1150 × 0.000 000 79 kg (1)

= 0.00091 kg m –1 [no u.e.] (1) 2

Formula in cell I3:

υ = µ/T

⇒ T = µ υ 2 (1)

⇒ I3 = H3 * H3 * G3

OR H3 Λ 2 * G3 (1) 2

Comment:

No + reason (e.g. 133 >> 47) (1)

OR

Yes + reason (e.g. 47, 64, 133 all same order of magnitude) (1)

More detail, e.g, f changes by factor 32, OR l by factor of 15. T only by factor 2.5⇒ similar Ts. (1)(1)

OR other sensible points. 3[11]

213. Why resistance changes:

Wire lengthens OR cross–sectional area OR diameter reduces (1)

Use of R = ρ l/A to explain [R and l, R ∝ 1/A (1) 2

Advantage:

A long length of wire OR small area OR multiple stretching (1) 1

Diagram:

Circuit with ammeter in series (1)voltmeter in parallel (with strain gauge) (1)

OR multimeter across strain gauge (1)(1)[Multimeter with power supply – 1 only] Max 2

Resistance:

R = ρ l /A

= 4.9 × 10–7 Ω m × 0.2 m/π × (2 × 10 –4 m /2)2 (1)[i.e. area = (1)] (1)

= 3.1Ω (1) 3[8]

214. Diagrams:

Diagram showing 2 waves π radians out of phase (1)

Adding to give (almost) zero amplitude (1)

Reference to destructive interference (1) Max 2

Wavelength of red light:

For example, red wavelength is 1.5 times blue wavelength (1)[OR red wavelength is 50% more than blue wavelength]

= 1.5 × 460 nm = 690 nm (1) 2

Dark bands :

Spacing = 4.0 mm (1) 1

Explanation of pattern:

Sunlight has a range of frequencies/colours (1)

Gaps between part of feather (act as slits) (1)

Different colours [OR gap width] in the sunlight diffracted by different amounts (1)

Red light bends more [OR blue less] hence coloured edges (1)[No colours linked to refraction] Max 3

[8]

215. Threshold wave:

Electron requires certain amount of energy to escape from surface (1)

This energy comes from one photon (1)

Use of E = hf (1)

(So photon needs) minimum frequency (1)

Hence maximum wavelength

OR use of E = hc/λ (1) Max 4

Work function:

f = c/λ = 3.0 × 108 / 700 × 10–9 m (1)

= 4.28 × 10 14 Hz (1)

E = hf = 6.63 ×10–34 J s × 4.28 × 10 14 Hz = 2.84 × 10–19 (J) [Allow e.c.f.] (1) 3

Circuit :

Circuit showing resistors only in series (1)

Potentials labelled (1)[Use of potential divider – allowed]Resistor values 1: 1: 1 OR 1:2 (1) Max 2

Suggestion:

Cosmic rays travel more slowly than light (1) 1[10]

216. Light from sky:

Light is polarised (1) 1

Change 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) 1

Radio 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]

217. Definition of symbols:

n = number of electrons/carriers per unit volume (per m3)ORelectron (or carrier) density (1)

υ = average (OR drift) velocity (OR speed) (1) 2

Ratio Value Explanation

x

y

n

n 1 Same material (1) (1)

x

y

l

l 1 Connected in series/Kirchoff’s 1st law/conservation of charge/current is the same (1) (1)

x

y

v

v 2 A is halved so ν double[Accept qualitative, e.g. A ↓ so v ↑, or goodanalogy] (1) (1)

6

[Accept e.g. ny = nx.....]

[No e.c.f ]

[NB Mark value first, without looking at explanation. If value correct, mark explanation. If value wrong, don’t mark explanation except: if υ y/υ x = ½ or 1:2, see if explanation is correct physics, and if so give (1). No e.c.f.]

[8]

218. Demonstration that resistance is 0.085 Ω :

R = ρ l/A (1)

= 1.7 ×10–8 Ω m ×20 m / (4.0 ×10–6 m2) (1) 2

Calculation of voltage drop:

V = 37 A × 0.085 Ω (1)

= 3.1 V × 2 = 6.3 V [Not if Vshower then found] (1) 2

[Only one conductor, leading to 3.1 V, gets 1st mark][Nothing if wires in parallel]

Explanation:

Lower resistance/R = 0.057 Ω /less voltage drop/new V=3

2old V (1)

Power dissipated in cable/energy wasted/wire not so hotOR more p.d/current/power to showerOR system more efficient (1) 2

[6]

219. Calculation of kinetic energy:

f = λ

1–8103 sm×(E = hf = 1.63 ×10–17 J) (1)

φ converted to J: 6.20 × 1.6 × 10–19 OR Photon energy converted toeV: 1.63 ÷ 1.6 × 10

(Subtract to obtain kinetic energy)

Kinetic energy = (1.5 – 1.56) × 10–17 J[OR 95.7/97.4 eV]

[Beware 1.6398 0/3; > 101 eV 0/3]

Demonstration of speed of electrons:

1.53 × 10–17 J = ½ × 9.11 × 10–31 kg × υ 2 (1)[e.c.f their kinetic energy in joules]

υ = 5.8 × 106 m s–1 (1)[If υ is not between 5 and 7 must comment to get mark]

[5]

220. Explanation: .

I= E/r + R (1) 1

Appropriate formula for cell E9:

C9 * D9 OR RI OR 1 Ω × 4 A (1) 1

Appropriate formula for cell F 11

D11 *E11 OR VI OR 3A × 6V OR C11 * D11 *D11OR RI2 OR 2 Ω × (3 A)2 (1) 1

Short circuit current:

6 A (1) 1

Explanation:

r and R in series → potential division (1) 1

as R ↑, r constant → R has greater share of 12 V (1) 1

OR other valid argument

Sketch graph of power against resistance:

P / W

0 2 1 0 R / Ω

1 8

18 (1)2 (1)Shape including asymptote (1) 3

Comment:

Maximum when R = r (1)

in accordance with maximum power theorem (1)

OR P → 0 as R → ∞ (1) Max 2[11]

221. Explanation of words:

Coherent

Same frequency and constant phase relationship (1) 1

Standing wave

Any two points from:

Superposition/interference

Two (or more) wavetrains passing through each other

Having equal A, f, λ

+ system of nodes and antinodes (1) (1) 2

Position of one antinode marked on diagram

Correctly marked A (in centre of rings – hot zone) (1) 1

Wavelength demonstration:

λ = c/f = 3 × 108 /2.45 × 109 m

= 12.2 cm (1) 1

Path difference:

(22.1 + 14) – (20 + 10) cm

= 6.1 cm (1) 1

Explanation:

6.1 cm = ½ × λ (1) 1

Waves at X in antiphase/ destructive interference (1) 1

→ node (1) 1

Explanation of how two separate microwave frequencies overcomesuneven heating problem:

Different wavelengths (1) 1

So a path difference which gives destructive interference atone wavelength may not do so at another (1) 1

[11]

222. Explanation of line spectra:

Specific frequencies or wavelengths (1)

Detail, e.g. absorption/emission (1)

OR within narrow band of wavelengths 2

Explanation 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]

223. Why warm surface water floats:

Cold water is denser than warm water (1) 1

Explanation of why ultrasound waves reflect thermocline:

This is surface separating layers of different density (1) 1

Explanation of why submarine is difficult to detect:

Ultrasound from ship partially reflects upwards fromthermocline so little is transmitted (1)

Any reflected sonar from submarine partially reflectsdownwards from thermocline (1) 2

Explanation of why sonar cannot be used from a satellite:

Lack of medium to transmit sound waves from satellite 1

Calculation of time between emission and detection of radar pulse:

2s /c (1)

= 2 × 6.0 × 107 m ÷ 3.0 × 108 ms–1 = 0.4 s (1) 2

Calculation of minimum change in height of ocean:

Minimum observable distance

= ct = 3.0 × 108 m s–1 × 1.0 × 10–9 s = 0.30 m (1)

so change in ocean height = 0.15 m (1) 2

Possible problem:

Sensible answer eg (1)

atmospheric pressure could change ocean height

bulge not large enough compared with waves

tidal effects

whales 1[10]

224. Explanation:

Light hits glass–juice boundary at less than the critical angle (1)

And is refracted into the juice (1) 2

Marking angles on diagram:

the critical angle C – between ray and normal on prism/liquid face (1)

an incident angle i – between incident ray and normal atair/ glass or glass air interface (1)

a refracted angle r – between refracted ray and normalat air/glass or glass air interface (1) 3

Explanation of term critical angle:

The angle in the more (1)

dense medium where the refracted angle in the less dense medium is 90 (1) 2

Plot of results on grid:

[NB Axes are labelled on the grid]

Scales: y–axis (1)

x–axis (1)

Points correctly plotted (1) 4

Best fit line (curve expected) (1)

Refractive index found from graph:

Value = 1.400 ± 0.002 (1) 1[12]

225. (a) Mark the method before marking the circuit

Suitable circuit A

V

Short circuit option

V

A

What is measured

Set of readings of V and I

V and I Two sets of V and I

V and I

What is then done

Plot V against I Record V for open circuit

Substitute inV = E – Ir

Record V for open circuit

Finding E and r

E = interceptr = – gradient

E = open circuit voltager from V = E–Ir

Solve simultaneous equations

E=open circuit voltage r from r=E–Ir

-

Suitable circuit

V RA

R

Potentiometer

T oP o t

What is measured

V for known R I for known R Two sets of I and R

l for known R

What is then done

Record V for open circuit

Record V for open circuit

Substitute in E = I(R + r)

l’ for open circuit

Finding E and r E = open E = open Solve E from l’

circuit voltage r fromE/V = (R + r)/R

circuitvoltager fromE = I(R + r)

simultaneous equations

(calibrated)

R

r)(R

l

l +=′

Mark other procedures in a similar way 4

[Mark text, then tick for circuit if it does the job described.

If diagram alone, ask if it can do the job and give mark if yes]

(b) (i) p.d. across battery:

V = E − Ir

= 12.0 V − 3.0 A × 3.0 Ω (substitution)

= 3.0 V 2

(ii) Straight line from (0,12) to (3,3) (e.c.f.) 1

Current: 2.05 to 2.10 A 1[8]

[Allow correct intersection of their line (ignore shape), ± 0.05 A, of the characteristic with their graph, even if theirs is wrong. A line MUST be drawn for the last mark.]

226. Explanation of variation shown on the graph:

More electrons set free. Any one from: as temperature increases; thermal energy/vibration increases/resistance decreases/current increases 2

Resistance of thermistor:

V (across thermistor) = 1.20 VResistance ratio = voltage ratioR = 495 Ω

or

I = 0.80 V/330 Ω (substitution)= 0.002424 AV across thermistor = 1.20 VR = 1.20 V/0.002424 A= 495 Ω

or

I = 0.80 V/330 Ω= 0.002424 AR(total) = 2.0 V/0.002424 A

= 825 ΩR = 825 Ω − 330 Ω= 495 Ω 3

Explanation:

Thermistor resistance lowWhy: thermistor hotter/more current, power, charge carriersWhy v. small: thermistor takes smaller fraction of p.d. or ratio of p.d. 3

[8]

227. Circumstances under which two progressive waves produce a stationary wave:

Both transverse/longitudinal/same typeWaves have same frequency/wavelengthand travel/act in opposite directions/reflected back. Max 2 marks

Experiment using microwaves to produce stationary waves:

T r a n s m i t t e r M e t a l o rp l a t e b a c k w a r d s t r a n s m i t t e r

Adjust distance of transmitter/plateHow it could be shown that a stationary wave had been produced:Note readings on probe/detector/receiver form a series of maximum or minimum readings or zero 3

[5]

228.

M M M

RRC

C

One of compression C and one rarefaction R marked as above.Wavelength of wave = 11 - 11.6 cm (u.e.)

One of maximum displacement M marked as above [M, 5th, 6th, 7th ].Amplitude of wave = 8 (± 1 mm) [consequent mark]

[4]

229. Explanation:Photons/quanta

Photon releases / used electronEnergy/frequency of red < energy/frequency of ultra violetRed insufficient energy to release electrons so foil stays 4

Ultraviolet of greater intensity: foil/leaf collapses quicker/fasterRed light of greater intensity: no change/nothing 2

Observations if zinc plate and electroscope were positively charged:Foil rises or Foil stays same/nothing

as electrons released it becomes more Released electrons attracted back by

positive positive plate/more difficult to

release electrons 2[8]

230. Energy level diagram:

0 e V

– 1 . 5 1 e V

– 3 . 3 9 e V

– 1 3 . 6 V

− 13.6 → 0

− 1.51 → 0 AND − 3.39 → 0 ONLY 2

Why level labelled − 13.6 eV is called ground state:Correct reference to stability/lowest energy state/level of the electron/ atom/hydrogen 1

Transition which would result in emission of light of wavelength 660 nm:

Correct use of c =fλ or E = hc/λ or f = m10660

ms1039–

–18

××

Correct use of eV/J i.e. ÷ 1.6 × 10–19

∆ E = 1.88

Transition = 1.5 → 3.39

[May be a downward arrow on diagram] 4[7]

231. Completion of a correct circuit diagram:

Ammeter in series with lamp and supply [Ignore voltmeter position]

Voltmeter across lamp and ammeter [and maybe with ammeter 2

A

V+

–

0 - 1 2 V

Measurements:

Record voltmeter reading

Record corresponding ammeter reading [“corresponding” may beimplied]

Repeat for range of supply voltage settings [or currents] 3Labelled sketch:

I2 A

1 2 V V

Label axes I and V [with or without units]Graph line with correct curvature [overlook any tendency of the currentvalue to saturate]

Show 12 V, 2 A correctly [Allow 12 and 2 if units are labelled on axes] 3

[The second mark is lost if axes are not labelled, unless 2 A and 12 V arepresent, with the units, to make sense of the axes.]

[8]

232. Diagram of torch circuit:

The lamp will light

Correct circuit 2

[Circuit showing one cell only is allowed one mark only unless the cell is labelled 4.5 V. If a resistor is included, allow first mark only unless it is clearly labelled in some way as an internal resistance.]

3 . 5 V / 3

0 . 3 A

3 . 5 V

Voltage across each circuit component and current in lamp:Either 3.5 V/3 shown across the terminals of one cell or 3.5 V acrossall three cells3.5 V shown to be across the lamp0.3 A flowing in the lamp [i.e. an isolated 0.3 A near the lamp does notscore] 3

Calculation of internal resistance of one of the cells:

Lost volts = 4.5 V - 3.5 V or 1.5 V – 3

V5.3

or total resistance = (4.5 V)/0.3 A) = 15 KΩ

Internal resistance of one cell = [(1.0 V)/(0.3 A)] ÷ 3

or [(0.33 V) (0.3 A)] or lamp resistance = (3.5 V) / (0.3 A)11.7 Ω= 1.1 Ω or = (3.3Ω )/3 = 1.1 Ω 3

[Some of these latter marks can be read from the diagram if it is so

labelled][8]

233. Use of graph to estimate work function of the metal:

φ = (6.63 × 10–34 J s) (6.0 × 1014 Hz) – (some value)

Value in brackets: (1.6 × 10–19 × 0.5 J)

3.2 × 10–19 J or 2 eV 3

Addition to axes of graph A obtained when intensity of light increased:

A starts at –0.5

A → larger than /max

Addition to axes of graph B obtained when frequency of light increased:

B starts at less than – 0.5

B → same of lower than /max 4[7]

234. Description:

EitherTwo connected dippers just touching/above the water

OrDipping beam or single source (1)reaches two slits (1)

Vibrated electrically (1)Level tank/shallow water/sloping sides (1)

-

EitherIlluminate project on to screen

OrUse stroboscope (1)to freeze the pattern (1)

Max 5

Diagram:(i) Correct line A - centre line (1)(ii) Correct line B (above or below A) (1)(iii) Correct line C (between A and B) (1)

both B and C correct (1) 4[Total 9 marks]

235. Ionisation energy:

2810 eV (4.5 ×10–16 J) (1)

Calculation of maximum wavelength:Energy in eV chosen above converted to joules (1)Use of λ = c/f (1)Maximum wavelength = 4.4 × 10–10 m (1)

Part of electromagnetic spectrum:γ -ray / X-ray (1) 5

Calculation of the de Broglie wavelength:λ = h/p p identified as momentum (1)Either m or υ correctly substituted (1)Wavelength = 1.1 × 10–13 m (1) 3

[Total 8 marks]

236. Experiments on the photoelectric effect show that

• the kinetic energy of photoelectrons released depends upon the frequency of the incident light and not on its intensity,

• light below a certain threshold frequency cannot release photoelectrons.

How do these conclusions support a particle theory but not a wave theory of light?Particle theory: E = hf implied packets/photons (1)

One photon releases one electron giving it k.e. (1)

Increase f ⇒ greater k.e. electrons (1)

Lower f; finally ke = O ie no electrons released Waves (1)

Energy depends on intensity / (amplitude)2 (1)

More intense light should give greater k.e–NOT SEEN (1)

More intense light gives more electrons but no change in maximum kinetic energy (1)

Waves continuous ∴ when enough are absorbed electrons should be released–NOT SEEN (1)

(6 marks)

Calculate the threshold wavelength for a metal surface which has a work function of 6.2 eV.6.2eV × 1.6 × 10–19 C (1)

Use of Ehc=λ (1)

Threshold wavelength = 2.0 × 10–7 m (1)

To which part of the electromagnetic spectrum does this wavelength belong?UV ecf their λ (1)

(4 marks)[Total 10 marks]

237. The diagram below shows a loudspeaker which sends a note of constant frequency towards a

vertical metal sheet. As the microphone is moved between the loudspeaker and the metal sheet the amplitude of the vertical trace on the oscilloscope continually changes several times between maximum and minimum values. This shows that a stationary wave has been set up in the space between the loudspeaker and the metal sheet.

M e t a ls h e e tL o u d s p e a k e r

M i c r o p h o n e

T o o s c i l l o s c o p e( t i m e b a s e o f f )

S i g n a lg e n e r a t o r

How has the stationary wave been produced?by superposition/interference (1)

with a reflected wave/wave of same speed and wavelength in opposite direction (1)

(2 marks)

State how the stationary wave pattern changes when the frequency of the signal generator is doubled. Explain your answer.

Maxima/nodes/equivalent are closer together (1)since wavelength is halved (1)

(2 marks)

What measurements would you take, and how would you use them, to calculate the speed of sound in air?

Measure distance between minima/equivalent (1)

Repeat/take average (1)

Method of finding frequency (1)

λ = 2 × (node – node)/equivalent (1)

V = f × λ (1)

(Four marks maximum)

Other methods eligible for full marks.

(4 marks)

Suggest why the minima detected near the sheet are much smaller than those detected near the loudspeaker.

Near the sheet there is almost complete cancellation (1)

since incident and reflected waves are of almost equal amplitude (1)

(2 marks)[Total 10 marks]