Photoelectric effect — Paper 4 · A Level Physics 9702

7 The photoelectric effect may be summarised in terms of the word equation photon energy = work function energy + maximum kinetic energy of emitted electrons. (a) Explain (i) what is meant by a photon, … … … [2] (ii) why most electrons are emitted with kinetic energy less than the maximum. … … … [2] (b) Light of constant intensity is incident on a metal surface, causing electrons to be emitted. State and explain why the rate of emission of electrons changes as the frequency of the incident light is increased. … … … [2]

6 marks

Mark scheme: 7 (a) (i) quantum/packet/discrete amount of energy M1 electromagnetic mentioned A1 [2] (ii) max. k.e. corresponds to electron emitted from surface B1 energy is required to bring electron to surface B1 [2] (b) at higher frequency, fewer photons (per second) for same intensity M1 so rate of emission decreases A1 [2] (allow argument based on photoelectric efficiency)

This question in 9702/41 Oct/Nov 2006

7 (a) State three pieces of evidence provided by the photoelectric effect for a particulate For nature of electromagnetic radiation. Examiner’s Use 1. … … 2. … … 3. … … [3] (b) (i) Briefly describe the concept of a photon. … … … [2] (ii) Explain how lines in the emission spectrum of gases at low pressure provide evidence for discrete electron energy levels in atoms. … … … [2] (c) Three electron energy levels in atomic hydrogen are represented in Fig. 7.1. For Examiner’s Use increasing energy Fig. 7.1 The wavelengths of the spectral lines produced by electron transitions between these three energy levels are 486 nm, 656 nm and 1880 nm. (i) On Fig. 7.1, draw arrows to show the electron transitions between the energy levels that would give rise to these wavelengths. Label each arrow with the wavelength of the emitted photon. [3] (ii) Calculate the maximum change in energy of an electron when making transitions between these levels. energy = … J [3]

13 marks

Mark scheme: 7 (a) e.g. ‘instantaneous’ emission (of electrons) threshold frequency below which no emission (max) electron energy dependent on frequency (max) electron energy not dependent on intensity rate of emission (of electrons) depends on intensity (any three sensible suggestions, 1 each) B3 [3] (b) (i) ‘packet’ / quantum of energy M1 of electromagnetic energy / radiation A1 [2] (ii) discrete wavelengths mean photons have particular energies M1 energy of photon determined by energy change of (orbital) electron M1 so discrete energy levels A0 [2] (c) (i) three energy changes shown correctly B1 arrows ‘pointing’ in correct direction B1 wavelengths correctly identified B1 [3] (ii) chooses λ = 486 nm C1 ∆E = hc / λ C1 = (6.63 × 10–34 × 3.0 × 108) / (4.86 × 10–9) = 4.09 × 10–19 J (allow 2 s.f.) A1 [3]

This question in 9702/41 Oct/Nov 2008

8 (a) Explain why, for the photoelectric effect, the existence of a threshold frequency and a For very short emission time provide evidence for the particulate nature of electromagnetic Examiner’s radiation, as opposed to a wave theory. Use … … … … … … [4] (b) State and explain two relations in which the Planck constant h is the constant of proportionality. 1. … … … … 2. … … … … [6]

10 marks

Mark scheme: 8 (a) wave theory predicts any frequency would give rise to emission of electron M1 if exposure time is sufficiently long A1 photon has (specific value of) energy dependent on frequency M1 emission if energy greater than threshold / work function / energy to remove electron from surface A1 [4] (b) photon is packet/quantum of energy M1 of electromagnetic radiation A1 (photon) energy = h × frequency B1 [3] every particle has an (associated) wavelength B1 wavelength = h / p M1 where p is the momentum (of the particle) A1 [3]

This question in 9702/41 May/June 2009

7 (a) State an effect, one in each case, that provides evidence for For Examiner’s (i) the wave nature of a particle, Use … [1] (ii) the particulate nature of electromagnetic radiation. … [1] (b) Four electron energy levels in an atom are shown in Fig. 7.1. –0.87 × 10–19 J –1.36 × 10–19 J electron energy –2.42 × 10–19 J –5.44 × 10–19 J Fig. 7.1 (not to scale) An emission spectrum is associated with the electron transitions between these energy levels. For this spectrum, (i) state the number of lines, … [1] (ii) calculate the minimum wavelength. wavelength = … m [2]

5 marks

Mark scheme: 7 (a) (i) e.g. electron / particle diffraction B1 [1] (ii) e.g. photoelectric effect B1 [1] (b) (i) 6 A1 [1] (ii) change in energy = 4.57 × 10–19 J λ = hc / E C1 = (6.63 × 10–34 × 3.0 × 108) / (4.57 × 10–19) = 4.4 × 10–7 m A1 [2]

This question in 9702/41 Oct/Nov 2010

7 (a) State an effect, one in each case, that provides evidence for For Examiner’s (i) the wave nature of a particle, Use … [1] (ii) the particulate nature of electromagnetic radiation. … [1] (b) Four electron energy levels in an atom are shown in Fig. 7.1. –0.87 × 10–19 J –1.36 × 10–19 J electron energy –2.42 × 10–19 J –5.44 × 10–19 J Fig. 7.1 (not to scale) An emission spectrum is associated with the electron transitions between these energy levels. For this spectrum, (i) state the number of lines, … [1] (ii) calculate the minimum wavelength. wavelength = … m [2]

5 marks

Mark scheme: 7 (a) (i) e.g. electron / particle diffraction B1 [1] (ii) e.g. photoelectric effect B1 [1] (b) (i) 6 A1 [1] (ii) change in energy = 4.57 × 10–19 J λ = hc / E C1 = (6.63 × 10–34 × 3.0 × 108) / (4.57 × 10–19) = 4.4 × 10–7 m A1 [2]

This question in 9702/42 Oct/Nov 2010

8 (a) By reference to the photoelectric effect, state what is meant by the threshold frequency. For Examiner’s … Use … … [2] (b) The surface of a zinc plate has a work function of 5.8 × 10–19 J. In a particular laboratory experiment, ultraviolet light of wavelength 120 nm is incident on the zinc plate. A photoelectric current I is detected. In order to view the apparatus more clearly, a second lamp emitting light of wavelength 450 nm is switched on. No change is made to the ultraviolet lamp. Using appropriate calculations, state and explain the effect on the photoelectric current of switching on this second lamp. … … [4]

6 marks

Mark scheme: 8 (a) minimum frequency for electron to be emitted (from surface) M1 of electromagnetic radiation / light / photons A1 [2] (b) E = hc / λ or E = hf and c = fλ C1 either threshold wavelength = (6.63 × 10–34 × 3.0 × 108) / (5.8 × 10–19) = 340 nm or energy of 340 nm photon = 4.4 × 10–19 J or threshold frequency = 8.7 × 1014 Hz or 450 nm → 6.7 × 1014 Hz A1 appropriate comment comparing wavelengths / energies / frequencies B1 so no effect on photo-electric current B1 [4] GCE A LEVEL – October/November 2010 9702 43 Section B

This question in 9702/43 Oct/Nov 2010

7 Experiments are conducted to investigate the photoelectric effect. For Examiner’s (a) It is found that, on exposure of a metal surface to light, either electrons are emitted Use immediately or they are not emitted at all. Suggest why this observation does not support a wave theory of light. … … … … [3] (b) Data for the wavelength λ of the radiation incident on the metal surface and the maximum kinetic energy EK of the emitted electrons are shown in Fig. 7.1. λ / nm EK / 10–19 J 650 – 240 4.44 Fig. 7.1 (i) Without any calculation, suggest why no value is given for EK for radiation of wavelength 650 nm. … … [1] (ii) Use data from Fig. 7.1 to determine the work function energy of the surface. work function energy = … J [3] (c) Radiation of wavelength 240 nm gives rise to a maximum photoelectric current I. For The intensity of the incident radiation is maintained constant and the wavelength is now Examiner’s reduced. Use State and explain the effect of this change on (i) the maximum kinetic energy of the photoelectrons, … … … [2] (ii) the maximum photoelectric current I. … … … [2]

11 marks

Mark scheme: 7 (a) for a wave, electron can ‘collect’ energy continuously B1 for a wave, electron will always be emitted / electron will be emitted at all frequencies….. M1 after a sufficiently long delay A1 [3] (b) (i) either wavelength is longer than threshold wavelength or frequency is below the threshold frequency or photon energy is less than work function B1 [1] (ii) hc / λ = φ + EMAX C1 (6.63 × 10–34 × 3.0 × 108) / (240 × 10–9) = φ + 4.44 × 10–19 C1 φ = 3.8 × 10–19 J (allow 3.9 × 10–19 J) A1 [3] (c) (i) photon energy larger M1 so (maximum) kinetic energy is larger A1 [2] (ii) fewer photons (per unit time) M1 so (maximum) current is smaller A1 [2]

This question in 9702/41 May/June 2011

7 An explanation of the photoelectric effect includes the terms photon energy and work function For energy. Examiner’s Use (a) Explain what is meant by (i) a photon, … … … [2] (ii) work function energy. … … [1] (b) In an experiment to investigate the photoelectric effect, a student measures the wavelength λ of the light incident on a metal surface and the maximum kinetic energy 1 Emax of the emitted electrons. The variation with Emax of is shown in Fig. 7.1. λ 4 1 / 106 m–1 λ 3 2 1 0 –4 –3 –2 –1 0 1 2 3 4 Emax / 10–19 J Fig. 7.1 (i) The work function energy of the metal surface is Φ. State an equation, in terms of λ, Φ and Emax, to represent conservation of energy for the photoelectric effect. Explain any other symbols you use. … … … [2] (ii) Use your answer in (i) and Fig. 7.1 to determine For Examiner’s 1. the work function energy Φ of the metal surface, Use Φ = … J [2] 2. a value for the Planck constant. Planck constant = … J s [3]

10 marks

Mark scheme: 7 (a) (i) packet/quantum of energy M1 of electromagnetic radiation A1 [2] (ii) minimum energy to cause emission of an electron (from surface) B1 [1] (b) (i) hc/λ = Φ + Emax M1 c and h explained A1 [2] (ii) 1. either when 1/λ = 0, Φ = –Emax or evidence of use of x-axis intercept from graph or chooses point close to the line and substitutes values of 1/λ and Emax into hc/λ = Φ + Emax C1 Φ = 4.0 × 10–19 J (allow ±0.2 × 10–19 J) A1 [2] 2. either gradient of graph is 1/hc C1 gradient = 4.80 × 1024 → 5.06 × 1024 M1 h = 1/(gradient × 3.0 × 108) = 6.6 × 10–34 J s → 6.9 × 10–34 J s A1 or chooses point close to the line and substitutes values of 1/λ and Emax into hc/λ = Φ + Emax (C1) values of 1/λ and Emax are correct within half a square (M1) h = 6.6 × 10–34 J s → 6.9 × 10–34 J s (A1) [3] (Allow full credit for the correct use of any appropriate method) (Do not allow ‘circular’ calculations in part 2 that lead to the same value of Planck constant that was substituted in part 1)

This question in 9702/43 Oct/Nov 2011

7 The photoelectric effect may be represented by the equation For Examiner’s photon energy = work function energy + maximum kinetic energy of electron. Use (a) State what is meant by work function energy. … … [1] (b) The variation with frequency f of the maximum kinetic energy EK of photoelectrons emitted from the surface of sodium metal is shown in Fig. 7.1. 0.8 0.6 EK / eV 0.4 0.2 0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 f / 1014 Hz Fig. 7.1 Use the gradient of the graph of Fig. 7.1 to determine a value for the Planck constant h. Show your working. h = … J s [2] (c) The sodium metal in (b) has a work function energy of 2.4 eV. The sodium is replaced by For calcium which has a work function energy of 2.9 eV. Examiner’s Use On Fig. 7.1, draw a line to show the variation with frequency f of the maximum kinetic energy EK of photoelectrons emitted from the surface of calcium. [3]

6 marks

Mark scheme: 7 (a) minimum energy to remove an electron from the metal/surface B1 [1] (b) gradient = 4.17 × 10–15 (allow 4.1 → 4.3) C1 h = 4.15 × 10–15 × 1.6 × 10–19 or h = 4.1 to 4.3 × 10–15 eV s A1 = 6.6 × 10–34 J s A0 [2] (c) graph: straight line parallel to given line with intercept at any higher frequency B1 intercept at between 6.9 × 1014 Hz and 7.1 × 1014 Hz B1 [3] GCE AS/A LEVEL – May/June 2012 9702 42

This question in 9702/42 May/June 2012

8 (a) State what is meant by a photon. For Examiner’s … Use … … [2] (b) It has been observed that, where photoelectric emission of electrons takes place, there is negligible time delay between illumination of the surface and emission of an electron. State three other pieces of evidence provided by the photoelectric effect for the particulate nature of electromagnetic radiation. 1. … … 2. … … 3. … … [3] (c) The work function of a metal surface is 3.5 eV. Light of wavelength 450 nm is incident on the surface. Determine whether electrons will be emitted, by the photoelectric effect, from the surface. [3]

8 marks

Mark scheme: 8 (a) discrete quantity / packet / quantum of energy of electromagnetic radiation B1 energy of photon = Planck constant × frequency B1 [2] (b) threshold frequency (1) rate of emission is proportional to intensity (1) max. kinetic energy of electron dependent on frequency (1) max. kinetic energy independent of intensity (1) (any three, 1 each, max 3) B3 [3] (c) either E = hc/λ or hc/λ = eV C1 λ = 450 nm to give work function of 3.5 eV energy = 4.4 × 10–19 or 2.8 eV to give λ = 355 nm M1 2.8 eV < 3.5 eV so no emission 355 nm < 450 nm so no A1 [3] or work function = 3.5 eV threshold frequency = 8.45×1014 Hz C1 450 nm = 6.67×1014 Hz M1 6.67 × 1014 Hz < 8.45 × 1014 Hz A1 GCE AS/A LEVEL – October/November 2012 9702 41 Section B

This question in 9702/41 Oct/Nov 2012

8 (a) State what is meant by a photon. For Examiner’s … Use … … [2] (b) It has been observed that, where photoelectric emission of electrons takes place, there is negligible time delay between illumination of the surface and emission of an electron. State three other pieces of evidence provided by the photoelectric effect for the particulate nature of electromagnetic radiation. 1. … … 2. … … 3. … … [3] (c) The work function of a metal surface is 3.5 eV. Light of wavelength 450 nm is incident on the surface. Determine whether electrons will be emitted, by the photoelectric effect, from the surface. [3]

8 marks

Mark scheme: 8 (a) discrete quantity / packet / quantum of energy of electromagnetic radiation B1 energy of photon = Planck constant × frequency B1 [2] (b) threshold frequency (1) rate of emission is proportional to intensity (1) max. kinetic energy of electron dependent on frequency (1) max. kinetic energy independent of intensity (1) (any three, 1 each, max 3) B3 [3] (c) either E = hc/λ or hc/λ = eV C1 λ = 450 nm to give work function of 3.5 eV energy = 4.4 × 10–19 or 2.8 eV to give λ = 355 nm M1 2.8 eV < 3.5 eV so no emission 355 nm < 450 nm so no A1 [3] or work function = 3.5 eV threshold frequency = 8.45×1014 Hz C1 450 nm = 6.67×1014 Hz M1 6.67 × 1014 Hz < 8.45 × 1014 Hz A1 GCE AS/A LEVEL – October/November 2012 9702 42 Section B

This question in 9702/42 Oct/Nov 2012

7 Some data for the work function energy Φ and the threshold frequency f0 of some metal For surfaces are given in Fig. 7.1. Examiner’s Use metal Φ / 10–19 J f0 / 1014 Hz sodium 3.8 5.8 zinc 5.8 8.8 platinum 9.0 Fig. 7.1 (a) (i) State what is meant by the threshold frequency. … … … [2] (ii) Calculate the threshold frequency for platinum. threshold frequency = … Hz [2] (b) Electromagnetic radiation having a continuous spectrum of wavelengths between 300 nm and 600 nm is incident, in turn, on each of the metals listed in Fig. 7.1. Determine which metals, if any, will give rise to the emission of electrons. … … … … [2] (c) When light of a particular intensity and frequency is incident on a metal surface, electrons are emitted. State and explain the effect, if any, on the rate of emission of electrons from this surface for light of the same intensity and higher frequency. … … … … [3]

9 marks

Mark scheme: 7 (a) (i) lowest frequency of e.m. radiation M1 giving rise to emission of electrons (from the surface) A1 [2] (ii) E = hf C1 threshold frequency = (9.0 × 10–19) / (6.63 × 10–34) = 1.4 × 1015 Hz A1 [2] (b) either 300 nm ≡ 10 × 1015 Hz (and 600 nm ≡ 5.0 × 1014 Hz) or 300 nm ≡ 6.6 × 10–19 J (and 600 nm ≡ 3.3 × 10–19 J) or zinc λ0 = 340 nm, platinum λ0 = 220 nm (and sodium λ0 = 520 nm) M1 emission from sodium and zinc A1 [2] (c) each photon has larger energy M1 fewer photons per unit time M1 fewer electrons emitted per unit time A1 [3] GCE AS/A LEVEL – May/June 2013 9702 41

This question in 9702/41 May/June 2013

7 Some data for the work function energy Φ and the threshold frequency f0 of some metal For surfaces are given in Fig. 7.1. Examiner’s Use metal Φ / 10–19 J f0 / 1014 Hz sodium 3.8 5.8 zinc 5.8 8.8 platinum 9.0 Fig. 7.1 (a) (i) State what is meant by the threshold frequency. … … … [2] (ii) Calculate the threshold frequency for platinum. threshold frequency = … Hz [2] (b) Electromagnetic radiation having a continuous spectrum of wavelengths between 300 nm and 600 nm is incident, in turn, on each of the metals listed in Fig. 7.1. Determine which metals, if any, will give rise to the emission of electrons. … … … … [2] (c) When light of a particular intensity and frequency is incident on a metal surface, electrons are emitted. State and explain the effect, if any, on the rate of emission of electrons from this surface for light of the same intensity and higher frequency. … … … … [3]

9 marks

Mark scheme: 7 (a) (i) lowest frequency of e.m. radiation M1 giving rise to emission of electrons (from the surface) A1 [2] (ii) E = hf C1 threshold frequency = (9.0 × 10–19) / (6.63 × 10–34) = 1.4 × 1015 Hz A1 [2] (b) either 300 nm ≡ 10 × 1015 Hz (and 600 nm ≡ 5.0 × 1014 Hz) or 300 nm ≡ 6.6 × 10–19 J (and 600 nm ≡ 3.3 × 10–19 J) or zinc λ0 = 340 nm, platinum λ0 = 220 nm (and sodium λ0 = 520 nm) M1 emission from sodium and zinc A1 [2] (c) each photon has larger energy M1 fewer photons per unit time M1 fewer electrons emitted per unit time A1 [3] GCE AS/A LEVEL – May/June 2013 9702 43

This question in 9702/43 May/June 2013

7 (a) By reference to the photoelectric effect, explain For Examiner’s (i) what is meant by work function energy, Use … … … [2] (ii) why, even when the incident light is monochromatic, the emitted electrons have a range of kinetic energy up to a maximum value. … … … [2] (b) Electromagnetic radiation of frequency f is incident on a metal surface. The variation with frequency f of the maximum kinetic energy EMAX of electrons emitted from the surface is shown in Fig. 7.1. 4 EMAX / 10 –18 J 3 2 1 0 0 1 2 3 4 5 f / 1015 Hz Fig. 7.1 (i) Use Fig. 7.1 to determine the work function energy of the metal surface. For Examiner’s Use work function energy = … J [3] (ii) A second metal has a greater work function energy than that in (i). On Fig. 7.1, draw a line to show the variation with f of EMAX for this metal. [2] (iii) Explain why the graphs in (i) and (ii) do not depend on the intensity of the incident radiation. … … … [2]

11 marks

Mark scheme: 7 (a) (i) minimum photon energy B1 minimum energy to remove an electron (from the surface) B1 [2] (ii) either maximum KE is photon energy – work function energy or max KE when electron ejected from the surface B1 energies lower than max because energy required to bring electron to the surface B1 [2] (b) (i) threshold frequency = 1.0 × 1015 Hz (allow ±0.05 × 1015) C1 work function energy = hf0 C1 = 6.63 × 10–34 × 1.0 × 1015 = 6.63 × 10–19 J A1 [3] (allow alternative approaches based on use of co-ordinates of points on the line) (ii) sketch: straight line with same gradient M1 displaced to right A1 [2] (iii) intensity determines number of photons arriving per unit time B1 intensity determines number of electrons per unit time (not energy) B1 [2]

This question in 9702/43 Oct/Nov 2013

9 For a particular metal surface, it is observed that there is a minimum frequency of light below which photoelectric emission does not occur. This observation provides evidence for a particulate nature of electromagnetic radiation. (a) State three further observations from photoelectric emission that provide evidence for a particulate nature of electromagnetic radiation. 1. … … 2. … … 3. … … [3] (b) Some data for the variation with frequency f of the maximum kinetic energy EMAX of electrons emitted from a metal surface are shown in Fig. 9.1. 0.6 0.5 E MAX / eV 0.4 0.3 0.2 0.1 0 5.5 6.0 6.5 7.0 7.5 f / 1014 Hz Fig. 9.1 (i) Explain why emitted electrons may have kinetic energy less than the maximum at any particular frequency. … … … [2] (ii) Use Fig. 9.1 to determine 1. the threshold frequency, threshold frequency = … Hz [1] 2. the work function energy, in eV, of the metal surface. work function energy = … eV [3]

9 marks

Mark scheme: 9 (a) e.g. no time delay between illumination and emission max. (kinetic) energy of electron dependent on frequency max. (kinetic) energy of electron independent of intensity rate of emission of electrons dependent on/proportional to intensity (any three separate statements, one mark each, maximum 3) B3 [3] (b) (i) (photon) interaction with electron may be below surface B1 energy required to bring electron to surface B1 [2] GCE AS/A LEVEL – May/June 2014 9702 41 (ii) 1. threshold frequency = 5.8 × 1014 Hz A1 [1] 2. Φ = hf0 C1 = 6.63 × 10–34 × 5.8 × 1014 = 3.84 × 10–19 (J) C1 = (3.84 × 10–19)/(1.6 × 10–19) = 2.4 eV A1 [3] or hf = Φ + EMAX (C1) chooses point on line and substitutes values EMAX, f and h into equation with the units of the hf term converted from J to eV (C1) Φ = 2.4 eV (A1)

This question in 9702/41 May/June 2014

9 For a particular metal surface, it is observed that there is a minimum frequency of light below which photoelectric emission does not occur. This observation provides evidence for a particulate nature of electromagnetic radiation. (a) State three further observations from photoelectric emission that provide evidence for a particulate nature of electromagnetic radiation. 1. … … 2. … … 3. … … [3] (b) Some data for the variation with frequency f of the maximum kinetic energy EMAX of electrons emitted from a metal surface are shown in Fig. 9.1. 0.6 0.5 E MAX / eV 0.4 0.3 0.2 0.1 0 5.5 6.0 6.5 7.0 7.5 f / 1014 Hz Fig. 9.1 (i) Explain why emitted electrons may have kinetic energy less than the maximum at any particular frequency. … … … [2] (ii) Use Fig. 9.1 to determine 1. the threshold frequency, threshold frequency = … Hz [1] 2. the work function energy, in eV, of the metal surface. work function energy = … eV [3]

9 marks

Mark scheme: 9 (a) e.g. no time delay between illumination and emission max. (kinetic) energy of electron dependent on frequency max. (kinetic) energy of electron independent of intensity rate of emission of electrons dependent on/proportional to intensity (any three separate statements, one mark each, maximum 3) B3 [3] (b) (i) (photon) interaction with electron may be below surface B1 energy required to bring electron to surface B1 [2] GCE AS/A LEVEL – May/June 2014 9702 43 (ii) 1. threshold frequency = 5.8 × 1014 Hz A1 [1] 2. Φ = hf0 C1 = 6.63 × 10–34 × 5.8 × 1014 = 3.84 × 10–19 (J) C1 = (3.84 × 10–19)/(1.6 × 10–19) = 2.4 eV A1 [3] or hf = Φ + EMAX (C1) chooses point on line and substitutes values EMAX, f and h into equation with the units of the hf term converted from J to eV (C1) Φ = 2.4 eV (A1)

This question in 9702/43 May/June 2014

7 (a) By reference to the photoelectric effect, state what is meant by the threshold frequency. … … … [2] (b) Electrons are emitted from a metal surface when light of a particular wavelength is incident on the surface. Explain why the emitted electrons have a range of values of kinetic energy below a maximum value. … … … … [2] (c) The wavelength of the incident radiation is λ. The variation with 1/λ of the maximum kinetic energy EMAX of electrons emitted from a metal surface is shown in Fig. 7.1. 4.0 3.0 EMAX / 10–19 J 2.0 1.0 0 1.0 1.5 2.0 2.5 3.0 3.5 1 – / 106 m–1 h Fig. 7.1 (i) Use Fig. 7.1 to determine, without reference to the work function energy, the threshold frequency f0. f0 = … Hz [2] (ii) Use your answer in (i) to calculate the work function energy Φ. Φ = … J [2] (d) Caesium metal has a work function energy of 2.2 × 10–19 J. On the axes of Fig. 7.1, sketch a graph to show the variation with 1/λ of EMAX for caesium metal. [2]

10 marks

Mark scheme: 7 (a) minimum frequency of e.m. radiation/a photon (not “light”) M1 for emission of electrons from a surface A1 [2] (reference to light/UV rather than e.m. radiation, allow 1/2) (b) EMAX corresponds to electron emitted from surface B1 electron (below surface) requires energy to bring it to surface, so less than EMAX B1 [2] (c) (i) 1/λ0 = 1.85 × 106 (allow 1.82 to 1.88) C1 f0 = c / λ0 = 3.00 × 108 × 1.85 × 106 = 5.55 × 1014 Hz A1 [2] (ii) Φ = hf0 = 6.63 × 10–34 × 5.55 × 1014 (allow ECF from (c)(i)) C1 = 3.68 × 10–19 J A1 [2] (d) sketch: straight line with same gradient M1 intercept between 1.0 and 1.5 A1 [2]

This question in 9702/41 Oct/Nov 2015

7 (a) By reference to the photoelectric effect, state what is meant by the threshold frequency. … … … [2] (b) Electrons are emitted from a metal surface when light of a particular wavelength is incident on the surface. Explain why the emitted electrons have a range of values of kinetic energy below a maximum value. … … … … [2] (c) The wavelength of the incident radiation is λ. The variation with 1/λ of the maximum kinetic energy EMAX of electrons emitted from a metal surface is shown in Fig. 7.1. 4.0 3.0 EMAX / 10–19 J 2.0 1.0 0 1.0 1.5 2.0 2.5 3.0 3.5 1 – / 106 m–1 h Fig. 7.1 (i) Use Fig. 7.1 to determine, without reference to the work function energy, the threshold frequency f0. f0 = … Hz [2] (ii) Use your answer in (i) to calculate the work function energy Φ. Φ = … J [2] (d) Caesium metal has a work function energy of 2.2 × 10–19 J. On the axes of Fig. 7.1, sketch a graph to show the variation with 1/λ of EMAX for caesium metal. [2]

10 marks

Mark scheme: 7 (a) minimum frequency of e.m. radiation/a photon (not “light”) M1 for emission of electrons from a surface A1 [2] (reference to light/UV rather than e.m. radiation, allow 1/2) (b) EMAX corresponds to electron emitted from surface B1 electron (below surface) requires energy to bring it to surface, so less than EMAX B1 [2] (c) (i) 1/λ0 = 1.85 × 106 (allow 1.82 to 1.88) C1 f0 = c / λ0 = 3.00 × 108 × 1.85 × 106 = 5.55 × 1014 Hz A1 [2] (ii) Φ = hf0 = 6.63 × 10–34 × 5.55 × 1014 (allow ECF from (c)(i)) C1 = 3.68 × 10–19 J A1 [2] (d) sketch: straight line with same gradient M1 intercept between 1.0 and 1.5 A1 [2]

This question in 9702/42 Oct/Nov 2015

8 Light of wavelength λ is incident on a metal surface having a work function energy Φ. Photoelectrons of maximum kinetic energy EMAX are emitted from the surface. (a) State an equation relating Φ, EMAX and λ. Explain any other symbols you use. … … … [2] (b) The variation with 1/λ of EMAX is shown in Fig. 8.1. EMAX 0 1/ h Fig. 8.1 (i) By reference to your answer in (a), explain why the gradient of the line does not depend on the metal surface. … … … [2] (ii) The work function energy of sodium is 2.28 eV. Determine the minimum wavelength λ0 at which EMAX is zero. λ0 = … m [3]

7 marks

Mark scheme: 8 (a) hc / λ = Φ + EMAX M1 h = Planck constant, c = speed of light / e.m. radiation A1 [2] (b) (i) gradient of line is hc M1 h and c are both constants A1 [2] (ii) Φ = 2.28 × 1.6 × 10–19 C1 = 3.65 × 10–19 (J) hc / λ0 = 3.65 × 10–19 λ0 = (6.63 × 10–34 × 3.0 × 108) / (3.65 × 10–19) C1 = 5.45 × 10–7 m A1 [3]

This question in 9702/43 Oct/Nov 2015

11 (a) With reference to the photoelectric effect, state what is meant by the threshold frequency. … … … [2] (b) Electromagnetic radiation of wavelength λ is incident on a metal surface. Electrons of maximum kinetic energy EMAX are emitted. (i) On Fig. 11.1, sketch the variation with 1/λ of EMAX. EMAX 0 0 1 /h Fig. 11.1 [2] (ii) State an equation relating the gradient of the graph drawn on Fig. 11.1 to the Planck constant h. Explain any symbols you use. … … [1] (iii) Explain why, for any particular wavelength of electromagnetic radiation, most of the electrons are emitted with kinetic energies less than the maximum value EMAX. … … … … [2] (iv) Light of a particular wavelength is incident on a metal surface and gives rise to a photoelectric current. The wavelength is reduced. The intensity of the light is kept constant. State and explain the effect, if any, on the photoelectric current. … … … … … [3] [Total: 10]

10 marks

Mark scheme: 11 (a) minimum frequency for electron(s) to be emitted (from surface) M1 reference to frequency of electromagnetic radiation / photon A1 or frequency causing emission of electron(s) from surface with zero kinetic energy (M1) reference to frequency of electromagnetic radiation / photon (A1) [2] (b) (i) positive intercept on (1 / λ)-axis (when extrapolated) B1 straight line with positive gradient B1 [2] (ii) gradient = hc where c is the speed of light B1 [1] (iii) maximum kinetic energy when electron emitted from surface B1 energy is required to bring an electron to the surface B1 [2] (iv) each photon has more energy M1 fewer photons per unit time M1 fewer electrons per unit time / less current A1 [3]

This question in 9702/42 Feb/March 2016

12 (a) State an effect, one in each case, that provides evidence for (i) the wave nature of a particle, … [1] (ii) the particulate nature of electromagnetic radiation. … [1] (b) Four electron energy levels in an isolated atom are shown in Fig. 12.1. –0.54 eV –0.85 eV energy –1.51 eV –3.40 eV Fig. 12.1 For the emission spectrum associated with these energy levels, (i) on Fig. 12.1, mark with an arrow the transition that gives rise to the shortest wavelength, [1] (ii) show that the wavelength of the transition in (i) is 4.35 × 10–7 m. [2] (c) (i) State what is meant by the de Broglie wavelength. … … … [2] (ii) Calculate the speed of an electron having a de Broglie wavelength equal to the wavelength in (b)(ii). speed = … m s–1 [2] [Total: 9]

9 marks

Mark scheme: 12 (a) (i) electron diffraction/electron microscope (allow other sensible suggestions) B1 [1] (ii) photoelectric effect/Compton scattering (allow other sensible suggestions) B1 [1] (b) (i) arrow clear from –0.54 eV to –3.40 eV B1 [1] (ii) E = hc / λ or E = hf and c = fλ C1 λ= (6.63 × 10–34 × 3.00 × 108) / [(3.40 – 0.54) × 1.60 × 10–19] = 4.35 × 10–7 m A1 [2] (c) (i) wavelength associated with a particle M1 that is moving/has momentum/has speed/has velocity A1 [2] (ii) λ= h / mv v = (6.63 × 10–34) / (9.11 × 10–31 × 4.35 × 10–7) C1 = 1.67 × 103 m s–1 A1 [2]

This question in 9702/42 Oct/Nov 2016

10 (a) Explain what is meant by the photoelectric effect. … … … [2] (b) One wavelength of electromagnetic radiation emitted from a mercury vapour lamp is 436 nm. Calculate the photon energy corresponding to this wavelength. energy = … J [2] (c) Light from the lamp in (b) is incident, separately, on the surfaces of caesium and tungsten metal. Data for the work function energies of caesium and tungsten metal are given in Fig. 10.1. metal work function energy / eV caesium 1.4 tungsten 4.5 Fig. 10.1 Calculate the threshold wavelength for photoelectric emission from (i) caesium, threshold wavelength = … nm [2] (ii) tungsten. threshold wavelength = … nm [1] (d) Use your answers in (c) to state and explain whether the radiation from the mercury lamp of wavelength 436 nm will give rise to photoelectric emission from each of the metals. caesium: … … tungsten: … … [2] [Total: 9]

9 marks

Mark scheme: 10 (a) electromagnetic radiation/photons incident on a surface B1 causes emission of electrons (from the surface) B1 [2] (b) E = hc / λ = (6.63 × 10–34 × 3.00 × 108) / (436 × 10–9) C1 = 4.56 × 10–19 J (4.6 × 10–19 J) A1 [2] (c) (i) Φ = hc / λ0 λ0 = (6.63 × 10–34 × 3.00 × 108) / (1.4 × 1.60 × 10–19) C1 = 890 nm A1 [2] (ii) λ0 = (6.63 × 10–34 × 3.00 × 108) / (4.5 × 1.60 × 10–19) = 280 nm A1 [1] (d) caesium: wavelength of photon less than threshold wavelength (or v.v.) or λ0 = 890 nm > 436 nm so yes A1 tungsten: wavelength of photon greater than threshold wavelength (or v.v.) or λ0 = 280 nm < 436 nm so no A1 [2]

This question in 9702/43 Oct/Nov 2016

10 (a) Briefly describe two phenomena associated with the photoelectric effect that cannot be explained using a wave theory of light. 1. … … 2. … … [2] (b) The maximum energy EMAX of electrons emitted from a metal surface when illuminated by light of wavelength λ is given by the expression 1 1 EMAX = λ – hc ( λ0) where h is the Planck constant and c is the speed of light. (i) Identify the symbol λ0. … [1] 1 (ii) The variation with of EMAX for the metal surface is shown in Fig. 10.1. λ 4 EMAX / 10–19 J 3 2 1 0 1.5 2.0 2.5 3.0 3.5 4.0 1 / 106 m–1 λ Fig. 10.1 1. Use Fig. 10.1 to determine the magnitude of λ0. λ0 = … m [1] 2. Use the gradient of Fig. 10.1 to determine a value for the Planck constant h. h = … J s [3] (c) The metal surface in (b) becomes oxidised. Photoelectric emission is still observed but the work function energy is increased. 1 On Fig. 10.1, draw a line to show the variation with of EMAX for the oxidised surface. [2] λ [Total: 9]

9 marks

Mark scheme: 10(a) two from: • frequency below which electrons not ejected • maximum energy of electron depends on frequency • maximum energy of electrons does not depend on intensity • instantaneous emission of electrons B2 10(b)(i) (λ0 is the) threshold wavelength or wavelength corresponding to threshold frequency or maximum wavelength for emission of electrons B1 10(b)(ii)1. intercept = 1 / λ0 = 2.2 × 106m–1 λ0 = 4.5 × 10–7 m or 450 nm A1 10(b)(ii)2. gradient = hc C1 gradient = 2.0 × 10–25 or correct substitution into gradient formula C1 h = (2.0 × 10–25) / (3.0 × 108) = 6.7 × 10–34 J s A1 10(c) line: same gradient B1 straight line, positive gradient, intercept at greater than 2.2 × 106 when candidate’s line extrapolated B1

This question in 9702/42 May/June 2017

10 (a) A metal surface is illuminated with light of a single wavelength λ. On Fig. 10.1, sketch the variation with λ of the maximum kinetic energy EMAX of the electrons emitted from the surface. On your graph mark, with the symbol λ0, the threshold wavelength. EMAX 0 λ Fig. 10.1 [3] (b) A neutron is moving in a straight line with momentum p. The de Broglie wavelength associated with this neutron is λ. On Fig. 10.2, sketch the variation with momentum p of the de Broglie wavelength λ. λ 0 0 p Fig. 10.2 [2] [Total: 5]

5 marks

Mark scheme: 10(a) B1 graph line with λ always < λ0 B1 negative gradient with correct concave curvature B1 10(b) curve with negative gradient and correct concave curvature M1 not touching either axis A1

This question in 9702/42 Oct/Nov 2017

11 (a) (i) Explain what is meant by a photon. … … … [2] (ii) By reference to intensity of light, state one piece of evidence provided by the photoelectric effect for a particulate nature of light. … … [1] (b) Some electron energy levels in a solid are illustrated in Fig. 11.1. conduction band forbidden band valence band Fig. 11.1 A semiconductor material has a very high resistance in darkness. Light incident on the semiconductor material causes its resistance to decrease. Explain the resistance of the semiconductor material in different light conditions. … … … … … … [5] [Total: 8]

8 marks

Mark scheme: 11(a)(i) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 11(a)(ii) (maximum) energy of emitted electrons is independent of intensity or no emission of electrons below the threshold frequency regardless of intensity or no emission of electrons when photon energy is less than work function (energy) regardless of intensity B1 11(b) in darkness: conduction band empty so high resistance B1 in daylight: electrons in valence band absorb photons B1 in daylight: electrons ‘jump’ to conduction band B1 this leaves holes in valence band B1 more charge carriers in daylight so resistance decreases B1

This question in 9702/41 May/June 2018

10 (a) Describe the photoelectric effect. … … … [2] (b) Data for the work function energy Φ of two metals are shown in Fig. 10.1. Φ/ J sodium 3.8 × 10–19 zinc 5.8 × 10–19 Fig. 10.1 Light of wavelength 420 nm is incident on the surface of each of the metals. (i) State what is meant by a photon. … … … [2] (ii) Calculate the energy of a photon of the incident light. energy = … J [2] (iii) State whether photoelectric emission will occur from each of the metals. sodium: … zinc: … [1] [Total: 7]

7 marks

Mark scheme: 10(a) emission of electron B1 when electromagnetic radiation incident (on surface) B1 10(b)(i) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 10(b)(ii) E = hc / λ C1 = (6.63 × 10–34 × 3.00 × 108) / (420 × 10–9) = 4.7 × 10–19 J A1 10(b)(iii) sodium: yes zinc: no B1

This question in 9702/42 May/June 2018

11 (a) (i) Explain what is meant by a photon. … … … [2] (ii) By reference to intensity of light, state one piece of evidence provided by the photoelectric effect for a particulate nature of light. … … [1] (b) Some electron energy levels in a solid are illustrated in Fig. 11.1. conduction band forbidden band valence band Fig. 11.1 A semiconductor material has a very high resistance in darkness. Light incident on the semiconductor material causes its resistance to decrease. Explain the resistance of the semiconductor material in different light conditions. … … … … … … [5] [Total: 8]

8 marks

Mark scheme: 11(a)(i) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 11(a)(ii) (maximum) energy of emitted electrons is independent of intensity or no emission of electrons below the threshold frequency regardless of intensity or no emission of electrons when photon energy is less than work function (energy) regardless of intensity B1 11(b) in darkness: conduction band empty so high resistance B1 in daylight: electrons in valence band absorb photons B1 in daylight: electrons ‘jump’ to conduction band B1 this leaves holes in valence band B1 more charge carriers in daylight so resistance decreases B1

This question in 9702/43 May/June 2018

11 (a) State three pieces of evidence provided by the photoelectric effect for a particulate nature of electromagnetic radiation. 1. … … 2. … … 3. … … [3] (b) The work function energies of some metals are shown in Fig. 11.1. work function energy / eV sodium 2.4 calcium 2.9 zinc 3.6 silver 4.3 Fig. 11.1 Each metal is irradiated with electromagnetic radiation of wavelength 380 nm. (i) Calculate the energy, in eV, of a photon of electromagnetic radiation of wavelength 380 nm. energy = … eV [3] (ii) Determine which metals will give rise to the emission of photoelectrons. Explain your answer. … … [2] (c) Photons of wavelength 380 nm are incident normally on a metal surface at a rate of 7.6 × 1014 s–1. All the photons are absorbed in the surface and no photoelectrons are emitted. Calculate the force exerted on the metal surface by the incident photons. force = … N [3] [Total: 11]

11 marks

Mark scheme: 11(a) Any three points from: • (max) energy of emitted electrons depends on frequency • (max) energy of emitted electrons does not depend on intensity • rate of emission of electrons depends on intensity (at constant frequency) • existence of frequency below which no emission of electrons • instantaneous emission of electrons • increasing the frequency at constant intensity decreases the rate of emission of electrons B3 11(b)(i) photon energy = hc / λ C1 = (6.63 × 10–34 × 3.0 × 108) / (380 × 10–9) ( = 5.23 × 10–19 J) C1 = 3.3 eV A1 11(b)(ii) photon energy must be greater than work function (energy) B1 so sodium and calcium B1 11(c) λ = h / p C1 p = (6.63 × 10–34) / (380 × 10–9) = 1.74 × 10–27 N s C1 force = 1.74 × 10–27 × 7.6 × 1014 = 1.3 × 10–12 N A1

This question in 9702/41 May/June 2019

11 (a) State three pieces of evidence provided by the photoelectric effect for a particulate nature of electromagnetic radiation. 1. … … 2. … … 3. … … [3] (b) The work function energies of some metals are shown in Fig. 11.1. work function energy / eV sodium 2.4 calcium 2.9 zinc 3.6 silver 4.3 Fig. 11.1 Each metal is irradiated with electromagnetic radiation of wavelength 380 nm. (i) Calculate the energy, in eV, of a photon of electromagnetic radiation of wavelength 380 nm. energy = … eV [3] (ii) Determine which metals will give rise to the emission of photoelectrons. Explain your answer. … … [2] (c) Photons of wavelength 380 nm are incident normally on a metal surface at a rate of 7.6 × 1014 s–1. All the photons are absorbed in the surface and no photoelectrons are emitted. Calculate the force exerted on the metal surface by the incident photons. force = … N [3] [Total: 11]

11 marks

Mark scheme: 11(a) Any three points from: • (max) energy of emitted electrons depends on frequency • (max) energy of emitted electrons does not depend on intensity • rate of emission of electrons depends on intensity (at constant frequency) • existence of frequency below which no emission of electrons • instantaneous emission of electrons • increasing the frequency at constant intensity decreases the rate of emission of electrons B3 11(b)(i) photon energy = hc / λ C1 = (6.63 × 10–34 × 3.0 × 108) / (380 × 10–9) ( = 5.23 × 10–19 J) C1 = 3.3 eV A1 11(b)(ii) photon energy must be greater than work function (energy) B1 so sodium and calcium B1 11(c) λ = h / p C1 p = (6.63 × 10–34) / (380 × 10–9) = 1.74 × 10–27 N s C1 force = 1.74 × 10–27 × 7.6 × 1014 = 1.3 × 10–12 N A1

This question in 9702/43 May/June 2019

11 (a) With reference to the photoelectric effect, state what is meant by work function energy. … … … [2] (b) The work function energy of a clean metal surface is 5.5 × 10–19 J. Electromagnetic radiation of wavelength 280 nm is incident on the metal surface. The metal is in a vacuum. (i) Calculate: 1. the photon energy photon energy = … J [2] 2. the maximum speed vMAX of the electrons emitted from the surface. vMAX = … m s–1 [3] (ii) Explain why most of the emitted electrons will have a speed lower than vMAX. … … [1] (c) The electromagnetic radiation incident on the metal surface may change in intensity or in frequency. Complete Fig. 11.1 by inserting either ‘increases’ or ‘decreases’ or ‘no change’ to describe the effects of the changes shown on the maximum speed and on the rate of emission of electrons. maximum speed of rate of emission change electrons of electrons reduced intensity at constant frequency … … increased frequency at constant intensity … … Fig. 11.1 [4] [Total: 12]

12 marks

Mark scheme: 11(a) energy (of photon) required to remove electron M1 from a surface or reference to minimum energy or reference to zero kinetic energy A1 11(b)(i) 1. photon energy = hc / λ C1 = (6.63 × 10–34 × 3.00 × 108) / (280 × 10–9) = 7.1 × 10–19 J A1 2. electron energy = (7.1 – 5.5) × 10–19 J C1 ½ × 9.11 × 10–31 × v2 = (7.1 – 5.5) × 10–19 C1 v = 5.9 × 105 m s–1 A1 11(b)(ii) energy is required to bring electron to the surface B1 11(c) no change decreases increases decreases B4

This question in 9702/41 Oct/Nov 2019

11 (a) With reference to the photoelectric effect, state what is meant by work function energy. … … … [2] (b) The work function energy of a clean metal surface is 5.5 × 10–19 J. Electromagnetic radiation of wavelength 280 nm is incident on the metal surface. The metal is in a vacuum. (i) Calculate: 1. the photon energy photon energy = … J [2] 2. the maximum speed vMAX of the electrons emitted from the surface. vMAX = … m s–1 [3] (ii) Explain why most of the emitted electrons will have a speed lower than vMAX. … … [1] (c) The electromagnetic radiation incident on the metal surface may change in intensity or in frequency. Complete Fig. 11.1 by inserting either ‘increases’ or ‘decreases’ or ‘no change’ to describe the effects of the changes shown on the maximum speed and on the rate of emission of electrons. maximum speed of rate of emission change electrons of electrons reduced intensity at constant frequency … … increased frequency at constant intensity … … Fig. 11.1 [4] [Total: 12]

12 marks

Mark scheme: 11(a) energy (of photon) required to remove electron M1 from a surface or reference to minimum energy or reference to zero kinetic energy A1 11(b)(i) 1. photon energy = hc / λ C1 = (6.63 × 10–34 × 3.00 × 108) / (280 × 10–9) = 7.1 × 10–19 J A1 2. electron energy = (7.1 – 5.5) × 10–19 J C1 ½ × 9.11 × 10–31 × v2 = (7.1 – 5.5) × 10–19 C1 v = 5.9 × 105 m s–1 A1 11(b)(ii) energy is required to bring electron to the surface B1 11(c) no change decreases increases decreases B4

This question in 9702/43 Oct/Nov 2019

10 (a) By reference to the photoelectric effect, explain what is meant by work function energy. … … … [2] (b) In an experiment, electromagnetic radiation of frequency f is incident on a metal surface. The results in Fig. 10.1 show the variation with frequency f of the maximum kinetic energy EMAX of electrons emitted from the surface. 1.4 EMAX /10–19 J 1.2 1.0 0.8 0.6 0.4 0.2 0 5.0 5.5 6.0 6.5 7.0 7.5 f / 1014 Hz Fig. 10.1 (i) Determine the work function energy in J of the metal used in the experiment. work function energy = … J [2] (ii) The work function energy in eV for some metals is given in Table 10.1. Table 10.1 metal work function / eV tungsten 4.49 magnesium 3.68 potassium 2.26 Determine the metal used in the experiment. Show your working. … … [1] (c) The intensity of the electromagnetic radiation for one particular frequency in (b) is increased. State and explain the change, if any, in: (i) the maximum kinetic energy of the emitted electrons … … [1] (ii) the rate of emission of photoelectrons. … … [1] [Total: 7]

7 marks

Mark scheme: 10(a) energy of a photon required to remove an electron B1 either: energy to remove electron from a surface or: minimum energy to remove electron or: energy to remove electron with zero kinetic energy B1 10(b)(i) Correct read off from graph of f as 5.45 × 1014 Hz when EMAX = 0 5.45 × 1014 × 6.63 × 10–34 C1 = 3.6 × 10–19 J A1 10(b)(ii) 3.6 × 10–19 / 1.6 × 10–19 = 2.3 eV so potassium A1 10(c)(i) each photon has same energy so no change B1 10(c)(ii) more photons (per unit time) so (rate of emission) increases B1

This question in 9702/42 Feb/March 2020

11 (a) Electromagnetic radiation is incident on a metal surface. It is observed that there is a minimum frequency of electromagnetic radiation below which emission of electrons does not occur. This observation provides evidence for a particulate nature of electromagnetic radiation. State two other observations associated with photoelectric emission that provide evidence for a particulate nature of electromagnetic radiation. 1. … … 2. … … [2] (b) The maximum kinetic energy EMAX of electrons emitted from a metal surface is determined for different wavelengths λ of the electromagnetic radiation incident on the surface. 1 The variation with of EMAX is shown in Fig. 11.1. λ 0.6 EMAX / eV 0.4 0.2 0 1.9 2.0 2.1 2.2 2.3 2.4 1 / 106 m–1 λ Fig. 11.1 (i) Use Fig. 11.1 to determine the threshold frequency f0. f0 = … Hz [2] (ii) Use the gradient of the line on Fig. 11.1 to determine a value for the Planck constant h. Explain your working. h = … J s [4] (c) The electromagnetic radiation is now incident on a metal with a larger work function energy than the metal in (b). 1 On Fig. 11.1, sketch the variation with of EMAX. [2] λ [Total: 10]

10 marks

Mark scheme: 11(a) any two points from: • (maximum) kinetic energy of electrons is independent of intensity • maximum kinetic energy of electrons depends on frequency • no time delay (between illumination and emission) B2 11(b)(i) (for EMAX = 0,) 1 / λ0 = 1.93 × 106 (m–1) C1 f0 = 3.00 × 108 × 1.93 × 106 = 5.8 × 1014 Hz A1 11(b)(ii) hc / λ = Φ + EMAX C1 hc = gradient C1 gradient = e.g. [(0.40 – 0.20) × 1.60 × 10–19] / [(2.25 – 2.09) × 106] (working needed) (= 2.0 × 10–25) M1 h = (2.0 × 10–25) / (3.00 × 108) = 6.7 × 10–34 J s (both working and answer needed) A1 11(c) straight line with same gradient as the original B1 straight line with x-axis intercept greater than 1.93 × 106 m–1 B1

This question in 9702/42 Oct/Nov 2020

12 (a) Electromagnetic radiation of a single constant frequency is incident on a metal surface. This causes an electron to be emitted. Explain why the maximum kinetic energy of the electron is independent of the intensity of the incident radiation. … … … … … [3] (b) Ultraviolet radiation of wavelength 250 nm is incident on the surface of a sheet of zinc. The maximum kinetic energy of the emitted electrons is 1.4 eV. Determine, in eV: (i) the energy of a photon of the ultraviolet radiation energy = … eV [3] (ii) the work function energy of the surface of the zinc. energy = … eV [2] [Total: 8]

8 marks

Mark scheme: 12(a) • frequency determines energy of photon • intensity determines number of photons (per unit time) • intensity does not determine energy of a photon Any two points, 1 mark each B2 kinetic energy (of the electron) depends on the energy of one photon B1 12(b)(i) E = hc / λ or E = hf and c = fλ C1 E = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1 (= 7.96 × 10–19 J) = 5.0 eV A1 12(b)(ii) EMAX = photon energy – work function C1 work function = 5.0 – 1.4 = 3.6 eV A1

This question in 9702/41 May/June 2021

5 (a) An isolated metal sphere of radius r is charged so that the electric potential at its surface is V0. On Fig. 5.1, sketch the variation with distance x from the centre of the sphere of the electric potential. Your graph should extend from x = 0 to x = 3r. 1.0 V0 electric potential 0.5 V0 0 0 r 2r 3r x Fig. 5.1 [3] (b) Photons having wavelength λ are incident on a metal surface. The maximum wavelength for which there is emission of electrons is λ 0. λ 0 For photons of wavelength , the maximum kinetic energy of the emitted electrons is EMAX. 2 On Fig. 5.2, sketch the variation with wavelength λ of the maximum kinetic energy for values λ 0 of wavelength between λ = and λ = λ 0. 3 3 EMAX energy 2 EMAX EMAX 0 0 λ λ λ 0 0 0 3 2 λ Fig. 5.2 [3] (c) A pure sample of a radioactive isotope contains N0 nuclei. The half-life of the isotope is T12. The product of the radioactive decay is stable. The variation with time t of the number N of nuclei of the radioactive isotope is shown in Fig. 5.3. N0 number N0 2 N 0 0 T time t Fig. 5.3 On Fig. 5.3: ● label, on the time axis, the time t = 1.0T12 and the time t = 2.0T12 ● sketch the variation with time t of the number of nuclei of the decay product for time t = 0 to time t = T. [3] [Total: 9]

9 marks

Mark scheme: 5(a) from x = 0 to x = r: horizontal line at V = 1.0V0 B1 from x = r to x = 3r: curve with negative gradient of decreasing magnitude starting at (r, 1.0V0) B1 line passing through (2r, ½V0) and (3r, ⅓V0) B1 5(b) line with negative gradient from λ = ⅓λ0 to λ = λ0 B1 line passing through (λ0, 0) B1 curve with negative gradient of decreasing magnitude passing through (½λ0, EMAX) and (⅓λ0, 2EMAX) B1 5(c) 1.0T½ shown at ½N0 and 2.0T½ shown at ¼N0 B1 line starting at (0, 0) and reaching (T, N0–N) B1 line starting at (0, 0) and reaching original curve at (1.0T½, ½N0) B1

This question in 9702/42 May/June 2021

12 (a) Electromagnetic radiation of a single constant frequency is incident on a metal surface. This causes an electron to be emitted. Explain why the maximum kinetic energy of the electron is independent of the intensity of the incident radiation. … … … … … [3] (b) Ultraviolet radiation of wavelength 250 nm is incident on the surface of a sheet of zinc. The maximum kinetic energy of the emitted electrons is 1.4 eV. Determine, in eV: (i) the energy of a photon of the ultraviolet radiation energy = … eV [3] (ii) the work function energy of the surface of the zinc. energy = … eV [2] [Total: 8]

8 marks

Mark scheme: 12(a) • frequency determines energy of photon • intensity determines number of photons (per unit time) • intensity does not determine energy of a photon Any two points, 1 mark each B2 kinetic energy (of the electron) depends on the energy of one photon B1 12(b)(i) E = hc / λ or E = hf and c = fλ C1 E = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1 (= 7.96 × 10–19 J) = 5.0 eV A1 12(b)(ii) EMAX = photon energy – work function C1 work function = 5.0 – 1.4 = 3.6 eV A1

This question in 9702/43 May/June 2021

10 (a) State an experimental phenomenon that provides evidence for: (i) the particulate nature of electromagnetic radiation … [1] (ii) the wave nature of matter. … [1] (b) A particle of matter moves with momentum p. (i) State the equation that gives the effective wavelength λ of the particle. State the name of any other symbols used. [2] (ii) State the name given to the wavelength of the moving particle. … [1] (c) Electrons are accelerated from rest through a potential difference (p.d.) of 4.8 kV. (i) Show that the final speed of the electrons is 4.1 × 107 m s–1. [2] (ii) Calculate the effective wavelength of a beam of electrons moving at the speed in (c)(i). wavelength = … m [2] [Total: 9]

9 marks

Mark scheme: 10(a)(i) photoelectric effect B1 10(a)(ii) electron diffraction B1 10(b)(i) λ = h / p M1 h is the Planck constant A1 10(b)(ii) de Broglie (wavelength) B1 10(c)(i) ½mv2 = eV C1 ½ × 9.11 × 10–31 × v2 = 1.60 × 10–19 × 4800 so v = 4.1 × 107 m s–1 A1 10(c)(ii) λ = h / mv = 6.63 × 10–34 / (9.11 × 10–31 × 4.1 × 107) C1 = 1.8 × 10–11 m A1

This question in 9702/41 Oct/Nov 2021

9 (a) State what is meant by: (i) the photoelectric effect … … … [2] (ii) work function energy. … … [1] (b) A polished calcium plate in a vacuum is investigated by illuminating the surface with light. It is found that no photoelectric current is produced when the frequency of the light is less than 6.93 × 1014 Hz. (i) State the name of the frequency below which no photoelectric current is produced. … [1] (ii) Explain how the photon model of electromagnetic radiation accounts for this phenomenon. … … … … [3] (iii) Calculate the work function energy, in eV, of calcium. work function energy = … eV [2] [Total: 9]

9 marks

Mark scheme: 9(a)(i) emission of electrons (from a metal surface) B1 when electromagnetic radiation is incident (on electrons) B1 9a(ii) minimum energy required for an electron to leave surface B1 9(b)(i) threshold (frequency) B1 9(b)(ii) • photons are (discrete) packets of energy • energy of photons depends on frequency (of EM radiation) • electrons can only absorb a single photon (of energy) Any two points, 1 mark each B2 emission only possible if photon energy is at least the work function B1 9(b)(iii) work function = hf0 = 6.63 × 10–34 × 6.93 × 1014 C1 = 4.59 × 10–19 (J) = 4.59 × 10–19 / 1.60 × 10–19 (eV) = 2.87 eV A1

This question in 9702/42 Oct/Nov 2021

10 (a) State an experimental phenomenon that provides evidence for: (i) the particulate nature of electromagnetic radiation … [1] (ii) the wave nature of matter. … [1] (b) A particle of matter moves with momentum p. (i) State the equation that gives the effective wavelength λ of the particle. State the name of any other symbols used. [2] (ii) State the name given to the wavelength of the moving particle. … [1] (c) Electrons are accelerated from rest through a potential difference (p.d.) of 4.8 kV. (i) Show that the final speed of the electrons is 4.1 × 107 m s–1. [2] (ii) Calculate the effective wavelength of a beam of electrons moving at the speed in (c)(i). wavelength = … m [2] [Total: 9]

9 marks

Mark scheme: 10(a)(i) photoelectric effect B1 10(a)(ii) electron diffraction B1 10(b)(i) λ = h / p M1 h is the Planck constant A1 10(b)(ii) de Broglie (wavelength) B1 10(c)(i) ½mv2 = eV C1 ½ × 9.11 × 10–31 × v2 = 1.60 × 10–19 × 4800 so v = 4.1 × 107 m s–1 A1 10(c)(ii) λ = h / mv = 6.63 × 10–34 / (9.11 × 10–31 × 4.1 × 107) C1 = 1.8 × 10–11 m A1

This question in 9702/43 Oct/Nov 2021

7 (a) State what is meant by a photon. … … … [2] (b) Electromagnetic radiation of a varying frequency f and constant intensity I is used to illuminate a metal surface. At certain frequencies, electrons are emitted from the surface of the metal. The variation with f of the maximum kinetic energy EMAX of the emitted electrons is shown in Fig. 7.1. 4.0 EMAX / 10–19 J 3.0 2.0 1.0 0 0 2 4 6 8 10 12 f / 1014 Hz Fig. 7.1 (i) State the name of this phenomenon. … [1] (ii) Describe three conclusions that can be drawn from the graph in Fig. 7.1. The conclusions may be qualitative or quantitative. 1 … … 2 … … 3 … … [3] (c) The experiment in (b) is repeated twice, each time making one change. State, with a reason, how the graph obtained would compare with Fig. 7.1 when: (i) a different metal is used, but keeping the intensity I of the radiation the same … … … [2] (ii) the same metal is used, but with electromagnetic radiation of intensity 2I. … … … [2] [Total: 10]

10 marks

Mark scheme: 7(a) quantum of energy M1 of electromagnetic radiation A1 7(b)(i) photoelectric effect B1 7(b)(ii)  there is a frequency below which no electrons are emitted or threshold frequency = 5.4  1014 Hz  work function of the metal = 3.6  10–19 J (or 2.2 eV)  EMAX increases (linearly) with (increasing) frequency  gradient of the line is the Planck constant or gradient of the line is 6.7  10–34 J s Any three bullet points, 1 mark each B3 7(c)(i) different threshold frequency B1 (line has) same gradient but different intercept B1 7(c)(ii) photons have same energy B1 line unchanged B1

This question in 9702/41 May/June 2022

8 (a) State one piece of experimental evidence for: (i) the particulate nature of electromagnetic radiation … [1] (ii) the wave nature of matter. … [1] (b) (i) Calculate the de Broglie wavelength λ of an alpha-particle moving at a speed of 6.2 × 107 m s–1. λ = … m [3] (ii) The speed v of the alpha-particle in (b)(i) is gradually reduced to zero. On Fig. 8.1, sketch the variation with v of λ. λ 0 0 6.2 v / 107 m s–1 Fig. 8.1 [2] (c) Suggest an explanation for why people are not observed to diffract when they walk through a doorway. … … … [1] [Total: 8]

8 marks

Mark scheme: 8(a)(i) photoelectric effect B1 8(a)(ii) electron diffraction B1 8(b)(i)  = h / p C1 p = 4  1.66  10–27  6.2  107 ( = 4.1  10–19 N s) C1  = 6.63  10–34 / 4.1  10–19 = 1.6  10–15 m A1 8(b)(ii) line with negative gradient throughout B1 curve asymptotic to both axes with non-zero  at v = 6.2  107 m s–1 B1 8(c) (de Broglie) wavelength negligible compared with width of doorway B1

This question in 9702/42 May/June 2022

7 (a) State what is meant by a photon. … … … [2] (b) Electromagnetic radiation of a varying frequency f and constant intensity I is used to illuminate a metal surface. At certain frequencies, electrons are emitted from the surface of the metal. The variation with f of the maximum kinetic energy EMAX of the emitted electrons is shown in Fig. 7.1. 4.0 EMAX / 10–19 J 3.0 2.0 1.0 0 0 2 4 6 8 10 12 f / 1014 Hz Fig. 7.1 (i) State the name of this phenomenon. … [1] (ii) Describe three conclusions that can be drawn from the graph in Fig. 7.1. The conclusions may be qualitative or quantitative. 1 … … 2 … … 3 … … [3] (c) The experiment in (b) is repeated twice, each time making one change. State, with a reason, how the graph obtained would compare with Fig. 7.1 when: (i) a different metal is used, but keeping the intensity I of the radiation the same … … … [2] (ii) the same metal is used, but with electromagnetic radiation of intensity 2I. … … … [2] [Total: 10]

10 marks

Mark scheme: 7(a) quantum of energy M1 of electromagnetic radiation A1 7(b)(i) photoelectric effect B1 7(b)(ii)  there is a frequency below which no electrons are emitted or threshold frequency = 5.4  1014 Hz  work function of the metal = 3.6  10–19 J (or 2.2 eV)  EMAX increases (linearly) with (increasing) frequency  gradient of the line is the Planck constant or gradient of the line is 6.7  10–34 J s Any three bullet points, 1 mark each B3 7(c)(i) different threshold frequency B1 (line has) same gradient but different intercept B1 7(c)(ii) photons have same energy B1 line unchanged B1

This question in 9702/43 May/June 2022

8 (a) State what is meant by the work function energy of a metal. … … … [2] (b) Ultraviolet radiation of frequency 1.36 × 1015 Hz is incident, in a vacuum, on a metal surface. The power of the radiation incident on the surface is 8.36 mW. Photoelectrons are emitted with a maximum kinetic energy of 3.09 × 10–19 J. (i) Determine the number of photons incident on the surface per unit time. number per unit time = … s–1 [2] (ii) Calculate the work function energy Φ of the metal. Φ = … J [2] (c) The frequency of the radiation incident on the surface in (b) is increased while the power remains constant. State and explain the effect of this change on: (i) the maximum kinetic energy of the photoelectrons … … … [2] (ii) the rate of emission of photoelectrons. … … … [2] [Total: 10]

10 marks

Mark scheme: 8(a) photon energy (to remove electron) B1 minimum energy to remove electron B1 or energy to remove electron from surface or energy to remove electron with zero kinetic energy 8(b)(i) photon energy = hf C1 number per unit time = 8.36  10–3 / (1.36  1015  6.63  10–34) A1 = 9.27  1015 s–1 8(b)(ii) hf = + EMAX C1  = (1.36  1015  6.63  10–34) – (3.09  10–19) A1 = 5.93  10–19 J 8(c)(i) greater photon energy (and same work function) M1 so maximum kinetic energy is increased A1 8(c)(ii) (greater photon energy and same power so) lower number of photons (per unit time) M1 (each electron absorbs one photon) so lower rate of emission A1

This question in 9702/41 Oct/Nov 2022

8 (a) State what is meant by the work function energy of a metal. … … … [2] (b) Ultraviolet radiation of frequency 1.36 × 1015 Hz is incident, in a vacuum, on a metal surface. The power of the radiation incident on the surface is 8.36 mW. Photoelectrons are emitted with a maximum kinetic energy of 3.09 × 10–19 J. (i) Determine the number of photons incident on the surface per unit time. number per unit time = … s–1 [2] (ii) Calculate the work function energy Φ of the metal. Φ = … J [2] (c) The frequency of the radiation incident on the surface in (b) is increased while the power remains constant. State and explain the effect of this change on: (i) the maximum kinetic energy of the photoelectrons … … … [2] (ii) the rate of emission of photoelectrons. … … … [2] [Total: 10]

10 marks

Mark scheme: 8(a) photon energy (to remove electron) B1 minimum energy to remove electron B1 or energy to remove electron from surface or energy to remove electron with zero kinetic energy 8(b)(i) photon energy = hf C1 number per unit time = 8.36  10–3 / (1.36  1015  6.63  10–34) A1 = 9.27  1015 s–1 8(b)(ii) hf = + EMAX C1  = (1.36  1015  6.63  10–34) – (3.09  10–19) A1 = 5.93  10–19 J 8(c)(i) greater photon energy (and same work function) M1 so maximum kinetic energy is increased A1 8(c)(ii) (greater photon energy and same power so) lower number of photons (per unit time) M1 (each electron absorbs one photon) so lower rate of emission A1

This question in 9702/43 Oct/Nov 2022

8 (a) State what is meant by a photon. … … … [2] (b) When the surface of a metal plate is illuminated with electromagnetic radiation, electrons are sometimes emitted from the metal. (i) State the name of this phenomenon. … [1] (ii) It is observed that this phenomenon occurs only when the frequency of the electromagnetic radiation is greater than a certain minimum value, regardless of the intensity of the radiation. Explain how this observation provides evidence for the existence of photons. … … … … … [3] (c) Fig. 8.1 shows the variation of the maximum kinetic energy of the emitted electrons in (b) with the frequency of the incident radiation. maximum kinetic energy 0 0 frequency Fig. 8.1 State the name of the quantity represented by: (i) the gradient of the line in Fig. 8.1 … [1] (ii) the y-intercept of the extrapolated line in Fig. 8.1. … [1] [Total: 8]

8 marks

Mark scheme: 8(a) packet / quantum of energy M1 of electromagnetic radiation A1 8(b)(i) photoelectric effect B1 8(b)(ii) • electron needs a minimum energy to escape B3 or electron emitted if energy in packet is enough • energy must be absorbed in packets that are related to frequency • intensity relates to number of packets (not to energy in packet) • electron absorbs only a single whole packet Any three points, 1 mark each 8(c)(i) Planck constant B1 8(c)(ii) – work function (energy) B1

This question in 9702/42 Oct/Nov 2023

8 A polished sheet of magnesium in a vacuum emits electrons when it is illuminated by ultraviolet radiation. (a) State the name of this phenomenon. … [1] (b) For emission of electrons to occur, the frequency of the ultraviolet radiation must be at least 8.8 × 1014 Hz. (i) Calculate the work function energy of magnesium. work function energy = … J [2] (ii) For ultraviolet radiation with a frequency of 11 × 1014 Hz, calculate the maximum speed of the emitted electrons. maximum speed = … m s–1 [3] (c) The frequency f of the ultraviolet radiation incident on the magnesium sheet is varied between 8.0 × 1014 Hz and 11 × 1014 Hz. On Fig. 8.1, sketch the variation with f of the maximum kinetic energy EMAX of the emitted electrons. Use the space below for any working that you need. 2.0 1.5 EMAX / 10–19 J 1.0 0.5 0 8.0 8.5 9.0 9.5 10.0 10.5 11.0 f / 1014 Hz Fig. 8.1 [3] [Total: 9]

9 marks

Mark scheme: 8(a) photoelectric effect B1 8(b)(i) E = hf C1 work function = 6.63  10–34  8.8  1014 A1 = 5.8  10–19 J 8(b)(ii) hf =  + ½ mvMAX2 C1 6.63  10–34  11  1014 = (5.8  10–19) + (½  9.11  10–31  vMAX2) C1 vMAX = 5.7  105 m s–1 A1 8(c) EMAX shown as zero from f = 8.0 to 8.8 and non-zero from f = 8.8 to 11 B1 all non-zero EMAX shown as a single straight line with a positive gradient B1 line passing through (11, 1.45) B1

This question in 9702/41 Oct/Nov 2024

8 A polished sheet of magnesium in a vacuum emits electrons when it is illuminated by ultraviolet radiation. (a) State the name of this phenomenon. … [1] (b) For emission of electrons to occur, the frequency of the ultraviolet radiation must be at least 8.8 × 1014 Hz. (i) Calculate the work function energy of magnesium. work function energy = … J [2] (ii) For ultraviolet radiation with a frequency of 11 × 1014 Hz, calculate the maximum speed of the emitted electrons. maximum speed = … m s–1 [3] (c) The frequency f of the ultraviolet radiation incident on the magnesium sheet is varied between 8.0 × 1014 Hz and 11 × 1014 Hz. On Fig. 8.1, sketch the variation with f of the maximum kinetic energy EMAX of the emitted electrons. Use the space below for any working that you need. 2.0 1.5 EMAX / 10–19 J 1.0 0.5 0 8.0 8.5 9.0 9.5 10.0 10.5 11.0 f / 1014 Hz Fig. 8.1 [3] [Total: 9]

9 marks

Mark scheme: 8(a) photoelectric effect B1 8(b)(i) E = hf C1 work function = 6.63  10–34  8.8  1014 A1 = 5.8  10–19 J 8(b)(ii) hf =  + ½ mvMAX2 C1 6.63  10–34  11  1014 = (5.8  10–19) + (½  9.11  10–31  vMAX2) C1 vMAX = 5.7  105 m s–1 A1 8(c) EMAX shown as zero from f = 8.0 to 8.8 and non-zero from f = 8.8 to 11 B1 all non-zero EMAX shown as a single straight line with a positive gradient B1 line passing through (11, 1.45) B1

This question in 9702/43 Oct/Nov 2024

9 (a) State what is meant by the photoelectric effect. … … … [2] (b) The photoelectric effect is investigated in two stages using the circuit shown in Fig. 9.1. metal plate electromagnetic X vacuum radiation, frequency f P V Y polished metal plate A Fig. 9.1 The polished metal plate Y is illuminated with electromagnetic radiation of frequency f and constant power. In stage 1 of the investigation, frequency f is set to a constant value of 2.5 × 1015 Hz. The current I in the ammeter is varied by adjusting the potentiometer P. Fig. 9.2 shows the variation of I with the voltmeter reading V. There is a value VS of V at which the current just falls to zero. In stage 2 of the investigation, stage 1 is repeated for different values of frequency. As frequency f is varied, the voltmeter reading VS at which the current just falls to zero is measured. Fig. 9.3 shows the variation of VS with f. 4 8 I / mA VS / V 2 4 0 0 0 2 4 6 0 1 2 3 V / V f / 1015 Hz Fig. 9.2 Fig. 9.3 (i) Explain, with reference to photons, why VS depends on the frequency of the incident electromagnetic radiation. … … … … … … [3] (ii) State three quantitative conclusions that can be drawn from the results in Fig. 9.2 and Fig. 9.3. Use the space for any working. 1 … … 2 … … 3 … … [3] [Total: 8]

8 marks

Mark scheme: 9(a) emission of electrons (from a metal surface) B1 when electromagnetic radiation is incident (on surface / electrons) B1 9(b)(i) current falls to zero when applied voltage equals energy per unit charge of emitted electrons B1 energy of photon depends on frequency B1 maximum energy of electron depends on energy of photon B1 9(b)(ii) Any three points from: B3 • threshold frequency = 1.5 × 1015 Hz • threshold wavelength = 2.0 × 10–7 m • work function = 6.2 eV (or 9.9 × 10–19 J) • Planck constant = 6.6 × 10–34 J s (not 6.63 × 10–34 J s) • number per unit time (of photons / electrons) = 1.7 × 1016 s–1 (in stage 1) • power of incident radiation = 0.028 W (in stage 1)

This question in 9702/42 May/June 2025

9 (a) State what is meant by the photoelectric effect. … … … [2] (b) The photoelectric effect is investigated using two clean metal plates. One plate is made from metal X and the other is made from metal Y. Metal X has work function energy Φ. Metal Y has work function energy 2Φ. Metal X has threshold frequency F. State expressions, in terms of either or both of Φ and F, for (i) the threshold frequency of metal Y threshold frequency = … [1] (ii) the Planck constant. Planck constant = … [1] (c) The maximum kinetic energy EK of photoelectrons is determined for each of the plates in (b) for different frequencies f of incident radiation. On Fig. 9.1, sketch the variation of EK with f for each plate. Label your lines X and Y to identify which line relates to which plate. 2Φ EK Φ 0 0 F 2F 3F f –Φ –2Φ Fig. 9.1 [4] [Total: 8]

8 marks

Mark scheme: 9(a) emission of electrons (from a metal surface) M1 when electromagnetic radiation is incident (on surface) A1 9(b)(i) threshold frequency = 2F A1 9(b)(ii) Planck constant =  / F A1 9(c) two diagonal straight lines with positive gradient in the positive EK region only, starting at non-zero values of f B1 line labelled X passing through (F, 0) and line labelled Y passing through (2F,0) and extending to 3F or +2 B1 two diagonal straight lines with equal gradients B1 X line extrapolates back to (0, –) and Y line extrapolates back to (0, –2) B1

This question in 9702/44 Oct/Nov 2025