Energy and momentum of a photon — 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

5 (a) (i) Explain what is meant by a photon. … … [1] (ii) Show that the photon energy of light of wavelength 350 nm is 5.68 × 10–19 J. [1] (iii) State the value of the ratio energy of photon of light of wavelength 700 nm . energy of photon of light of wavelength 350 nm ratio = …………….. [1] (b) Two beams of monochromatic light have similar intensities. The light in one beam has wavelength 350 nm and the light in the other beam has wavelength 700 nm. The two beams are incident separately on three different metal surfaces. The work function of each of these surfaces is shown in Fig. 5.1. metal work function / eV tungsten 4.49 magnesium 3.68 potassium 2.26 Fig. 5.1 (i) Explain what is meant by the work function of the surface. … … … [2] Examiner’s Use (ii) State which combination, if any, of monochromatic light and metal surface could give rise to photo-electric emission. Give a quantitative explanation of your answer. … … … … … … [3]

8 marks

Mark scheme: 5 (a) (i) packet/discrete quantity/quantum (of energy) of e.m. radiation B1 [1] (ii) either E = (6.63 × 10–34 × 3 × 108)/(350 × 10–9) or E = (6.63 × 10–34 × 8.57 × 1014) M1 E = 5.68 × 10–19 J A0 [1] (iii) 0.5 B1 [1] (b) (i) energy of photon M1 to cause emission of electron from surface either with zero k.e or photon energy is minimum A1 [2] (ii) correct conversion eV → J or J → eV seen once B1 photon energy must be greater than work function C1 350 nm wavelength and potassium metal A1 [3]

This question in 9702/41 May/June 2007

8 A positron ( +1e0 ) is a particle that has the same mass as an electron and has a charge of For +1.6 × 10–19 C. Examiner’s A positron will interact with an electron to form two γ-ray photons. Use +1e0 + –1e0 → 2 γ Assuming that the kinetic energy of the positron and the electron is negligible when they interact, (a) suggest why the two photons will move off in opposite directions with equal energies, … … … … … … [3] (b) calculate the energy, in MeV, of one of the γ-ray photons. energy = … MeV [3]

6 marks

Mark scheme: 8 (a) since momentum before combining is zero B1 momenta must be equal and opposite after B1 equal momenta so photon energies equal B1 [3] (b) E = mc2 C1 = 9.1 × 10–31 × (3.0 × 108)2 = 8.19 × 10–14 (J) C1 = (8.19 × 10–14) / (1.6 × 10–13) = 0.51 MeV A1 [3] GCE A/AS LEVEL – May/June 2008 9702 04 Section B

This question in 9702/41 May/June 2008

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) Explain how a line emission spectrum leads to an understanding of the existence of For discrete electron energy levels in atoms. Examiner’s Use … … … … [3] (b) Some of the lines of the emission spectrum of atomic hydrogen are shown in Fig. 7.1. 410 434 486 656 wavelength / nm Fig. 7.1 The photon energies associated with some of these lines are shown in Fig. 7.2. wavelength / nm photon energy / 10–19 J 410 4.85 434 4.58 486 …………… 656 3.03 Fig. 7.2 (i) Complete Fig. 7.2 by calculating the photon energy for a wavelength of 486 nm. [2] (ii) Energy levels of a single electron in a hydrogen atom are shown in Fig. 7.3. For Examiner’s Use –0.60 –0.87 –1.36 –2.42 –5.45 energy / 10–19 J –21.80 Fig. 7.3 (not to scale) Use data from (i) to show, on Fig. 7.3, the transitions associated with each of the four spectral lines shown in Fig. 7.1. Show each transition with an arrow. [2]

7 marks

Mark scheme: 7 (a) each line corresponds to a (specific) photon energy … B1 photon emitted when electron changes its energy level … B1 discrete energy changes so discrete levels … B1 [3] (b) (i) E = hc / λ …(allow ratio ideas) … C1 = (6.63 × 10-34 × 3.0 × 108) / (486 × 10-9) = 4.09 × 10-19 J … A1 [2] (ii) four transitions to/from – 5.45 × 10-19 J level … B1 all transitions shown from higher to lower energy (level) … B1 [2] [Total: 7]

This question in 9702/42 Oct/Nov 2009

8 A π0 meson is a sub-atomic particle. For A stationary π0 meson, which has mass 2.4 × 10–28 kg, decays to form two γ-ray photons. Examiner’s The nuclear equation for this decay is Use π0 γ + γ. (a) Explain why the two γ-ray photons have the same energy. … … … [2] (b) Determine, for each γ-ray photon, (i) the energy, in joule, energy = … J [2] (ii) the wavelength, wavelength = … m [2] (iii) the momentum. For Examiner’s Use momentum = … N s [2]

8 marks

Mark scheme: 8 (a) momentum conservation hence momenta of photons are equal (but opposite) M1 same momentum so same energy A1 [2] (b) (i) (∆)E = (∆)mc2 C1 = 1.2 × 10–28 × (3.0 × 108)2 = 1.08 × 10–11 J A1 [2] (ii) E = hc / λ λ = (6.63 × 10–34 × 3.0 × 108) / (1.08 × 10–11) C1 = 1.84 × 10–14 m A1 [2] (iii) λ = h / p p = (6.63 × 10–34) / (1.84 × 10–14) C1 = 3.6 × 10–20 N s A1 [2] Section B

This question in 9702/41 May/June 2010

8 (a) Explain what is meant by a photon. For Examiner’s … Use … … … [3] (b) An emission spectrum is seen as a series of differently coloured lines on a black background. Suggest how this observation provides evidence for discrete electron energy levels in atoms. … … … … [2]

5 marks

Mark scheme: 8 (a) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 (allow 1 mark for ‘packet of electromagnetic radiation’) energy = Planck constant × frequency (seen here or in b) B1 [3] (b) each (coloured) line corresponds to one wavelength/frequency B1 energy = Planck constant × frequency implies specific energy change between energy levels B1 so discrete levels A0 [2] GCE AS/A LEVEL – May/June 2012 9702 41

This question in 9702/41 May/June 2012

8 (a) Explain what is meant by a photon. For Examiner’s … Use … … … [3] (b) An emission spectrum is seen as a series of differently coloured lines on a black background. Suggest how this observation provides evidence for discrete electron energy levels in atoms. … … … … [2]

5 marks

Mark scheme: 8 (a) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 (allow 1 mark for ‘packet of electromagnetic radiation’) energy = Planck constant × frequency (seen here or in b) B1 [3] (b) each (coloured) line corresponds to one wavelength/frequency B1 energy = Planck constant × frequency implies specific energy change between energy levels B1 so discrete levels A0 [2] GCE AS/A LEVEL – May/June 2012 9702 43

This question in 9702/43 May/June 2012

11 High-speed electrons are incident on a metal target. The spectrum of the emitted X-ray For radiation is shown in Fig. 11.1. Examiner’s Use intensity 0 wavelength Fig. 11.1 (a) Explain why (i) there is a continuous distribution of wavelengths, … … … [2] (ii) there is a sharp cut-off at short wavelength. … … … [2] (b) State (i) what is meant by the hardness of an X-ray beam, … … … [2] (ii) how hardness is controlled. … … … [2] (c) (i) Suggest why, when producing an X-ray image, long-wavelength X-ray radiation For poses a greater hazard to health than short-wavelength radiation. Examiner’s Use … … [1] (ii) Suggest how this hazard is minimised. … … [1]

10 marks

Mark scheme: 11 (a) (i) e.m. radiation produced whenever charged particle is accelerated M1 electrons hitting target have distribution of accelerations A1 [2] (ii) either wavelength shorter/shortest for greater/greatest acceleration or λmin = hc/ Emax or minimum wavelength for maximum energy B1 all electron energy given up in one collision/converted to single photon B1 [2] (b) (i) hardness measures the penetration of the beam C1 greater hardness, greater penetration A1 [2] (ii) controlled by changing the anode voltage C1 higher anode voltage, greater penetration/hardness A1 [2] (c) (i) long-wavelength radiation more likely to be absorbed in the body/less likely to penetrate through body B1 [1] (ii) (aluminium) filter/metal foil placed in the X-ray beam B1 [1]

This question in 9702/43 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

8 Light of wavelength 590 nm is incident normally on a surface, as illustrated in Fig. 8.1. light surface wavelength 590 nm Fig. 8.1 The power of the light is 3.2 mW. The light is completely absorbed by the surface. (a) Calculate the number of photons incident on the surface in 1.0 s. number = … [3] (b) Use your answer in (a) to determine (i) the total momentum of the photons arriving at the surface in 1.0 s, momentum = … kg m s−1 [3] (ii) the force exerted on the surface by the light. force = … N [1]

7 marks

Mark scheme: 8 (a) photon energy R hc / λ R (6.63 × 10–34 × 3.0 × 108) / (590 × 10–9) C1 R 3.37 × 10–19 J C1 number R (3.2 × 10–3) / (3.37 × 10–19) = = = R 9.5 × 1015 (allow 9.4 × 1015) A1 [3] (b) (i) p R h / λ C1 = = = R (6.63 × 10–34) / (590 × 10–9) = = = R 1.12 × 10–27 kg m s–1 C1 total momentum R 9.5 × 1015 × 1.12 × 10–27 = = = R 1.06 × 10–11 kg m s–1 A1 [3] (ii) force R 1.06 × 10–11 N A1 [1]

This question in 9702/42 May/June 2014

8 White light is incident on a cloud of cool hydrogen gas, as illustrated in Fig. 8.1. cool hydrogen gas incident emergent white light light Fig. 8.1 The spectrum of the light emerging from the gas cloud is found to contain a number of dark lines. (a) Explain why these dark lines occur. … … … … … [3] (b) Some electron energy levels in a hydrogen atom are illustrated in Fig. 8.2. –0.38 –0.55 –0.85 –1.51 energy / eV –3.41 Fig. 8.2 One dark line is observed at a wavelength of 435 nm. (i) Calculate the energy, in eV, of a photon of light of wavelength 435 nm. energy = … eV [4] (ii) On Fig. 8.2, draw an arrow to indicate the energy change that gives rise to this dark line. [1]

8 marks

Mark scheme: 8 (a) photon ‘absorbed’ by electron B1 photon has energy equal to difference in energy of two energy levels B1 electron de-excites emitting photon (of same energy) in any direction B1 [3] (b) (i) E = hc / λ C1 = (6.63 × 10–34 × 3 × 108) / (435 × 10–9) C1 = 4.57 × 10–19 J (allow 2 s.f.) C1 = (4.57 × 10–19) / (1.6 × 10–19) (eV) = 2.86 eV (allow 2 s.f.) A1 [4] (ii) arrow pointing in either direction between –3.41 eV and –0.55 eV B1 [1]

This question in 9702/41 Oct/Nov 2014

8 (a) State what is meant by a photon. … … … [2] (b) A beam of light is incident normally on a metal surface, as illustrated in Fig. 8.1. light beam metal surface area of cross-section 1.3 × 10–5 m2 Fig. 8.1 The beam of light has cross-sectional area 1.3 × 10−5 m2 and power 2.7 × 10−3 W. The light has wavelength 570 nm. The light energy is absorbed by the metal and no light is reflected. (i) Show that a photon of this light has an energy of 3.5 × 10−19 J. [1] (ii) Calculate, for a time of 1.0 s, 1. the number of photons incident on the surface, number = … [2] 2. the change in momentum of the photons. change in momentum = … kg m s−1 [3] (c) Use your answer in (b)(ii) to calculate the pressure that the light exerts on the metal surface. pressure = … Pa [2]

10 marks

Mark scheme: 8 (a) discrete amount / packet / quantum of energy M1 of electromagnetic radiation / EM radiation A1 [2] (b) (i) E = hc / λ = (6.63 × 10–34 × 3.0 × 108) / (570 × 10–9) = 3.49 × 10–19 J A1 [1] (ii) 1. number = (2.7 × 10–3) / (3.5 × 10–19) C1 = 7.7 × 1015 A1 [2] 2. momentum of photon = h / λ C1 = (6.63 × 10–34) / (570 × 10–9) = 1.16 × 10–27 kg m s–1 C1 change in momentum = 1.16 × 10–27 × 7.7 × 1015 = 8.96 × 10–12 kg m s–1 A1 [3] (allow E = pc route to 9 × 10–12) (c) pressure = (change in momentum per second) / area C1 = (8.96 × 10–12) / (1.3 × 10–5) = 6.9 × 10–7 Pa A1 [2] 14 6

This question in 9702/43 Oct/Nov 2014

8 A photon of wavelength 6.50 × 10−12 m is incident on an isolated stationary electron, as illustrated in Fig. 8.1. deflected photon wavelength 6.84 × 10–12 m incident photon e wavelength 6.50 × 10–12 m electron mass me Fig. 8.1 The photon is deflected elastically by the electron of mass me. The wavelength of the deflected photon is 6.84 × 10−12 m. (a) Calculate, for the incident photon, (i) its momentum, momentum = … N s [2] (ii) its energy. energy = … J [2] (b) The angle θ through which the photon is deflected is given by the expression h Δλ = (1 – cos θ) mec where Δλ is the change in wavelength of the photon, h is the Planck constant and c is the speed of light in free space. (i) Calculate the angle θ. θ = … ° [2] (ii) Use energy considerations to suggest why Δλ must always be positive. … … … … [3]

9 marks

Mark scheme: 8 (a) (i) p = h / λ = (6.63 × 10–34) / (6.50 × 10–12) C1 = 1.02 × 10–22 N s A1 [2] (ii) E = hc / λ or E = pc = (6.63 × 10–34 × 3.00 × 108) / (6.50 × 10–12) C1 = 3.06 × 10–14 J A1 [2] (b) (i) 0.34 × 10–12 = (6.63 × 10–34) / (9.11 × 10–31 × 3.0 × 108) × (1 – cos θ) C1 θ = 30.7° A1 [2] (ii) deflected electron has energy M1 this energy is derived from the incident photon A1 deflected photon has less energy, longer wavelength (so ∆λ always positive) B1 [3]

This question in 9702/41 May/June 2015

6 (a) Explain what is meant by a photon. … … … [2] (b) An X-ray photon of energy 3.06 × 10–14 J is incident on an isolated stationary electron, as illustrated in Fig. 6.1. deflected photon wavelength 6.80 × 10–12 m incident photon e energy 3.06 × 10–14 J Fig. 6.1 The photon is deflected elastically by the electron through angle θ. The deflected photon has a wavelength of 6.80 × 10–12 m. (i) On Fig. 6.1, draw an arrow to indicate a possible initial direction of motion of the electron after the photon has been deflected. [1] (ii) Calculate 1. the energy of the deflected photon, photon energy = … J [2] 2. the speed of the electron after the photon has been deflected. speed = … m s–1 [3] (c) Explain why the magnitude of the final momentum of the electron is not equal to the change in magnitude of the momentum of the photon. … … … [2]

10 marks

Mark scheme: 6 (a) packet/quantum/discrete amount of energy M1 of electromagnetic energy/radiation/waves A1 [2] (b) (i) arrow below axis and pointing to right B1 [1] (ii) 1. E = hc / λ = (6.63 × 10–34 × 3.0 × 108) / (6.80 × 10–12) C1 = 2.93 × 10–14 J (accept 2 s.f.) A1 [2] 2. energy of electron = (3.06 – 2.93) × 10–14 = 1.3 × 10–15 J C1 speed = (2E / m ) C1 = 5.4 × 107 m s–1 A1 [3] (c) momentum is a vector quantity B1 either must consider momentum in two directions or direction changes so cannot just consider magnitude B1 [2]

This question in 9702/42 May/June 2015

8 A photon of wavelength 6.50 × 10−12 m is incident on an isolated stationary electron, as illustrated in Fig. 8.1. deflected photon wavelength 6.84 × 10–12 m incident photon e wavelength 6.50 × 10–12 m electron mass me Fig. 8.1 The photon is deflected elastically by the electron of mass me. The wavelength of the deflected photon is 6.84 × 10−12 m. (a) Calculate, for the incident photon, (i) its momentum, momentum = … N s [2] (ii) its energy. energy = … J [2] (b) The angle θ through which the photon is deflected is given by the expression h Δλ = (1 – cos θ) mec where Δλ is the change in wavelength of the photon, h is the Planck constant and c is the speed of light in free space. (i) Calculate the angle θ. θ = … ° [2] (ii) Use energy considerations to suggest why Δλ must always be positive. … … … … [3]

9 marks

Mark scheme: 8 (a) (i) p = h / λ = (6.63 × 10–34) / (6.50 × 10–12) C1 = 1.02 × 10–22 N s A1 [2] (ii) E = hc / λ or E = pc = (6.63 × 10–34 × 3.00 × 108) / (6.50 × 10–12) C1 = 3.06 × 10–14 J A1 [2] (b) (i) 0.34 × 10–12 = (6.63 × 10–34) / (9.11 × 10–31 × 3.0 × 108) × (1 – cos θ) C1 θ = 30.7° A1 [2] (ii) deflected electron has energy M1 this energy is derived from the incident photon A1 deflected photon has less energy, longer wavelength (so ∆λ always positive) B1 [3]

This question in 9702/43 May/June 2015

12 High-energy electrons collide with a metal target, producing X-ray photons. The variation with wavelength of the intensity of the X-ray beam is illustrated in Fig. 12.1. intensity 0 wavelength Fig. 12.1 (a) Explain why there is (i) a continuous distribution of wavelengths, … … … … [3] (ii) a sharp cut-off at short wavelength, … … … [2] (iii) a series of peaks superimposed on the continuous distribution of wavelengths. … … [1] (b) In the X-ray imaging of body structures, longer wavelength photons are frequently filtered out of the X-ray beam. (i) State how this filtering is achieved. … … [1] (ii) Suggest the reason for this filtering. … … [1] [Total: 8]

8 marks

Mark scheme: 12 (a) (i) (X–ray) photon produced when electron/charged particle is stopped/accelerated (suddenly) B1 range of accelerations (in target) M1 hence distribution of wavelengths A1 [3] (ii) electron gives all its energy to one photon B1 electron stopped in single collision B1 [2] (iii) de-excitation of (orbital) electrons in target/anode/metal B1 [1] (b) (i) aluminium sheet/filter/foil (placed in beam from tube) B1 [1] (ii) (long wavelength X-rays) do not pass through the body B1 [1]

This question in 9702/42 May/June 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) State what is meant by a photon. … … [2] (b) Light in a beam has a continuous spectrum that lies within the visible region. The photons of light have energies ranging from 1.60 eV to 2.60 eV. The beam passes through some hydrogen gas. It then passes through a diffraction grating and an absorption spectrum is observed. (i) All of the light absorbed by the hydrogen is re-emitted. Explain why dark lines are still observed in the absorption spectrum. … … [1] (ii) Some of the energy levels of an electron in a hydrogen atom are illustrated in Fig. 10.1. –0.54 –0.85 –1.51 energy / eV –3.40 –13.60 Fig. 10.1 (not to scale) The dark lines in the absorption spectrum are the result of electron transitions between energy levels. On Fig. 10.1, draw arrows to show the initial electron transitions between energy levels that could give rise to dark lines in the absorption spectrum. [2] (iii) Calculate the shortest wavelength of the light in the beam. wavelength = … m [3] [Total: 8]

8 marks

Mark scheme: 10(a) packet / quantum of energy M1 of electromagnetic radiation A1 10(b)(i) light is re-emitted in all directions / only part of the re-emitted light is in the direction of the beam B1 10(b)(ii) an arrow between –3.40 eV and –1.51 eV and an arrow between –3.40 eV and –0.85 eV B1 all arrows shown point ‘upwards’ B1 10(b)(iii) E = hc / λ or E = hf and c = fλ C1 2.60 × 1.60 × 10–19 = (6.63 × 10–34 × 3.00 × 108) / λ C1 λ = 4.8 × 10–7 m A1

This question in 9702/42 Feb/March 2017

11 (a) State what is meant by a photon. … … [1] (b) Indium-123 (12349In) is radioactive. A nucleus of indium-123 emits a γ-ray photon of energy 1.1 MeV. Determine, for this γ-radiation, (i) the frequency, frequency = … Hz [2] (ii) the momentum of a photon. momentum = … N s [2] (c) The indium-123 nucleus is stationary before emission of the γ-ray photon. Use your answer in (b)(ii) to estimate the recoil speed of the nucleus after emission of the photon. speed = … m s−1 [2] [Total: 7]

7 marks

Mark scheme: 11(a) packet/quantum of energy of electromagnetic/EM radiation B1 11(b)(i) E = hf 1.1 × 106 × 1.60 × 10–19 = 6.63 × 10–34 × f C1 f = 2.7 × 1020 (2.65 × 1020) Hz A1 11(b)(ii) p = h / λ = hf / c = (6.63 × 10–34 × 2.65 × 1020) / (3.00 × 108) or p = E / c = (1.1 × 1.60 × 10–13) / (3.00 × 108) C1 p = 5.9 × 10–22 (5.87 × 10–22) N s A1 11(c) 123 × 1.66 × 10–27 × v = 5.87 × 10–22 C1 v = 2.9 × 103 m s–1 A1

This question in 9702/41 Oct/Nov 2017

11 (a) State what is meant by a photon. … … [1] (b) Indium-123 (12349In) is radioactive. A nucleus of indium-123 emits a γ-ray photon of energy 1.1 MeV. Determine, for this γ-radiation, (i) the frequency, frequency = … Hz [2] (ii) the momentum of a photon. momentum = … N s [2] (c) The indium-123 nucleus is stationary before emission of the γ-ray photon. Use your answer in (b)(ii) to estimate the recoil speed of the nucleus after emission of the photon. speed = … m s−1 [2] [Total: 7]

7 marks

Mark scheme: 11(a) packet/quantum of energy of electromagnetic/EM radiation B1 11(b)(i) E = hf 1.1 × 106 × 1.60 × 10–19 = 6.63 × 10–34 × f C1 f = 2.7 × 1020 (2.65 × 1020) Hz A1 11(b)(ii) p = h / λ = hf / c = (6.63 × 10–34 × 2.65 × 1020) / (3.00 × 108) or p = E / c = (1.1 × 1.60 × 10–13) / (3.00 × 108) C1 p = 5.9 × 10–22 (5.87 × 10–22) N s A1 11(c) 123 × 1.66 × 10–27 × v = 5.87 × 10–22 C1 v = 2.9 × 103 m s–1 A1

This question in 9702/43 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 stationary isolated nucleus emits a γ-ray photon of energy 0.51MeV. (a) State what is meant by a photon. … … … [2] (b) For the γ-ray photon, calculate (i) its wavelength, wavelength = … m [2] (ii) its momentum. momentum = … N s [2] (c) (i) For this nucleus, determine the change in mass Δm during the decay that gives rise to the energy of the γ-ray photon. Δm = … kg [2] (ii) Explain why, after the decay, the nucleus is no longer stationary. … … … [1] [Total: 9]

9 marks

Mark scheme: 11(a) discrete amount/quantum/packet of energy M1 of electromagnetic radiation A1 11(b)(i) energy = hc / λ C1 λ = (6.63 × 10–34 × 3.00 × 108) / (0.51 × 106 × 1.60 × 10–19) = 2.4 × 10–12 m A1 11(b)(ii) p = h / λ = (6.63 × 10–34) / (2.44 × 10–12) or p = E / c = (0.51 × 1.60 × 10–13) / (3.00 × 108) C1 p = 2.7 × 10–22 N s A1 11(c)(i) E = c2∆m C1 ∆m = (0.51 × 1.60 × 10–13) / (3.00 × 108)2 = 9.1 × 10–31 kg A1 11(c)(ii) (momentum is conserved so) nucleus must have momentum in opposite direction to photon B1

This question in 9702/41 Oct/Nov 2018

11 (a) State what is meant by a photon. … … … [2] (b) Describe the appearance of a visible line emission spectrum, as seen using a diffraction grating. … … … … [2] (c) The lowest electron energy levels in an isolated hydrogen atom are shown in Fig. 11.1. – 0.54 – 0.38 – 0.85 –1.50 – 3.40 energy / eV –13.6 Fig. 11.1 (not to scale) (i) An electron is initially at the energy level –0.85 eV. State the total number of different wavelengths that may be emitted as the electron de-excites (loses energy). number = … [1] (ii) Photons resulting from electron de-excitation from the –0.85 eV energy level are incident on the surface of a sample of platinum. Platinum has a work function energy of 5.6 eV. Determine 1. the maximum kinetic energy, in eV, of a photoelectron emitted from the surface of the platinum, maximum energy = … eV [2] 2. the wavelength of the photon producing the photoelectron in (ii) part 1. wavelength = … m [3] [Total: 10]

10 marks

Mark scheme: 11(a) discrete amount/quantum/packet of energy M1 of electromagnetic radiation A1 11(b) mostly dark/dark background B1 coloured lines B1 11(c)(i) 6 A1 11(c)(ii) 1. maximum photon energy = 13.6 – 0.85 (= 12.75 eV) C1 maximum kinetic energy = (13.6 – 0.85) – 5.6 = 7.2 eV A1 2. energy = hc / λ C1 λ = (6.63 × 10–34 × 3.00 × 108) / [(13.6 – 0.85) × 1.60 × 10–19] C1 = 9.8 × 10–8 m A1

This question in 9702/42 Oct/Nov 2018

11 A stationary isolated nucleus emits a γ-ray photon of energy 0.51MeV. (a) State what is meant by a photon. … … … [2] (b) For the γ-ray photon, calculate (i) its wavelength, wavelength = … m [2] (ii) its momentum. momentum = … N s [2] (c) (i) For this nucleus, determine the change in mass Δm during the decay that gives rise to the energy of the γ-ray photon. Δm = … kg [2] (ii) Explain why, after the decay, the nucleus is no longer stationary. … … … [1] [Total: 9]

9 marks

Mark scheme: 11(a) discrete amount/quantum/packet of energy M1 of electromagnetic radiation A1 11(b)(i) energy = hc / λ C1 λ = (6.63 × 10–34 × 3.00 × 108) / (0.51 × 106 × 1.60 × 10–19) = 2.4 × 10–12 m A1 11(b)(ii) p = h / λ = (6.63 × 10–34) / (2.44 × 10–12) or p = E / c = (0.51 × 1.60 × 10–13) / (3.00 × 108) C1 p = 2.7 × 10–22 N s A1 11(c)(i) E = c2∆m C1 ∆m = (0.51 × 1.60 × 10–13) / (3.00 × 108)2 = 9.1 × 10–31 kg A1 11(c)(ii) (momentum is conserved so) nucleus must have momentum in opposite direction to photon B1

This question in 9702/43 Oct/Nov 2018

11 (a) State what is meant by a photon. … … … [2] (b) Calculate the energy, in eV, of a photon of light of wavelength 540 nm. energy = … eV [3] (c) The outermost electron energy bands of a semiconductor material are illustrated in Fig. 11.1. conduction band forbidden band valence band Fig. 11.1 The width of the forbidden band is 1.1 eV. Explain why, when photons of light, each of energy 2.1 eV, are incident on the semiconductor material, its resistance decreases. … … … … … … [4] [Total: 9]

9 marks

Mark scheme: 11(a) quantum / packet / discrete amount of energy M1 of electromagnetic radiation A1 11(b) E = hc / λ C1 = (6.63 × 10–34 × 3.0 × 108) / (540 × 10–9) C1 = (3.68 × 10–19) / (1.6 × 10–19) = 2.3 eV A1 11(c) Any 4 from: photon absorbed by electron in valence band (1) photon energy > energy of forbidden band (1) electron promoted to conduction band (1) hole left in valence band (1) more charge carriers so lower resistance (1) B4

This question in 9702/42 Feb/March 2019

11 (a) State what is meant by a photon. … … … [2] (b) A stationary cobalt-60 (6027Co) nucleus emits a γ-ray photon of energy 1.18 MeV. (i) Calculate the wavelength of the photon. wavelength = … m [2] (ii) Show that the momentum of the photon is 6.3 × 10–22 N s. [2] (c) Use information in (b)(ii) to determine the recoil speed of the cobalt-60 nucleus when the γ-ray photon is emitted. speed = … m s–1 [2] [Total: 8]

8 marks

Mark scheme: 11(a) packet/quantum of energy M1 of electromagnetic radiation A1 11(b)(i) E = hc / λ C1 1.18 × 1.60 × 10–13 = (6.63 × 10–34 × 3.00 × 108) / λ λ = 1.05 × 10–12 m A1 11(b)(ii) λ = h / p or E = pc C1 p = (6.63 × 10–34) / (1.05 × 10–12) or p = (1.18 × 1.60 × 10–13) / (3.00 × 108) leading to p = 6.3 × 10–22 N s B1 11(c) 6.3 × 10–22 = 60 × 1.66 × 10–27 × v C1 v = 6.3 × 103 m s–1 A1

This question in 9702/42 May/June 2019

11 A photon of wavelength 540 nm collides with an isolated stationary electron, as illustrated in Fig. 11.1. electron incident photon wavelength 540 nm deflected photon wavelength 544 nm Fig. 11.1 The photon is deflected elastically by the electron. The wavelength of the deflected photon is 544 nm. (a) (i) State what is meant by a photon. … … … [2] (ii) On Fig. 11.1, draw an arrow to indicate the approximate direction of motion of the deflected electron. [1] (b) Calculate: (i) the momentum of the deflected photon momentum = … N s [2] (ii) the energy transferred to the deflected electron. energy = … J [2] (c) Another photon of wavelength 540 nm collides with an isolated stationary electron. Explain why it is not possible for the deflected photon to have a wavelength less than 540 nm. … … … [2] [Total: 9]

9 marks

Mark scheme: 11(a)(i) quantum of energy M1 of electromagnetic radiation A1 11(a)(ii) arrow (on Fig. 11.1) pointing upwards and to the right B1 11(b)(i) λ = h / p C1 p = (6.63 × 10–34) / (544 × 10–9) = 1.22 × 10–27 N s A1 11(b)(ii) energy = hc / λ C1 = 6.63 × 10–34 × 3.00 × 108 × (540–1 – 544–1) × 109 = 2.7 × 10–21 J A1 11(c) (smaller wavelength corresponds to) greater photon energy B1 any one point from: • (deflected) photon loses energy (so not possible) • (deflected) photon would need to gain energy (so not possible) • electron would need to lose energy (so not possible) • initially electron energy is zero (so not possible) B1

This question in 9702/41 Oct/Nov 2020

11 A photon of wavelength 540 nm collides with an isolated stationary electron, as illustrated in Fig. 11.1. electron incident photon wavelength 540 nm deflected photon wavelength 544 nm Fig. 11.1 The photon is deflected elastically by the electron. The wavelength of the deflected photon is 544 nm. (a) (i) State what is meant by a photon. … … … [2] (ii) On Fig. 11.1, draw an arrow to indicate the approximate direction of motion of the deflected electron. [1] (b) Calculate: (i) the momentum of the deflected photon momentum = … N s [2] (ii) the energy transferred to the deflected electron. energy = … J [2] (c) Another photon of wavelength 540 nm collides with an isolated stationary electron. Explain why it is not possible for the deflected photon to have a wavelength less than 540 nm. … … … [2] [Total: 9]

9 marks

Mark scheme: 11(a)(i) quantum of energy M1 of electromagnetic radiation A1 11(a)(ii) arrow (on Fig. 11.1) pointing upwards and to the right B1 11(b)(i) λ = h / p C1 p = (6.63 × 10–34) / (544 × 10–9) = 1.22 × 10–27 N s A1 11(b)(ii) energy = hc / λ C1 = 6.63 × 10–34 × 3.00 × 108 × (540–1 – 544–1) × 109 = 2.7 × 10–21 J A1 11(c) (smaller wavelength corresponds to) greater photon energy B1 any one point from: • (deflected) photon loses energy (so not possible) • (deflected) photon would need to gain energy (so not possible) • electron would need to lose energy (so not possible) • initially electron energy is zero (so not possible) B1

This question in 9702/43 Oct/Nov 2020

12 (a) State what is meant by a photon. … … … [2] (b) A stationary nucleus of samarium-157 (15762 Sm) emits a gamma-ray (γ-ray) photon of energy 0.57 MeV. Determine, for one γ-ray photon: (i) its wavelength wavelength = … m [2] (ii) its momentum. momentum = … N s [2] (c) (i) Using your answer to (b)(ii), determine the speed of the samarium-157 nucleus after emission of the photon. speed = … m s−1 [2] (ii) By reference to your answer in (c)(i), explain quantitatively why the speed of the samarium-157 nucleus may be assumed to be negligible compared with the speed of the photon. … … [1] [Total: 9]

9 marks

Mark scheme: 12(a) quantum of energy M1 of electromagnetic radiation A1 12(b)(i) energy = hc / λ or energy = hf and f = c / λ C1 0.57 × 106 × 1.60 × 10–19 = (6.63 × 10–34 × 3.00 × 108) / λ λ = 2.2 × 10–12 m A1 Question Answer Marks 12(b)(ii) p = h / λ C1 = (6.63 × 10–34) / (2.2 × 10–12) = 3.0 × 10–22 N s A1 or p = E / c (C1) = (0.57 × 106 × 1.60 × 10–19) / (3.00 × 108) = 3.0 × 10–22 N s (A1) 12(c)(i) mass (of Sm-157 nucleus) = 157 × 1.66 × 10–27 or mass (of Sm-157 nucleus) = 0.157 / (6.02 × 1023) C1 recoil speed = (3.00 × 10–22) / (157 × 1.66 × 10–27) = 1.2 × 103 m s–1 A1 12(c)(ii) (1.2 ×) 103 m s–1 is much less than (3.0 ×) 108 m s–1 B1

This question in 9702/42 May/June 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

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) (i) Show that the momentum p of a photon of electromagnetic radiation with wavelength λ is given by h p = λ where h is the Planck constant. [2] (ii) Use the expression in (a)(i) to show that a photon in free space that has a momentum of 9.5 × 10–28 N s is a photon of red light. [1] (b) A beam of red light of intensity 160 W m–2 is incident normally on a plane mirror, as shown in Fig. 8.1. The momentum of each photon in the beam is 9.5 × 10–28 N s. plane mirror beam of red light, intensity 160 W m–2 Fig. 8.1 All of the light is reflected by the mirror in the opposite direction to its original path. The cross-sectional area of the beam is 2.5 × 10–6 m2. (i) Show that the number of photons incident on the mirror per unit time is 1.4 × 1015 s–1. [2] (ii) Use the information in (b)(i) to determine the pressure exerted by the light beam on the mirror. pressure = … Pa [3] (c) The beam of red light in (b) is now replaced with a beam of blue light of the same intensity. Suggest and explain whether the pressure exerted on the mirror by the beam of blue light is less than, the same as, or greater than the pressure exerted by the beam of red light. … … … … [2] [Total: 10]

10 marks

Mark scheme: 8(a)(i) p = E / c M1 E = hc / and completion of algebra leading to p = h /  A1 8(a)(ii) wavelength = (6.63  10–34) / (9.5  10–28) = 700  10–9 m so red B1 8(b)(i) power = intensity  area C1 number per unit time = (160  2.5  10–6) / (9.5  10–28  3.00  108) = 1.4  1015 s–1 A1 8(b)(ii) pressure = force / area C1 force = rate of change of momentum C1 = 2  9.5  10–28  1.4  1015 pressure = (2  9.5  10–28  1.4  1015) / (2.5  10–6) A1 = 1.1  10–6 Pa 8(c) photons have greater momentum B1 or fewer photons per unit time greater photon momentum but smaller number of photons (per unit time) so pressure is the same B1

This question in 9702/41 Oct/Nov 2023

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) (i) Show that the momentum p of a photon of electromagnetic radiation with wavelength λ is given by h p = λ where h is the Planck constant. [2] (ii) Use the expression in (a)(i) to show that a photon in free space that has a momentum of 9.5 × 10–28 N s is a photon of red light. [1] (b) A beam of red light of intensity 160 W m–2 is incident normally on a plane mirror, as shown in Fig. 8.1. The momentum of each photon in the beam is 9.5 × 10–28 N s. plane mirror beam of red light, intensity 160 W m–2 Fig. 8.1 All of the light is reflected by the mirror in the opposite direction to its original path. The cross-sectional area of the beam is 2.5 × 10–6 m2. (i) Show that the number of photons incident on the mirror per unit time is 1.4 × 1015 s–1. [2] (ii) Use the information in (b)(i) to determine the pressure exerted by the light beam on the mirror. pressure = … Pa [3] (c) The beam of red light in (b) is now replaced with a beam of blue light of the same intensity. Suggest and explain whether the pressure exerted on the mirror by the beam of blue light is less than, the same as, or greater than the pressure exerted by the beam of red light. … … … … [2] [Total: 10]

10 marks

Mark scheme: 8(a)(i) p = E / c M1 E = hc / and completion of algebra leading to p = h /  A1 8(a)(ii) wavelength = (6.63  10–34) / (9.5  10–28) = 700  10–9 m so red B1 8(b)(i) power = intensity  area C1 number per unit time = (160  2.5  10–6) / (9.5  10–28  3.00  108) = 1.4  1015 s–1 A1 8(b)(ii) pressure = force / area C1 force = rate of change of momentum C1 = 2  9.5  10–28  1.4  1015 pressure = (2  9.5  10–28  1.4  1015) / (2.5  10–6) A1 = 1.1  10–6 Pa 8(c) photons have greater momentum B1 or fewer photons per unit time greater photon momentum but smaller number of photons (per unit time) so pressure is the same B1

This question in 9702/43 Oct/Nov 2023

8 (a) State what is meant by a photon. … … … [2] (b) Fig. 8.1 shows a tube in which X-rays are produced at a metal target. X Y particles filament vacuum glass tube metal target Fig. 8.1 Particles are accelerated from the filament to the target by a constant high voltage applied across the terminals X and Y. (i) State the name of the particles. … [1] (ii) On Fig. 8.1, use + and – signs to label terminals X and Y to indicate the polarity of the high voltage. [1] (c) For an accelerating voltage of 32 kV in Fig. 8.1, determine: (i) the maximum energy, in MeV, of an X-ray photon produced at the target maximum photon energy = … MeV [1] (ii) the maximum momentum of an X-ray photon produced at the target maximum photon momentum = … N s [2] (iii) the minimum wavelength of X-rays produced at the target. minimum wavelength = … m [3] (d) Explain why X-rays can be used to produce images of internal body structures that have good contrast. … … … … … [3] [Total: 13]

13 marks

Mark scheme: 8(a) packet / quantum of energy M1 of electromagnetic radiation A1 8(b)(i) electron(s) B1 8(b)(ii) X labelled – and Y labelled + B1 8(c)(i) 0.032 MeV A1 8(c)(ii) momentum = E / c C1 momentum = (0.032 × 1.60  10–13) / (3.00  108) = 1.7  10–23 N s A1 8(c)(iii) E = hf and  = c / f C1  = hc / E = (6.63  10–34 × 3.00  108) / (0.032  1.60 × 10–13) C1  = 3.9  10–11 m A1 8(d) discussion of bone and soft tissue B1 discussion of different attenuation (coefficients) or discussion differences in penetration / transmission / absorption B1 transmitted intensities (by bone and tissue) are very different (leading to good contrast images) B1

This question in 9702/41 May/June 2024

8 (a) State what is meant by a photon. … … … [2] (b) Fig. 8.1 shows a tube in which X-rays are produced at a metal target. X Y particles filament vacuum glass tube metal target Fig. 8.1 Particles are accelerated from the filament to the target by a constant high voltage applied across the terminals X and Y. (i) State the name of the particles. … [1] (ii) On Fig. 8.1, use + and – signs to label terminals X and Y to indicate the polarity of the high voltage. [1] (c) For an accelerating voltage of 32 kV in Fig. 8.1, determine: (i) the maximum energy, in MeV, of an X-ray photon produced at the target maximum photon energy = … MeV [1] (ii) the maximum momentum of an X-ray photon produced at the target maximum photon momentum = … N s [2] (iii) the minimum wavelength of X-rays produced at the target. minimum wavelength = … m [3] (d) Explain why X-rays can be used to produce images of internal body structures that have good contrast. … … … … … [3] [Total: 13]

13 marks

Mark scheme: 8(a) packet / quantum of energy M1 of electromagnetic radiation A1 8(b)(i) electron(s) B1 8(b)(ii) X labelled – and Y labelled + B1 8(c)(i) 0.032 MeV A1 8(c)(ii) momentum = E / c C1 momentum = (0.032 × 1.60  10–13) / (3.00  108) = 1.7  10–23 N s A1 8(c)(iii) E = hf and  = c / f C1  = hc / E = (6.63  10–34 × 3.00  108) / (0.032  1.60 × 10–13) C1  = 3.9  10–11 m A1 8(d) discussion of bone and soft tissue B1 discussion of different attenuation (coefficients) or discussion differences in penetration / transmission / absorption B1 transmitted intensities (by bone and tissue) are very different (leading to good contrast images) B1

This question in 9702/43 May/June 2024

9 Electrons in a vacuum are accelerated from rest through a potential difference (p.d.) V to form a beam. The electrons each have mass m and charge q. The beam is incident on a graphite crystal that acts as a diffraction grating. After passing through the crystal, the beam reaches a fluorescent screen. An interference pattern is observed on this screen. (a) Explain what this observation shows about the nature of electrons. … … … [1] (b) Determine an expression, in terms of m, q and V, for the momentum p of an electron in the beam. p = … [3] (c) The p.d. through which the electrons are accelerated is now increased to a greater value. Describe and explain the effect of this change on the interference pattern observed. … … … [2] (d) The electrons are now accelerated through different values of V, resulting in pairs of corresponding values for p and the de Broglie wavelength λ. 1 (i) On Fig. 9.1, sketch the variation of p with λ. p 0 0 1 λ Fig. 9.1 [2] (ii) State the name of the quantity represented by the gradient of the line in Fig. 9.1. … [1] [Total: 9]

9 marks

Mark scheme: 9(a) diffraction is characteristic of wave behaviour so shows that electrons can behave like waves B1 9(b) qV = ½mv2 C1 p = mv C1 p = m  √(2qV / m) A1 = √(2qVm) 9(c) (electrons have) greater momentum so smaller (de Broglie) wavelength B1 fringes become closer together B1 9(d)(i) straight line with positive gradient B1 line with positive gradient passing through the origin B1 9(d)(ii) Planck constant B1

This question in 9702/42 Oct/Nov 2024

8 (a) State what is meant by a photon. … … … [2] (b) A laser emits red light of a single wavelength. The light is produced when electrons move from a higher energy level to a lower energy level. The difference in energy between the two levels is 1.96 eV. (i) Calculate the wavelength of the light. wavelength = … m [3] (ii) The power of the beam emitted by the laser is 1.0 × 10–2 W. Calculate the number of photons emitted per unit time by the laser. number per unit time = … s–1 [1] (iii) The photons are incident normally on a surface. Half of the number of photons are absorbed by the surface, and half are reflected. Determine the average force exerted by the beam of photons on the surface. average force = … N [4] [Total: 10]

10 marks

Mark scheme: 8(a) • quantum of energy M1 • of electromagnetic radiation A1 8(b)(i) ()E = hc /  C1  = (6.63  10–34  3.00  108) / (1.96  1.60  10–19) C1 = 6.3  10–7 m A1 8(b)(ii) number per unit time = power / energy per photon A1 = (1.0  10–2) / (1.96  1.60  10–19) = 3.2  1016 s–1 8(b)(iii) either: force = rate of change of momentum C1 or: F = p / t p = E / c C1 half the photons have change in momentum p, the other half have change in momentum 2p C1 F = [(1.96  1.60  10–19) / (3.00  108)]  3.2  1016  [(2 + 1) / 2] A1 = 5.0  10–11 N

This question in 9702/42 Feb/March 2025

8 (a) State what is meant by a photon. … … … [2] 238 (b) A stationary nucleus of uranium-238 ( 92U) undergoes alpha decay to produce a nucleus 234 of thorium-234 ( 90Th). The kinetic energy of the emitted alpha particle is 4.200 MeV. A gamma-ray photon is also emitted during the decay. Assume that the rebound kinetic energy of the thorium nucleus is negligible. Table 8.1 shows the masses of the nuclides involved in the decay reaction. The mass of the uranium-238 nuclide is missing. Table 8.1 nuclide nuclide mass / u 4 4.000 407 2α 234 233.915 174 90Th 238 92U The total energy released in the decay of the nucleus of uranium-238 is 4.274 MeV. (i) Calculate the mass, in u, of the uranium-238 nuclide. Give your answer to five decimal places. mass = … u [3] (ii) Determine a value for the wavelength of the gamma radiation emitted during the decay of the uranium-238 nucleus. wavelength = … m [3] (iii) In practice, the rebound kinetic energy of the thorium nucleus is not negligible. Explain, without further calculation, how your answer in (b)(ii) compares with the true wavelength of gamma radiation emitted during the decay of the uranium-238 nucleus. … … … [1] (c) Gamma radiation emitted during the decay of a sample of uranium-238 has a single wavelength. decay by beta emission, and also emit gamma radiation in the Nuclei of cobalt-60 (6027Co) process. Suggest why there is not a single wavelength for the gamma radiation emitted during the decay of a sample of cobalt-60. … … … [2] [Total: 11]

11 marks

Mark scheme: 8(a) packet / quantum of energy M1 of electromagnetic radiation A1 8(b)(i) E = c2m C1 m = (4.274  106  1.60  10–19) / (1.66  10–27  (3.00  108)2) C1 ( = 0.00458 u) m = 233.915174 + 4.000407 + 0.00458 A1 = 237.92016 u 8(b)(ii) E = hc /  C1 or E = hf and c = f (4.274 – 4.200)  1.60  10–13 = (6.63  10–34  3.00  108) /  C1 = 1.7  10–11 m A1 8(b)(iii) (true) energy of gamma photon is smaller so (true) wavelength is larger B1 8(c) (anti)neutrinos are emitted during beta decay B1 particles emitted during beta decay carry varying amounts of energy, so energy of gamma photon is also variable (between B1 decays)

This question in 9702/41 Oct/Nov 2025

8 (a) State what is meant by a photon. … … … [2] 238 (b) A stationary nucleus of uranium-238 ( 92U) undergoes alpha decay to produce a nucleus 234 of thorium-234 ( 90Th). The kinetic energy of the emitted alpha particle is 4.200 MeV. A gamma-ray photon is also emitted during the decay. Assume that the rebound kinetic energy of the thorium nucleus is negligible. Table 8.1 shows the masses of the nuclides involved in the decay reaction. The mass of the uranium-238 nuclide is missing. Table 8.1 nuclide nuclide mass / u 4 4.000 407 2α 234 233.915 174 90Th 238 92U The total energy released in the decay of the nucleus of uranium-238 is 4.274 MeV. (i) Calculate the mass, in u, of the uranium-238 nuclide. Give your answer to five decimal places. mass = … u [3] (ii) Determine a value for the wavelength of the gamma radiation emitted during the decay of the uranium-238 nucleus. wavelength = … m [3] (iii) In practice, the rebound kinetic energy of the thorium nucleus is not negligible. Explain, without further calculation, how your answer in (b)(ii) compares with the true wavelength of gamma radiation emitted during the decay of the uranium-238 nucleus. … … … [1] (c) Gamma radiation emitted during the decay of a sample of uranium-238 has a single wavelength. decay by beta emission, and also emit gamma radiation in the Nuclei of cobalt-60 (6027Co) process. Suggest why there is not a single wavelength for the gamma radiation emitted during the decay of a sample of cobalt-60. … … … [2] [Total: 11]

11 marks

Mark scheme: 8(a) packet / quantum of energy M1 of electromagnetic radiation A1 8(b)(i) E = c2m C1 m = (4.274  106  1.60  10–19) / (1.66  10–27  (3.00  108)2) C1 ( = 0.00458 u) m = 233.915174 + 4.000407 + 0.00458 A1 = 237.92016 u 8(b)(ii) E = hc /  C1 or E = hf and c = f (4.274 – 4.200)  1.60  10–13 = (6.63  10–34  3.00  108) /  C1 = 1.7  10–11 m A1 8(b)(iii) (true) energy of gamma photon is smaller so (true) wavelength is larger B1 8(c) (anti)neutrinos are emitted during beta decay B1 particles emitted during beta decay carry varying amounts of energy, so energy of gamma photon is also variable (between B1 decays)

This question in 9702/43 Oct/Nov 2025