Quantum physics

12(a) • frequency determines energy of photon B2 • intensity determines number of photons (per unit time) • intensity does not determine energy of a photon Any two points, 1 mark each kinetic energy (of the electron) depends on the energy of one photon B1 12(b)(i) E = hc / λ C1 or E = hf and c = fλ E = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1 (= 7.96 × 10–19 J) A1 = 5.0 eV 12(b)(ii) E = photon energy – work function C1 MAX work function = 5.0 – 1.4 A1 = 3.6 eV © UCLES 2021 Page 18 of 18

12(a) • frequency determines energy of photon B2 • intensity determines number of photons (per unit time) • intensity does not determine energy of a photon Any two points, 1 mark each kinetic energy (of the electron) depends on the energy of one photon B1 12(b)(i) E = hc / λ C1 or E = hf and c = fλ E = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1 (= 7.96 × 10–19 J) A1 = 5.0 eV 12(b)(ii) E = photon energy – work function C1 MAX work function = 5.0 – 1.4 A1 = 3.6 eV © UCLES 2021 Page 18 of 18

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 C1 = 6.63 × 10–34 / (9.11 × 10–31 × 4.1 × 107) = 1.8 × 10–11 m A1 © UCLES 2021 Page 14 of 15

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 B2 • energy of photons depends on frequency (of EM radiation) • electrons can only absorb a single photon (of energy) Any two points, 1 mark each emission only possible if photon energy is at least the work function B1 9(b)(iii) work function = hf = 6.63 × 10–34 × 6.93 × 1014 C1 0 = 4.59 × 10–19 (J) A1 = 4.59 × 10–19 / 1.60 × 10–19 (eV) = 2.87 eV © UCLES 2021 Page 15 of 19

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 C1 = 6.63 × 10–34 / (9.11 × 10–31 × 4.1 × 107) = 1.8 × 10–11 m A1 © UCLES 2021 Page 14 of 15

8(a) h h M1 λ = or λ = p mv where h is the Planck constant and A1 p is the momentum (of particle) / mv is the momentum (of particle) / m is the mass (of particle) and v is the velocity (of particle) 8(b)(i) (electron) diffraction B1 8(b)(ii) moving electrons behave like waves B1 8(b)(iii) spacing between atoms ≈ wavelength of electron B1 or diameter of atom ≈ wavelength of electron 8(b)(iv) Any one of: M1 • wavelength has decreased • electron had greater momentum so (accelerating) p.d. was increased A1 © UCLES 2022 Page 12 of 17

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 B3 or threshold frequency = 5.4  1014 Hz  work function of the metal = 3.6  10–19 J (or 2.2 eV)  E increases (linearly) with (increasing) frequency MAX  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 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 © UCLES 2022 Page 13 of 16

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 B3 or threshold frequency = 5.4  1014 Hz  work function of the metal = 3.6  10–19 J (or 2.2 eV)  E increases (linearly) with (increasing) frequency MAX  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 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 © UCLES 2022 Page 13 of 16

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 =  + E C1 MAX  = (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 © UCLES 2022 Page 13 of 15

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 =  + E C1 MAX  = (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 © UCLES 2022 Page 13 of 15

7(a) photon absorbed (by electron) and electron excited B1 photon energy equal to difference in (energy of two) energy levels B1 photon energy relates to a single wavelength / single frequency B1 electron de-excites and emits photon in any direction B1 7(b) hc C1 =E  uses 658nm C1 6.63 1 0–34  3.00 1 08 A1 = – E – (–3.40 × 1.60 × 10–19) 1 658 1 0–9 E = –2.42  10–19J 1 © UCLES 2023 Page 15 of 18

7(a) wavelength associated with a moving particle B1 7(b)(i) (electron) diffraction B1 7(b)(ii) beam spreads out indicating diffraction B1 or light and dark regions indicate an interference pattern electron beam is behaving as a wave B1 7(c)(i) central blob and concentric rings B1 rings closer together (than previously) B1 7(c)(ii) (greater p.d. so) electrons to have greater momentum B1 greater momentum so decrease in (de Broglie) wavelength B1 lower (de Broglie) wavelength (for same grating spacing in crystal) causes: B1 smaller diffraction angle or smaller angle of intensity maxima (for each order) or decrease in fringe spacing in diffraction pattern © UCLES 2023 Page 13 of 16

8(a) transition (emits) (one) photon with energy equal to the difference in energy between the two levels B1 frequency of radiation corresponds to energy of photon B1 8(b)(i) line to the left of the pair in Fig. 8.2, labelled A B1 larger gap between line A and the nearest of the pair in Fig. 8.2 than between the lines in the pair B1 8(b)(ii) line to the left of both the pair in Fig. 8.2 and line A, labelled B B1 larger gap between line B and line A than between line A and the nearest one of the pair in Fig. 8.2 B1 8(c) E = hf C1 E = E + h(f + f ) A1 3 1 A B © UCLES 2023 Page 13 of 15

7(a) wavelength associated with a moving particle B1 7(b)(i) (electron) diffraction B1 7(b)(ii) beam spreads out indicating diffraction B1 or light and dark regions indicate an interference pattern electron beam is behaving as a wave B1 7(c)(i) central blob and concentric rings B1 rings closer together (than previously) B1 7(c)(ii) (greater p.d. so) electrons to have greater momentum B1 greater momentum so decrease in (de Broglie) wavelength B1 lower (de Broglie) wavelength (for same grating spacing in crystal) causes: B1 smaller diffraction angle or smaller angle of intensity maxima (for each order) or decrease in fringe spacing in diffraction pattern © UCLES 2023 Page 13 of 16

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 © UCLES 2023 Page 14 of 16

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 © UCLES 2023 Page 13 of 15

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 © UCLES 2023 Page 14 of 16

7(a) p = E / c C1 = (3.11  10–19) / (3.00  108) A1 = 1.04  10–27 N s 7(b)(i) E = hf and c = f so C1 energy of one photon = hc /  350  10–3 = N  (6.63  10–34  3.00  108) / (640  10–9) N = 1.1  1018 A1 7(b)(ii) F = (change in) momentum / time M1 Clear use of p = E / c and t = E / P to complete the algebra and arrive at the final equation: A1 e.g. F = [E / c] / [E / P] = P / c 7(c)(i) maximum wavelength (of electromagnetic radiation) that causes electrons to be emitted (from surface of metal) B1 7(c)(ii) work function = 2.26  1.60  10–19 (J) C1 E = hc /  so (2.26  1.60  10–19) = (6.63  10–34  3.00  108) /  0  = 5.50  10–7 m A1 0 Question Answer Marks

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 mv 2 C1 MAX 6.63  10–34  11  1014 = (5.8  10–19) + (½  9.11  10–31  v 2) C1 MAX v = 5.7  105 m s–1 A1 MAX 8(c) E shown as zero from f = 8.0 to 8.8 and non-zero from f = 8.8 to 11 B1 MAX all non-zero E shown as a single straight line with a positive gradient B1 MAX line passing through (11, 1.45) B1 © Cambridge University Press & Assessment 2024 Page 13 of 15

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 © Cambridge University Press & Assessment 2024 Page 13 of 14

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 mv 2 C1 MAX 6.63  10–34  11  1014 = (5.8  10–19) + (½  9.11  10–31  v 2) C1 MAX v = 5.7  105 m s–1 A1 MAX 8(c) E shown as zero from f = 8.0 to 8.8 and non-zero from f = 8.8 to 11 B1 MAX all non-zero E shown as a single straight line with a positive gradient B1 MAX line passing through (11, 1.45) B1 © Cambridge University Press & Assessment 2024 Page 13 of 15

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  10–7m 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  1016s–1 © Cambridge University Press & Assessment 2025 Page 12 of 14 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  10–11N Question Answer Marks

8(a) wavelength associated with a moving particle B1 8(b)  = h / p C1 = (6.63 × 10–34) / (9.11 × 10–31 × 4.9 × 107) A1 = 1.5 × 10–11 m 8(c) similarity: any one point from: B1 • same mass • same magnitude of charge • both leptons difference: any one point from: B1 • electron has negative charge, positron has positive charge • positron is anti-particle of electron • electron is a particle, positron is an anti-particle 8(d)(i) (pair) annihilation B1 8(d)(ii) their mass gets converted into energy B1 (their mass–energy) becomes the energy of the gamma photons B1 8(d)(iii) they travel in opposite directions to conserve momentum B1 8(d)(iv) kinetic energy = ½ × 9.11 × 10–31 × (4.9 × 107)2 = 1.1 × 10–15 J A1 © Cambridge University Press & Assessment 2025 Page 16 of 19 8(d)(v) E = mc2 C1 E = hc /  C1 or E = hf and c = f (1.1 × 10–15) + (9.11 × 10–31 × (3.00 × 108)2) = (6.63 × 10–34 × 3.00 × 108) /  A1  = 2.39 × 10–12 m © Cambridge University Press & Assessment 2025 Page 17 of 19

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) © Cambridge University Press & Assessment 2025 Page 17 of 18

8(a) wavelength associated with a moving particle B1 8(b)  = h / p C1 = (6.63 × 10–34) / (9.11 × 10–31 × 4.9 × 107) A1 = 1.5 × 10–11 m 8(c) similarity: any one point from: B1 • same mass • same magnitude of charge • both leptons difference: any one point from: B1 • electron has negative charge, positron has positive charge • positron is anti-particle of electron • electron is a particle, positron is an anti-particle 8(d)(i) (pair) annihilation B1 8(d)(ii) their mass gets converted into energy B1 (their mass–energy) becomes the energy of the gamma photons B1 8(d)(iii) they travel in opposite directions to conserve momentum B1 8(d)(iv) kinetic energy = ½ × 9.11 × 10–31 × (4.9 × 107)2 = 1.1 × 10–15 J A1 © Cambridge University Press & Assessment 2025 Page 16 of 19 8(d)(v) E = mc2 C1 E = hc /  C1 or E = hf and c = f (1.1 × 10–15) + (9.11 × 10–31 × (3.00 × 108)2) = (6.63 × 10–34 × 3.00 × 108) /  A1  = 2.39 × 10–12 m © Cambridge University Press & Assessment 2025 Page 17 of 19

8(a) Any three points from: B3 • electrons moving between levels emit a single photon • energy of photon = difference between energy levels • energy of photon depends on frequency • discrete frequencies (in spectrum) so differences between electron energies must be discrete • discrete differences between electron energies means energy levels must be discrete 8(b)(i) energy = – (13.6  1.60  10–19) A1 = – 2.18  10–18 J 8(b)(ii) E = hf C1 = (6.63  10–34  2.47  1015) / (1.60  10–19) = 10.2 eV A1 8(b)(iii) n = 2 energy level = – 3.4 eV A1 n = 3 energy difference = 12.1 eV A1 n = 4 energy difference = 12.8 eV A1 n = 3 energy level = – 1.5 eV and n = 4 energy level = – 0.8 eV A1 © Cambridge University Press & Assessment 2025 Page 15 of 17