Energy levels in atoms and line spectra — Paper 4 · A Level Physics 9702

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

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

7 (a) Explain how the line spectrum of hydrogen provides evidence for the existence of For discrete electron energy levels in atoms. Examiner’s Use … … … … … [3] (b) Some electron energy levels in atomic hydrogen are illustrated in Fig. 7.1. –0.85 eV –1.50 eV energy A B –3.40 eV Fig. 7.1 Two possible electron transitions A and B giving rise to an emission spectrum are shown. These electron transitions cause light of wavelengths 654 nm and 488 nm to be emitted. (i) On Fig. 7.1, draw an arrow to show a third possible transition. [1] (ii) Calculate the wavelength of the emitted light for the transition in (i). wavelength = … m [3] (c) The light in a beam has a continuous spectrum of wavelengths from 400 nm to 700 nm. For The light is incident on some cool hydrogen gas, as illustrated in Fig. 7.2. Examiner’s Use incident emergent cool hydrogen gas light light Fig. 7.2 Using the values of wavelength in (b), state and explain the appearance of the spectrum of the emergent light. … … … … … … [4]

11 marks

Mark scheme: 7 (a) each line represents photon of specific energy M1 photon emitted as a result of energy change of electron M1 specific energy changes so discrete levels A1 [3] (b) (i) arrow from –0.85 eV level to –1.5 eV level B1 [1] (ii) ∆E = hc /λ C1 = (1.5 – 0.85) × 1.6 × 10–19 C1 = 1.04 × 10–19 J λ = (6.63 × 10–34 × 3.0 × 108)/(1.04 × 10–19) = 1.9 × 10–6 m A1 [3] (c) spectrum appears as continuous spectrum crossed by dark lines B1 two dark lines B1 electrons in gas absorb photons with energies equal to the excitation energies M1 light photons re-emitted in all directions A1 [4]

This question in 9702/41 Oct/Nov 2011

7 (a) Explain how the line spectrum of hydrogen provides evidence for the existence of For discrete electron energy levels in atoms. Examiner’s Use … … … … … [3] (b) Some electron energy levels in atomic hydrogen are illustrated in Fig. 7.1. –0.85 eV –1.50 eV energy A B –3.40 eV Fig. 7.1 Two possible electron transitions A and B giving rise to an emission spectrum are shown. These electron transitions cause light of wavelengths 654 nm and 488 nm to be emitted. (i) On Fig. 7.1, draw an arrow to show a third possible transition. [1] (ii) Calculate the wavelength of the emitted light for the transition in (i). wavelength = … m [3] (c) The light in a beam has a continuous spectrum of wavelengths from 400 nm to 700 nm. For The light is incident on some cool hydrogen gas, as illustrated in Fig. 7.2. Examiner’s Use incident emergent cool hydrogen gas light light Fig. 7.2 Using the values of wavelength in (b), state and explain the appearance of the spectrum of the emergent light. … … … … … … [4]

11 marks

Mark scheme: 7 (a) each line represents photon of specific energy M1 photon emitted as a result of energy change of electron M1 specific energy changes so discrete levels A1 [3] (b) (i) arrow from –0.85 eV level to –1.5 eV level B1 [1] (ii) ∆E = hc /λ C1 = (1.5 – 0.85) × 1.6 × 10–19 C1 = 1.04 × 10–19 J λ = (6.63 × 10–34 × 3.0 × 108)/(1.04 × 10–19) = 1.9 × 10–6 m A1 [3] (c) spectrum appears as continuous spectrum crossed by dark lines B1 two dark lines B1 electrons in gas absorb photons with energies equal to the excitation energies M1 light photons re-emitted in all directions A1 [4]

This question in 9702/42 Oct/Nov 2011

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

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 9702/41 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

7 (a) The emission spectrum of atomic hydrogen consists of a number of discrete wavelengths. For Explain how this observation leads to an understanding that there are discrete electron Examiner’s energy levels in atoms. Use … … … … [2] (b) Some electron energy levels in atomic hydrogen are illustrated in Fig. 7.1. –0.54 eV –0.85 eV –1.5 eV energy –3.4 eV Fig. 7.1 The longest wavelength produced as a result of electron transitions between two of the For energy levels shown in Fig. 7.1 is 4.0 × 10–6 m. Examiner’s Use (i) On Fig. 7.1, 1. draw, and mark with the letter L, the transition giving rise to the wavelength of 4.0 × 10–6 m, [1] 2. draw, and mark with the letter S, the transition giving rise to the shortest wavelength. [1] (ii) Calculate the wavelength for the transition you have shown in (i) part 2. wavelength = … m [3] (c) Photon energies in the visible spectrum vary between approximately 3.66 eV and 1.83 eV. Determine the energies, in eV, of photons in the visible spectrum that are produced by transitions between the energy levels shown in Fig. 7.1. photon energies … eV [2]

9 marks

Mark scheme: 7 (a) each wavelength is associated with a discrete change in energy M1 discrete energy change / difference implies discrete levels A1 [2] (b) (i) 1. arrow from –0.54 eV to –0.85 eV, labelled L B1 [1] 2. arrow from –0.54 eV to –3.4 eV , labelled S B1 [1] (two correct arrows, but only one label – allow 2 marks) (two correct arrows, but no labels – allow 1 mark) (ii) E = hc / λ C1 (3.4 – 0.54) × 1.6 × 10–19 = (6.63 × 10–34 × 3.0 × 108) / λ C1 λ = 4.35 × 10–7 m A1 [3] (c) –1.50 → –3.4 = 1.9 eV –0.85 → –3.4 = 2.55 eV (allow 2.6 eV) –0.54 → –3.4 = 2.86 eV (allow 2.9 eV) 3 correct, 2 marks with –1 mark for each additional energy 2 correct, 1 mark but no marks if any additional energy differences B2 [2] GCE AS/A LEVEL – May/June 2013 9702 42

This question in 9702/42 May/June 2013

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 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 9702/42 Oct/Nov 2014

13 Copper-66 is a radioactive isotope. When a nucleus of copper-66 decays, the emissions include a β− particle and a γ-ray photon. The count rate produced from a sample of the isotope copper-66 is measured using a detector and counter, as illustrated in Fig. 13.1. to counter detector radioactive sample shielding Fig. 13.1 (a) State three reasons why the activity of the sample of copper-66 is not equal to the measured count rate. 1. … … 2. … … 3. … … [3] (b) In a time of 42.0 minutes, the count rate from the sample of copper-66 is found to decrease from 3.62 × 104 Bq to 1.21 × 102 Bq. Calculate the half-life of copper-66. half-life = … minutes [2] (c) The γ-ray photons emitted from radioactive nuclei have specific energies, dependent on the nucleus emitting the photons. By comparison with emission line spectra, suggest what can be deduced about energy levels in nuclei. … … [1] [Total: 6]

6 marks

Mark scheme: 13 (a) e.g. background count (rate)/radiation multiple possible counts from each decay radiation emitted in all directions dead-time of counter (daughter) product unstable/also emits radiation self-absorption of radiation in sample or absorption in air/detector window three sensible suggestions, 1 each B3 [3] (b) A = A0 exp(– ln 2 × t / T½) 1.21 × 102 = 3.62 × 104 exp(– ln2 × 42.0 / T½) or 1.21 × 102 = 3.62 × 104 exp(–λ × 42.0) C1 T½ = 5.1 minutes (306 s) A1 [2] (c) discrete energy levels (in nuclei) B1 [1]

This question in 9702/41 May/June 2016

13 Copper-66 is a radioactive isotope. When a nucleus of copper-66 decays, the emissions include a β− particle and a γ-ray photon. The count rate produced from a sample of the isotope copper-66 is measured using a detector and counter, as illustrated in Fig. 13.1. to counter detector radioactive sample shielding Fig. 13.1 (a) State three reasons why the activity of the sample of copper-66 is not equal to the measured count rate. 1. … … 2. … … 3. … … [3] (b) In a time of 42.0 minutes, the count rate from the sample of copper-66 is found to decrease from 3.62 × 104 Bq to 1.21 × 102 Bq. Calculate the half-life of copper-66. half-life = … minutes [2] (c) The γ-ray photons emitted from radioactive nuclei have specific energies, dependent on the nucleus emitting the photons. By comparison with emission line spectra, suggest what can be deduced about energy levels in nuclei. … … [1] [Total: 6]

6 marks

Mark scheme: 13 (a) e.g. background count (rate)/radiation multiple possible counts from each decay radiation emitted in all directions dead-time of counter (daughter) product unstable/also emits radiation self-absorption of radiation in sample or absorption in air/detector window three sensible suggestions, 1 each B3 [3] (b) A = A0 exp(– ln 2 × t / T½) 1.21 × 102 = 3.62 × 104 exp(– ln2 × 42.0 / T½) or 1.21 × 102 = 3.62 × 104 exp(–λ × 42.0) C1 T½ = 5.1 minutes (306 s) A1 [2] (c) discrete energy levels (in nuclei) B1 [1]

This question in 9702/43 May/June 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) 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 beam of light consists of a continuous range of wavelengths from 420 nm to 740 nm. The light passes through a cloud of cool gas, as shown in Fig. 11.1. incident light emergent light cool gas wavelengths 420 nm – 740 nm Fig. 11.1 (a) The spectrum of the light emerging from the cloud of cool gas is viewed using a diffraction grating. Explain why this spectrum contains a number of dark lines. … … … … … … [4] (b) Some of the electron energy levels of the atoms in the cloud of gas are represented in Fig. 11.2. – 0.38 eV – 0.54 eV – 0.85 eV – 1.5 eV energy – 3.4 eV – 13.6 eV Fig. 11.2 (not to scale) (i) Light of wavelength 420 nm has a photon energy of 2.96 eV. Calculate the photon energy, in eV, of light of wavelength 740 nm. photon energy = … eV [2] (ii) Use data from (i) and your answer in (i) to show, on Fig. 11.2, the changes in energy levels giving rise to the dark lines in (a). [2] [Total: 8]

8 marks

Mark scheme: 11(a) electrons (in gas atoms/molecules) interact with photons B1 photon energy causes electron to move to higher energy level/to be excited B1 photon energy = difference in energy of (electron) energy levels B1 when electrons de-excite, photons emitted in all directions (so dark line) B1 11(b)(i) photon energy ∝ 1 / λ C1 energy = 1.68 eV A1 or E = hc / λ E = 6.63 × 10–34 × 3.0 × 108 / (740 × 10–9) = 2.688 × 10–19 J (C1) energy = 1.68 eV (A1) 11(b)(ii) 3.4 eV → 1.5 eV 3.4 eV → 0.85 eV 3.4 eV → 0.54 eV all correct and none incorrect 2/2 2 correct and 1 incorrect or only 2 correctly drawn 1/2 B2

This question in 9702/41 May/June 2017

11 A beam of light consists of a continuous range of wavelengths from 420 nm to 740 nm. The light passes through a cloud of cool gas, as shown in Fig. 11.1. incident light emergent light cool gas wavelengths 420 nm – 740 nm Fig. 11.1 (a) The spectrum of the light emerging from the cloud of cool gas is viewed using a diffraction grating. Explain why this spectrum contains a number of dark lines. … … … … … … [4] (b) Some of the electron energy levels of the atoms in the cloud of gas are represented in Fig. 11.2. – 0.38 eV – 0.54 eV – 0.85 eV – 1.5 eV energy – 3.4 eV – 13.6 eV Fig. 11.2 (not to scale) (i) Light of wavelength 420 nm has a photon energy of 2.96 eV. Calculate the photon energy, in eV, of light of wavelength 740 nm. photon energy = … eV [2] (ii) Use data from (i) and your answer in (i) to show, on Fig. 11.2, the changes in energy levels giving rise to the dark lines in (a). [2] [Total: 8]

8 marks

Mark scheme: 11(a) electrons (in gas atoms/molecules) interact with photons B1 photon energy causes electron to move to higher energy level/to be excited B1 photon energy = difference in energy of (electron) energy levels B1 when electrons de-excite, photons emitted in all directions (so dark line) B1 11(b)(i) photon energy ∝ 1 / λ C1 energy = 1.68 eV A1 or E = hc / λ E = 6.63 × 10–34 × 3.0 × 108 / (740 × 10–9) = 2.688 × 10–19 J (C1) energy = 1.68 eV (A1) 11(b)(ii) 3.4 eV → 1.5 eV 3.4 eV → 0.85 eV 3.4 eV → 0.54 eV all correct and none incorrect 2/2 2 correct and 1 incorrect or only 2 correctly drawn 1/2 B2

This question in 9702/43 May/June 2017

10 Some of the electron energy bands in a semiconductor material at the absolute zero of temperature are shown in Fig. 10.1. conduction band (empty) forbidden band valence band (filled) Fig. 10.1 Use band theory to explain why, as the temperature of the semiconductor material rises, the electrical resistance of the sample of material decreases. … … … … … … … … … [5]

5 marks

Mark scheme: 10 Any five points from: • as temperature rises electrons gain energy • electrons enter conduction band • (positively charged) holes left in valence band • more charge carriers (so resistance decreases) • (as temperature rises,) lattice vibrations increase • effect of increase in number of electrons or holes or charge carriers outweighs effect of increased lattice vibrations (so resistance decreases) B5

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

10 Some of the electron energy bands in a semiconductor material at the absolute zero of temperature are shown in Fig. 10.1. conduction band (empty) forbidden band valence band (filled) Fig. 10.1 Use band theory to explain why, as the temperature of the semiconductor material rises, the electrical resistance of the sample of material decreases. … … … … … … … … … [5]

5 marks

Mark scheme: 10 Any five points from: • as temperature rises electrons gain energy • electrons enter conduction band • (positively charged) holes left in valence band • more charge carriers (so resistance decreases) • (as temperature rises,) lattice vibrations increase • effect of increase in number of electrons or holes or charge carriers outweighs effect of increased lattice vibrations (so resistance decreases) B5

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

10 (a) White light passes through a cloud of cool low-pressure gas, as illustrated in Fig. 10.1. cool gas white emergent light light Fig. 10.1 For light that has passed through the gas, its continuous spectrum is seen to contain a number of darker lines. Use the concept of discrete electron energy levels to explain the existence of these darker lines. … … … … … … [4] (b) The uppermost electron energy bands in a solid are illustrated in Fig. 10.2. conduction band (CB) forbidden band (FB) valence band (VB) Fig. 10.2 Use band theory to explain the dependence on light intensity of the resistance of a light-dependent resistor (LDR). … … … … … … … [5] [Total: 9]

9 marks

Mark scheme: 10(a) • photon gives energy to electron (in an inner shell) or electron (in an inner shell) absorbs a photon • electron moves (from lower) to higher energy level • energy (of photon) is equal to difference in energy levels • electron de-excites giving off photon (of same energy) • photons emitted in all directions Any four points, 1 mark each B4 10(b) (in light) photons gives energy to electrons in VB or (in light) electrons in VB absorb photons B1 electron crosses FB/jumps to CB B1 (positive) holes left/created in VB B1 low intensity: few electrons in CB/most electrons in VB or high intensity: more photons so more electrons in CB or electron-hole pairs are charge carriers B1 more charge carriers results in lower resistance B1

This question in 9702/41 May/June 2020

10 (a) White light passes through a cloud of cool low-pressure gas, as illustrated in Fig. 10.1. cool gas white emergent light light Fig. 10.1 For light that has passed through the gas, its continuous spectrum is seen to contain a number of darker lines. Use the concept of discrete electron energy levels to explain the existence of these darker lines. … … … … … … [4] (b) The uppermost electron energy bands in a solid are illustrated in Fig. 10.2. conduction band (CB) forbidden band (FB) valence band (VB) Fig. 10.2 Use band theory to explain the dependence on light intensity of the resistance of a light-dependent resistor (LDR). … … … … … … … [5] [Total: 9]

9 marks

Mark scheme: 10(a) • photon gives energy to electron (in an inner shell) or electron (in an inner shell) absorbs a photon • electron moves (from lower) to higher energy level • energy (of photon) is equal to difference in energy levels • electron de-excites giving off photon (of same energy) • photons emitted in all directions Any four points, 1 mark each B4 10(b) (in light) photons gives energy to electrons in VB or (in light) electrons in VB absorb photons B1 electron crosses FB/jumps to CB B1 (positive) holes left/created in VB B1 low intensity: few electrons in CB/most electrons in VB or high intensity: more photons so more electrons in CB or electron-hole pairs are charge carriers B1 more charge carriers results in lower resistance B1

This question in 9702/43 May/June 2020

9 (a) Fig. 9.1 shows the visible part of the emission spectrum from hydrogen gas in a laboratory on the Earth. The numbers indicate the wavelength, in nm, represented by each line. 411 435 488 658 Fig. 9.1 (i) Explain how the emission spectrum provides evidence for the existence of discrete energy levels for the electron in a hydrogen atom. … … … … [3] (ii) Fig. 9.2 shows five of the energy levels in the hydrogen atom. The wavelengths of radiation shown in Fig. 9.1 relate to transitions to the – 3.400 eV level in Fig. 9.2. – 0.378 eV – 0.544 eV – 0.850 eV energy X – 3.400 eV Fig. 9.2 (not to scale) Show that the energy level X is –1.51 eV. [3] (b) The same part of the emission spectrum from hydrogen as in (a), observed in light from stars in a distant galaxy, is shown in Fig. 9.3. The numbers indicate the wavelengths in nm. 429 454 509 686 Fig. 9.3 The spectrum shows the same pattern as Fig. 9.1 but with different wavelengths. (i) State the name of the phenomenon that gives rise to the change in the wavelengths. … [1] (ii) State what this phenomenon shows about the motion of the galaxy. … [1] (iii) Use one of the lines in Fig. 9.1, and the corresponding line in Fig. 9.3, to determine the speed of the distant galaxy relative to the observer. speed = … m s–1 [3] (c) The galaxy in (b) is known to be a distance of 5.7 × 1024 m from the Earth. Use your answer in (b)(iii) to determine a value for the Hubble constant H0. H0 = … s–1 [2] [Total: 13]

13 marks

Mark scheme: 9(a)(i) • energy of photon has a corresponding frequency B3 • change in electron energy level emits a single photon • photon energy = difference in energy levels • discrete frequencies must have come from discrete energy gaps • discrete energy changes imply discrete energy levels Any three points, 1 mark each 9(a)(ii) transition (to – 3.400 eV) from X corresponds to 658 nm line C1 E1 – E2 = hc /  C1 E1 – (– 3.400) = (6.63  10–34  3.00  108) / (658  10–9  1.60  10–19) A1 and so E1 = –1.51 eV (full substitution and answer needed) 9(b)(i) redshift B1 9(b)(ii) moving away (from observer) B1 9(b)(iii)  / = v / c C1 e.g. for 658 nm line:  = 686 – 658 ( = 28 nm) (other lines may be used) 28 / 658 = v / (3.00  108) (other lines may be used) C1 v = 1.3  107 m s–1 A1 9(c) v = H0d C1 H0 = (1.3  107) / (5.7  1024) A1 = 2.3  10–18 s–1

This question in 9702/42 Oct/Nov 2022

7 (a) A beam of white light passes through a cloud of cool gas. The spectrum of the transmitted light is viewed and contains a number of dark lines. Explain why these dark lines occur. … … … … … … … … … [4] (b) Some energy levels for the electron in an isolated hydrogen atom are illustrated in Fig. 7.1. n = 6 n = 5 n = 4 n = 3 energy n = 2 Fig. 7.1 Table 7.1 shows the wavelengths of photons that are emitted in the transitions to n = 2 from the other energy levels shown in Fig. 7.1. Table 7.1 wavelength / nm 412 435 488 658 The energy associated with the energy level n = 2 is – 3.40 eV. Calculate the energy, in J, of energy level n = 3. energy = … J [3] [Total: 7]

7 marks

Mark scheme: 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 658 nm C1 6.63 1 0 –34  3.00 1 0 8 A1 –9 = – E1 – (–3.40 × 1.60 × 10–19) 658 1 0 E1 = –2.42  10–19 J

This question in 9702/42 Feb/March 2023

8 Fig. 8.1 shows the lowest four energy levels of an electron in an isolated atom. n = 4 n = 3 n = 2 increasing energy n = 1 Fig. 8.1 Fig. 8.2 shows the lines in the emission spectrum of the atom that correspond to the transitions of the electron from n = 3 to n = 1 and from n = 4 to n = 1. increasing frequency Fig. 8.2 (a) Explain, with reference to photons, why there is a single frequency of electromagnetic radiation that corresponds to each of these transitions. … … … [2] (b) (i) On Fig. 8.2, draw a line that corresponds to the transition of the electron from n = 2 to n = 1. Label this line A. [2] (ii) On Fig. 8.2, draw a line that corresponds to the transition of the electron from n = 3 to n = 2. Label this line B. [2] (c) The frequency of radiation represented by line A is fA. The frequency of radiation represented by line B is fB. The energy of the ground state (n = 1) is E1. Determine an expression, in terms of fA, fB, E1 and the Planck constant h, for the energy E3 of the energy level n = 3. E3 = … [2] [Total: 8]

8 marks

Mark scheme: 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 E3 = E1 + h(fA + fB) A1

This question in 9702/42 May/June 2023

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 Fig. 8.1 shows the three lowest-frequency lines in the part of the emission spectrum for hydrogen that relates to electron transitions to the ground state (level n = 1). 3.09 2.92 2.47 Fig. 8.1 (not to scale) The numbers represent the frequencies, in 1015 Hz, associated with the spectral lines. (a) Use the photon model of electromagnetic radiation to explain how the existence of spectral lines in the emission spectrum provides evidence for discrete electron energy levels in the hydrogen atom. … … … … … [3] (b) The energy of the ground state (level n = 1) in a hydrogen atom is –13.6 eV. (i) Calculate the energy, in J, of the ground state. energy = … J [1] (ii) Show that the energy difference between levels n = 1 and n = 2 is 10.2 eV. [2] (iii) Complete Table 8.1 to show the energy differences from the ground state, and the energies of the levels up to n = 4, in the hydrogen atom. Use the space for any working. Table 8.1 (energy difference level energy / eV from n = 1) / eV n = 4 n = 3 n = 2 10.2 n = 1 0.0 –13.6 [4] [Total: 10]

10 marks

Mark scheme: 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

This question in 9702/42 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