Mass defect and nuclear binding energy — Paper 4 · A Level Physics 9702

7 (a) Explain what is meant by the binding energy of a nucleus. … … [1] (b) Fig. 7.1 shows the variation with nucleon number (mass number) A of the binding energy per nucleon EB of nuclei. E B 0 0 A Fig. 7.1 One particular fission reaction may be represented by the nuclear equation 235 1 141 92 92U + 0n → 56Ba + 36Kr + 310n. (i) On Fig. 7.1, label the approximate positions of 1. the uranium (235 nucleus with the symbol U, 92U) 2. the barium (141 nucleus with the symbol Ba, 56Ba) 92 3. the krypton ( 36Kr) nucleus with the symbol Kr. [2] (ii) The neutron that is absorbed by the uranium nucleus has very little kinetic energy. Explain why this fission reaction is energetically possible. … … … [2] Examiner’s Use (c) Barium-141 has a half-life of 18 minutes. The half-life of Krypton-92 is 3.0 s. In the fission reaction of a mass of Uranium-235, equal numbers of barium and krypton nuclei are produced. Estimate the time taken after the fission of the sample of uranium for the ratio number of Barium-141 nuclei number of Krypton-92 nuclei to be approximately equal to 8. time = … s [3]

8 marks

Mark scheme: 7 (a) energy required to (completely) separate the nucleons (in a nucleus) … B1 [1] (b) (i) U labelled near right-hand end of line … B1 Ba and Kr in approximately correct positions … B1 [2] (ii) binding energy is A × EB … B1 either binding energy of U < binding energy of (Ba + Kr) or EB of U < EB of (Ba + Kr) … B1 [2] (c) Krypton-92 reduced to 1/8 in 9 s … M1 in 9 s, very little decay of Barium-141 … M1 so, approximately 9 s … A1 [3] OR λKr = 0.231 or λBa = 6.42 × 10-4 (M1) 8 = e-λB × t/e-λK × t (C1) t = 9.0 s (A1) GCE A/AS LEVEL – October/November 2007 9702 04 Section B

This question in 9702/41 Oct/Nov 2007

8 The controlled reaction between deuterium (2 H) and tritium (3 H) has involved ongoing 1 1 For research for many years. The reaction may be summarised as Examiner’s Use 2 H + 3 H 4 + 1 n + Q 1 1 2He 0 where Q = 17.7 MeV. Binding energies per nucleon are shown in Fig. 8.1. binding energy per nucleon / MeV 2 H 1.12 1 1 n – 0 42He 7.07 Fig. 8.1 (a) Suggest why binding energy per nucleon for the neutron is not quoted. … … [1] (b) Calculate the mass defect, in kg, of a helium 42He nucleus. mass defect = … kg [3] (c) (i) State the name of the type of reaction illustrated by this nuclear equation. … [1] (ii) Determine the binding energy per nucleon, in MeV, of tritium (3 H). 1 binding energy per nucleon = … MeV [3]

8 marks

Mark scheme: 8 (a) neutron is a single nucleon / particle … B1 [1] (b) binding energy = 4 × 7.07 × 1.6 × 10-13 … C1 = 4.52 × 10-12 J binding energy = c2 ∆m … C1 4.52 × 10-12 = (3.0 × 108)2 × ∆m ∆m = 5.03 × 10-29 kg … A1 [3] (c) (i) fusion ……(do not allow fussion) … B1 [1] (ii) (2 × 1.12) + 3x = 28.28 … C1 …... –17.7 … C1 x = 2.78 MeV per nucleon … A1 [3] (use of +17.7 gives x = 14.6 MeV, allow 1 mark only) [Total: 8] GCE A/AS LEVEL – October/November 2009 9702 41 Section B

This question in 9702/41 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 In some power stations, nuclear fission is used as a source of energy. For Examiner’s (a) State what is meant by nuclear fission. Use … … … [2] (b) The nuclear fission reaction produces neutrons. In the power station, the neutrons may be absorbed by rods made of boron-10. Complete the nuclear equation for the absorption of a single neutron by a boron-10 nucleus with the emission of an a-particle. 10 … 5B + … 3Li + … [3] (c) Suggest why, when neutrons are absorbed in the boron rods, the rods become hot as a result of this nuclear reaction. … … … … [3]

8 marks

Mark scheme: 8 (a) splitting of a heavy nucleus (not atom/nuclide) M1 into two (lighter) nuclei of approximately same mass A1 [2] (b) 01 n 4 2 He (allow 42 α ) M2 7 3 Li A1 [3] (c) emitted particles have kinetic energy B1 range of particles in the control rods is short / particles stopped in rods / lose kinetic energy in rods B1 kinetic energy of particles converted to thermal energy B1 [3] GCE AS/A LEVEL – October/November 2010 9702 41 Section B

This question in 9702/41 Oct/Nov 2010

2 (a) State the basic assumptions of the kinetic theory of gases. For Examiner’s … Use … … … … … [4] (b) Use equations for the pressure of an ideal gas to deduce that the average translational kinetic energy <EK> of a molecule of an ideal gas is given by the expression 3 R <EK> = T 2 NA where R is the molar gas constant, NA is the Avogadro constant and T is the thermodynamic temperature of the gas. [3] 2 (c) A deuterium nucleus 1H and a proton collide. A nuclear reaction occurs, represented by the equation 2 1 3 1H + 1p 2He + c. (i) State and explain whether the reaction represents nuclear fission or nuclear fusion. … … … [2] (ii) For the reaction to occur, the minimum total kinetic energy of the deuterium nucleus For and the proton is 2.4 × 10–14 J. Examiner’s Assuming that a sample of a mixture of deuterium nuclei and protons behaves as Use an ideal gas, calculate the temperature of the sample for this reaction to occur. temperature = … K [3] (iii) Suggest why the assumption made in (ii) may not be valid. … … [1]

13 marks

Mark scheme: 2 (a) atoms / molecules / particles behave as elastic (identical) spheres (1) volume of atoms / molecules negligible compared to volume of containing vessel (1) time of collision negligible to time between collisions (1) no forces of attraction or repulsion between atoms / molecules (1) atoms / molecules / particles are in (continuous) random motion (1) (any four, 1 each) B4 [4] (b) pV = 31 Nm<c2> and pV = nRT or pV = NkT B1 1 Nm<c2> = nRT or = NkT and <EK> = ½m<c2> B1 3 n = N/NA or k = R/NA B1 3 <EK> = × R/NA × T A0 [3] 2 (c) (i) reaction represents either build-up of nucleus from light nuclei or build-up of heavy nucleus from nuclei M1 so fusion reaction A1 [2] (ii) proton and deuterium nucleus will have equal kinetic energies B1 1.2 × 10–14 = 32 × 8.31 / (6.02 × 1023) × T C1 T = 5.8 × 108 K A1 [3] (use of E = 2.4 × 10–14 giving 1.16 × 109 K scores 1 mark) (iii) either inter-molecular / atomic / nuclear forces exist or proton and deuterium nucleus are positively charged / repel B1 [1] GCE A LEVEL – October/November 2010 9702 43

This question in 9702/43 Oct/Nov 2010

8 (a) The variation with nucleon number A of the binding energy per nucleon BE of nuclei is For shown in Fig. 8.1. Examiner’s Use BE 0 A Fig. 8.1 On Fig. 8.1, mark the approximate positions of (i) iron-56 (label this point Fe), [1] (ii) zirconium-97 (label this point Zr), [1] (iii) hydrogen-2 (label this point H). [1] (b) (i) State what is meant by nuclear fission. … … … [2] (ii) By reference to Fig. 8.1, explain how fission is energetically possible. … … … … [2]

7 marks

Mark scheme: 8 (a) (i) Fe shown near peak A1 [1] (ii) Zr shown about half-way along plateau A1 [1] (iii) H shown at less than 0.4 of maximum height A1 [1] (b) (i) heavy / large nucleus breaks up / splits M1 into two nuclei / fragments of approximately equal mass A1 [2] (ii) binding energy of nucleus = BE × A B1 binding energy of parent nucleus is less than sum of binding energies of fragments B1 [2] GCE AS/A LEVEL – May/June 2011 9702 41 Section B

This question in 9702/41 May/June 2011

8 (a) State what is meant by the binding energy of a nucleus. For Examiner’s … Use … … [2] (b) Show that the energy equivalence of 1.0 u is 930 MeV. [3] (c) Data for the masses of some particles and nuclei are given in Fig. 8.1. mass / u proton 1.0073 neutron 1.0087 deuterium (21H) 2.0141 zirconium (9740Zr) 97.0980 Fig. 8.1 Use data from Fig. 8.1 and information from (b) to determine, in MeV, (i) the binding energy of deuterium, binding energy = … MeV [2] (ii) the binding energy per nucleon of zirconium. For Examiner’s Use binding energy per nucleon = … MeV [3]

10 marks

Mark scheme: 8 (a) energy required to separate nucleons in a nucleus M1 to infinity A1 [2] (b) 1u = 1.66 × 10–27 kg E = mc2 C1 = 1.66 × 10–27 × (3.0 × 108)2 M1 = 1.49 × 10–10 J = (1.49 × 10–10) / (1.6 × 10–13) M1 = 930 MeV A0 [3] (c) (i) ∆m = 2.0141u – (1.0073 + 1.0087)u = –1.9 × 10–3 u C1 binding energy = 1.9 × 10–3 × 930 =1.8 MeV A1 [2] (ii) ∆m = (57 × 1.0087u) + (40 × 1.0073u) – 97.0980u C1 = (–)0.69 u binding energy per nucleon = (0.69 × 930) / 97 C1 = 6.61 MeV A1 [3] GCE AS/A LEVEL – May/June 2011 9702 42 Section B

This question in 9702/42 May/June 2011

8 (a) State what is meant by the binding energy of a nucleus. For Examiner’s … Use … … [2] (b) Show that the energy equivalence of 1.0 u is 930 MeV. [3] (c) Data for the masses of some particles and nuclei are given in Fig. 8.1. mass / u proton 1.0073 neutron 1.0087 deuterium (21H) 2.0141 zirconium (9740Zr) 97.0980 Fig. 8.1 Use data from Fig. 8.1 and information from (b) to determine, in MeV, (i) the binding energy of deuterium, binding energy = … MeV [2] (ii) the binding energy per nucleon of zirconium. For Examiner’s Use binding energy per nucleon = … MeV [3]

10 marks

Mark scheme: 8 (a) energy required to separate nucleons in a nucleus M1 to infinity A1 [2] (b) 1u = 1.66 × 10–27 kg E = mc2 C1 = 1.66 × 10–27 × (3.0 × 108)2 M1 = 1.49 × 10–10 J = (1.49 × 10–10) / (1.6 × 10–13) M1 = 930 MeV A0 [3] (c) (i) ∆m = 2.0141u – (1.0073 + 1.0087)u = –1.9 × 10–3 u C1 binding energy = 1.9 × 10–3 × 930 =1.8 MeV A1 [2] (ii) ∆m = (57 × 1.0087u) + (40 × 1.0073u) – 97.0980u C1 = (–)0.69 u binding energy per nucleon = (0.69 × 930) / 97 C1 = 6.61 MeV A1 [3] GCE AS/A LEVEL – May/June 2011 9702 43 Section B

This question in 9702/43 May/June 2011

8 When a neutron is captured by a uranium-235 nucleus, the outcome may be represented by For the nuclear equation shown below. Examiner’s Use 235 1 95 139 1 0 U + n Mo + La + x n + 7 e 92 0 42 57 0 –1 (a) (i) Use the equation to determine the value of x. x = … [1] 0 (ii) State the name of the particle represented by the symbol e. –1 … [1] (b) Some data for the nuclei in the reaction are given in Fig. 8.1. mass / u binding energy per nucleon / MeV 235 uranium-235 ( U) 235.123 92 95 molybdenum-95 ( Mo) 94.945 8.09 42 139 lanthanum-139 ( La) 138.955 7.92 57 1 proton ( p) 1.007 1 1 neutron ( n) 1.009 0 Fig. 8.1 Use data from Fig. 8.1 to (i) determine the binding energy, in u, of a nucleus of uranium-235, binding energy = … u [3] (ii) show that the binding energy per nucleon of a nucleus of uranium-235 is 7.18 MeV. For Examiner’s Use [3] (c) The kinetic energy of the neutron before the reaction is negligible. Use data from (b) to calculate the total energy, in MeV, released in this reaction. energy = … MeV [2]

10 marks

Mark scheme: 8 (a) (i) x = 2 A1 [1] (ii) either beta particle or electron B1 [1] (b) (i) mass of separate nucleons = {(92 × 1.007) + (143 × 1.009)} u C1 = 236.931 u C1 binding energy = 236.931 u – 235.123 u = 1.808 u A1 [3] GCE AS/A LEVEL – October/November 2012 9702 43 (ii) E = mc2 C1 energy = 1.808 × 1.66 × 10–27 × (3.0 × 108)2 = 2.7 × 10–10 J C1 binding energy per nucleon = (2.7 × 10–10) / (235 × 1.6 × 10–13) M1 = 7.18 MeV A0 [3] (c) energy released = (95 × 8.09) + (139 × 7.92) – (235 × 7.18) C1 = 1869.43 – 1687.3 = 182 MeV A1 [2] (allow calculation using mass difference between products and reactants) Section B

This question in 9702/43 Oct/Nov 2012

8 (a) State what is meant by a nuclear fusion reaction. For Examiner’s … Use … … [2] (b) One nuclear reaction that takes place in the core of the Sun is represented by the equation 2 H + 1 H 3 He + energy. 1 1 2 Data for the nuclei are given in Fig. 8.1. mass / u proton 1 H 1.00728 1 deuterium 2 H 2.01410 1 helium 3 He 3.01605 2 Fig. 8.1 (i) Calculate the energy, in joules, released in this reaction. energy = … J [3] (ii) The temperature in the core of the Sun is approximately 1.6 × 107 K. Suggest why such a high temperature is necessary for this reaction to take place. … … … [2]

7 marks

Mark scheme: 8 (a) two (light) nuclei combine M1 to form a more massive nucleus A1 [2] (b) (i) ∆m = (2.01410 u + 1.00728 u) – 3.01605 u = 5.33 × 10–3 u C1 energy = c2 × ∆m C1 = 5.33 × 10–3 × 1.66 × 10–27 × (3.00 × 108)2 = 8.0 × 10–13 J A1 [3] (ii) speed/kinetic energy of proton and deuterium must be very large B1 so that the nuclei can overcome electrostatic repulsion B1 [2] Section B

This question in 9702/41 May/June 2013

8 (a) Explain why the mass of an α-particle is less than the total mass of two individual For protons and two individual neutrons. Examiner’s Use … … … [2] (b) An equation for one possible nuclear reaction is 42He + 147N 178O + 11p. Data for the masses of the nuclei are given in Fig. 8.1. mass / u proton 11p 1.00728 helium-4 42He 4.00260 nitrogen-14 147N 14.00307 oxygen-17 178O 16.99913 Fig. 8.1 (i) Calculate the mass change, in u, associated with this reaction. mass change = … u [2] (ii) Calculate the energy, in J, associated with the mass change in (i). energy = … J [2] (iii) Suggest and explain why, for this reaction to occur, the helium-4 nucleus must have For a minimum speed. Examiner’s Use … … … [2]

8 marks

Mark scheme: 8 (a) energy is given out / released on formation of the α-particle (or reverse argument) M1 either E = mc2 so mass is less or reference to mass-energy equivalence A1 [2] (b) (i) mass change = 18.00567 u – 18.00641 u C1 = 7.4 × 10–4 u (sign not required) A1 [2] (ii) energy = c2∆m = (3.0 × 108)2 × 7.4 × 10–4 × 1.66 × 10–27 C1 = 1.1 × 10–13 J A1 [2] (allow use of u = 1.67 × 10–27 kg) (allow method based on 1u equivalent to 930 MeV to 933 MeV) (iii) either mass of products greater than mass of reactants M1 this mass/energy provided as kinetic energy of the helium-4 nucleus A1 or both nuclei positively charged (M1) energy required to overcome electrostatic repulsion (A1) [2] GCE AS/A LEVEL – May/June 2013 9702 42 Section B

This question in 9702/42 May/June 2013

8 (a) State what is meant by a nuclear fusion reaction. For Examiner’s … Use … … [2] (b) One nuclear reaction that takes place in the core of the Sun is represented by the equation 2 H + 1 H 3 He + energy. 1 1 2 Data for the nuclei are given in Fig. 8.1. mass / u proton 1 H 1.00728 1 deuterium 2 H 2.01410 1 helium 3 He 3.01605 2 Fig. 8.1 (i) Calculate the energy, in joules, released in this reaction. energy = … J [3] (ii) The temperature in the core of the Sun is approximately 1.6 × 107 K. Suggest why such a high temperature is necessary for this reaction to take place. … … … [2]

7 marks

Mark scheme: 8 (a) two (light) nuclei combine M1 to form a more massive nucleus A1 [2] (b) (i) ∆m = (2.01410 u + 1.00728 u) – 3.01605 u = 5.33 × 10–3 u C1 energy = c2 × ∆m C1 = 5.33 × 10–3 × 1.66 × 10–27 × (3.00 × 108)2 = 8.0 × 10–13 J A1 [3] (ii) speed/kinetic energy of proton and deuterium must be very large B1 so that the nuclei can overcome electrostatic repulsion B1 [2] Section B

This question in 9702/43 May/June 2013

8 (a) State what is meant by nuclear binding energy. For Examiner’s … Use … … [2] (b) The variation with nucleon number A of the binding energy per nucleon BE is shown in Fig. 8.1. BE 0 0 A Fig. 8.1 When uranium-235 (23592U) absorbs a slow-moving neutron, one possible nuclear reaction is 23592U + 10n 9542Mo + 13957La + 210n + 7 –1β0 + energy. (i) State the name of this type of nuclear reaction. … [1] (ii) On Fig. 8.1, mark the position of 1. the uranium-235 nucleus (label this position U), [1] 2. the molybdenum-95 (9542Mo) nucleus (label this position Mo), [1] 3. the lanthanum-139 (13957La) nucleus (label this position La). [1] (iii) The masses of some particles and nuclei are given in Fig. 8.2. For Examiner’s Use mass / u β-particle 5.5 × 10–4 neutron 1.009 proton 1.007 uranium-235 235.123 molybdenum-95 94.945 lanthanum-139 138.955 Fig. 8.2 Calculate, for this reaction, 1. the change, in u, of the rest mass, change in mass = … u [2] 2. the energy released, in MeV, to three significant figures. energy = … MeV [3]

11 marks

Mark scheme: 8 (a) energy to separate nucleons (in a nucleus) M1 separate to infinity A1 [2] (b) (i) fission B1 [1] (ii) 1. U: near right-hand end of line B1 [1] 2. Mo: to right of peak, less than 1/3 distance from peak to U B1 [1] 3. La: 0.4 → 0.6 of distance from peak to U B1 [1] GCE A LEVEL – October/November 2013 9702 41 (iii) 1. right-hand side, mass = 235.922 u C1 mass change = 0.210 u A1 [2] 2. energy = mc2 C1 energy = 0.210 × 1.66 × 10–27 × (3.0 × 108)2 energy = 3.1374 × 10–11 J C1 energy = 196 MeV (need 3 s.f.) A1 [3] (use of 1 u = 934 MeV, allow 3/3; use of 1 u = 930 MeV or 932 MeV, allow 2/3) (use of 1.67 × 10–27 not 1.66 ×10–27 scores max. 2/3) Section B

This question in 9702/41 Oct/Nov 2013

8 (a) State what is meant by nuclear binding energy. For Examiner’s … Use … … [2] (b) The variation with nucleon number A of the binding energy per nucleon BE is shown in Fig. 8.1. BE 0 0 A Fig. 8.1 When uranium-235 (23592U) absorbs a slow-moving neutron, one possible nuclear reaction is 23592U + 10n 9542Mo + 13957La + 210n + 7 –1β0 + energy. (i) State the name of this type of nuclear reaction. … [1] (ii) On Fig. 8.1, mark the position of 1. the uranium-235 nucleus (label this position U), [1] 2. the molybdenum-95 (9542Mo) nucleus (label this position Mo), [1] 3. the lanthanum-139 (13957La) nucleus (label this position La). [1] (iii) The masses of some particles and nuclei are given in Fig. 8.2. For Examiner’s Use mass / u β-particle 5.5 × 10–4 neutron 1.009 proton 1.007 uranium-235 235.123 molybdenum-95 94.945 lanthanum-139 138.955 Fig. 8.2 Calculate, for this reaction, 1. the change, in u, of the rest mass, change in mass = … u [2] 2. the energy released, in MeV, to three significant figures. energy = … MeV [3]

11 marks

Mark scheme: 8 (a) energy to separate nucleons (in a nucleus) M1 separate to infinity A1 [2] (b) (i) fission B1 [1] (ii) 1. U: near right-hand end of line B1 [1] 2. Mo: to right of peak, less than 1/3 distance from peak to U B1 [1] 3. La: 0.4 → 0.6 of distance from peak to U B1 [1] GCE A LEVEL – October/November 2013 9702 42 (iii) 1. right-hand side, mass = 235.922 u C1 mass change = 0.210 u A1 [2] 2. energy = mc2 C1 energy = 0.210 × 1.66 × 10–27 × (3.0 × 108)2 energy = 3.1374 × 10–11 J C1 energy = 196 MeV (need 3 s.f.) A1 [3] (use of 1 u = 934 MeV, allow 3/3; use of 1 u = 930 MeV or 932 MeV, allow 2/3) (use of 1.67 × 10–27 not 1.66 ×10–27 scores max. 2/3) Section B

This question in 9702/42 Oct/Nov 2013

10 (a) Explain what is meant by the binding energy of a nucleus. … … … [2] (b) Data for the masses of some particles are given in Fig. 10.1. mass / u proton 1.00728 neutron 1.00867 tritium (31H) nucleus 3.01551 polonium (21084Po) nucleus 209.93722 Fig. 10.1 The energy equivalent of 1.0 u is 930 MeV. (i) Calculate the binding energy, in MeV, of a tritium (31H) nucleus. binding energy = … MeV [3] (ii) The total mass of the separate nucleons that make up a polonium-210 (21084Po) nucleus is 211.70394 u. Calculate the binding energy per nucleon of polonium-210. binding energy per nucleon = … MeV [3] (c) One possible fission reaction is 23592U + 10n 14156Ba + 9236Kr + 310n . By reference to binding energy, explain, without any calculation, why this fission reaction is energetically possible. … … … [2]

10 marks

Mark scheme: 10 (a) energy required to separate the nucleons (in a nucleus) M1 to infinity A1 [2] (allow reverse statement) (b) (i) ∆m = (2 × 1.00867) + 1.00728 – 3.01551 C1 = 9.11 × 10–3 u C1 binding energy = 9.11 × 10–3 × 930 = 8.47 MeV A1 [3] (allow 930 to 934 MeV so answer could be in range 8.47 to 8.51 MeV) (allow 2 s.f.) (ii) ∆m = 211.70394 – 209.93722 = 1.76672 u C1 binding energy per nucleon = (1.76672 × 930)/210 C1 = 7.82 MeV A1 [3] (allow 930 to 934 MeV so answer could be in range 7.82 to 7.86 MeV) (allow 2 s.f.) (c) total binding energy of barium and krypton M1 is greater than binding energy of uranium A1 [2] Section B

This question in 9702/41 May/June 2014

10 (a) Explain what is meant by the binding energy of a nucleus. … … … [2] (b) Data for the masses of some particles are given in Fig. 10.1. mass / u proton 1.00728 neutron 1.00867 tritium (31H) nucleus 3.01551 polonium (21084Po) nucleus 209.93722 Fig. 10.1 The energy equivalent of 1.0 u is 930 MeV. (i) Calculate the binding energy, in MeV, of a tritium (31H) nucleus. binding energy = … MeV [3] (ii) The total mass of the separate nucleons that make up a polonium-210 (21084Po) nucleus is 211.70394 u. Calculate the binding energy per nucleon of polonium-210. binding energy per nucleon = … MeV [3] (c) One possible fission reaction is 23592U + 10n 14156Ba + 9236Kr + 310n . By reference to binding energy, explain, without any calculation, why this fission reaction is energetically possible. … … … [2]

10 marks

Mark scheme: 10 (a) energy required to separate the nucleons (in a nucleus) M1 to infinity A1 [2] (allow reverse statement) (b) (i) ∆m = (2 × 1.00867) + 1.00728 – 3.01551 C1 = 9.11 × 10–3 u C1 binding energy = 9.11 × 10–3 × 930 = 8.47 MeV A1 [3] (allow 930 to 934 MeV so answer could be in range 8.47 to 8.51 MeV) (allow 2 s.f.) (ii) ∆m = 211.70394 – 209.93722 = 1.76672 u C1 binding energy per nucleon = (1.76672 × 930)/210 C1 = 7.82 MeV A1 [3] (allow 930 to 934 MeV so answer could be in range 7.82 to 7.86 MeV) (allow 2 s.f.) (c) total binding energy of barium and krypton M1 is greater than binding energy of uranium A1 [2] Section B

This question in 9702/43 May/June 2014

9 One likely means by which nuclear fusion may be achieved on a practical scale is the D-T reaction. (a) State what is meant by nuclear fusion. … … [1] (b) In the D-T reaction, a deuterium (21H) nucleus fuses with a tritium (31H) nucleus to form a helium-4 (42He) nucleus. The nuclear equation for the reaction is 21H + 31H 42He + 10n + energy Some data for this reaction are given in Fig. 9.1. mass / u deuterium (21H) 2.01356 tritium (31H) 3.01551 helium-4 (42He) 4.00151 neutron (10n) 1.00867 Fig. 9.1 (i) Calculate the energy, in MeV, equivalent to 1.00 u. Explain your working. energy = … MeV [3] (ii) Use data from Fig. 9.1 and your answer in (i) to determine the energy released in this D-T reaction. energy = … MeV [2] (iii) Suggest why, for the D-T reaction to take place, the temperature of the deuterium and the tritium must be high. … … … [2]

8 marks

Mark scheme: 9 (a) ‘light’ nuclei combine to form ‘heavier’ nuclei B1 [1] (b) (i) either energy = c2∆m or energy = (3.00 × 108)2 × 1.66 × 10–27 C1 energy = 1.494 × 10–10 J C1 = (1.494 × 10–10) / (1.60 × 10–13) = 934 MeV (3 s.f.) A1 [3] (ii) ∆m = (2.01356 + 3.01551) – (4.00151 + 1.00867) = 5.02907 – 5.01018 = 0.01889 u C1 energy = 0.01889 × 934 = 17.6 MeV (allow 2 s.f.) A1 [2] (iii) high temperature means high speeds / kinetic energy of nuclei B1 D and T nuclei collide despite repelling one another B1 [2] Section B

This question in 9702/41 Oct/Nov 2014

9 One likely means by which nuclear fusion may be achieved on a practical scale is the D-T reaction. (a) State what is meant by nuclear fusion. … … [1] (b) In the D-T reaction, a deuterium (21H) nucleus fuses with a tritium (31H) nucleus to form a helium-4 (42He) nucleus. The nuclear equation for the reaction is 21H + 31H 42He + 10n + energy Some data for this reaction are given in Fig. 9.1. mass / u deuterium (21H) 2.01356 tritium (31H) 3.01551 helium-4 (42He) 4.00151 neutron (10n) 1.00867 Fig. 9.1 (i) Calculate the energy, in MeV, equivalent to 1.00 u. Explain your working. energy = … MeV [3] (ii) Use data from Fig. 9.1 and your answer in (i) to determine the energy released in this D-T reaction. energy = … MeV [2] (iii) Suggest why, for the D-T reaction to take place, the temperature of the deuterium and the tritium must be high. … … … [2]

8 marks

Mark scheme: 9 (a) ‘light’ nuclei combine to form ‘heavier’ nuclei B1 [1] (b) (i) either energy = c2∆m or energy = (3.00 × 108)2 × 1.66 × 10–27 C1 energy = 1.494 × 10–10 J C1 = (1.494 × 10–10) / (1.60 × 10–13) = 934 MeV (3 s.f.) A1 [3] (ii) ∆m = (2.01356 + 3.01551) – (4.00151 + 1.00867) = 5.02907 – 5.01018 = 0.01889 u C1 energy = 0.01889 × 934 = 17.6 MeV (allow 2 s.f.) A1 [2] (iii) high temperature means high speeds / kinetic energy of nuclei B1 D and T nuclei collide despite repelling one another B1 [2] Section B

This question in 9702/42 Oct/Nov 2014

9 (a) State what is meant by the binding energy of a nucleus. … … … [2] (b) Data for two isotopes of uranium are given in Fig. 9.1. isotope binding energy per nucleon / MeV binding energy / MeV uranium-235 7.59 … uranium-238 … 1802 Fig. 9.1 (i) State what is meant by isotopes. … … … [2] (ii) Complete Fig. 9.1. [2] (c) Uranium-235 has a half-life of 7.1 × 108 years. (i) Show that the decay constant λ of uranium-235 is 3.1 × 10−17 s−1. [1] (ii) A sample of uranium-235 has an activity of 5.0 × 103 Bq. Calculate the mass of the sample. mass = … g [3]

10 marks

Mark scheme: 9 (a) energy required to separate the nucleons (in a nucleus) or energy required to separate the protons and neutrons in a nucleus M1 (or energy released when nucleons combine (to form a nucleus)/energy released when protons and neutrons combine to form a nucleus) either completely or to infinity A1 [2] (either free protons and neutrons or from infinity) (b) (i) either different forms of same element or nuclei having same number of M1 protons with different numbers of neutrons A1 [2] (ii) 1784 MeV (accept min. 3 s.f.) A1 7.57 MeV A1 [2] (c) (i) λ = ln 2 / (7.1 × 108 × 365 × 24 × 3600) = 3.1 × 10–17 s–1 B1 [1] (ii) A = λN 5000 = 3.1 × 10–17 × N C1 N = 1.61 × 1020 mass = 235 × (1.61 × 1020) / (6.02 × 1023) C1 = 0.063 g (accept min. 2 s.f.) A1 [3] Section B

This question in 9702/43 Oct/Nov 2015

14 Phosphorus-30 (3015P) was the first artificial radioactive nuclide to be produced in a laboratory. This was achieved by bombarding aluminium-27 (2713Al) with α-particles. A partial nuclear equation to represent this reaction is Φ 2713Al + 30 α → 15P + (a) State the full nuclear notation for (i) the α-particle, … [1] (ii) the particle represented by the symbol Φ. … [1] (b) Data for the rest masses of the particles in the reaction are given in Fig. 14.1. particle mass / u 2713Al 26.98153 α 4.00260 3015P 29.97830 Φ 1.00867 Fig. 14.1 Calculate, for this reaction, (i) the change in the total rest mass of the particles, mass change = … u [2] (ii) the energy, in joule, equivalent to the mass change calculated in (i). energy = … J [2] (c) With reference to your answer in (b)(i), comment on the energy of the α-particle such that the reaction can take place. … … … … [2] [Total: 8]

8 marks

Mark scheme: 14 (a) (i) 42 He or 42 α B1 [1] (ii) 01n B1 [1] (b) (i) ∆m = (29.97830 +1.00867) – (26.98153 + 4.00260) C1 = 30.98697 – 30.98413 = 2.84 × 10–3 u C1 [2] (ii) E = c2∆m or mc2 C1 = (3.0 × 108)2 × 2.84 × 10–3 × 1.66 × 10–27 = 4.2 × 10–13 J A1 [2] (c) mass of products is greater than mass of Al plus α or reaction causes (net) increase in (rest) mass (of the system) B1 α-particle must have at least this amount of kinetic energy B1 [2]

This question in 9702/42 Oct/Nov 2016

12 (a) Define the binding energy of a nucleus. … … … [2] (b) A stationary nucleus of uranium-238 (23982U) decays to form a nucleus of thorium-234 (239 40Th). An α-particle and a gamma-ray photon are emitted. The equation representing the decay is 23 9 82U 23940Th + 42He + 00γ The masses of the nuclei are given in Fig. 12.1. nucleus mass / u uranium-238 238. 05076 thorium-234 234.04357 helium-4 4.00260 Fig. 12.1 (i) State the relationship between the binding energies of the nuclei that is consistent with this reaction being energetically possible. … … [1] (ii) Calculate, for this reaction, 1. the change, in u, of the mass, change of mass = … u [1] 2. the total energy, in J, released. energy = … J [2] (iii) State and explain whether the energy of the gamma-ray photon is equal to the energy released in the reaction. … … … [2] [Total: 8]

8 marks

Mark scheme: 12(a) either (minimum) energy required / work done to separate the nucleons (in a nucleus) M1 to infinity A1 or energy released when nucleons come together (to form a nucleus) (M1) from infinity (A1) 12(b)(i) (total) binding energy of thorium and helium (nuclei) greater than binding energy of uranium (nucleus) B1 12(b)(ii)1 change in mass = 238.05076 – (234.04357 + 4.00260) = 4.59 × 10–3 u A1 12(b)(ii)2 either E = mc 2 = 4.59 × 10–3 × 1.66 × 10–27 × (3.00 × 108)2 C1 = 6.9 × 10–13 J A1 or 1u = 931 MeV E = 4.59 × 10–3 × 931 × 106 × 1.6 × 10–19 (C1) = 6.8 × 10–13 J (A1) 12(b)(iii) Th nucleus / He nucleus / product nucleus has kinetic energy M1 energy of gamma photon must be less than energy released A1

This question in 9702/42 Feb/March 2017

12 One possible nuclear reaction that takes place in a nuclear reactor is given by the equation 23592U + 10n 9542Mo + 13957La + 210n + x –1e0 Data for the nuclei and particles are given in Fig. 12.1. nucleus or particle mass / u 23592U 235.123 9542Mo 94.945 13957La 138.955 10n 1.00863 –1e0 5.49 × 10–4 Fig. 12.1 (a) Determine, for this nuclear reaction, the value of x. x = … [1] (b) (i) Show that the energy equivalent to 1.00 u is 934 MeV. [3] (ii) Calculate the energy, in MeV, released in this reaction. Give your answer to three significant figures. energy = … MeV [3] (c) Suggest the forms of energy into which the energy calculated in (b)(ii) is transformed. … … … … [2] [Total: 9]

9 marks

Mark scheme: 12(a) x = 7 A1 12(b)(i) E = mc2 C1 = 1.66 × 10–27 × (3.0 × 108)2 = 1.494 × 10–10 J C1 division by 1.6 × 10–13 clear to give 934 MeV A1 12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10–4) or ∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10–4) C1 = 0.21053 u C1 energy = 0.21053 × 934 = 197 MeV A1 12(c) kinetic energy of nuclei/particles/products/fragments B1 γ–ray photon energy B1

This question in 9702/41 May/June 2017

12 One nuclear reaction that can take place in a nuclear reactor may be represented, in part, by the equation 23592 U + 10 n 9542 Mo + 13957 La + 210 n + …………. + energy Data for a nucleus and some particles are given in Fig. 12.1. nucleus or particle mass / u 13957 La 138.955 10 n 1.00863 11 p 1.00728 –1 e0 5.49 × 10–4 Fig. 12.1 (a) Complete the nuclear reaction shown above. [1] (b) (i) Show that the energy equivalent to 1.00 u is 934 MeV. [3] (ii) Calculate the binding energy per nucleon, in MeV, of lanthanum-139 (13957 La). binding energy per nucleon = … MeV [3] Question 12 continues on the next page. (c) State and explain whether the binding energy per nucleon of uranium-235 (23592 U) will be greater, equal to or less than your answer in (b)(ii). … … … … [3] [Total: 10]

10 marks

Mark scheme: 12(a) 7 e 0 1 − A1 12(b)(i) E = mc2 C1 = 1.66 × 10–27 × (3.00 × 108)2 M1 = 1.494 × 10–10 J division by 1.60 × 10–13 clear to give 934 MeV A1 12(b)(ii) ∆m = (82 × 1.00863u) + (57 × 1.00728u) – 138.955u = (–) 1.16762 (u) C1 energy = 1.16762 × 934 C1 energy per nucleon = (1.16762 × 934) / 139 = 7.85 MeV A1 12(c) above A = 56, binding energy per nucleon decreases as A increases B1 U-235 has larger nucleon number M1 so less (binding energy per nucleon) A1 or fission takes place with uranium (B1) fission reaction releases energy (M1) binding energy per nucleon less (for uranium than for products) (A1)

This question in 9702/42 May/June 2017

12 One possible nuclear reaction that takes place in a nuclear reactor is given by the equation 23592U + 10n 9542Mo + 13957La + 210n + x –1e0 Data for the nuclei and particles are given in Fig. 12.1. nucleus or particle mass / u 23592U 235.123 9542Mo 94.945 13957La 138.955 10n 1.00863 –1e0 5.49 × 10–4 Fig. 12.1 (a) Determine, for this nuclear reaction, the value of x. x = … [1] (b) (i) Show that the energy equivalent to 1.00 u is 934 MeV. [3] (ii) Calculate the energy, in MeV, released in this reaction. Give your answer to three significant figures. energy = … MeV [3] (c) Suggest the forms of energy into which the energy calculated in (b)(ii) is transformed. … … … … [2] [Total: 9]

9 marks

Mark scheme: 12(a) x = 7 A1 12(b)(i) E = mc2 C1 = 1.66 × 10–27 × (3.0 × 108)2 = 1.494 × 10–10 J C1 division by 1.6 × 10–13 clear to give 934 MeV A1 12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10–4) or ∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10–4) C1 = 0.21053 u C1 energy = 0.21053 × 934 = 197 MeV A1 12(c) kinetic energy of nuclei/particles/products/fragments B1 γ–ray photon energy B1

This question in 9702/43 May/June 2017

3 (a) A mass is undergoing simple harmonic motion with amplitude x0. The maximum velocity of the mass has magnitude v0. On Fig. 3.1, show the variation with displacement x of the velocity v of the mass. v v0 0 −x0 0 x0 x −v0 Fig. 3.1 [2] (b) A straight stiff wire carries a constant current in a region of uniform magnetic flux density. The angle θ between the direction of the current and the direction of the magnetic field is varied. The maximum force on the wire is F0. On Fig. 3.2, show the variation with angle θ of the force F on the wire for values of θ between 0° and 90°. F0 F 0 0 90 θ/° Fig. 3.2 [2] (c) A sinusoidal supply has frequency 250 Hz and r.m.s. potential difference 2.8 V. On the axes of Fig. 3.3, show quantitatively the variation with time t of the voltage V for one cycle of the varying voltage. 8 V / V 6 4 2 00 1 2 3 4 5 t / ms −2 −4 −6 −8 Fig. 3.3 [2] (d) One particular fission reaction may be represented by the equation 23 9 52U + 10n 14516Ba + 9326Kr + 310n The variation with nucleon number A of the binding energy per nucleon BE is shown in Fig. 3.4. BE 0 0 A Fig. 3.4 On Fig. 3.4, mark on the line the position of (i) the nucleus 23952U (label this point U), (ii) the nucleus 14516Ba (label this point Ba), (iii) the nucleus 9326Kr (label this point Kr). [2] [Total: 8]

8 marks

Mark scheme: 3(a) reasonably shaped circle or oval surrounding the origin B1 closed loop passing through (0,±v0) and (±x0,0) B1 3(b) line from (0,0) to (90, F0) B1 curve with decreasing positive gradient, zero gradient at θ = 90 B1 3(c) reasonable sinusoidal wave, one cycle, period 4.0 ms B1 amplitude at 4.0 V B1 3(d) U near right-hand end of line with Ba between U and peak of graph B1 Ba on right hand side of peak and Kr between Ba and peak of graph B1

This question in 9702/42 Feb/March 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

2 (a) State what is meant by an ideal gas. … … … [2] (b) An ideal gas comprised of single atoms is contained in a cylinder and has a volume of 1.84 × 10–2 m3 at a pressure of 2.12 × 107 Pa. The mass of gas in the cylinder is 3.20 kg. (i) Determine, to three significant figures, the root-mean-square (r.m.s.) speed of the atoms of the gas. r.m.s. speed = … m s–1 [3] (ii) The temperature of the gas in the cylinder is 22 °C. Determine, to three significant figures, 1. the amount, in mol, of the gas, amount = … mol [2] 2. the mass of one atom of the gas. mass = … kg [2] (c) Use your answer in (b)(ii) part 2 to determine the nucleon number A of an atom of the gas. A = … [1] [Total: 10]

10 marks

Mark scheme: 2(a) M1 symbols p,V and T explained A1 2(b)(i) pV = ⅓ Nm<c2> and M = Nm (and so) p = ⅓ρ <c2> C1 2.12 × 107 = ⅓ × [3.20 / (1.84 × 10–2)] × <c2> C1 cr.m.s. = 605 m s–1 A1 2(b)(ii) 1. pV = nRT and T = (22 + 273) K C1 n = (2.12 × 107 × 1.84 × 10–2) / (8.31 × 295) = 159 mol A1 2. mass = 3.20 / (159 × 6.02 × 1023) or mass = [2 × (3 / 2) × 1.38 × 10–23 × 295] / 6052 C1 mass = 3.34 × 10–26 kg A1 2(c) A = (3.34 × 10–26) / (1.66 × 10–27) = 20 A1

This question in 9702/42 Oct/Nov 2018

12 (a) State what is meant by nuclear fusion and nuclear fission. nuclear fusion: … … … nuclear fission: … … … [3] (b) A nuclear reaction which may, in the future, be used for the generation of electrical energy is 2 3 4 1 H + 1 H 2 He + x . (i) Name the particle x. … [1] (ii) Data for the binding energy per nucleon EB of some nuclei are given in Fig. 12.1. binding energy per nucleon EB / 10–13 J deuterium 21 H 1.7813 tritium 31 H 4.5285 helium 42 He 11.3290 Fig. 12.1 1. State the binding energy per nucleon of x. binding energy per nucleon = … J 2. Calculate the energy change that takes place in this reaction. energy change = … J [3] (iii) Use your answer in (ii) part 2 to determine the energy release when 2.0 g of deuterium ( 21 H ) reacts with 3.0 g of tritium ( 31 H ). energy = … J [1] [Total: 8]

8 marks

Mark scheme: 12(a) fusion: two nuclei combine to form a (single) nucleus B1 fission: a (single) large nucleus divides to form (smaller) nuclei B1 Any one from: • fusion is initiated by (very) high temperatures • fission is initiated by neutron bombardment • resulting nuclei in fission are of similar size • (both processes) release energy • binding energy per nucleon increases • total binding energy increases • fission involves release of neutrons B1 12(b)(i) neutron B1 12(b)(ii) 1. zero A1 2. (4 × 11.3290 × 10–13) – (2 × 1.7813 × 10–13) – (3 × 4.5285 × 10–13) C1 energy change = 45.316 × 10–13 – 17.148 × 10–13 = 2.82 × 10–12 J A1 12(b)(iii) 1.0 mol or NA nuclei of each energy = 2.817 × 10–12 × 6.02 × 1023 = 1.7 × 1012 J 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

12 The incomplete nuclear equation for one possible reaction that takes place in the core of a nuclear reactor is 23592U + 10n 13957La + 9542Mo + 210n + … (a) (i) State the name given to this type of nuclear reaction. … [1] (ii) Complete the nuclear equation. [2] (b) The mass defect for the reaction is 0.223 u. (i) Calculate the energy, in J, equivalent to 0.223 u. energy = … J [2] (ii) Suggest two forms of the energy released in this reaction. 1. … … 2. … … [2] [Total: 7]

7 marks

Mark scheme: 12(a)(i) fission B1 12(a)(ii) either 0–1e or 0–1β M1 7 A1 12(b)(i) energy = c2 ∆m = 0.223 × 1.66 × 10–27 × (3.00 × 108)2 C1 = 3.33 × 10–11 J A1 Question Answer Marks 12(b)(ii) Any 2 from: kinetic energy of products gamma photons neutrinos B2

This question in 9702/42 Feb/March 2019

12 (a) A sample of a radioactive isotope contains N nuclei of the isotope at time T. At time (T + ΔT ), the sample contains (N – ΔN ) nuclei of the isotope. The time interval ΔT is short. Use the symbols N, ΔN, T and ΔT to give expressions for: (i) the average activity of the sample during the time ΔT … [1] (ii) the probability of decay of a nucleus in the time ΔT … [1] (iii) the decay constant λ of the isotope. … [1] (b) The isotope polonium-208 (20884 Po) is radioactive and decays to form lead-204 ( 20482 Pb). The nuclear equation for this decay is 208 84 Po 20482 Pb + 42 He. Data for nuclear masses are given in Fig. 12.1. mass / u 4 2 He 4.002 603 204 82 Pb 203.973 043 208 84 Po 207.981 245 Fig. 12.1 (i) Determine, for the decay of one nucleus of polonium-208: 1. the change, in u, of the mass mass change = … u [1] 2. the total energy, in pJ, released. energy = … pJ [3] (ii) The polonium-208 nucleus is initially stationary. The initial kinetic energy of the 4 2 He nucleus (α-particle) is found to be less than the energy calculated in (i) part 2. Suggest two possible reasons for this difference. 1. … … 2. … … [2] [Total: 9]

9 marks

Mark scheme: 12(a)(i) B1 12(a)(ii) ∆N / N B1 12(a)(iii) ∆N / (N ∆T) B1 12(b)(i) 1. mass change = 5.60 × 10–3 u A1 2. energy = (∆)mc2 C1 = 5.6 × 10–3 × 1.66 × 10–27 × (3.0 × 108)2 ( = 8.36 × 10–13 J) C1 = 0.84 pJ A1 12(b)(ii) kinetic energy (of recoil) of lead (nucleus) B1 energy of γ-ray photon B1

This question in 9702/41 May/June 2019

12 (a) State what is meant by the binding energy of a nucleus. … … … [2] (b) Some masses are shown in Fig. 12.1. mass / u proton (11p) 1.007 neutron (10n) 1.009 lanthanum-141 (14157La) nucleus 140.911 Fig. 12.1 Calculate the binding energy of a nucleus of lanthanum-141. binding energy = … J [4] (c) The nuclide lanthanum-141 (14157La) has a half-life of 3.9 hours. Initially, a radioactive source contains only lanthanum-141. The initial activity of the source is A0. (i) Calculate the time for the activity of the lanthanum-141 to be reduced to 0.40A0. time = … hours [3] (ii) Suggest why the total activity of the radioactive source measured at the time calculated in (i) may be greater than 0.40A0. … … [1] [Total: 10]

10 marks

Mark scheme: 12(a) energy required to separate the nucleons (in a nucleus) M1 to infinity A1 or energy released when nucleons come together (to form nucleus) (M1) from infinity (A1) 12(b) mass defect = 140.911 – (57 × 1.007) – (84 × 1.009) C1 = 140.911 – 142.155 = (–)1.244 (u) C1 energy = c2(∆)m C1 = (3.00 × 108)2 × 1.244 × 1.66 × 10–27 = 1.9 × 10–10 J A1 12(c)(i) A = A0e–λt and ln 2 = λt½ C1 0.40 = exp(–ln 2 × t / 3.9) C1 or (0.5)n = 0.40 (C1) n = 1.32 and t = 1.32 × 3.9 (C1) t = 5.2 hours A1 12(c)(ii) daughter product may be radioactive or random nature of decay B1

This question in 9702/42 May/June 2019

12 (a) A sample of a radioactive isotope contains N nuclei of the isotope at time T. At time (T + ΔT ), the sample contains (N – ΔN ) nuclei of the isotope. The time interval ΔT is short. Use the symbols N, ΔN, T and ΔT to give expressions for: (i) the average activity of the sample during the time ΔT … [1] (ii) the probability of decay of a nucleus in the time ΔT … [1] (iii) the decay constant λ of the isotope. … [1] (b) The isotope polonium-208 (20884 Po) is radioactive and decays to form lead-204 ( 20482 Pb). The nuclear equation for this decay is 208 84 Po 20482 Pb + 42 He. Data for nuclear masses are given in Fig. 12.1. mass / u 4 2 He 4.002 603 204 82 Pb 203.973 043 208 84 Po 207.981 245 Fig. 12.1 (i) Determine, for the decay of one nucleus of polonium-208: 1. the change, in u, of the mass mass change = … u [1] 2. the total energy, in pJ, released. energy = … pJ [3] (ii) The polonium-208 nucleus is initially stationary. The initial kinetic energy of the 4 2 He nucleus (α-particle) is found to be less than the energy calculated in (i) part 2. Suggest two possible reasons for this difference. 1. … … 2. … … [2] [Total: 9]

9 marks

Mark scheme: 12(a)(i) B1 12(a)(ii) ∆N / N B1 12(a)(iii) ∆N / (N ∆T) B1 12(b)(i) 1. mass change = 5.60 × 10–3 u A1 2. energy = (∆)mc2 C1 = 5.6 × 10–3 × 1.66 × 10–27 × (3.0 × 108)2 ( = 8.36 × 10–13 J) C1 = 0.84 pJ A1 12(b)(ii) kinetic energy (of recoil) of lead (nucleus) B1 energy of γ-ray photon B1

This question in 9702/43 May/June 2019

12 One possible nuclear reaction that takes place is 23 9 52U + 10n 9452Mo + 1357La9 + 210n + 7–10e Data for nuclei in this reaction are given in Fig. 12.1. total mass of separate binding energy nucleus mass / u mass defect / u nucleons / u per nucleon / MeV 9 4 52Mo 94.906 95.765 0.859 8.443 13 57La9 138.906 140.125 1.219 8.189 23 9 52U 235.044 236.909 1.865 … Fig. 12.1 (a) Show that the energy equivalent to a mass of 1.00 u is 934 MeV. [2] (b) (i) Use data from Fig. 12.1 to calculate the binding energy per nucleon of a nucleus of uranium-235 (23952U). Complete Fig. 12.1. [2] (ii) The nucleon number of an isotope of the element rutherfordium is 267. State whether the binding energy per nucleon of this isotope will be greater than, equal to or less than the binding energy per nucleon of uranium-235. … [1] (c) Calculate the total energy, in MeV, released in this nuclear reaction. energy = … MeV [2] (d) The nuclei in 1.2 × 10–7 mol of uranium-235 all undergo this reaction in a time of 25 ms. Calculate the average power release during the time of 25 ms. power = … W [3] [Total: 10]

10 marks

Mark scheme: 12(a) M1 E = (1.49 × 10–10) / (1.60 × 10–19) = 9.34 × 108 = 934 MeV A1 or binding energy = 8.443 × 95 [or equivalent using La-139 nucleus] (M1) binding energy / mass defect = (8.443 × 95) / 0.859 = 934 MeV (A1) 12(b)(i) binding energy = 1.865 × 934 (= 1741.91 MeV) C1 binding energy per nucleon = 1741.91 / 235 = 7.41 (MeV) A1 12(b)(ii) less (than) B1 12(c) energy = {(1.219 + 0.859) – 1.865)} × 934 or energy = (95 × 8.443) + (139 × 8.189) – (235 × 7.412) C1 = 199 MeV A1 12(d) number of reactions = 1.2 × 10–7 × 6.02 × 1023 = 7.22 × 1016 C1 energy release (for one reaction) = 199 × 1.60 × 10–13 (= 3.18 × 10–11 J) C1 power = (7.22 × 1016 × 3.18 × 10–11) / (25 × 10–3) = 9.2 × 107 W A1

This question in 9702/41 Oct/Nov 2019

12 One possible nuclear reaction that takes place is 23 9 52U + 10n 9452Mo + 1357La9 + 210n + 7–10e Data for nuclei in this reaction are given in Fig. 12.1. total mass of separate binding energy nucleus mass / u mass defect / u nucleons / u per nucleon / MeV 9 4 52Mo 94.906 95.765 0.859 8.443 13 57La9 138.906 140.125 1.219 8.189 23 9 52U 235.044 236.909 1.865 … Fig. 12.1 (a) Show that the energy equivalent to a mass of 1.00 u is 934 MeV. [2] (b) (i) Use data from Fig. 12.1 to calculate the binding energy per nucleon of a nucleus of uranium-235 (23952U). Complete Fig. 12.1. [2] (ii) The nucleon number of an isotope of the element rutherfordium is 267. State whether the binding energy per nucleon of this isotope will be greater than, equal to or less than the binding energy per nucleon of uranium-235. … [1] (c) Calculate the total energy, in MeV, released in this nuclear reaction. energy = … MeV [2] (d) The nuclei in 1.2 × 10–7 mol of uranium-235 all undergo this reaction in a time of 25 ms. Calculate the average power release during the time of 25 ms. power = … W [3] [Total: 10]

10 marks

Mark scheme: 12(a) M1 E = (1.49 × 10–10) / (1.60 × 10–19) = 9.34 × 108 = 934 MeV A1 or binding energy = 8.443 × 95 [or equivalent using La-139 nucleus] (M1) binding energy / mass defect = (8.443 × 95) / 0.859 = 934 MeV (A1) 12(b)(i) binding energy = 1.865 × 934 (= 1741.91 MeV) C1 binding energy per nucleon = 1741.91 / 235 = 7.41 (MeV) A1 12(b)(ii) less (than) B1 12(c) energy = {(1.219 + 0.859) – 1.865)} × 934 or energy = (95 × 8.443) + (139 × 8.189) – (235 × 7.412) C1 = 199 MeV A1 12(d) number of reactions = 1.2 × 10–7 × 6.02 × 1023 = 7.22 × 1016 C1 energy release (for one reaction) = 199 × 1.60 × 10–13 (= 3.18 × 10–11 J) C1 power = (7.22 × 1016 × 3.18 × 10–11) / (25 × 10–3) = 9.2 × 107 W A1

This question in 9702/43 Oct/Nov 2019

12 (a) Explain what is meant by the binding energy of a nucleus. … … … [2] (b) The following nuclear reaction takes place: 23592 U + 10 n 14455 Cs + 90x Rb + y10 n (i) Determine the values of x and y. x = … y = … [1] (ii) State the name of this type of nuclear reaction. … [1] (iii) Compare the binding energy per nucleon of uranium-235 with the binding energy per nucleon of caesium-144. … … [1] (c) Yttrium-90 decays into zirconium-90, a stable isotope. A sample initially consists of pure yttrium-90. Calculate the time, in days, when the ratio of the number of yttrium-90 nuclei to the number of zirconium-90 nuclei would be 2.0. The half-life of yttrium-90 is 2.7 days. time = … days [3] [Total: 8]

8 marks

Mark scheme: 12(a) (minimum) energy required to separate the nucleons M1 to infinity A1 12(b)(i) 37 2 B1 12(b)(ii) fission B1 12(b)(iii) binding energy per nucleon smaller for U than for Cs B1 12(c) Current ratio 2 Y to 1 Zr, so initially 3 Y 2 = 3 e–λt λ = 0.693 / 2.7 C1 ln(2 / 3) = – (ln 2 / 2.7)t C1 t = 1.6 days A1 or (½)n = 2 / 3 (C1) n = 0.585 (C1) time = 0.585 × 2.7 = 1.6 days (A1)

This question in 9702/42 Feb/March 2020

11 An electron, at rest, has mass me and charge –q. A positron is a particle that, at rest, has mass me and charge +q. A positron interacts with an electron. The electron and the positron may be considered to be at rest. The outcome of this interaction is that the electron and the positron become two gamma-ray (γ-ray) photons, each having the same energy. (a) Calculate, for one of the γ-ray photons: (i) the photon energy, in J energy = … J [2] (ii) its momentum. momentum = … N s [2] (b) State and explain the direction, relative to each other, in which the γ-ray photons are emitted. … … … … [2] [Total: 6]

6 marks

Mark scheme: 11(a)(i) E = mc2 C1 = 9.11 × 10–31 × (3.0 × 108)2 = 8.2 × 10–14 J A1 11(a)(ii) p = h / λ and E = hc / λ or E = pc C1 p = (8.2 × 10–14) / (3.0 × 108) = 2.7 × 10–22 N s A1 11(b) total momentum (before and after interaction) is zero or momentum must be conserved (in the interaction) or momentum of the photons must be equal and opposite B1 (photons emitted in) opposite directions B1

This question in 9702/41 May/June 2020

12 (a) State what is meant by the mass defect of a nucleus. … … … [2] (b) Some masses are shown in Table 12.1. Table 12.1 mass / u proton 11p 1.007 276 neutron 10n 1.008 665 helium‑4 (42He) nucleus 4.001 506 Show that: (i) the energy equivalence of 1.00 u is 934 MeV [2] (ii) the binding energy per nucleon of a helium-4 nucleus is 7.09 MeV. [2] (c) Isotopes of hydrogen have binding energies per nucleon of less than 3 MeV. Suggest why a nucleus of helium‑4 does not spontaneously break down to become nuclei of hydrogen. … … … [2] [Total: 8] To avoid the issue of disclosure of answer‑related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download

8 marks

Mark scheme: 12(a) difference between mass of nucleus and mass of (constituent) nucleons M1 where nucleons are separated to infinity A1 12(b)(i) E = mc2 C1 = 1.66 × 10–27 × (3.00 × 108)2 / (1.60 × 10–13) = 934 MeV A1 12(b)(ii) mass defect = 2 × (1.007276 + 1.008665) – 4.001506 ( = 0.030376) B1 binding energy per nucleon = (0.030376 × 934) / 4 = 7.09 MeV A1 12(c) binding energy per nucleon is much greater M1 so would require a large amount of energy to separate the nucleons in helium A1 or amount of energy released in forming hydrogen isotopes (M1) is less than energy required to break apart helium nucleus (A1)

This question in 9702/42 May/June 2020

11 An electron, at rest, has mass me and charge –q. A positron is a particle that, at rest, has mass me and charge +q. A positron interacts with an electron. The electron and the positron may be considered to be at rest. The outcome of this interaction is that the electron and the positron become two gamma-ray (γ-ray) photons, each having the same energy. (a) Calculate, for one of the γ-ray photons: (i) the photon energy, in J energy = … J [2] (ii) its momentum. momentum = … N s [2] (b) State and explain the direction, relative to each other, in which the γ-ray photons are emitted. … … … … [2] [Total: 6]

6 marks

Mark scheme: 11(a)(i) E = mc2 C1 = 9.11 × 10–31 × (3.0 × 108)2 = 8.2 × 10–14 J A1 11(a)(ii) p = h / λ and E = hc / λ or E = pc C1 p = (8.2 × 10–14) / (3.0 × 108) = 2.7 × 10–22 N s A1 11(b) total momentum (before and after interaction) is zero or momentum must be conserved (in the interaction) or momentum of the photons must be equal and opposite B1 (photons emitted in) opposite directions B1

This question in 9702/43 May/June 2020

12 (a) (i) Define nuclear binding energy. … … … [2] (ii) Explain what is meant by a nuclear fission reaction. … … … [2] (b) A student suggests that one possible nuclear reaction is 5626Fe + 10n 209F + 3717Cl. The binding energy per nucleon of a nucleus varies with the nucleon number. Use this variation to explain why the reaction would not result in an overall release of energy. … … … … … [3] [Total: 7]

7 marks

Mark scheme: 12(a)(i) energy required to separate nucleons (of nucleus) M1 to infinity A1 12(a)(ii) a (single) large nucleus divides to form (smaller) nuclei B1 any one point from: • initiated by neutron bombardment • resulting nuclei are of similar size • binding energy per nucleon increases • total binding energy increases • neutrons released • combined mass of smaller nuclei is less than mass of large nucleus B1 12(b) binding energy per nucleon is a maximum at around A = 56 B1 products of splitting a 56Fe nucleus must have a lower total binding energy B1 (reaction would require) a net input of energy B1

This question in 9702/42 Oct/Nov 2020

12 (a) State what is meant by luminosity of a star. … … [1] (b) The luminosity of the Sun is 3.83 × 1026 W. The distance between the Earth and the Sun is 1.51 × 1011 m. Calculate the radiant flux intensity F of the Sun at the Earth. Give a unit with your answer. F = … unit … [2] (c) Use data from (b) to calculate the mass that is converted into energy every second in the Sun. mass = … kg [1] (d) The radius of the Sun is 6.96 × 108 m. Show that the temperature T of the surface of the Sun is 5770 K. [1] (e) The wavelength λmax of light for which the maximum rate of emission occurs from the Sun is 5.00 × 10–7 m. The temperature of the surface of the star Sirius is 9940 K. Use information from (d) to determine the wavelength of light for which the maximum rate of emission occurs from Sirius. wavelength = … m [2] [Total: 7]

7 marks

Mark scheme: 12(a) total power of radiation emitted (by the star) B1 12(b) 2 L F = 4 d π 26 112 3.83 10 = 4 1.51 10 × × π × × C1 2 = 1340 W m− A1 Question Answer Marks 12(c) 2 E m c = 26 82 3.83 10 = 3.00 10 × × 9 = 4.26 10 kg × A1 12(d) 2 4 L = 4 T r πσ 26 8 82 4 3.83 10 = 4 5.67 10 6.96 10 T − × × π× × × × × leading to T = 5770 K B1 12(e) (max) 1 T ∝ λ 7 5.00 10 9940 5770 λ − × = C1 7 2.90 10 m − λ = × A1

This question in 9702/42 Feb/March 2022

8 (a) (i) State what is meant by nuclear binding energy. … … … [2] (ii) On Fig. 8.1, sketch a line to show the variation with nucleon number A of the binding energy per nucleon E of a nucleus. E 0 0 250 A Fig. 8.1 [2] (b) In one type of nuclear process, deuterium (21H) undergoes the reaction 21H + 21H 32He + 10n. (i) State the name of this type of nuclear process. … [1] (ii) Explain, with reference to your line in (a)(ii), why this reaction results in the release of energy. … … … [2] (c) Table 8.1 shows the masses of the particles involved in the reaction in (b). Table 8.1 particle mass / u 10n 1.008 665 21H 2.014 102 32He 3.016 029 Calculate the energy released when 1.00 mol of deuterium undergoes the reaction. energy = … J [5] [Total: 12]

12 marks

Mark scheme: 8(a)(i) energy required to separate the nucleons (in the nucleus) M1 to infinity A1 8(a)(ii) curve starting close to the origin and forming a single peak B1 peak shown to left of centre, with steep line on LHS of peak and shallow line on RHS of peak B1 8(b)(i) fusion B1 8(b)(ii) both particles have low A values or both particles are at left-hand end of graph B1 He-3 has higher binding energy (per nucleon) than H-2 B1 8(c) m = [(2  2.014102) – (3.016029 + 1.008665)] u ( = 0.00351 u) C1 E = mc2 C1 = 0.00351  1.66  10–27  (3.00  108)2 ( = 5.24  10–13 J) C1 1.00 mol of deuterium forms 0.500 mol of helium-3 C1 total energy = 0.500  6.02  1023  5.24  10–13 = 1.58  1011 J A1

This question in 9702/41 May/June 2022

8 (a) (i) State what is meant by nuclear binding energy. … … … [2] (ii) On Fig. 8.1, sketch a line to show the variation with nucleon number A of the binding energy per nucleon E of a nucleus. E 0 0 250 A Fig. 8.1 [2] (b) In one type of nuclear process, deuterium (21H) undergoes the reaction 21H + 21H 32He + 10n. (i) State the name of this type of nuclear process. … [1] (ii) Explain, with reference to your line in (a)(ii), why this reaction results in the release of energy. … … … [2] (c) Table 8.1 shows the masses of the particles involved in the reaction in (b). Table 8.1 particle mass / u 10n 1.008 665 21H 2.014 102 32He 3.016 029 Calculate the energy released when 1.00 mol of deuterium undergoes the reaction. energy = … J [5] [Total: 12]

12 marks

Mark scheme: 8(a)(i) energy required to separate the nucleons (in the nucleus) M1 to infinity A1 8(a)(ii) curve starting close to the origin and forming a single peak B1 peak shown to left of centre, with steep line on LHS of peak and shallow line on RHS of peak B1 8(b)(i) fusion B1 8(b)(ii) both particles have low A values or both particles are at left-hand end of graph B1 He-3 has higher binding energy (per nucleon) than H-2 B1 8(c) m = [(2  2.014102) – (3.016029 + 1.008665)] u ( = 0.00351 u) C1 E = mc2 C1 = 0.00351  1.66  10–27  (3.00  108)2 ( = 5.24  10–13 J) C1 1.00 mol of deuterium forms 0.500 mol of helium-3 C1 total energy = 0.500  6.02  1023  5.24  10–13 = 1.58  1011 J A1

This question in 9702/43 May/June 2022

10 Carbon-15 (156 C) is an isotope of carbon that undergoes radioactive decay to nitrogen-15 (157 N), which is a stable isotope of nitrogen. Radioactive decay is both a random and a spontaneous process. (a) State what is meant by: (i) random … … [1] (ii) spontaneous. … … [1] (b) A small sample of carbon-15 decays. The mass M of carbon-15 in the sample decreases with time t. Fig. 10.1 shows the variation with t of the value of ln (M / 10–16 g). – 4 0 2 4 6 8 10 12 t / s – 5 In (M / 10–16 g) – 6 – 7 – 8 Fig. 10.1 (i) State how Fig. 10.1 demonstrates that radioactive decay is random. … … [1] (ii) On Fig. 10.1, draw the straight line of best fit. [1] (iii) Show that the decay constant λ of carbon-15 is given by the magnitude of the gradient of your line in (b)(ii). [1] (iv) Use your line in (b)(ii) to determine λ. Give a unit with your answer. λ = … unit … [2] (v) Use your answer in (b)(iv) to calculate the half-life of carbon-15. half-life = … s [1] (c) The equation for the decay of carbon-15 can be written as 156C 157N + –1β0 + 00ν. State and explain how the mass of the products of the decay must compare with the mass of the carbon-15 nucleus. … … … [2] [Total: 10]

10 marks

Mark scheme: 10(a)(i) cannot predict when a (particular) nucleus will decay B1 or cannot predict which nucleus will decay next 10(a)(ii) not affected by external / environmental factors B1 10(b)(i) line fluctuates B1 or trend is a straight line 10(b)(ii) straight line of best fit drawn on Fig. 10.1 B1 10(b)(iii) M = M0 exp (–t) B1 so ln M = ln M0 – t so gradient = –(and magnitude of gradient = ) 10(b)(iv) gradient = (–) (8.0 – 4.8) / (11.6 – 0) (allow any correct pair of values from Fig. 10.1) C1 = 0.28 s–1 A1 10(b)(v) half-life = 0.693 /  A1 = 0.693 / 0.28 = 2.5 s 10(c) (for reaction to occur,) energy is released B1 energy release comes from fall in mass so total mass of products must be less (than mass of carbon-15) B1

This question in 9702/41 Oct/Nov 2022

10 Carbon-15 (156 C) is an isotope of carbon that undergoes radioactive decay to nitrogen-15 (157 N), which is a stable isotope of nitrogen. Radioactive decay is both a random and a spontaneous process. (a) State what is meant by: (i) random … … [1] (ii) spontaneous. … … [1] (b) A small sample of carbon-15 decays. The mass M of carbon-15 in the sample decreases with time t. Fig. 10.1 shows the variation with t of the value of ln (M / 10–16 g). – 4 0 2 4 6 8 10 12 t / s – 5 In (M / 10–16 g) – 6 – 7 – 8 Fig. 10.1 (i) State how Fig. 10.1 demonstrates that radioactive decay is random. … … [1] (ii) On Fig. 10.1, draw the straight line of best fit. [1] (iii) Show that the decay constant λ of carbon-15 is given by the magnitude of the gradient of your line in (b)(ii). [1] (iv) Use your line in (b)(ii) to determine λ. Give a unit with your answer. λ = … unit … [2] (v) Use your answer in (b)(iv) to calculate the half-life of carbon-15. half-life = … s [1] (c) The equation for the decay of carbon-15 can be written as 156C 157N + –1β0 + 00ν. State and explain how the mass of the products of the decay must compare with the mass of the carbon-15 nucleus. … … … [2] [Total: 10]

10 marks

Mark scheme: 10(a)(i) cannot predict when a (particular) nucleus will decay B1 or cannot predict which nucleus will decay next 10(a)(ii) not affected by external / environmental factors B1 10(b)(i) line fluctuates B1 or trend is a straight line 10(b)(ii) straight line of best fit drawn on Fig. 10.1 B1 10(b)(iii) M = M0 exp (–t) B1 so ln M = ln M0 – t so gradient = –(and magnitude of gradient = ) 10(b)(iv) gradient = (–) (8.0 – 4.8) / (11.6 – 0) (allow any correct pair of values from Fig. 10.1) C1 = 0.28 s–1 A1 10(b)(v) half-life = 0.693 /  A1 = 0.693 / 0.28 = 2.5 s 10(c) (for reaction to occur,) energy is released B1 energy release comes from fall in mass so total mass of products must be less (than mass of carbon-15) B1

This question in 9702/43 Oct/Nov 2022

9 (a) Define mass defect. … … … [2] (b) Table 9.1 shows the mass defects of three nuclei. Table 9.1 nucleus mass defect / u 21H 0.002 388 31H 0.009 105 42He 0.030 377 The nuclear fusion process in a particular star is described by 21H + 31H 42He + X where X is a particle that has no mass defect. (i) State the name of particle X. … [1] (ii) Show that the energy released when one nucleus of 42He is formed in this fusion reaction is 2.8 × 10–12 J. [3] (c) The star in (b) has a radius of 2.3 × 109 m and a luminosity of 1.4 × 1028 W. All the energy released from the formation of 42He is radiated away from the star. All the energy that is radiated from the star has been released in the formation of 42He. Determine: (i) the mass of 42He produced per unit time by the fusion process mass per unit time = … kg s–1 [3] (ii) the surface temperature of the star. temperature = … K [2] [Total: 11]

11 marks

Mark scheme: 9(a) difference between mass of nucleus and (total) mass of nucleons M1 when infinitely separated A1 9(b)(i) neutron B1 9(b)(ii) E = m c2 C1 m = (0.030377 – 0.002388 – 0.009105)u ( = 0.018884u) C1 energy release = (0.030377 – 0.002388 – 0.009105)  1.66  10–27  (3.00  108)2 = 2.8  10–12 J A1 9(c)(i) number of atoms per unit time = (1.4  1028) / (2.8  10–12) ( = 5.0  1039 s–1) C1 mass of one atom = 4  1.66  10–27 or (4  10–3) / (6.02  1023) ( = 6.64  10–27 kg) C1 mass per unit time = 6.64  10–27  5.0  1039 = 3.3  1013 kg s–1 A1 9(c)(ii) L = 4σr2T4 1.4  1028 = 4  5.67  10–8  (2.3  109)2  T4 C1 T = 7800 K A1

This question in 9702/42 May/June 2023

9 (a) State what is meant by nuclear fusion. … … … [2] (b) On Fig. 9.1, sketch the variation of binding energy per nucleon with nucleon number A for values of A between 1 and 250. binding energy per nucleon 0 1 250 A Fig. 9.1 [2] (c) On your line in Fig. 9.1, label: (i) a point X that could represent a nucleus that undergoes alpha-decay [1] (ii) a point Y that could represent a nucleus that undergoes nuclear fusion. [1] (d) A nucleus Z undergoes nuclear fission to form strontium-93 ( 9338Sr) and xenon-139 (13954Xe) according to 10n + Z 9338Sr + 13954Xe + 210n. Table 9.1 shows the binding energies of the strontium-93 and xenon-139 nuclei. Table 9.1 nucleus binding energy / J × 10–10 9338Sr 1.25 × 10–10 13954Xe 1.81 The fission of 1.00 mol of Z releases 1.77 × 1013 J of energy. Determine the binding energy per nucleon, in MeV, of Z. binding energy per nucleon = … MeV [4] [Total: 10]

10 marks

Mark scheme: 9(a) (two small) nuclei join together M1 to form one larger nucleus A1 9(b) line with a peak at A  56 B1 line with steep initial positive gradient on the left of peak and shallower negative gradient at all points to the right of peak B1 and line does not return to 0 binding energy 9(c)(i) X shown at value of A to the right of the peak B1 9(c)(ii) Y shown at value of A close to 1 B1 9(d) energy from 1 nucleus = (1.77  1013) / (6.02  1023) C1 ( = 2.94  10–11 J) binding energy of Z = [(1.25 + 1.81)  10–10] – 2.94  10–11 C1 ( = 2.77  10–10 J) nucleon number of Z = 93 + 139 + 2 – 1 C1 ( = 233) binding energy per nucleon = (2.77  10–10) / (233  1.60  10–13) A1 = 7.43 MeV

This question in 9702/41 Oct/Nov 2023

9 (a) State what is meant by nuclear fusion. … … … [2] (b) On Fig. 9.1, sketch the variation of binding energy per nucleon with nucleon number A for values of A between 1 and 250. binding energy per nucleon 0 1 250 A Fig. 9.1 [2] (c) On your line in Fig. 9.1, label: (i) a point X that could represent a nucleus that undergoes alpha-decay [1] (ii) a point Y that could represent a nucleus that undergoes nuclear fusion. [1] (d) A nucleus Z undergoes nuclear fission to form strontium-93 ( 9338Sr) and xenon-139 (13954Xe) according to 10n + Z 9338Sr + 13954Xe + 210n. Table 9.1 shows the binding energies of the strontium-93 and xenon-139 nuclei. Table 9.1 nucleus binding energy / J × 10–10 9338Sr 1.25 × 10–10 13954Xe 1.81 The fission of 1.00 mol of Z releases 1.77 × 1013 J of energy. Determine the binding energy per nucleon, in MeV, of Z. binding energy per nucleon = … MeV [4] [Total: 10]

10 marks

Mark scheme: 9(a) (two small) nuclei join together M1 to form one larger nucleus A1 9(b) line with a peak at A  56 B1 line with steep initial positive gradient on the left of peak and shallower negative gradient at all points to the right of peak B1 and line does not return to 0 binding energy 9(c)(i) X shown at value of A to the right of the peak B1 9(c)(ii) Y shown at value of A close to 1 B1 9(d) energy from 1 nucleus = (1.77  1013) / (6.02  1023) C1 ( = 2.94  10–11 J) binding energy of Z = [(1.25 + 1.81)  10–10] – 2.94  10–11 C1 ( = 2.77  10–10 J) nucleon number of Z = 93 + 139 + 2 – 1 C1 ( = 233) binding energy per nucleon = (2.77  10–10) / (233  1.60  10–13) A1 = 7.43 MeV

This question in 9702/43 Oct/Nov 2023

9 (a) State what is meant by the binding energy of a nucleus. … … … [2] (b) Table 9.1 shows the masses of two sub-atomic particles and a polonium-212 (21284Po) nucleus. Table 9.1 mass / u proton 1.007 276 neutron 1.008 665 polonium-212 nucleus 211.942 749 For the polonium-212 nucleus, determine: (i) the mass defect Δm, in kg Δm = … kg [3] (ii) the binding energy binding energy = … J [2] (iii) the binding energy per nucleon. binding energy per nucleon = … J [1] (c) (i) On Fig. 9.1, sketch the variation with nucleon number A of binding energy per nucleon for values of A from 1 to 250. binding energy per nucleon 0 1 250 A Fig. 9.1 [2] (ii) On your line in Fig. 9.1, draw an X to show the approximate position of polonium-212. [1] (iii) Polonium-212 is radioactive and undergoes alpha-decay. Suggest and explain, with reference to Fig. 9.1, why the alpha-decay of polonium-212 results in a release of energy. … … … [2] [Total: 13]

13 marks

Mark scheme: 9(a) energy required to separate (all) the nucleons (in the nucleus) M1 to infinity A1 9(b)(i) m = {[(84  1.007276) + (128  1.008665)] – 211.942749} (u) ( = 1.778 u) C1 = 1.778  1.66  10–27 (kg) C1 = 2.95  10–27 kg A1 9(b)(ii) E = ()mc2 C1 binding energy = 2.95  10–27  (3.00  108)2 = 2.66  10–10 J A1 9(b)(iii) binding energy per nucleon = (2.66  10–10) / 212 = 1.25  10–12 J A1 9(c)(i) line rising to a single peak that is to the left of the ‘9’ in the Fig. 9.1 label and then continually decreasing B1 steep positive gradient on the left of the peak and shallow negative gradient on the right B1 9(c)(ii) X shown on the line at a value of A that is to the right of the left-hand edge of the ‘A’ in the axis label, and to the left of ‘2’ in the 250 label B1 9(c)(iii) nucleus formed (as a result of the decay) has a lower nucleon number B1 (nucleus formed has a) greater binding energy per nucleon B1

This question in 9702/42 May/June 2024

10 (a) Radioactive decay is both random and spontaneous. (i) State what is meant by random. … … [1] (ii) State what is meant by spontaneous. … … [1] (iii) State one piece of evidence for the random nature of decay. … … [1] (b) (i) Describe the differences between nuclear fission and nuclear fusion. … … … … … [3] (ii) Explain, with reference to the variation of binding energy per nucleon with nucleon number, why the processes of nuclear fission and nuclear fusion both result in a release of energy. … … … [2] [Total: 8]

8 marks

Mark scheme: 10(a)(i) cannot predict when a particular nucleus will decay B1 or cannot predict which nucleus will decay next 10(a)(ii) (decay is) not affected by external (environmental) factors B1 10(a)(iii) fluctuations in (measured) count rate B1 10(b)(i) • large nuclei undergo fission whereas small nuclei undergo fusion B3 • fission involves one nucleus splitting into two (or more) (smaller) nuclei • fusion involves two nuclei joining together to form one (larger) nucleus • fission is (usually) initiated by neutron bombardment • fusion is (usually) initiated by (very) high temperatures Any three points, 1 mark each 10(b)(ii) binding energy per nucleon is greatest for intermediate nucleon numbers B1 (may be shown on sketch graph with axes labelled ‘binding energy per nucleon’ and ‘nucleon number’) both fusion and fission involve an increase in binding energy (per nucleon) B1

This question in 9702/42 Oct/Nov 2024

9 Polonium‑193 (19384Po) is an unstable nuclide. A nucleus of polonium‑193 decays to a nucleus of lead‑189 (18982Pb) by emitting an alpha‑particle. (a) Radioactive decay is both random and spontaneous. State what is meant by: (i) random … … [1] (ii) spontaneous. … … [1] (b) Define half‑life. … … … [1] (c) Data for the binding energy per nucleon of the particles involved in the decay of a nucleus of polonium‑193 are given in Table 9.1. Table 9.1 particle binding energy per nucleon / eV 19384Po 7.774 18982Pb 7.826 4α 7.074 2 Determine the energy, in eV, released when a nucleus of polonium‑193 decays into a nucleus of lead‑189. energy = … eV [2] (d) A pure sample of polonium‑193 contains N0 nuclei. After a time t the sample contains N nuclei of polonium‑193. The variation of ln (N / N0) with t is shown in Fig. 9.1. t / ms 0 0.2 0.4 0.6 0.8 1.0 0 –0.2 –0.4 –0.6 In (N / N0) –0.8 –1.0 –1.2 –1.4 Fig. 9.1 (i) State the name of the quantity that is represented by the magnitude of the gradient of the line in Fig. 9.1. … [1] (ii) Use Fig. 9.1 to determine the half‑life, in ms, of polonium‑193. half‑life = … ms [2] (e) Positron emission tomography (PET scanning) uses a radioactive tracer. (i) State what happens to the positrons emitted by the tracer. … … [1] (ii) Explain why a tracer with a half‑life of approximately 2 hours is a suitable tracer to use. … … [1] [Total: 10]

10 marks

Mark scheme: 9(a)(i) either: cannot predict when a (particular) nucleus will decay B1 or: cannot predict which nucleus will decay next 9(a)(ii) not affected by external / environmental factors B1 9(b) time for activity to halve B1 9(c) energy = (189  7.826) + (4  7.074) – (193  7.774) C1 = 7.03 eV A1 9(d)(i) decay constant A1 9(d)(ii) decay constant / magnitude of gradient = 1.4 / 0.84 C1 half-life = ln2 / (1.4 / 0.84) A1 = 0.42 ms 9(e)(i) positrons collide with electrons and annihilate B1 9(e)(ii) long enough to have time to conduct investigation, not so long as to cause patient unnecessary exposure to radiation B1

This question in 9702/42 Feb/March 2025

11 The deuterium nucleus (21H) has a mass defect of 0.002 388 u. The helium-4 nucleus (42He) has a mass defect of 0.030 377 u. Helium-4 is formed from deuterium in a nuclear reaction that can be represented by the equation 3 21H 42He + 11p + 10n. (a) (i) State the name of this type of nuclear reaction. … [1] (ii) Show that the energy released when one nucleus of helium-4 is formed from deuterium is 3.47 × 10–12 J. [3] (b) A star has a radius of 6.96 × 108 m. Helium-4 is produced in this star, from deuterium, at a mass rate of 7.34 × 1011 kg s–1. All the energy released from this process is radiated away from the star. All the energy that is radiated from the star is released by this process. (i) Calculate the luminosity of the star. luminosity = … W [3] (ii) Use your answer in (b)(i) to determine the surface temperature of the star. temperature = … K [2] [Total: 9]

9 marks

Mark scheme: 11(a)(i) (nuclear) fusion B1 11(a)(ii) ∆m = [0.030377 – (3  0.002388)] u B1 ( = 0.023213 u) E = ∆m c2 C1 = 0.023213  1.66  10–27  (3.00  108)2 = 3.47  10–12 J A1 11(b)(i) mass of 1 mol of helium-4 = 4 g C1 or mass of 1 helium atom = 4 u N rate = (6.02  1023  7.34  1011) / (4  10–3) C1 or N rate = (7.34  1011) / (4  1.66  10–27) luminosity = 1.10  1038  3.47  10–12 A1 = 3.83  1026 W 11(b)(ii) L = 4r2T4 C1 3.83  1026 = 4  5.67  10–8  (6.96  108)2  T4 T = 5770 K A1

This question in 9702/44 May/June 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

9 (a) State what is meant by the mass defect of a nucleus. … … … [2] (b) The nuclear fusion reaction for the formation of helium-4 from deuterium is represented by 21H + 21H 42He. Table 9.1 shows the masses of the nuclides involved in this reaction. Table 9.1 nuclide nuclide mass / u 21H 2.013 553 42He 4.001 505 Calculate the energy released in the formation of 1.00 mol of helium-4. energy = … J [4] (c) The star Sirius has a radius of 1.19 × 109 m and loses mass due to nuclear fusion at a rate of 1.09 × 1011 kg s–1. Assume that the power of the radiation emitted by the star is equal to the power released by this process. (i) Determine a value for the luminosity of Sirius. Give a unit with your answer. luminosity = … unit … [2] (ii) Use your answer in (c)(i) to determine the surface temperature of Sirius. surface temperature = … K [2] (d) Explain how cosmologists use standard candles to estimate the distance of a galaxy from the Earth. … … … … … [3] [Total: 13]

13 marks

Mark scheme: 9(a) difference between mass of nucleus and mass of (constituent) nucleons M1 when nucleons are separated to infinity A1 9(b) m = (2  2.013553) – (4.001505) (u) C1 ( = 0.025601 u) E = c2m C1 energy from one He-4 nucleus= 0.025601  1.66  10–27  (3.00  108)2 C1 (= 3.82  10–12 J) energy to form 1.00 mol= 3.82  10–12  6.02  1023 A1 = 2.30  1012 J 9(c)(i) L = 1.09  1011  (3.00  108)2 C1 = 9.81  1027 W A1 9(c)(ii) L = 4 r2T4 C1 9.81  1027 = 4  5.67  10–8  (1.19  109)2  T4 T = 9930 K A1 9(d) standard candles have known luminosity B1 radiant flux intensity (from star) measured (on the Earth) B1 distance found from F = L / (4d2) B1

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