Cambridge A Level Physics 9702 — 2025 May/June Paper 2 · Variant 4

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level DC (JP) 357182 © UCLES 2025 PHYSICS 9702/24 Paper 2 AS Level Structured Questions May/June 2025 1 hour 15 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You may use a calculator. ● You should show all your working and use appropriate units. INFORMATION ● The total mark for this paper is 60. ● The number of marks for each question or part question is shown in brackets [ ]. * 1 4 1 4 7 6 0 1 8 8 * ,










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2 9702/24/M/J/25 © UCLES 2025 Data acceleration of free fall g = 9.81 m s–2 speed of light in free space c = 3.00 × 108 m s–1 elementary charge e = 1.60 × 10–19 C unified atomic mass unit 1 u = 1.66 × 10–27 kg rest mass of proton mp = 1.67 × 10–27 kg rest mass of electron me = 9.11 × 10–31 kg Avogadro constant NA = 6.02 × 1023 mol–1 molar gas constant R = 8.31 J K–1 mol–1 Boltzmann constant k = 1.38 × 10–23 J K–1 gravitational constant G = 6.67 × 10–11 N m2 kg–2 permittivity of free space ε0 = 8.85 × 10–12 F m–1 (4 1 0 rf = 8.99 × 109 m F–1) Planck constant h = 6.63 × 10–34 J s Stefan–Boltzmann constant σ = 5.67 × 10–8 W m–2 K–4 Formulae uniformly accelerated motion s = ut + 2 1 at 2 v 2 = u 2 + 2as hydrostatic pressure ∆p = ρg∆h upthrust F = ρgV Doppler effect for sound waves fo = v v f v s s ! electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel R 1 = R 1 1 + R 1 2 + ... * 0000800000002 * ,










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3 9702/24/M/J/25 © UCLES 2025 [Turn over 1 (a) Define the moment of a force. … … [1] (b) A trapdoor has a hinge at end A, as shown in Fig. 1.1. horizontal trapdoor hinge 80 cm 75 N 42° F B A Fig. 1.1 (not to scale) The trapdoor has length 80 cm and weight 75 N. The mass of the trapdoor is uniformly distributed along its length. A force F acts at right angles to the trapdoor at end B so that the trapdoor is held in equilibrium at an angle of 42° to the horizontal. (i) State the principle of moments. … … … [2] (ii) Calculate the component of the weight that is perpendicular to the trapdoor. component of weight = … N [1] (iii) Calculate the magnitude of the force F. F = … N [2] [Total: 6] * 0000800000003 * ,










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4 9702/24/M/J/25 © UCLES 2025 2 An object of constant mass moves in a straight line. The variation with time t of the momentum p of the object is shown in Fig. 2.1. –2 –1 0 1 2 3 4 0 8 12 2 6 10t / s p / kg m s–1 Fig. 2.1 (a) Define momentum. … … [1] (b) Calculate the change in momentum of the object from time t = 0 to t = 12 s. change in momentum = … kg m s–1 [1] (c) Calculate the magnitude of the resultant force acting on the object. force = … N [2] * 0000800000004 * ,










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5 9702/24/M/J/25 © UCLES 2025 [Turn over (d) Describe the variation of the speed of the object from time t = 0 to t = 8.0 s. … … [1] (e) By reference to Fig. 2.1, explain why the resultant force acting on the object during the first 8.0 s of its motion cannot be due to air resistance. … … … … [2] (f) At time t = 0 the displacement d of the object is zero. On Fig. 2.2, sketch the variation of d with time t from t = 0 to t = 12 s. Numerical values of d are not required. 0 4 0 8 12 2 6 10 t / s d Fig. 2.2 [3] [Total: 10] * 0000800000005 * ,










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6 9702/24/M/J/25 © UCLES 2025 3 The lower end of a vertical spring is fixed to a horizontal surface, as shown in Fig. 3.1. block, mass 5.5 kg spring horizontal surface Fig. 3.1 The mass of the spring is negligible. A block of mass 5.5 kg drops vertically onto the spring and is brought to rest as the spring is compressed. (a) The block has kinetic energy 110 J as it makes contact with the spring. Calculate the speed of the block as it makes contact with the spring. speed = … m s–1 [2] (b) The gravitational potential energy of the block decreases by 20 J as the spring is compressed to its maximum compression x0. Show that x0 is 0.37 m. [2] (c) Assume that, as the spring compresses, all of the energy lost by the block is converted into the elastic potential energy of the spring. Use the data from (a) and (b) to determine the maximum elastic potential energy of the spring. Show your working. maximum elastic potential energy = … J [1] * 0000800000006 * ,










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7 9702/24/M/J/25 © UCLES 2025 [Turn over (d) The variation of the force F acting on the spring with the compression x of the spring is shown in Fig. 3.2. x x0 F0 F 0 0 Fig. 3.2 Use the information in (b) and your answer in (c) to show that the maximum force F0 exerted on the spring by the block is 700 N. [2] (e) Use the information in (d) to determine, for the instant that the block is first brought to rest by the spring, the magnitude of: (i) the resultant force acting on the block resultant force = … N [2] (ii) the acceleration of the block. acceleration = … m s–2 [2] [Total: 11] * 0000800000007 * ,










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8 9702/24/M/J/25 © UCLES 2025 4 (a) A source oscillates with frequency f to produce a progressive wave of wavelength λ. The source takes time t to produce n complete oscillations. (i) State what is meant by a progressive wave. … … [1] (ii) State expressions, in terms of some or all of f, λ and n, for: ● the distance moved by a wavefront in time t distance = … ● time t. time t = … [2] (iii) Use your answers in (ii) to determine an expression for the speed v of the wave in terms of f and λ. [1] (b) Two identical microwave sources X and Y emit waves in phase. The sources are separated by a distance of 30 cm, as shown in Fig. 4.1. Q P Y X 72 cm 30 cm Fig. 4.1 (not to scale) The intensity of the microwaves is to be investigated at points P and Q. Line PQ is parallel to line XY. Distance XP is equal to distance YP. Distance YQ is 72 cm and angle XYQ is 90°. The wavelength of the microwaves is 4.0 cm. * 0000800000008 * ,










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9 9702/24/M/J/25 © UCLES 2025 [Turn over (i) Calculate the frequency, in GHz, of the microwaves. frequency = … GHz [2] (ii) Show that the difference between the path lengths XQ and YQ is 6 cm. [1] (iii) State and explain what may be deduced about the intensity of the microwaves at point Q. … … … … … [3] (iv) A microwave detector is positioned at P and connected to a cathode-ray oscilloscope (CRO). The controls of the CRO are adjusted so that a waveform is shown on the screen. Describe the changes to the amplitude of the waveform as the detector is moved from P to Q. … … … … [2] [Total: 12] * 0000800000009 * ,










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10 9702/24/M/J/25 © UCLES 2025 5 (a) (i) State and explain the effect, if any, on the resistance of a filament wire in a lamp as the current in the wire decreases. … … [1] (ii) On Fig. 5.1, sketch the I–V characteristic of a filament lamp. V I 0 0 Fig. 5.1 [2] (b) A battery of electromotive force (e.m.f.) E and negligible internal resistance is connected in parallel with two filament lamps A and B, as shown in Fig. 5.2. A 18 W E 1.5 A 3.3 A B Fig. 5.2 The current in the battery is 3.3 A and the current in lamp A is 1.5 A. The power dissipated in lamp A is 18 W. * 0000800000010 * ,










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11 9702/24/M/J/25 © UCLES 2025 [Turn over (i) Calculate the e.m.f. E of the battery. E = … V [2] (ii) The filament wire of lamp B has a cross-sectional area of 1.4 × 10–9 m2. The number of free (conduction) electrons per unit volume in the metal of the filament wire is 3.4 × 1028 m–3. Calculate the average drift speed of the free electrons in the filament wire of lamp B. average drift speed = … m s–1 [3] [Total: 8] * 0000800000011 * ,









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12 9702/24/M/J/25 © UCLES 2025 6 A battery of electromotive force (e.m.f.) 6.0 V and negligible internal resistance is connected in series with a variable resistor and a uniform resistance wire XY, as shown in Fig. 6.1. R 6.0 V uniform wire of resistance 8.0 Ω X Y 2.00 m Fig. 6.1 Wire XY has length 2.00 m and resistance 8.0 Ω. The resistance R of the variable resistor is adjusted so that the potential difference across wire XY is 2.4 V. (a) Determine R. R = … Ω [2] (b) Explain why the potential difference V between any two points on wire XY is proportional to the distance L between those points. … … … … [2] * 0000800000012 * ,









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13 9702/24/M/J/25 © UCLES 2025 [Turn over (c) A cell of e.m.f. E and internal resistance r is connected to the circuit, as shown in Fig. 6.2. R 6.0 V E r X P Y Fig. 6.2 Resistance R is unchanged. The movable connection P is positioned on wire XY so that the galvanometer reading is zero. Distance XP is 1.24 m. (i) Calculate E. E = … V [2] (ii) The value of R is now decreased. State and explain the change that must be made to the position of P on wire XY so that the galvanometer reads zero again. … … … … [2] [Total: 8] * 0000800000013 * ,









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14 9702/24/M/J/25 © UCLES 2025 BLANK PAGE * 0000800000014 * ,









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15 9702/24/M/J/25 © UCLES 2025 7 (a) State the names of two different leptons. 1 … 2 … [2] (b) In the following list, underline all the particles that are hadrons. antineutrino beta-plus meson neutron [1] (c) By reference to quark composition, show that the charge of a proton is +1.6 × 10–19 C. [2] [Total: 5] * 0000800000015 * ,









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16 9702/24/M/J/25 © UCLES 2025 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. 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 at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. * 0000800000016 * ,









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This document consists of 14 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level PHYSICS 9702/24 Paper 2 AS Level Structured Questions May/June 2025 MARK SCHEME Maximum Mark: 60 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the May/June 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 14 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 14 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind. Science-Specific Marking Principles 1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks should not be awarded if the keywords are used incorrectly. 2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored. 3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection). 4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme where necessary and any exceptions to this general principle will be noted. 5 ‘List rule’ guidance For questions that require n responses (e.g. State two reasons …): • The response should be read as continuous prose, even when numbered answer spaces are provided. • Any response marked ignore in the mark scheme should not count towards n. • Incorrect responses should not be awarded credit but will still count towards n. • Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should be treated as a single incorrect response. • Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 14 6 Calculation specific guidance Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show your working’. For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values. For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient (a) to a value between 1 and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme. Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded. Exceptions to this general principle will be noted in the mark scheme. 7 Guidance for chemical equations Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme. State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 14 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning Information missing or insufficient for credit Arithmetic error Benefit of the doubt given Contradiction in response, mark not awarded Incorrect point or mark not awarded Error carried forward applied Ignore the response Mandatory mark not awarded Power of ten error Blank page seen Error in number of significant figures

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 14 Annotation Meaning Transcription error Correct point or mark awarded Incorrect physics

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 14 Abbreviations / Alternative and acceptable answers for the same marking point. ( ) Bracketed content indicates words which do not need to be explicitly seen to gain credit but which indicate the context for an answer. The context does not need to be seen but if a context is given that is incorrect then the mark should not be awarded. ___ Underlined content must be present in answer to award the mark. This means either the exact word or another word that has the same technical meaning. Mark categories B marks These are independent marks, which do not depend on other marks. For a B mark to be awarded, the point to which it refers must be seen specifically in the candidate’s answer. M marks These are mandatory marks upon which A marks later depend. For an M mark to be awarded, the point to which it refers must be seen specifically in the candidate’s answer. If a candidate is not awarded an M mark, then the later A mark cannot be awarded either. C marks These are compensatory marks which can be awarded even if the points to which they refer are not written down by the candidate, providing subsequent working gives evidence that they must have known them. For example, if an equation carries a C mark and the candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then the C mark is awarded. If a correct answer is given to a numerical question, all of the preceding C marks are awarded automatically. It is only necessary to consider each of the C marks in turn when the numerical answer is not correct. A marks These are answer marks. They may depend on an M mark or allow a C mark to be awarded by implication.

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 14 Question Answer Marks 1(a) force × perpendicular distance (of line of action of force to a point) or product of force and perpendicular distance (to a point) B1 1(b)(i) in (rotational) equilibrium B1 sum / total of CW moments about a point = sum / total of ACW moments about the (same) point. B1 1(b)(ii) component of weight = 75 × cos 42 = 56 N A1 1(b)(iii) F × 80 or 56 × 40 or F × 0.8 or 56 × 0.4 F × 80 = 56 × 40 C1 F = 28 N A1 Question Answer Marks 2(a) product of mass and velocity B1 2(b) change in momentum = (–1.4) – (+2.8) = (–) 4.2 kg m s–1 A1 2(c) F = p / ()t or F = gradient = 4.2 / 12 C1 = 0.35 N A1 2(d) constant / uniform (rate of) decrease (of speed to zero). B1

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 14 Question Answer Marks 2(e) • The (resultant) force is constant / does not decrease • air resistance would vary / decrease / not constant • the force is not zero when speed / velocity is zero / at 8 s Any two of the above 3 marking points (1 mark each, max 2) B2 2(f) line from origin with decreasing positive gradient B1 gradient changes from positive to negative at 8.0 s B1 after t = 8.0 s the line has a negative gradient of increasing magnitude and a positive value of d at t = 12 s B1 Question Answer Marks 3(a) E (K) = 2 1 2 mv 2 1 110 5.5 2 v =   C1 v = 6.3 m s–1 A1 3(b) ()E(P) / 20 = mg()h OR ()E(P) / 20 = mgx0 C1 (x0 =) 20 / 5.5 × 9.81 = 0.37 (m) Allow ()h for x0 A1 3(c) [max (E(P))] = 110 + 20 = 130 J A1

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 14 Question Answer Marks 3(d) (max) E(P) / 130 = 1 2 F(0)x(0) C1 (F0 =) 2 × 130 / 0.37 = 700 (N) A1 or (max) E(P) / 130 = 1 / 2k x(0)2 (F0 =) k x(0) (C1) (F0 =) 2 × 130 × 0.37 / (0.37)2 = 700 (N) (A1) 3(e)(i) (weight) = 5.5 × 9.81 C1 resultant force = 700 – (5.5 × 9.81) = 650 N A1 3(e)(ii) F = ma C1 a = 650 / 5.5 or (700 / 5.5) – 9.81 a = 120 m s–2 A1

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 14 Question Answer Marks 4(a)(i) (a wave that) transfers / propagates energy B1 4(a)(ii) distance = n B1 time t = n / f B1 4(a)(iii) (v = distance / time) (v) = n / (n / f) so (v) = f B1 4(b)(i) f = 3 × 108 / 4 × 10–2 = 7.5 × 109 (Hz) = 7.5 × 109 / 109 (GHz) C1 = 7.5 GHz A1 4(b)(ii) (difference in path lengths, XQ – YQ) 1 2 2 2 72 30   = +   – 72 = 6 (cm) A1 Or calculate angle Q using tan and then cos to obtain XQ of 78 tan  = 30 / 72 so  = 22.6 XQ = 72 / cos 22.6 = 78 (cm) path difference 78 – 72 = 6 (cm) (A1) Or calculate angle X using tan and then cos to obtain XQ of 78 tan  = 72 / 30 so  = 67.4 XQ = 72 / sin 67.4 = 78 (cm) path difference 78 – 72 = 6 (cm) (A1) 4(b)(iii) path difference (= 6 cm / 4 cm)  = 1.5 M1 (so) phase difference (at Q) = 540° or 180 M1 (so) intensity is minimum A1

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 14 Question Answer Marks 4(b)(iv) amplitude changes from maximum (at P) to minimum (at Q) C1 amplitude changes from maximum (at P) to minimum to maximum to minimum (at Q) A1 Question Answer Marks 5(a)(i) temperature decreases, resistance decreases B1 5(a)(ii) line from origin with decreasing gradient drawn in first quadrant or line from origin with decreasing gradient drawn in third quadrant M1 line drawn in the third or first quadrant of similar composition (straight line from origin followed by correct curve) and similar size as compared by eye to the first line A1 5(b)(i) E = P / I = 18 / 1.5 C1 = 12 V A1 5(b)(ii) I = 3.3 – 1.5 = 1.8 C1 I = Anvq 1.8 = 1.4 × 10–9 × 3.4 × 1028 × v × 1.6 × 10–19 C1 v = 0.24 m s–1 A1

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 14 Question Answer Marks 6(a) 2.4 / 6.0 = 8 / (R + 8) C1 R = 12  A1 or I = 2.4 / 8 = 0.30 R = (6 – 2.4) / 0.3 or R = (6 / 0.3) – 8 (C1) R = 12  (A1) 6(b) V = I ( L / A) or V = I R and R = ( L / A) M1 I, , A are constant (so V  L) A1 6(c)(i) VXP / VXY = LXP / LXY E / 2.4 = 1.24 / 2.00 C1 E = 1.5 V A1 6(c)(ii) p.d. across XY / wire increases / p.d. across XP increases M1 so P moved towards X / away from Y / to the left A1

9702/24 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 14 Question Answer Marks 7(a) • (electron) neutrino • (electron) antineutrino • electron • positron Any two, 1 mark each B2 7(b) only meson and neutron are underlined B1 7(c) (quark composition is) up, up, down / uud (charge =) 2 2 1 3 3 3 e e e + − = (+1)e = (+) 1.6 × 10–19 (C) B1 or (charge =) 1.07 × 10–19 + 1.07 × 10–19 – 5.33 × 10–20 = (+) 1.6 × 10–19 (C) A1