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GCE MAY 2008 : A2 2 Electromagnetism and Nuclear Physics

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Centre Number 71 Candidate Number ADVANCED General Certificate of Education 2008 Physics assessing Module 5: Electromagnetism and Nuclear Physics A2Y21 Assessment Unit A2 2 [A2Y21] THURSDAY 29 MAY, MORNING TIME 1 hour 30 minutes. INSTRUCTIONS TO CANDIDATES Write your Centre Number and Candidate Number in the spaces provided at the top of this page. Answer all six questions. Write your answers in the spaces provided in this question paper. INFORMATION FOR CANDIDATES The total mark for this paper is 90. Quality of written communication will be assessed in question 5(b)(ii). Figures in brackets printed down the right-hand side of pages indicate the marks awarded to each question. Your attention is drawn to the Data and Formulae Sheet which is inside this question paper. You may use an electronic calculator. Question 6 contributes to the synoptic assessment requirement of the Specification. You are advised to spend about 45 minutes in answering questions 1 5, and about 45 minutes in answering question 6. 4666 For Examiner s use only Question Number 1 2 3 4 5 6 Total Marks Marks If you need the values of physical constants to answer any questions in this paper, they may be found on the Data and Formulae Sheet. Examiner Only Marks Remark Answer all six questions 1 An uncharged capacitor C1 of capacitance 200 F is connected in series with a resistor R of value 470 k , a 1.50 V cell and a switch S as shown in Fig. 1.1. R C1 S 1.5 V Fig. 1.1 The switch is closed and the capacitor is charged. (a) (i) On Fig. 1.2 sketch a graph to show the variation of the energy E stored in the capacitor with voltage V as the capacitor is charged. E 0 0 V Fig. 1.2 [1] (ii) Calculate the energy Emax stored in the fully charged capacitor. Emax = ____________ J 4666 [2] 2 [Turn over (b) The switch S is opened and then a second, uncharged, capacitor C2 of value 400 F is connected as shown in Fig. 1.3. Examiner Only Marks Remark C2 R C1 S 1.5 V Fig. 1.3 (i) Calculate the voltage across the combination of capacitors after they have been joined and the switch is still open. Voltage = ___________________ V [3] (ii) Calculate the total energy stored in the combination of capacitors. Total energy stored = _____________ J 4666 3 [2] [Turn over (c) The answers to (a)(ii) and (b)(ii) should show that the energy stored in the combination of capacitors is considerably less than that originally stored in C1 alone. Explain this difference in energy. Examiner Only Marks Remark __________________________________________________________ __________________________________________________________ _______________________________________________________ [2] 4666 4 [Turn over BLANK PAGE (Questions continue overleaf) 4666 5 [Turn over 2 (a) State Faraday s law of electromagnetic induction. Examiner Only Marks Remark __________________________________________________________ __________________________________________________________ _______________________________________________________ [2] (b) Before the introduction of CDs and cassettes, one way of storing recorded music was to implant the data in a continuous narrow groove on the surface of a circular vinyl record. This method is now showing some return to popularity. The record is placed on a turntable which is rotated at a constant angular velocity. A stylus or needle is then placed in the groove and variations on the sides of the groove cause the needle and attached coil to oscillate in a horizontal direction in a magnetic field provided by a permanent magnet. As a result, an e.m.f. is induced in the coil. Fig. 2.1 shows a simplified plan view of this arrangement. coil N S pivot stylus direction of oscillation Fig. 2.1 Plan View 4666 6 [Turn over Fig. 2.2 shows the variation of the angular position of the coil with time t as the stylus tracks through a section of the groove. Examiner Only Marks Remark / 0.18 0.12 0.06 0 1.0 2.0 t/ms 0.06 0.12 0.18 Fig. 2.2 Explain why the movement of the stylus creates an e.m.f. __________________________________________________________ __________________________________________________________ _______________________________________________________ [1] (c) The maximum e.m.f. in the coil is induced at the times marked A (i.e. 0, 1.0 ms and 2.0 ms) in Fig. 2.2. (i) Making reference to the graph in Fig. 2.2 explain briefly why the induced e.m.f. is greatest at these times. ______________________________________________________ ______________________________________________________ ____________________________________________________ [2] 4666 7 [Turn over (ii) The magnitude of the gradient of the tangent to the curve in Fig. 2.2 at the times marked A is 600 degrees per second. This is the magnitude of the angular velocity of the coil at the times marked A. Examiner Only Marks Remark Show that the angular velocity of 600 degrees per second is equal to an angular velocity of 10.5 radians per second. ____________ [1] (iii) It can be shown that the magnitude of the e.m.f. EA generated by the coil at the times marked A is given by EA = BAn A Equation 2.1 where B is the magnetic flux density between the polepieces of the magnet, A is the area of the coil, n is the number of turns in the coil and A is the magnitude of the angular velocity of the coil at the times marked A. The magnetic flux density between the polepieces of the magnet is 14 mT. The coil, which has 8 turns, has an area of 20 mm2. Use Equation 2.1 to calculate the magnitude of the e.m.f. EA generated by the coil at the times marked A. E.m.f. EA = ____________________ V 4666 8 [2] [Turn over BLANK PAGE (Questions continue overleaf) 4666 9 [Turn over 3 The mass spectrograph is a device that uses the principle of the deflection of charged particles in a magnetic field. A plan view of a mass spectrograph is shown in Fig. 3.1. A beam of ions of charge +q and mass m, travelling at a constant speed v, enters a uniform magnetic field of flux density B applied throughout the shaded region which is evacuated. When the beam emerges from the slit, it moves in a semicircular path and leaves a trace on the photographic plate. Examiner Only Marks Remark Positive ion beam Photographic plate Slit Magnetic field Fig. 3.1 (a) (i) State the name of the rule which you would use to determine the direction in which the magnetic field must be applied for the ions to move as shown. ____________________________________________________ [1] (ii) In what direction must the magnetic field be applied to give the deflection shown in Fig. 3.1? Parallel to page Into page Out of page Place an in the box you consider to be the correct direction. [1] 4666 10 [Turn over (iii) Explain why the path of the ions in the magnetic field is semicircular. Examiner Only Marks Remark ______________________________________________________ ______________________________________________________ ____________________________________________________ [2] (b) Find an expression in terms of B, q, m and v for the radius r of the semicircle followed by the ions. r = ________________ [2] (c) The specific charge of an ion is the ratio of the charge of the ion to the mass of the ion. Calculate the radius of the semicircular path if the ions have a specific charge of +5.0 108 C kg 1 and are travelling at a speed of 4.5 105 m s 1. The value of B is 2.3 10 3 T. Radius = _________ cm [2] (d) On Fig. 3.1, sketch the path you would expect if the speed of the ions was increased. Label this path 2. [1] 4666 11 [Turn over 4 Fig. 4.1 shows an arrangement for investigating the penetrating power of different types of radiation. A radioactive source is placed at P and sheets of different materials are placed close to the source. A radiation detector is placed against the material. All air gaps are negligible. Examiner Only Marks Remark material P radiation detector radioactive source Fig. 4.1 In the absence of the source, the count rate of the detector is very small, being the background count rate Rb. The following table summarises the results of a series of experiments with the source in position. Expt Material at A Count rate 1 None R1 where R1>>Rb 2 Thin cardboard R2 where R1>>R2>>Rb 3 Sheet of aluminium 10 mm thick Slightly less than R2 (The symbol >> means very much greater than .) (a) Deduce the type or types of radiation emitted from the source. Explain your reasoning. __________________________________________________________ __________________________________________________________ __________________________________________________________ __________________________________________________________ _______________________________________________________ [4] 4666 12 [Turn over 0 (b) The nuclide 2833 Bi is an alpha particle emitter. An initial measurement of the activity of a sample of this isotope gives a count rate of 1200 counts s 1. After an interval of 24 hours the count rate falls to 290 counts s 1. Both these readings have been corrected for background radiation. Examiner Only Marks Remark 0 (i) Show that the decay constant of 2833 Bi is about 1.6 10 5 s 1. [2] (ii) Calculate the half life of this nuclide. Half life = _____________ hours [1] 0 (iii) Calculate the number N of 2833 Bi nuclei in the sample when the 1. count rate was 1200 s N = __________________ 4666 [2] 13 [Turn over In part (b)(ii) of this question you should answer in continuous prose. You will be assessed on the quality of your written communication. 5 Examiner Only Marks Remark (a) Explain what is meant by the nuclear binding energy of a nucleus. __________________________________________________________ _______________________________________________________ [2] (b) (i) On the axes in Fig. 5.1, sketch a graph to show how the binding energy per nucleon depends on the mass number of the nucleus. 12 Binding energy per 8 nucleon/ MeV per nucleon 4 0 0 50 100 150 200 250 Mass number Fig. 5.1 [2] (ii) Use the graph you have drawn in Fig. 5.1 to write an explanation of why energy may be released in fission reactions. ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ____________________________________________________ [3] Quality of written communication 4666 [2] 14 [Turn over BLANK PAGE (Questions continue overleaf) 4666 15 [Turn over 6 Comprehension question This question contributes to the synoptic assessment requirements of the Specification. In your answer, you will be expected to bring together and apply principles and contexts from different areas of physics, and use the skills of physics, in the particular situation described. You are advised to spend about 45 minutes in answering this question. Read the passage carefully and answer the questions which follow. Extreme diamonds 1 Diamond is an extremely strong structure because each of the carbon atoms is bonded to four others. Other crystals can show the same structure, for example silicon, which is in the same group in the periodic table. It is the structure of diamond and the strength of the bonds which make it so hard. The stable form of carbon at atmospheric pressure is, in fact, graphite, not diamond. 5 The four C s Four things tell you about the quality of a diamond: colour, carat, clarity and cut. Colour Not all diamonds are colourless. Colour is caused by impurities in the crystal. A lot of nitrogen impurities in a diamond give it a yellow colour. The colour of transparent diamonds is rated on a scale from icy white (the highest) to pale yellow (the lowest). We think of diamonds as being colourless the coloured ones, especially red, are rare. 10 Hardness A hard material is defined as one that is not readily scratched or indented. The Mohs scale of hardness ranges from talc at 1 (the softest) up to diamond at 10 (the hardest). 15 Any mineral will scratch all of those which are below it or equal to it and none of those above. Carat The mass of a diamond is measured in carats. One carat (1 ct) = 200.0 milligrams. Clarity Clarity levels indicate whether the diamond has any flaws. The most common flaw is a visible foreign substance, or imperfection, inside the diamond that is called an inclusion. These natural imperfections date from the time when the stone was formed. 20 Cut 25 Diamond has a high refractive index, which means that diamonds can be cut to reflect the light and sparkle more than other gems. Three-quarters of diamonds sold today are round brilliant cut to reflect as much light as possible out of the front face. This is shown in Fig. 6.1. 4666 16 [Turn over i1 i2 Fig. 6.1 Round brilliant cut diamond 30 The future Diamond has another interesting property. If you were to touch a large diamond it would feel cold, as metal does. The heat is conducted quickly away from your hand. This means that it can be used in situations where the environment will get very hot and the diamond can be used to conduct the heat away quickly without causing any damage. Where is this property needed? In the next generation of microprocessors. The current fast microprocessors run at about 200 C. To go faster would mean even higher temperatures, and manufacturers are almost at the limit of what is possible. With diamond wafers replacing silicon microchips, much higher temperatures could be used and heat could be conducted away much faster. In the future, we could be using diamond processors which run at speeds that would heat up and liquefy today s silicon chips. There is another problem. Most diamonds are electrical insulators. In nature only rare blue-grey diamonds can conduct electricity due to impurities in the crystal. To allow other diamonds to conduct electricity, a way of implanting non-carbon atoms into diamond is required. 35 40 45 Examiner Only Marks Remark (a) Write a few words, or a short sentence, to show the meaning of the following words or phrases as they are used in the passage. (i) structure (line 3), ______________________________________________________ ____________________________________________________ [1] (ii) colourless (line 9), ______________________________________________________ ____________________________________________________ [1] (iii) hardness (line 15), ______________________________________________________ ___________________________________________________ [1] 4666 17 [Turn over (iv) clarity (line 21), Examiner Only Marks Remark ______________________________________________________ ___________________________________________________ [1] (v) inclusion (line 23), ______________________________________________________ ___________________________________________________ [1] (vi) refractive index (line 26), ______________________________________________________ ___________________________________________________ [1] (vii) conducted (lines 32, 34), ______________________________________________________ ___________________________________________________ [1] (viii) microprocessor (line 36), ______________________________________________________ ___________________________________________________ [1] (ix) fast (line 36), ______________________________________________________ ___________________________________________________ [2] (x) wafers (line 38), ______________________________________________________ ___________________________________________________ [1] (xi) electrical insulator (line 42). ______________________________________________________ ___________________________________________________ [1] 4666 18 [Turn over (b) The refractive index of diamond is 2.42. Examiner Only Marks Remark (i) Calculate the speed of light in diamond. Speed of light in diamond = _______________m s 1 [2] (ii) Calculate the critical angle for diamond. Critical angle = _________ [2] (iii) Fig. 6.1 shows a way of cutting a diamond so that light is reflected to the front face. Explain how this happens. ______________________________________________________ ____________________________________________________ [2] (iv) The refractive index of window glass is 1.52. Explain why a diamond which is round brilliant cut would reflect more light than a similar round brilliant cut of window glass. ______________________________________________________ ______________________________________________________ ____________________________________________________ [3] (v) One way of measuring the refractive index of diamond is to immerse some small crystals in oil which has a higher refractive index than diamond at room temperatures. The refractive index of this oil at different temperatures is known. The oil is gently heated until the diamond crystals are no longer visible. Suggest why this would happen. ______________________________________________________ ______________________________________________________ ____________________________________________________ [3] 4666 19 [Turn over (c) An electron of an atom of nitrogen trapped in a crystal of diamond might absorb a photon of light and be excited. It would then fall back to a lower energy level and emit a photon corresponding to an energy level difference for the nitrogen atom. This is in the yellow region of the electromagnetic spectrum. This results in a yellow diamond. Examiner Only Marks Remark (i) The wavelength of the yellow light is 575 nm. Calculate the energy in eV of the photon emitted by the nitrogen atom. Photon energy = ___________ eV [4] (ii) A blue diamond is caused by the presence of the impurity boron. Explain why the diamond is called blue . ______________________________________________________ ____________________________________________________ [2] (d) (i) The largest gem quality diamond that has ever been found was discovered in South Africa in 1905 and named the Cullinan diamond. It had a mass of 3106 carats before it was cut. Calculate the mass, in grams, of the Cullinan diamond. Give your answer to an appropriate number of significant figures. Mass = _______________ g [3] (ii) The density of diamond is 3.515 g cm 3. Calculate the volume of the Cullinan diamond. Volume = 4666 cm3 [3] 20 [Turn over (iii) The Cullinan diamond was cut into 105 gems. The largest of these, the Great Star of Africa, has a mass of 530 carats and is part of the Crown Jewels. What percentage of the original Cullinan diamond is the Great Star of Africa? Percentage = __________________ % [2] (e) The specific heat capacity of diamond is 503 J kg 1 C 1. Calculate the heat required to raise the temperature of a diamond of mass 20.0 mg from room temperature of 20 C to its melting point temperature of 3550 C. Heat required = ___________________ J [4] (f) In its rough state, a diamond is fairly unremarkable in appearance. Such diamonds have dull, battered external surfaces. To reveal the hidden beauty of a rough diamond, the dull surface must be removed. (i) What material is used to remove the dull surface of a diamond when in its rough state? ____________________________________________________ [1] (ii) Explain why this material is suitable. ______________________________________________________ ______________________________________________________ ____________________________________________________ [2] THIS IS THE END OF THE QUESTION PAPER 4666 21 Examiner Only Marks Remark 530-056-1 GCE Physics (Advanced Subsidiary and Advanced) Data and Formulae Sheet Values of constants speed of light in a vacuum c = 3.00 108 m s 1 permeability of a vacuum 0 = 4 10 7 H m 1 permittivity of a vacuum 0 = 8.85 10 12 F m 1 1 = 8.99 109 F 1 m 4 0 ( ) elementary charge e = 1.60 10 19 C the Planck constant h = 6.63 10 34 J s unified atomic mass unit 1 u = 1.66 10 27 kg mass of electron me = 9.11 10 31 kg mass of proton mp = 1.67 10 27 kg molar gas constant R = 8.31 J K 1 mol 1 the Avogadro constant NA = 6.02 1023 mol 1 the Boltzmann constant k = 1.38 10 23 J K 1 gravitational constant G = 6.67 10 11 N m2 kg 2 acceleration of free fall on the Earth s surface g = 9.81 m s 2 electron volt 1 eV = 1.60 10 19 J A2Y2INS 4666.02 USEFUL FORMULAE The following equations may be useful in answering some of the questions in the examination: Thermal physics Mechanics Momentum-impulse relation mv mu = Ft for a constant force Average kinetic energy of a molecule 1 m<c2> 2 Power P = Fv Kinetic theory pV = 1 Nm <c2> 3 Conservation of energy 1 mv 2 2 1 mu 2 = Fs 2 for a constant force Simple harmonic motion Displacement x = x0 cos t or x = x0 sin t Velocity v = x 0 2 x 2 Simple pendulum T = 2 l / g Loaded helical spring T = 2 m / k Medical physics Sound intensity level/dB = 10 lg10(I/I0) Sound intensity difference/dB = 10 lg10(I2/I1) Resolving power sin = / D Waves Capacitors Capacitors in parallel 11 1 1 = + + C C1 C 2 C 3 C = C1 + C2 + C3 Time constant = RC Capacitors in series Electromagnetism Magnetic flux density due to current in (i)i long straight (i)i solenoid B= (ii) long straight (i)i conductor B= = ay/d Diffraction grating 0I 2 a A.c. generator E = E0 sin t = BAN sin t Stress and Strain Hooke s law F = kx Strain energy E = <F > x (= 1 Fx = 1 kx 2 2 2 if Hooke s law is obeyed) Electricity Vout = R1Vin/(R1 + R2) A = N A = A0e t t1 = 0.693/ 2 Photoelectric effect 1 mv2 = max 2 de Broglie equation 1/u + 1/v = 1/ f Radioactive decay Half life Light 4666.02 l Alternating currents d sin = n Potential divider 0NI Particles and photons Two-slit interference Lens formula = 3 kT 2 = h /p Particle Physics Nuclear radius 1 r = r0 A3 hf hf0

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Additional Info : Gce Physics May 2008 Assessment Unit A2 2, Module 5: Electromagnetism and Nuclear Physics
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