PHYC10004 SCHOOL OF PHYSICS SESSION 1 NOVEMBER 20, 2020 EXAMINATION QUESTION PAPER 1 Write your ID at the top of each page of your solutions STUDENT ID SUBJECT NAME: PHYSICS 2 PHYSICAL SCIENCE AND TECHNOLOGY SUBJECT CODE: UPLOADING TIME: PHYC10004 3 hours (180 MINS) NOT including reading or uploading time. Estimate 1 hour (60 MINS) Canvas Quiz, 2 hours (120 MINS) Long Answer UP TO 30 MINS (15 MINS FOR CANVAS QUIZ, 15 MINS FOR LONG ANSWER) 30 MINS (LONG ANSWER) TOTAL MARKS: 150 marks (50 marks Quiz, 100 marks Long Answer) NUMBER OF QUESTIONS: 29 QUIZ QUESTIONS AND 8 GRADESCOPE QUESTIONS TOTAL EXAM DURATION: READING TIME: AUTHORISED MATERIALS 1. 2. 3. 4. 5. Writing materials, stationery, rulers, and other implements Calculator Textbooks, lecture notes Resources on the Learning Management System (LMS) for PHYC10004 Mobile phone with camera or other apparatus for scanning your solutions INSTRUCTIONS TO STUDENTS 1. The conduct of the exam is governed by the rules of the University of Melbourne 2. Apart from the LMS and the requirements of scanning and uploading your solutions, access to the internet and communication with others, except the examiners, is not permitted during the exam 3. Write your solutions on blank plain or ruled sheets of paper 4. Start each question on a new page 5. Write your student id number at the top of each page of your solutions 6. Write your start time and finish time on the first page of your solutions and write your signature on the next line to certify this is a true record consistent with the honour code 7. At the conclusion of the writing time submit your solutions as a single PDF using the appropriate app 8. To ensure your submission is not late, you must access the exam at the designated time 9. Note that ALL answers must be submitted for all parts of the exam within the 240 minute window. Answers submitted after this time will not be accepted. Leave sufficient time to upload all of your solutions. 10. Evidence of technical difficulties should be recorded and reported as soon as possible. This will be considered for late submissions. 11. Any questions or technical difficulties can be addressed to; I. ph-firstyear-exam@unimelb.edu.au II. Exam Support Link on the LMS III. Lecturers Chris Chantler and Roger Rassool Question 1 – Gauss 25 Marks Consider Gauss’s Law for electric fields. a) Briefly (about 1 page) explain what the symbols in the equation for Gauss’ Law mean, especially how the integral is defined [see Formula sheets]. Also explain what charge q is referring to on the right-hand side. b) A point positive charge, +Q, is placed at the centre of a spherical cavity of radius R formed in a large block of conducting material (assume the conducting material goes from r=R to r= infinity). Use Gauss’ Law to find the magnitude and direction of the electric field at a distance i. ii. iii. r = R/2 (inside the cavity) r = R (at the edge between the cavity and the conducting material) r > R from the centre of the cavity (in the conductor). c) Two long concentric conducting cylinders of radius a and b, respectively, each carry equal and opposite charges, with a linear charge density λ C m-1. The cylinders are of negligible thickness. Assuming that b > a, use Gauss's law i. to show that E vanishes for r < a and r > b ii. to derive E for a < r < b, iii. to describe the direction of the electric field in the space between the cylinders [7 + (3 + 3 + 3) + (3 + 4 + 2)] = 25 marks] 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 1 of 12 Question 2 Rupert Bear and the Castle Trap 25 marks Rupert Bear is one of the great UK institutions and one of the first popular text comics, with the first comic in 1920. Rupert is a bear, living with his parents in an idyllic English town of Nutwood, but his adventures take him all over the world (just like physics without COVID!). It is a precursor of the Belgian and French TinTin [from 1929] and the French Asterix comics [from 1959] and there are still annuals every year. In this adventure Rupert meets up with his friend Bingo, the brainy pup, to help rescue the kidnapped Squire and foil some crooks with a metal-tracer machine. Without electricity and magnetism, they could not have been successful. a) We needed to rewrite Ampere’s Law to include a time-dependent term. Explain the revised equation, what displacement current means, how it relates to a time-varying electric flux. Inside a material, what values should permeability and permittivity have? b) Faraday’s Law showed another time-dependent effect. Explain what it is and all elements of the equation. c) Time-dependent magnetic flux can arise from several alternate experimental components changing with time. Explain three. 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 2 of 12 d) Bingo finds the trowel Rupert’s father has lost his trowel in the garden, so Bingo and Rupert find it using the metal-tracer. Explain how the metal detector can detect the metal of a trowel buried just under the surface. From lectures, and demonstrations, discuss the characteristic of the trowel which the metal detector can detect, knowing that the metal detector has a big magnetic field which is either static or moving. e) Rupert and the metal detector The metal detector can detect iron (in a trowel). Is it likely to be more sensitive to silver and gold if the object is of the same size. Why? 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 3 of 12 f) The French Trouve made a metal detector in 1874, and Alexander Graham Bell (the telephone guy) made one in 1881. They were used in Admiral Byrd’s second Antarctic expedition in 1933 and could detect objects 8 feet underground. They were used by the Allies to detect mines in the second world war. Many prospectors look for gold, jewellery or archaeological object like ancient coins or bullets with Bingo’s metaltracer. Sometimes these lie well below the surface. How can the machine be sensitive enough to detect these? g) By analogy with Bingo’s metal-tracer, explain the demonstration where a bar magnet was falling through an aluminium pipe. What happened, and why? [6 + 6 + 3 + 3 + 3 + 2 + 2 = 25 marks] 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 4 of 12 Question 3 Blackbody Radiation 8 marks A glowing candle, giving off a warm “yellowish” glow, can be considered a blackbody radiator of T=800 K. In comparison the sun which can also be modelled as a black body radiator of temperature of T=5500 K produces a much “whiter” spectrum of light. a) With the aid of a sketched graph, compare the blackbody emission spectrum of the sun and a candle. Label your axis and highlight the key points of difference. b) Assuming the glowing radiating part of the candle has an area of approximately 10-3 m², estimate the power output of the candle. Outline any assumptions you have made in your calculation. The ideal spectrum of visible sunlight is shown in the diagram below. Upon closer examination of the solar spectrum as measured at the surface of the earth reveals very fine structure and specific dark bands. c) Briefly explain one likely cause of these variations in your answer include a specific statement about the actual process which is taking place to cause these differences. [ 3 + 3 + 2 = 8 Marks] Question 4 Heisenberg Uncertainty Principle 10 marks a) Explain what is meant by the Heisenberg Uncertainty Principle (βpβx > h/4π) In the Bohr model, an electron in the ground state of a hydrogen atom is considered to have a kinetic energy of 13.56 eV. b) Show that the momentum of this electron is approximately 2 x10-24 kg.m.s-1. c) Use the Uncertainty Principle to estimate the diameter of the hydrogen atom. Make sure you explain the reasoning behind your estimate. [ 3 + 3 + 4 = 10 Marks] 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 5 of 12 Question 5 Bohr Model 7 marks Data: Z=1 for H, Hydrogen, Z=2 for He, Helium and Z=3 for Li, Lithium. According to the Bohr model, the energies of a hydrogen like atom or an ion which has one electron orbiting a positively charged nucleus are given by ππ 2 ππππ 4 1 πΈπΈππ = − 2 2 . 2 8ππππ β ππ where ‘n’ is the principal quantum number. An energy diagram of the quantum states for the H atom is shown in the figure above. a) Write down the value of the of the term: ππ 2 ππππ 4 8ππππ2 β2 1 . ππ2 (Include units in your answer) When H atoms are excited in a gas discharge tube, they may exist briefly in a state with n ≥ 2. As these atoms then de-excite to a lower quantum state they will emit radiation. Recall from lectures that the Balmer series, which is emitted when electrons in an excited state drop to the n = 2 state contains four lines that are in the visible region of the spectrum. b) What is the lowest energy of a photon of visible light emitted in the Balmer series of hydrogen. Atoms of helium have Z = 2. If one of the electrons in this atom is removed, it is called a singly ionised He ion, namely He+. c) What is the binding energy of the He+? Explain how you arrived at your answer. [ 2 + 2 + 3 = 7 Marks] 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 6 of 12 Question 6 Ionisation 8 marks The figure below shows the binding energy i.e. the energy required to remove the least strongly bound electron of the light elements up to F, (Fluorine, Z=9). a) Briefly explain the general trends indicated by the straight lines on this graph Include in your answer i. The reasons for the large difference in ionization energies of He and Li ii. The reason that the ionization energies gradually increase from Li to F. b) Estimate the ionization energy of the next element in the series Ne (Neon, Z =10). c) Provide an approximate value for the ionization energy of Na (Sodium Z=11) and explain your reasoning. [ (2 + 2) + 2 + 2 = 8 Marks] 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 7 of 12 Question 7 X-rays 11 marks The spectrum of x-rays produced when an intense beam of energic electrons is directed onto a Molybdenum (Mo) target is shown below. Note that there are two distinct peaks, one at 63 pm, the other at 72 pm. a) Briefly explain how the scattering of these electrons can produce the features of the spectrum. Include in your explanation a reference to the formation of the distinct peaks and the sharp cut-off at around 35 pm. b) Calculate the energy of the incident electrons. c) Assuming that the two peaks are the results of K transitions in Mo, estimate the separation in the energies of the L and M shell in Mo. d) Consider now the case where the energy of incident electrons is now doubled. On a clearly labelled diagram with axes similar to that above, sketch the new spectrum of x-rays you would expect these electrons to produce. [ 3 + 1 + 3 + 4 = 11 Marks] 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 8 of 12 Question 8 Characteristic X-rays 6 marks Moseley observed that characteristics x-rays from an element were unique. By carefully studying a large number of elements, he showed a linear relationship between the square root of the frequency of emission and an element’s position in the periodic table. Ultimately, Moseley showed that the energy of the πΎπΎπΌπΌ x-rays produced by any element satisfy the empirical relation: πΈπΈπΎπΎπΌπΌ = βππ = 13.56ππππ(ππ − 1)2 οΏ½ ~ ππππππππππ. × ππ 2 1 1 − 2οΏ½ 2 1 2 With reference to the Bohr model, explain how one may interpret the terms in this equation. For full marks, make sure you include a rationale for the “constant × (Z-1)2” term and the quantities in the square brackets. [6 Marks] 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 9 of 12 End of Exam Formula Sheet over page 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 10 of 12 Formula Sheet and Useful Data for PHYC10004 (may be detached) Page 1 of 2 Constants Mass of electron, me 9.109 x 10-31 kg = 0.00055 u Mass of proton, mp 1.673 x 10-27 kg = 1.00728 u Mass of neutron, mn 1.675 x 10-27 kg = 1.00866 u Electron charge magnitude, e 1.6 x 10-19 C Permeability of free space, µo 4π x 10-7 T m A-1 Permittivity of free space, εo 8.85 x 10-12 C2 N-1 m-2 Bohr radius, aB 0.529 x 10-10 m Universal gas constant, R 8.31 J K-1 mol-1 Avogadro’s number, NA 6.02 x 1023 mol-1 Specific heat capacity of ice 2220 J K-1 kg-1 Specific heat capacity of water 4187 J K-1 kg-1 Atomic mass unit u 1.6605 x 10-27 kg = 931.49 MeV/c2 -19 1 eV = 1.6 x 10 J Speed of light in vacuum, c 299 792 458 m s-1 Gravitational acceleration, g 9.8 m s-2 Planck’s constant, h = 6.63 x 10-34 J s = 4.14 x 10-15 eV s Rydberg constant, R 1.097373 x 107 m-1 Boltzmann constant, k 1.38 x 10-23 J K-1 Stefan-Boltzmann constant, σ 5.67 x 10-8 W m-2 K-4 1 atm = 101325 Pa = 760 mm Hg Latent heat of fusion for water 333 kJ kg-1 Density of water 1000.0 kg/m3 Formulae Electromagnetism ππ1 ππ2 ππ = ππΜ 4ππππππ ππ 2 ππ = −βππ οΏ½ π¬π¬. πππ¨π¨ = ππππππππ ππππ ππ = ππππ πΆπΆ = ππ ππ ππ = π π ππ0 ππ = ππππ × π©π© π΅π΅ = ππππ ππππ ππ0 ππππππ × πποΏ½ 4ππ ππ 2 ππΦπΈπΈ πππ·π· = ππππ ππππ πππ©π© = ππ ππ ππ π¬π¬ = πποΏ½ 4ππππππ ππ 2 π¬π¬ = ∞ ππ(ππ) = οΏ½ π¬π¬. ππππ = ππ ππ 4ππππππ ππ ππ = πππ π ππ = ππ × π¬π¬ ππ = −ππ. π¬π¬ ππππ ππππ = ππππ ππππ π΄π΄ππππ πΆπΆ = ππ ππππ ππ = = πππππππ£π£ππ ππππ ππ = ππ1 βππ = − οΏ½ π¬π¬. ππππ ππ2 ππ 2 ππ = ππππ = = ππ 2 π π π π 1 1 2 1 ππ 2 ππ = ππππ = πΆπΆππ = 2 2 2 πΆπΆ 1 πππΈπΈ = 2π π ππππ πΈπΈ 2 ππ = ππππ × π©π© ΦπΈπΈ = οΏ½ π¬π¬. πππ¨π¨ ππ = οΏ½ π¬π¬. ππππ = − ππΦπ΅π΅ ππππ ππ = ππ × (ππππππππ ππππ ππππππππ) ππ(ππ) ππ οΏ½ π©π©. πππ¨π¨ = 0 Φπ΅π΅ = οΏ½ π©π©. πππ¨π¨ οΏ½ π©π©. ππππ = ππππ ππππππππ + ππππ πππ·π· 2020 PHYC10004 Physics 2: Physical Science and Technology Exam ππ = ππ × π©π© ππ = −ππ. π©π© Page 11 of 12 Formula Sheet and Useful Data for PHYC10004 Page 2 of 2 Fluids and Thermal Energy π π = π΄π΄1 π£π£1 = π΄π΄2 π£π£2 ππππ = ππππππ 1 ππ + ππππβ + πππ£π£ 2 = ππππππππππ 2 ππ = πππ΄π΄π΄π΄π΄π΄ + ππππβ ππ = ππππ πΎπΎ = ππππππππππ ππ ππ πΎπΎ = ππππ πππ£π£ = 2 π π ππ ππππππ 2 ππ = πππΆπΆππ Δππ ππ = ππ ππππ = πππ£π£ + π π ππ = − οΏ½ ππππππ = −ππβππ Modern Physics ππππππππ ππ = 2898 ππππ πΎπΎ ππ = ππππππππ 4 β ππ β2 2 πΈπΈππ = ππ 8πππΏπΏ2 ππ = ππππ 4 ππ 2 13.6ππ 2 πΈπΈππ = − 2 2 2 = − ππ2 8ππ0 β ππ 1 1 2 − 2οΏ½ ππ2 ππ1 πΏπΏ = οΏ½ππ(ππ + 1)β πΏπΏπ§π§ = ππππ β ππ = ππππ οΏ½ ππ(π‘π‘) = ππ0 ππ −ππππ ππ ππ ππ = −ππππππln π’π’ππ (ππ) = πππ΅π΅ = ππππ ππππ ππ = ππππ log e 2 ππ1⁄2 ππ π π ππ (ππ) = π’π’ππ (ππ) 4 πΈπΈ = βππ = βππ ππ β (1 − cos ππ) ππππ 2ππ sin ππ = ππππ Δπ₯π₯Δππ ≥ 1 ππππππππππ = (1.2 × 10−15 ππ)π΄π΄3 ππ = ππ = ππππ ππ = ππππβππ βππ = 4ππππππ β2 2 ππ = πππ΅π΅ ππ2 ππππ 2 πππ§π§ = πππ π β 4√2ππ (ππ/ππ)ππ 2 3ππππ π£π£π π π π π π = οΏ½ ππ 8ππππβ 1 οΏ½ βππ οΏ½ 5 ππ ππ ππππππ − 1 ππππ β2 = 52.92 ππππ ππππππ 2 1 βππ = ππ + ππ βππ = πΎπΎππππππ + ππ ππππ = ππ = πΌπΌππππ β 2 ππππ = πππΌπΌππππ 1 ππ ππ 2 1 = ≈ 4ππππππ βππ 137 π΅π΅. πΈπΈ. = (π΄π΄ − ππ)ππππ + ππππππ − ππππππππππππππππ ππ 2 2020 PHYC10004 Physics 2: Physical Science and Technology Exam Page 12 of 12
