Particle Physics - General Physics - Solved Exam, Exams of Physics

Download Particle Physics - General Physics - Solved Exam and more Exams Physics in PDF only on Docsity! Exam Questions mass of proton = 1.6730 × 10-27 kg; mass of electron = 9.1 × 10–31 kg; mass of lithium nucleus = 1.1646 × 10-26 kg; mass of α-particle = 6.6443 × 10-27 kg; mass of neutron = 1.6749 × 10–27 kg; charge on electron = 1.6022 × 10–19 C; mass of pion = 2.4842 × 10–28 kg; speed of light, c = 2.9979 × 108 ms-1; Planck constant = 6.626 × 10-34 J s Particle Accelerators, Cockroft and Walton Experiment and E = mc2 1. [2009] In 1932 Cockcroft and Walton succeeded in splitting lithium nuclei by bombarding them with artificially accelerated protons using a linear accelerator. Each time a lithium nucleus was split a pair of alpha particles was produced. (i) How were the protons accelerated? (ii) How were the alpha particles detected? 2. [2005] High voltages can be used to accelerate alpha particles and protons but not neutrons. Explain why. 3. [2009] Most of the accelerated protons did not split a lithium nucleus. Explain why. 4. [2002] In 1932, Cockcroft and Walton carried out an experiment in which they used high-energy protons to split a lithium nucleus. Outline this experiment. 5. [2007] (i) Draw a labelled diagram to show how Cockcroft and Walton accelerated the protons. (ii) What is the velocity of a proton when it is accelerated from rest through a potential difference of 700 kV? 6. [2009] [2007] [2005][2002] Write a nuclear equation to represent the splitting of a lithium nucleus by a proton. 7. [2009] [2007][2002] Calculate the energy released in this reaction. 8. [2005][2009] Circular particle accelerators were later developed. Give an advantage of circular accelerators over linear accelerators. 9. [2004] In beta decay, a neutron decays into a proton with the emission of an electron. Write a nuclear equation for this decay. 10. [2004 Calculate the energy released during the decay of a neutron. The neutrino 11. [2008] The existence of the neutrino was proposed in 1930 but it was not detected until 1956. Give two reasons why it is difficult to detect a neutrino. 12. [2007] In beta decay it appeared that momentum was not conserved. How did Fermi’s theory of radioactive decay resolve this? 13. [2004] Momentum and energy do not appear to be conserved in beta decay. Explain how the existence of the neutrino, which was first named by Enrico Fermi, resolved this. Antimatter 14. [2007] Compare the properties of an electron with that of a positron. 15. [2007] What happens when an electron meets a positron? 16. [2003] Give one contribution made to Physics by Paul Dirac. Pair Production 17. [2005] In an accelerator, two high-speed protons collide and a series of new particles are produced, in addition to the two original protons. Explain why new particles are produced. 18. [2009] Cockcroft and Walton’s apparatus is now displayed at CERN in Switzerland, where very high energy protons are used in the Large Hadron Collider. In the Large Hadron Collider, two beams of protons are accelerated to high energies in a circular accelerator. The two beams of protons then collide producing new particles. Each proton in the beams has a kinetic energy of 2.0 GeV. (i) Explain why new particles are formed. (ii) What is the maximum net mass of the new particles created per collision? 19. [2008] (i) In a circular accelerator, two protons, each with a kinetic energy of 1 GeV, travelling in opposite directions, collide. After the collision two protons and three pions are emitted. What is the net charge of the three pions? Justify your answer. (ii) Calculate the combined kinetic energy of the particles after the collision. (iii) Calculate the maximum number of pions that could have been created during the collision. 20. [2003] The following reaction represents pair production: γ → e+ + e– Calculate the minimum frequency of the γ-ray photon required for this reaction to occur. 21. [2003] What is the effect on the products of a pair production reaction if the frequency of the γ-ray photon exceeds the minimum value? Pair Annihilation 22. [2006] During a nuclear interaction an antiproton collides with a proton. A huge collection of new particles was produced using circular accelerators. The quark model was proposed to put order on the new particles. List the six flavours of quark. 40. [2005] Give the quark composition of the proton. 41. [2003] Leptons, baryons and mesons belong to the “particle zoo”. Give (i) an example, (ii) a property, of each of these particles. Exam Solutions 1. (i) They were accelerated by the very large potential difference which existed between the top and the bottom (ii) They collide with a zinc sulphide screen, where they cause a flash and get detected by microscopes. 2. Alpha particles and protons are charged, neutrons are not. 3. The atom is mostly empty space so the protons passed straight through. 4. • Protons are produced and released at the top of the accelerator. • The protons get accelerated across a potential difference of 800 kVolt. • The protons collide with a lithium nucleus at the bottom, and as a result two alpha particles are produced. • The alpha particles move off in opposite directions at high speed. • They then collide with a zinc sulphide screen, where they cause a flash and get detected by microscopes. 5. (i) See diagram. (ii) P.E. = K.E. qV = ½ mv2 v2 = 2qV/m v2 = 2(1.6022 × 10-19)(7.00 × 105)/ 1.6726 × 10-27 v = 1.16 × 107 m s-1 6. 11H + 7 3Li → 4 2 4 2 HeHe + + K.E. 7. Loss in mass: Mass before = mass of proton (1.6726 × 10–27) + mass of lithium nucleus (1.1646 × 10–26) = 1.33186 × 10-26 kg Mass after = mass of two alpha particles = 2 × (6.6447 × 10–27) = 1.32894 × 10-26 kg Loss in mass = 1.33186 × 10-26 - 1.32894 × 10-26 = 2.92 × 10-29 kg E = mc2 = (2.92 × 10-29) (2.9979 × 108)2 = 2.6 × 10-12 J 8. Circular accelerators result in progressively increasing levels of speed/energy and occupy much less space than an equivalent linear accelerator. 9. 10. Mass lost = mass before – mass after = (mass of neutron) – [(mass of proton + electron)] = (1.6749 × 10–27) – [(1.6726 × 10–27 + 9.1094 × 10–31)] = 1.3891 × 10-30 kg E = mc2 = (1.3891 × 10-30)(2.9979 × 108)2 = 1.25 × 1013 J 11. No charge and very small mass. 12. Fermi (and Pauli) realised that another particle must be responsible for the missing momentum , which they called the neutrino. 13. Momentum and energy are conserved when the momentum and energy of the (associated) neutrino are taken into account. 14. Both have equal mass / charges equal / charges opposite (in sign) / matter and anti-matter 15. Pair annihilation occurs. 16. Dirac predicted antimatter. 17. The kinetic energy of the two protons gets converted into mass. 18. (i) When the protons collide into each other they lose their kinetic energy and it is this energy which gets converted into mass to form the new particles. (ii) Total energy = 4 GeV E = mc2 ⇒ m = E/ c2 ⇒ m = (4 × 109) (1.6 × 10-19)/(2.9979 × 108)2 ⇒ m = 7.121 × 10-27 kg 19. (i) Zero, because electric charge must be conserved. (ii) Energy equivalent of a pion:) E = mc2 E = (2.4842 )( 2.9979 × 108)2 E = 2.2327 × 10–11 J = 1.3935 × 108 eV For 3 pions E = 6.6980 × 10–11 J = 4.18047 × 108eV Energy after collision = (2 × 109) - (4.18047 × 108) = 1.58195 × 109 eV = 2.535 × 10– 10 J (iii)Number of pions = (1.58195 × 109) / 1.3935 × 108 = 11.3524 = 11 pions. Maximum number of pions = 3 + 11 = 14 pions. 20. E = (2)mc2 = hf 2(9.1 × 10–31)( 3.0 × 108)2 = (6.6 × 10–34)f ⇒ f = 2.5×1020 Hz 21. The electrons which were created would move off with greater speed. There may also be more particles produced. 22. (i) A photon is a discrete amount of electromagnetic radiation. (ii) m [= mass of proton + mass of antiproton ] = 2(1.673 × 10-27) = 3.346 × 10-27 kg E = mc2 = (3.346 × 10-27 )(2.998 × 108)2 = 3.0074 × 10-10 Energy for one photon = 1.5037 × 10-10 J E = hf ⇒ f = E/h / = 1.5037 × 10-10 / 6.626 × 10-34 = 2.2694 × 1023 Hz (iii)So that momentum is conserved. (iv) They move in opposite directions. (v) Total charge before = +1-1 = 0 Total charge after = 0 since photons have zero charge (vi) The energy of the photons is converted into matter. 23. (i) e+ + e- → 2γ (ii) Total charge on both sides is zero Momentum of positron + electron = momentum of photons 24. Strong (short range), Weak (short range), Gravitational (infinite range), Electromagnetic (infinite range). 25. Electromagnetic, strong, weak , gravitational 26. The strong nuclear force. 27. Strong: acts on nucleus/protons + neutrons/hadrons/baryons/mesons, short range Gravitational: attractive force, inverse square law/infinite range, all particles Electromagnetic: acts on charged particles, inverse square law/infinite range 28. Gravitational, Electromagnetic, Strong (nuclear), Weak (nuclear) 29. Strong