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FinalFa07Solutions

# FinalFa07Solutions - Final Examination Physics 115A Fall...

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Final Examination December 10, 2007 Physics 115A, Fall 2007 E. Abers Final Examination Solutions and Comments Question 1 [20 Points] Relativistic Kinematics Suppose a μ - particle (mass m μ ) at rest decays into an electron (mass m e ) and a photon. 1 Let ω be the angular frequency of the photon. Let p be the (magnitude of the) momentum of the electron. Part a) Write the energy conservation law in terms of m μ , m e , ω , p , and possibly planckover2pi1 and c . Part b) Write the momentum conservation law in terms of these quantities. Part c) What is p in terms of m μ , m e , and possibly planckover2pi1 and c (but not ω )? Don’t bother putting in the numbers in this problem. GO ON TO THE NEXT PAGE SOLUTION Part a) m μ c 2 = radicalbig m 2 e c 4 + p 2 c 2 + planckover2pi1 ω Part b) p = planckover2pi1 ω c Part c) m μ c 2 = radicalbig m 2 e c 4 + p 2 c 2 + cp m μ c 2 - cp = radicalbig m 2 e c 4 + p 2 c 2 m 2 μ c 4 - 2 m μ c 3 p = m 2 e c 4 p = c m 2 μ - m 2 e 2 m π 1 This decay has been looked for but never observed.

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2 Final Examination Question 2 [20 Points] The Hydrogen Atom A hydrogen atom at rest decays from the n = 3 level to the n = 1 level by emitting a photon. (This is the Lyman- β line.) Usually we ignore the recoil energy of the atom, the kinetic energy of the atom after the decay. Write a formula (not a numerical value) for the ratio of the recoil kinetic energy of the final atom to the energy of the emitted photon. Make any approximations you want that introduce errors of 1% or smaller. Use non- relativistic kinematics for all particles except the photon – the recoil energy is small. Note: The ratio of these two energies is dimensionless . Write your answer in the form Numerical constant × other dimensionless factors × α some power × parenleftbigg electron mass atomic mass parenrightbigg some power where α is the fine structure constant. GO ON TO THE NEXT PAGE SOLUTION The energy of the transition is E 3 - E 1 = α 2 mc 2 2 parenleftbigg 1 - 1 9 parenrightbigg = 4 9 α 2 mc 2 so energy conservation requires planckover2pi1 ω = 4 9 α 2 mc 2 - planckover2pi1 2 ω 2 2 Mc 2 (1) The last term is much smaller than planckover2pi1 ω , and so to a good approximation planckover2pi1 ω = 4 9 α 2 mc 2 and p = planckover2pi1 ω c = 4 9 α 2 mc The momentum of the photon is p = planckover2pi1 ω c By momentum conservation this is the momentum of the recoiling atom also. Its kinetic energy is therefore T = p 2 2 M = planckover2pi1 2 ω 2 2 Mc 2 where M is the atomic (not the electron) mass. The ratio of this energy to the photon energy is planckover2pi1 ω 2 Mc 2 = 1 2 Mc 2 4 9 α 2 mc 2 = 2 9 α 2 m M
Physics 115A, Fall 2007 3 Note: If you want to do better, you can solve the quadratic equation (1) for planckover2pi1 ω exactly, or find a power series for the frequency of the photon, for example by by successive approximations: planckover2pi1 ω = 4 9 α 2 mc 2 - 1 M 8 81 α 4 m 2 c 2 + · · · = 4 9 α 2 mc 2 1 + 2 9 α 2 m M + · · · « From the value of the ratio in equation (2) you see that the error is indeed very small.

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