Copy of 2011-spring - Illinois Institute of Technology Physics M.Sc Comprehensive and Ph.D Qualifying Examination PART I Thursday 4:00 8:00 PM General

# Copy of 2011-spring - Illinois Institute of Technology...

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Illinois Institute of Technology Physics M.Sc. Comprehensive and Ph.D. Qualifying Examination PART I Thursday January 13, 2011 4:00 - 8:00 PM General Instructions 1. Each problem is to be done on a separate piece of paper. Label the front of each page with your identifying code letter, the part number of the exam, and the number of the problem only; for example: A-I.6. Do not write your name on any material handed in for grading. 2. Any numerical data not specified in a problem should be found in the table of constants at the front of the exam. 3. DON’T PANIC: It is not expected that each student will solve every problem. Rather, it is advisable to do a thorough job on those problems that you do solve. 1 Physical Constants Speed of light in vacuum c = 2 . 998 × 10 8 m/s Planck’s constant h = 6 . 626 × 10 34 J · s planckover2pi1 = h/ 2 π = 1 . 055 × 10 34 J · s = 6 . 582 × 10 16 eV · s Permeability constant μ o = 4 π × 10 7 N/A 2 Permittivity constant 1 4 πǫ o = 8 . 988 × 10 9 N · m 2 /C 2 Fine structure constant α = e 2 4 πǫ o planckover2pi1 c = 7 . 30 × 10 3 = 1 137 Gravitational constant G = 6 . 67 × 10 11 m 3 /s 2 · kg Avogadro’s number N A = 6 . 023 × 10 23 mole 1 Boltzmann’s constant k = 1 . 381 × 10 23 J/K = 8 . 617 × 10 5 eV/K kT at room temperature k · 300 K = 0.0258 eV Universal gas constant R = 8 . 314 J/mole · K Stefan-Boltzmann constant σ = 5 . 67 × 10 8 W/m 2 · K 4 Electron charge magnitude e = 1 . 602 × 10 19 C Electron rest mass m e = 9 . 109 × 10 31 kg = 0.5110 MeV/c 2 Neutron rest mass m n = 1 . 675 × 10 27 kg = 939.6 MeV/c 2 Proton rest mass m p = 1 . 672 × 10 27 kg = 938.3 MeV/c 2 Deuteron rest mass m d = 3 . 343 × 10 27 kg = 1875.6 MeV/c 2 Atomic mass unit (C 12 = 12) u = 1 . 661 × 10 27 kg = 931.5 MeV/c 2 Mass of earth M E = 5 . 98 × 10 24 kg Radius of earth R E = 6 . 37 × 10 6 m Mass of sun M S = 1 . 99 × 10 30 kg Radius of sun R S = 6 . 96 × 10 8 m Gravitational acceleration at earth’s surface g = 9.81 m/s 2 Atmospheric pressure = 1 . 01 × 10 5 N/m 2 Radius of earth’s orbit = 1 . 50 × 10 11 m Radius of moon’s orbit = 3 . 84 × 10 8 m Conversion Factors 1 eV = 1 . 602 × 10 19 J 1 J = 6 . 242 × 10 18 eV 1 ˚ A = 10 10 m 1 Fermi = 10 15 m 1 barn (b) = 10 28 m 2 1 in = 2.54 cm 0 Celsius = 273.16 K 1 cal = 4.19 J 2 Problem 1: CP violation was discovered in the decays of Kaons by Cronin and Fitch in 1964, where they noticed K 0 L could decay to either three or two pions (a different CP eigenstate). CP violation was observed again in Kaons when is was observed the rate for K 0 L π e + ν e and K 0 L π + e ¯ ν e differ. Given m K 0 L = 500 MeV, m π ± = 140 MeV, and the electron and neutrino are effectively massless: (a) What is the maximum energy of the electron when a K 0 L decays at rest?  #### You've reached the end of your free preview.

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• Spring '08
• Laurent-Muehleisen
• Physics, Electron, Mass, Atomic mass unit, Illinois Institute of Technology, Physical constant
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