CHAPTER 44 - 1

# CHAPTER 44 - 1 - Note E=hf=hc =(6.63 10 3 34 Js(3 10 m/s(10...

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CHAPTER 44 – Elementary Particles Note: A useful expression for the energy of a photon in terms of its wavelength is E  =  hf  =  hc / λ  = (6.63 × 10 –34  J     s)(3 × 10 8  m/s)(10 9  nm/m)/(1.60 × 10 –19  J/eV) ; E  = (1.24 × 10 3  eV     nm)/  = (1.24 × 10 –12  MeV     m)/ . 1. The total energy of the proton is E  =  K  +  m p c 2  = 6.35 GeV + 0.938 GeV =       7.29 GeV . 2. The total energy of the electron is E  =  K  +  m e c 2  = 35 GeV + 0.511 MeV = 35 GeV. Because the mass is negligible, the momentum is  p  =  E / c . We find the wavelength from  =  h / p  =  hc / E   = (1.24 × 10 –12  MeV     m)/(35 GeV)(10 3  MeV/GeV) =       3.5 × 10 –17  m . 3. We find the magnetic field from the cyclotron frequency: f  =  qB /2p m ; 2.8 × 10 7  Hz = (1.60 × 10 –19  C) B /2p(1.67 × 10 –27  kg), which gives

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## This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at Berkeley.

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CHAPTER 44 - 1 - Note E=hf=hc =(6.63 10 3 34 Js(3 10 m/s(10...

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