CHAPTER 44 - 9 - 43. Because7.0TeVm c ,wehave p 2 K=E=pc.

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  ν e 43. Because 7.0 TeV »  m p c 2 , we have K  =  E  =  pc . We use the result from Problem 13: E (eV) =  Brc ; (7.0 TeV)(10 12  eV/TeV) =  B (4.25 × 10 3  m)(3.00 × 10 8  m/s), which gives  B  =        5.5 T . 44. ( a ) We assume the momentum of the electron-positron pair is zero.  The released energy is E  = 2 m e c 2  = 2(0.511 MeV) =       1.022 MeV . ( b ) We assume the momentum of the proton-antiproton pair is zero.  The released energy is E  = 2 m p c 2  = 2(938.3 MeV) =       1876.6 MeV . 45. ( a ) For the reaction p  + p   K +  +    Σ , the conservation laws are Charge:     – 1 + 1 = + 1 – 1; Spin: 0 +  !  = 0 +  ! ; Baryon number: 0 + 1 = 0 + 1; Lepton number: 0 + 0 = 0 + 0; Strangeness: 0 + 0 = + 1 – 1. Thus the reaction is       possible, through the strong interaction . ( b
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CHAPTER 44 - 9 - 43. Because7.0TeVm c ,wehave p 2 K=E=pc.

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