CHAPTER 44 - 3

CHAPTER 44 - 3 - 11 n=?E/2eV=(900GeV8.0GeV(10...

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11. The number of revolutions is n  = ? E /2 eV  = (900 GeV – 8.0 GeV)(10 3  MeV/GeV)/(1 e)(2.5 MV) = 3.57 × 10 5  rev. The total distance traveled is d  =  n 2p R  = (3.57 × 10 5 )2p(1.0 km) =       2.2 × 10 6  km . Very-high-energy protons will have a speed  v  ˜  c .  Thus the time is t  =  d / v  = (2.2 × 10 9  m)/(3.00 × 10 8  m/s) =       7.5 s . 12. Very-high-energy protons will have a speed  v  ˜  c .  Thus the time for one revolution is T  = 2p R / v  = 2p(1.0 × 10 3  m)/(3.00 × 10 8  m/s) = 2.09 × 10 –5  s. The number of turns is n  =  t / T  = (20 s)/(2.09 × 10 –5  s) = 9.57 × 10 5  turns. We find the energy provided on each turn from ? E / n  = (900 GeV – 150 GeV)/(9.57 × 10 5  turns) = 7.8 × 10 –4  GeV/turn =       0.78 MeV/turn . 13. In the relativistic limit

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CHAPTER 44 - 3 - 11 n=?E/2eV=(900GeV8.0GeV(10...

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