P29_031

# P29_031 - . 3 MeV. The radius of the orbit is given by r =...

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31. We approximate the total distance by the number of revolutions times the circumference of the orbit corresponding to the average energy. This should be a good approximation since the deuteron receives the same energy each revolution and its period does not depend on its energy. The deuteron accelerates twice in each cycle, and each time it receives an energy of qV =80 × 10 3 eV. Since its Fnal energy is 16 . 6 MeV, the number of revolutions it makes is n = 16 . 6 × 10 6 eV 2(80 × 10 3 eV) = 104 . Its average energy during the accelerating process is 8
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Unformatted text preview: . 3 MeV. The radius of the orbit is given by r = mv/qB , where v is the deuteron’s speed. Since this is given by v = p 2 K/m , the radius is r = m qB r 2 K m = 1 qB √ 2 Km . ±or the average energy r = p 2(8 . 3 × 10 6 eV)(1 . 60 × 10 − 19 J / eV)(3 . 34 × 10 − 27 kg) (1 . 60 × 10 − 19 C)(1 . 57 T) = 0 . 375 m . The total distance traveled is about n 2 πr = (104)(2 π )(0 . 375) = 2 . 4 × 10 2 m....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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