839_Physics ProblemsTechnical Physics

839_Physics ProblemsTechnical Physics - Chapter 29 *P29.59...

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Chapter 29 179 *P29.59 The electrons are all fired from the electron gun with the same speed v in UK if = qV mv = 1 2 2 −− = eV m v e af a f 1 2 2 v m e = 2 For φ small, cos is nearly equal to 1. The time T of passage of each electron in the chamber is given by dv T = Td m e = F H G I K J 2 12 Each electron moves in a different helix, around a different axis. If each completes just one revolution within the chamber, it will be in the right place to pass through the exit port. Its transverse velocity component vv = sin swings around according to Fm a ⊥⊥ = qv B mv r °= sin90 2 eB mv r mm T e ee == = ω π 2 T m eB d m F H G I K J 2 2 Then 2 2 B m e d V e F H G I K J = a f B d mV e e = F H G I K J 2 2 . *P29.60 Let v i represent the original speed of the alpha particle. Let v α and v p represent the particles’ speeds after the collision. We have conservation of momentum 4 4 pi p pp =+ and the relative velocity equation v ip −= − 0 . Eliminating v i , 44 4 v v −=+ αα 38 p = p = 3 8 . For the proton’s motion in the magnetic field, a = ev B R p sin90 2 eBR m v p p = . For the alpha particle, 29 0 4 2 ev B r p sin r eB p = 2 r m eB v m eB eBR m R p p p p = 2 3 8 2 3 8 3 4 . P29.61 Let
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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