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Drift tube lengths

  1. May 4, 2012 #1
    Lets say we had a 10-stage linac accelerator and we were using a 200 kV supply.
    Apparently the K.E at stage 2 (in the second tube) is twice that of the first tube (stage 1).
    why is this?
     
  2. jcsd
  3. May 4, 2012 #2
    In fact a better question is:

    If an electron, starting at 0 m/s is accelerated across a p.d of 300V its new KE is
    4.8 * 10-17J (E = QV).
    My question is, if I accelerate this electron (with this KE) through another p.d of 300V then, disregarding relativistic effects, will the new KE be
    4.8 * 10-17J *2 = 9.6*10-17J ?


    If not then what will it be?
     
  4. May 4, 2012 #3
    It is easier to do the calculations using eV (electron-volt) units. The electron mass in eV units is mc2 = 511,000 eV. The kinetic energy is
    [tex] KE=\frac{1}{2}mv^2=\frac{1}{2}\beta^2mc^2 [/tex]
    where v=βc and c = 3 x 108 meters per second. So
    [tex] \beta=\sqrt{\frac{2\cdot KE}{mc^2}} [/tex]
    So β = 0.03427, 0.04846, 0.05935, etc. for n=1, 2, 3, etc. and KE =n·300 eV.

    When the electron is inside a drift tube, it is completely shielded from the applied voltage. You have to use an ac voltage and a drift tube so that when the applied voltage is the wrong polarity, the drift tube shields the electron, and the electron only "sees" the correct polarity voltage.

    [added] If you use an ac frequency of f = 100 MHz, the drift tube lengths are [itex] L=\frac{\beta c}{2f}= [/itex] 0.05141, 0.07269, .08903 meters etc.
     
    Last edited: May 4, 2012
  5. May 5, 2012 #4
    OK so a good way to see it is as follows:

    tube 1 --> tube 2 - KE doubles
    tube 2 --> tube 3 - KE * 1.5

    so
    Tube 1 = L
    tube 2 = L*sqrt(2) = L-new
    tube 3 = L-new * sqrt(1.5)

    but this has answered by question so thanks a lot :)
     
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