Kinetic Energy of a proton

  1. I just want to make sure I'm on the right track with this problem. The problem states:

    "The Tevatron accelerator at the Fermi National Acclerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 GeV) in five stages (the numbers given in parentheses represent the total kinetic energy at the end of each stage):

    Cockcroft-Walton (750 KeV), Linac(400 MeV), Booseter (8 GeV), Main ring or injector (150 GeV), and finally the tevatron itself (1 TeV). What is the speed of the proton at the end of each stage?"

    I'm going to show what I did for the first one. I just want to know if I'm on the right track or not.

    My work:

    Let K denote the kinetic energy of the proton. Because the proton is going quite fast, relativistic energy equations will probably be needed so

    K = ([tex] (\gamma -1)mc^2 [/tex]

    and [tex] \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} [/tex]

    I will solve for velocity in terms of kinetic energy and mass.

    [tex] K = (\gamma -1)mc^2 [/tex]

    [tex]\frac{K}{mc^2} + 1 = \gamma [/tex]

    [tex]\frac{K + mc^2}{mc^2} = \frac{1}{\sqrt{1 - v^2/c^2}} [/tex]

    [tex] \frac{mc^2}{K+mc^2} = \sqrt{1 - v^2/c^2} [/tex]

    Let [tex] a \equiv mc^2 [/tex]


    [tex] \frac{a^2}{(K+a)^2} = 1 - v^2/c^2 [/tex]

    [tex] \beta = v/c [/tex]

    [tex] (\beta)^2 = 1 - \frac{a^2}{(K + a)^2} [/tex]

    [tex] \beta = \sqrt{1 - \frac{a^2}{(K + a)^2}}
    So, that's the algebra I used to get an expression for v (or Beta in this case). Now, the first stage ends with a Kinetic Energy of 750 KeV = 750*1000 eV = 750000 eV. Now, converting that to Joules is a simple task, 750000eV * 1.6022 * 10^(-19) Joules / eV = 1.202 * 10^-13

    Also, the value a above = mc^2 can be evaluated to equal 1.50 * 10^-10.


    [tex] \beta = \sqrt{1 - \frac{(1.50*10^-10)^2}{(1.202 * 10^(-13) + 1.50* 10^(-10))^2}} [/tex]

    I hope all that tex came out legibly. If that looks right, great. If it's wrong, just a hint or something would be great. Thanks
  2. jcsd
  3. That looks ok to me.
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