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Speed of Protons

  1. Apr 12, 2017 #1
    1. The problem statement, all variables and given/known data
    In a proton linear accelerator, protons are accelerated to have a kinetic energy of 530 MeV. What is the speed of these protons? (The rest mass of a proton is 1.67 × 10-27 kg.)

    2. Relevant equations
    E0=m0c^2
    E=E0/√1-(v^2)/(c^2)
    E = KE + E0

    3. The attempt at a solution
    Recognize that 1eV = 1.602E-19 Joules and 1MeV = 10E6 eV. I converted 530MeV to joules and it turned out to be 8.49E-11. I found the total energy using E = KE + E0 to be 2.35E-10. I then solved for velocity using E=E0/√1-(v^2)/(c^2) and it turned out to be 2.306E8 C
     
    Last edited: Apr 12, 2017
  2. jcsd
  3. Apr 12, 2017 #2

    phyzguy

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    You did something wrong. What was your value for E0 and E/E0 ?
     
  4. Apr 12, 2017 #3

    PeroK

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    Surely you ought to do this problem in MeV? That's why these units were invented!
     
  5. Apr 12, 2017 #4
    I found E0 = 1.503E-10 and E/E0 = 0.63957
     
  6. Apr 12, 2017 #5

    PeroK

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    The proton can't have less energy than its rest energy.

    I think you also need to try to move away from the plug-and-chug mentality. You should be looking to express speed in terms of energy and do one calculation at the end.
     
  7. Apr 12, 2017 #6
    The rest energy of the proton (E0) is 1.503E-10 and the total energy of the proton (E = KE + E0) is 2.35E-10, so I don't know what you mean when the proton can't have less energy than its rest energy.
     
  8. Apr 12, 2017 #7

    PeroK

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    You had:

    So, ##E = 0.6 E_0## is less than the rest energy.

    Don't you see how you are not being defeated by the physics, but the mass of complicated numbers. You must do this problem in MeV. It will be so much simpler.
     
  9. Apr 12, 2017 #8

    phyzguy

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    If E > E0, then how can E/E0 be less than 1?
     
  10. Apr 12, 2017 #9
    I apologize E0/E = 0.63957. It's funny because our professor told us to convert to joules first, so it makes it easier working with the units. If I did do the problem in MeV the approach would be the same right?
     
  11. Apr 12, 2017 #10

    PeroK

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    Let me explain the advantage of MeV. The mass of a proton is ##938 MeV/c^2##. That means that the rest energy of the proton is:

    ##E_0 = mc^2 = 938MeV##

    That's part of the advantage: the ##c^2## cancels out.

    The other suggestion is that if you start from:

    ##E = \gamma mc^2 = \gamma E_0##, then (as a simple exercise you can show):

    ##\frac{v}{c} = \sqrt{1 - (\frac{E_0}{E})^2}##

    Now, you just need one calculation and you don't have to struggle with numbers to 6 decimal places everywhere.
     
  12. Apr 12, 2017 #11
    I apologize, E/E0 yields 1.56 and it was E0/E that gives 0.639
     
  13. Apr 12, 2017 #12
    This is what I have been using and and I get 2.306E8 C as my final answer if I used joules. Next, keeping MeV is neat because it you get to work with smaller numbers, and using the numbers provided in the question I would get 2.307E8 C as my final answer. As a follow up if I wanted to find the mass of the proton (not rest mass), how would I go about doing that?

    Thanks once again.
     
  14. Apr 12, 2017 #13

    PeroK

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    I don't see the problem. ##E_0 = 938 MeV## and ##E = 530 + 938 MeV## and it's the same ratio as you eventually got using Joules. Joules are definitely inappropriate for atomic energies.

    The mass of the proton doesn't change. It's rest mass is its mass. Don't tell me your professor wants you to use "relativistic mass"?
     
  15. Apr 12, 2017 #14
    Haha. You guessed correctly, he wants us to use relativistic mass.
     
  16. Apr 12, 2017 #15

    PeroK

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  17. Apr 12, 2017 #16
    As a follow up question. If I want to find the mass of this proton how would I go about doing that. There is m = m0 / √1-(v2/c2). The speed of the proton turned out to be 2.306E8 C, but if I plugged that into the formula I would get a negative answer in the denominator. So, how can I find the mass of proton in order to find its momentum?
     
  18. Apr 12, 2017 #17

    phyzguy

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    You're still doing something waaaay wrong. If [itex] \frac{E_0}{E} = \sqrt{1-\frac{v^2}{c^2}} = 0.64 [/itex], then [itex] 1-\frac{v^2}{c^2} = 0.8[/itex]. This will give v/c between 0 and 1, not v = 2.3E8 c.
     
  19. Apr 12, 2017 #18
    E = E0 + KE

    Therefore,

    E = E0 + KE becomes
    E = 938MeV + 530MeV
    E = 1468MeV

    Now, let's use the equation E = E0 / √1-(v2/c2)

    E√1-(v2/c2) = E0

    √1-(v2/c2) = E0 / E

    1 - (v2/c2) = (E0/E)2

    v2/c2 = 1 - (E0/E)2

    Solve for V and you will get 2.3E8
     
  20. Apr 12, 2017 #19

    phyzguy

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    Sorry, I mis-typed it. So you mean 2.3E8 m/sec = 0.77 c? This looks correct. You kept typing 2.3E8 c, which is obviously impossible.
     
  21. Apr 12, 2017 #20

    PeroK

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    Yes, although you need the units of ##m/s##. Often you can leave the speed as a fraction of ##c##. In this case: ##v = 0.77c##.
     
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