1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need help converting units, momentum of quark question

  1. Jan 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A proton has a diameter of about 1fm. Estimate the minimum momentum in units of MeV/C of a quark confined in a proton.


    2. Relevant equations



    3. The attempt at a solution

    I used the uncertainty relation [itex]\Delta[/itex]P[itex]\Delta[/itex]X ~ [itex]\frac{\hbar}{2}[/itex]

    [itex]\Delta[/itex]P ~ [itex]\frac{\hbar}{2 x 1x10^{-15}}[/itex] = 5.276x10[itex]^{-20}[/itex] kgm/s

    I'm unsure how to put this into MeV/C

    thanks
     
  2. jcsd
  3. Jan 2, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You can convert kg in J/c^2 and that to MeV/c^2. Isolate a factor of c (based on m/s) and use this to cancel one c in the denominator.

    Alternatively, use ℏ given in MeV/c * length.
     
  4. Jan 2, 2013 #3
    Hmmmm I tried 5.276x10[itex]^{-20}[/itex] x (3x10[itex]^{8}[/itex])[itex]^{2}[/itex] = 4.7484x10[itex]^{-3}[/itex] Joules / c[itex]^{2}[/itex] Then to convert to eV I divided by 1.6x10[itex]^{-19}[/itex]

    And I ended up getting 2.96775x10[itex]^{16}[/itex]eV / C. the answer should be around 200 MeV/C
     
  5. Jan 2, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Did you consider m/s -> c?
    If I divide your value by 3*108, I get 108eV/c or ~100MeV/c.
     
  6. Jan 3, 2013 #5
    Hmm that seems more correct, did I do the sum incorrectly?
     
  7. Jan 3, 2013 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I don't see a sum, but a missing factor of 3*10^8 would explain the difference.
     
  8. Jan 3, 2013 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I suspect that Dunworry is British (or learned "British" English) and they, for some strange reason, use the word "sum" to refer to any arithmetic calcluation!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Need help converting units, momentum of quark question
Loading...