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Radius of proton given radius of electron

  1. Jun 11, 2007 #1
    1. The problem statement, all variables and given/known data

    http://img518.imageshack.us/img518/9337/untitled2gr1.jpg [Broken]

    3. The attempt at a solution

    (a) Using the right hand rule, my sketch looks like the following

    http://img340.imageshack.us/img340/5530/untitled3mk0.jpg [Broken]

    (b) F=qvbsin(theta)
    F=(1.6*10^-19)*(5*10^7)*(0.5)*sin(90deg)
    F= 4*10^-12 Newtons

    (c) Using F = (mv²)/r
    4*10^-12 = [(9.1*10^-31)*(5*10^7)²]/r
    r = 5.7*10^-4 m

    (d) I used the formula [Rp/Re] = [MpVp]/[MeVe], but the final answer is simply a radius which is 2000 times that of the electron. For some reason, I don't think it's right!
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 11, 2007 #2

    andrevdh

    User Avatar
    Homework Helper

    (d)

    [tex]F_e = F_p[/tex]

    therefore

    [tex]\frac{m_e v^2}{r_e} = \frac{m_p v^2}{r_p}[/tex]

    giving

    [tex]\frac{r_e}{m_e} = \frac{r_p}{m_p}[/tex]

    so that

    [tex]\frac{m_p}{m_e} r_e = r_p[/tex]

    or

    [tex]r_p = 2000\ r_e[/tex]
     
  4. Jun 11, 2007 #3
    hmm so I was right! :rofl:
     
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