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Simple Potential Energy of an electric force prob

  1. Oct 15, 2007 #1
    1. The problem statement, all variables and given/known data
    The unstable nucleus uranium-236 can be regarded as a uniformly charged sphere of charge Q = + 92e and radius [tex] R = 7.4 \times 10^{ - 15} {\rm m} [/tex]. In nuclear fission, this can divide into two smaller nuclei, each of [tex] {\frac{1}{2}} [/tex] the charge and [tex] {\frac{1}{2}} [/tex] the volume of the original uranium-236 nucleus. This is one of the reactions that occurred in the nuclear weapon that exploded over Hiroshima, Japan in August 1945. In a simple model for the fission process, immediately after the uranium-236 nucleus has undergone fission the "daughter" nuclei are at rest and just touching. Calculate the kinetic energy that each of the "daughter" nuclei will have when they are very far apart.

    2. Relevant equations

    [tex] U = \frac{1}{4\pi\epsilon_{0}}\frac{q * q_{0}}{r} [/tex]

    [tex] K = 1/2 m v^{2} [/tex]

    [tex] U_{1} + K_{1} = K_{2} + U_{2} [/tex]

    3. The attempt at a solution

    so I know for a fact because I got the answer correct(online hw) that the radius of the new spheres is 5.90*10^-15 m.
    I know that K1 is zero because they are at rest, and I know that U2 is zero because they are "far" apart. So I solve for U1=K2

    so for U1 I multiply the charges of +46E as stated in the problem divided by the radius time two, the distance between and then multiply that by the constant value 9*10^9

    [tex] 9*10^{9} \frac{(46*1.6*10^{-19})^{2}}{5.90*10^{-15}*2} [/tex]

    So I get 4.15*10^-11 J which is WRONG according to this computer... what the heck am I doin' wrong
  2. jcsd
  3. Oct 15, 2007 #2
    SOLVED! divide by two!
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