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Distance of alpha particle to gold nucleus

  1. Apr 17, 2010 #1
    1. The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32MeV to estimate the radius r of the gold nucleus.



    2. In the previous problem, I found r = 79e2 / 4pi*epsilon0mpv2



    3. 32MeV = 32*106eV = 5.1264*10-12 J

    I then did KE = 0.5mpv2, solving for mpv with KE = to the above value. This value is in the denominator of the equation for r given above. Doing the calculation I obtain 1.8*10-15 m, which is incorrect.
     
  2. jcsd
  3. Apr 18, 2010 #2

    Redbelly98

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    That doesn't seem right, for two reasons:

    1) 79e2 should instead be
    (charge of gold nucleus)x(charge of alpha particle)​
    The gold nucleus has 79e, and an alpha particle's charge is ___?

    2) The mpv2 kinetic energy term is missing the factor of 1/2. (This might be a simple typo on your part.) It's probably easier to simply replace this with "KE" or "5.1264*10-12J" however.

    That's in the right ballpark, so hopefully using the correct charge for an alpha particle, and making sure you include the kinetic energy properly, fixes things.
     
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