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Minimum kinetic energy of an Alpha particle confined to a nucleus

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    I've found the uncertainty of the kinetic energy through many means, and I'm confident that it's .052MeV.

    The easiest method that gave me this answer was:

    using http://puu.sh/bz0Fm/f45f952a89.png

    ((1.05*10^-34)^2 / 2(6.64*10^-27)(10^-14)^2 ) / 1.6*10^-19 = .052MeV

    So I know that the kinetic energy has an uncertainty of .052MeV and that its typical energy is 5MeV, so wouldn't the answer just be 5MeV - .052MeV = 4.948MeV?
  2. jcsd
  3. Sep 14, 2014 #2


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    In this type of question, the uncertainty in momentum is generally taken to be an estimate of the minimum momentum itself. Thus, the KE of .052 MeV would represent an estimate of the minimum KE of the alpha particle inside the nucleus. Did you try using .052 MeV as the answer?

    You also have to be careful in that different people will use somewhat different expressions for the right side of the uncertainty principle: h, h-bar, h-bar/2, etc.
  4. Sep 14, 2014 #3
    Wow, when I first put in my answer I put in .51 instead of .519, so it got marked as wrong :( I was right the whole time.

    There is this problem, though -- http://puu.sh/bz4AT/4545a75828.png [Broken]

    I can't even find anything like it in my text book, any tips just to get me started on it?

    Last edited by a moderator: May 6, 2017
  5. Sep 14, 2014 #4


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    Is that meant to be 0.0519 MeV?

    Are you familiar with the energy-time uncertainty principle? See http://pdg.web.cern.ch/pdg/cpep/unc_vir.html
    Last edited by a moderator: May 6, 2017
  6. Sep 14, 2014 #5
    Ok, yeah I've used that equation before and I got the correct answer, thanks!
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