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Minimum KE in a proton/neutron

  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data
    In a deuteron, a proton and a neutron are very weakly bound by the strong nuclear force with an average distance between the particles of about 5fm.
    Due to the uncertainty principle, the particles have a minimum momentum and hence a minimum kinetic energy of ...?

    Which answer is correct?


    2. Relevant equations



    3. The attempt at a solution


    ΔpΔx ~ (h-bar)/2
    Δp ~ (h-bar)/(2Δx)

    Then using:
    ΔE = (Δp)^2 / 2m
    ΔE = [ {(h-bar) / (2Δx)}^2 / 2m ]
    Using values of Δx=5x10^-15, m=1.67x10^-27
    ΔE = 3.3 x 10^-14 J

    Or should it be:
    Δp ~ (h-bar)/(Δx)

    Then using:
    ΔE = (Δp)^2 / 2m
    ΔE = [ {(h-bar) / (Δx)}^2 / 2m ]
    Using values of Δx=5x10^-15, m=1.67x10^-27
    ΔE = 1.3 x 10^-13 J
     
  2. jcsd
  3. Oct 21, 2012 #2
    I believe it's the former. Where did you get the idea to drop the 2?
     
  4. Oct 21, 2012 #3
    Because that 2nd version was given to us by the lecturer.

    The former attempt is mine.
     
  5. Oct 21, 2012 #4
    After further reading, [itex] \sigma x \sigma p \geq \frac{\hbar}{2} [/itex] is the modern form, which is most accurate.

    [itex] \sigma x \sigma p \geq \hbar [/itex] was Heisenbergs original formula. It is obviously still a true mathematical statement since the minimum uncertainty is greater than in the above formula, but it is not quite as accurate.

    I think your professor probably wants the original Heisenberg equation, since he decided to teach it that way. Either way, you can always write a note in next to the problem if you're worried about using the incorrect one.
     
  6. Oct 21, 2012 #5
    Thanks very much bossman. That's appreciated.

    I was unaware of that. I'll look into that now.

    At least now I know where the "problem" lies.
     
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