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Minimum KE in a proton/neutron

  • 21-10-2012 11:45am
    #1
    Smythe
    Registered Users, Registered Users 2 Posts: 434 ✭✭


    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 solution below is correct?


    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


Comments

  • #2
    krd
    Banned (with Prison Access) Posts: 3,455 ✭✭✭krd


    Smythe wrote: »
    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 solution below is correct?


    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

    A deuteron, a proton and a neutron walk into a bar.......And the barman says...

    Δp ~ (h-bar)/(2Δx) looks correct. Because even though there is uncertainty, losing that 2 would be pushing it.

    Δp ~ (h-bar)/(Δx) looks a bit much.


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