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Excitation levels and Nucleus Size

  1. Nov 16, 2007 #1
    How would you use excitation levels (or the lowest excitation level) to find the radius of a nucleus? And how would it differ to using the r=r0A^(1/3) formula?
     
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  3. Nov 16, 2007 #2
    Is it possible to use heisenberg again?
    eg.
    [tex]\Delta x \Delta p \approx \hbar[/tex]
    [tex]\Delta x \approx \frac{\hbar}{\sqrt{2mE}}[/tex]

    so subbing in the excitation level for E and the A value for m?
     
  4. Nov 16, 2007 #3

    malawi_glenn

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    Have you tried?
     
  5. Nov 17, 2007 #4
    Well, for:
    [tex]^{17}O[/tex] (oxygen)
    [tex]r=r_{0}A^{1/3}[/tex]
    [tex]r=1.3*17^{1/3}=3.34[/tex]

    and using the other way:
    [tex]\Delta x=\frac{\hbar}{\sqrt{2mE}}[/tex]
    Knowing that the first excitation level of 17-oxygen is 3055keV
    [tex]\Delta x=\frac{0.197}{\sqrt{2*17*3055}}[/tex]
    [tex]\Delta x=6.11*10^{-4}[/tex] units?? m I guess?

    Clearly not the same...
     
  6. Nov 17, 2007 #5

    malawi_glenn

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    Strange, I got 6.34*10^(-16) m for [tex] \Delta x [/tex].....

    The thing by using excitation energy is that you must know the energy of the ground state, 3055keV is just the delta E, not E of that level.
     
  7. Nov 17, 2007 #6
    Okay, so if the base level is 870keV, then the total excitation energy is 3055+870=3925... If we say 17u = 6*mp+8*mn+6*me then we arrive at:
    [tex]\Delta x = \frac{197 MeV fm}{\sqrt{2*(6*938.8+8*939.7+6*0.511)*3.925}}[/tex]
    [tex]\Delta x = \frac{197 MeV fm}{\sqrt{2*(6*938.8MeV+8*939.7MeV+6*0.511MeV)*3.925MeV}}[/tex]
    [tex]\Delta x = 0.613 fm[/tex]
    Far too small...
     
  8. Nov 17, 2007 #7

    malawi_glenn

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    Sounds very strange that the ground energy is smaller than its excitation energy.
    But any way, the higher the energy, the smaller deltaX.
     
  9. Nov 17, 2007 #8
    Okay, I think I get it.. the mass is only 938.8 or 939.7 because only 1 nucleon is excited, yeah? and the E is the actual value of the excitation, not the difference between excitation and ground state... so that:
    [tex]\Delta x = \frac{197 MeV fm}{\sqrt{2*938.8*3.055}}[/tex]
    [tex]\Delta x = 2.6fm[/tex]
    Which is still much smaller than 3.34... any reason for that?
     
  10. Nov 17, 2007 #9

    malawi_glenn

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    HUP gives lower limit.

    And also the deltaX should be the smallest cube-length that a particle with momenta (2mE)^½ can be contained in.

    If we could get exact numbers with HUP, then we would use it instead of scattering and so on.
     
    Last edited: Nov 18, 2007
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