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One simple question

  1. Nov 30, 2009 #1
    Hi all,
    in many books..they mention..quadrupole moment of nuclear with spin zero or half is zero..
    and the reason they give is spherical charge distribution (spherically symmetric.).. is that true?
    Or any other better explanation??
    thanks

    EDit: spherical charge distribution -does this mean-symmetric efg?
     
    Last edited: Nov 30, 2009
  2. jcsd
  3. Nov 30, 2009 #2
    you can give your post better names....

    you can derive that result for yourself very straightforward, the intepretation is that for nucleus with L = 0 (i.e. J = 0 or 1/2) el_quad = 0 and vice versa.

    spherical charge distrubution means that it means, the charger distrubution only depends on the radius.
     
  4. Nov 30, 2009 #3

    bcrowell

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    As a concrete example, consider 168Er, which is an even-even nucleus that is deformed (prolate). Its ground state has spin and parity 0+. Although the nucleus is deformed, the zero angular momentum of the ground state means that the ground-state wavefunction is a superposition of all possible orientations. Therefore the static quadruple moment <0+|Q|0+> vanishes. However, <0+|Q|2+>, so you do get collective E2 transitions from the first excited state with spin-parity 2+.
     
  5. Dec 1, 2009 #4
    Hi,
    okay..so something to do with Gordan coefficients..thanks for the hint..
    So Q=I(2I-1). Is this relation correct?..If it is correct..is there some book reference for that relation..just to understand further a book would be nice..
    thanks again
     
  6. Dec 1, 2009 #5
  7. Dec 1, 2009 #6
  8. Dec 1, 2009 #7
    hehe yeah
     
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