Understanding Quadrupole Moment of Zero/Half Spin Nuclear Charge

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Discussion Overview

The discussion revolves around the quadrupole moment of nuclei with zero or half spin, exploring the implications of spherical charge distribution and its relation to nuclear deformation. Participants examine theoretical interpretations and seek clarifications on the underlying principles.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question whether the quadrupole moment of nuclei with spin zero or half is indeed zero due to spherical charge distribution, seeking further explanations.
  • One participant suggests that the result can be derived straightforwardly, linking it to the angular momentum of the nucleus.
  • A specific example of the even-even nucleus 168Er is provided, illustrating that despite its deformation, the zero angular momentum leads to a vanishing static quadrupole moment.
  • Another participant inquires about the relation Q=I(2I-1) and requests references for further understanding.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of quadrupole moments in relation to nuclear spin and charge distribution. The discussion remains unresolved regarding the completeness of explanations and the validity of specific claims.

Contextual Notes

Some assumptions regarding the definitions of spherical charge distribution and angular momentum may not be fully articulated, and the implications of nuclear deformation are not universally agreed upon.

Rajini
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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:
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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.
 
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+.
 
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
 
Here you have it

http://www.phys.washington.edu/users/savage/Class_560/lec560_4/node4.html
 
ansgar said:
Here you have it

http://www.phys.washington.edu/users/savage/Class_560/lec560_4/node4.html

Thanks..Last line is funny!
 
Rajini said:
Thanks..Last line is funny!

hehe yeah
 

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