## one simple question

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?

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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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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+.

## one simple question

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...0_4/node4.html

 Quote by ansgar Here you have it http://www.phys.washington.edu/users...0_4/node4.html
Thanks..Last line is funny!

 Quote by Rajini Thanks..Last line is funny!
hehe yeah