Relation between nuclear shape and it spin number

[Renato Murback]

TL;DR

What is the relation between the nuclear shape (or charge distribution) and the nuclear spin?

I'm beginner in NMR studies and now I'm studying quadrupole interaction. In this case, spins larger than 1/2 have a non-spherical charge distribution and this asymmetry interacts with external electric fields.
But the question is: What is the relation between the nuclear shape (or charge distribution) and the nuclear spin? I know spin 1/2 has a spherical shape, and larger spins have differents shapes.... But why? How nuclear spins implies the nuclear shape?
Thank you very much for your attention

 

[berkeman]

Welcome to PF, Renato.

Paging @Dale

 

[Vanadium 50]

spin 1/2 = magnetic dipole.
spin 1 = electric quadrupole
spin 3/2 = magnetic octupole
etc.

See the Wigner-Eckhart Theorem.

 

[Renato Murback]

how is these spin numbers related with the nuclear shape? what is the relation between them?

 

[Vanadium 50]

If you go to Google, select Images, and type in Nuclear Quadrupole (e.g.) you will get plenty of pictures showing you the shape.

 

[Renato Murback]

I don't want to know the shape. I want to know the physical relation between the shape and the spin number. How does spin implies the shape? Or how does nuclear shape implies the spin number?

 

[Vanadium 50]

I told you the "physical reason". It's the Wigner-Eckhart Theorem.

 

[Renato Murback]

I'll learn about that. Thank you

 

[Vanadium 50]

While not a proof, here is an example that shows how Wigner-Eckhart works. Suppose I want the quadrupole moment of a spin-½ system. That's <½| ℚ | ½> where ℚ is the quadrupole operator. ℚ | ½> = a | 3/2 > + b | 5/2 > where a and b are numbers. Messy to calculate, but just numbers. But since <1/2 | 3/2 > and <1/2 | 5/2 >are both zero, we don't need to: <½| ℚ | ½> = 0. A spin-½ system cannot have a quadrupole moment.