How Would a Change in Electron Spin Alter the Identity of Noble Gases?

dalarev
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Homework Statement



Suppose that electrons had a spin number of \frac{3}{2} instead of \frac{1}{2}

That is, they have ould magnetic spin states of S_{z} = \frac{-3}{2}\hbar , \frac{-1}{2}\hbar , \frac{+1}{2}\hbar , \frac{+3}{2}\hbar

Which elements would be the "new" noble gases in this case?

Homework Equations



"real" electron spin = + 1/2, - 1/2.

Noble gases are in the group of Helium (Neon, Argon, etc.)

The Attempt at a Solution



I figured it was going to be easy because, looking at the dependence of the other 3 quantum numbers on each other, it would reveal the solution quickly. The dependence I'm referring to is n = 1, 2, 3, ...
l = 1, 2, n-1... and so on..

I can't find a dependence this clear with the spin number though.
 
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Hint: what is so special about the nobel gases?
 
malawi_glenn said:
Hint: what is so special about the nobel gases?

l = n-1, always, for noble gases. Where does m_{s} come into play?
 
The total number of m states gives the total number of electrons which may share the same n, l states. A noble gas is an element where all of its electron shells are completely full...
 
Ben Niehoff said:
The total number of m states gives the total number of electrons which may share the same n, l states. A noble gas is an element where all of its electron shells are completely full...

... and instead of TWO electrons filling each N-L orbital, ...
 
dalarev said:
l = n-1, always, for noble gases. Where does m_{s} come into play?


How do you build up the electron stucture for helium & neon
Does a pattern reveal?
 
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