# Suppose Electron spin != 1/2

1. 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?

2. Homework Equations

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

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

3. 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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malawi_glenn
Homework Helper
Hint: what is so special about the nobel gases?

Hint: what is so special about the nobel gases?
l = n-1, always, for noble gases. Where does m$$_{s}$$ come into play?

Ben Niehoff
Gold Member
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...

Redbelly98
Staff Emeritus
Homework Helper
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, ...

malawi_glenn
l = n-1, always, for noble gases. Where does m$$_{s}$$ come into play?