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Suppose Electron spin != 1/2

  • Thread starter dalarev
  • Start date
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1. Homework Statement

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

That is, they have ould magnetic spin states of S[tex]_{z}[/tex] = [tex]\frac{-3}{2}[/tex][tex]\hbar[/tex] , [tex]\frac{-1}{2}[/tex][tex]\hbar[/tex] , [tex]\frac{+1}{2}[/tex][tex]\hbar[/tex] , [tex]\frac{+3}{2}[/tex][tex]\hbar[/tex]

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.
 

Answers and Replies

malawi_glenn
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Hint: what is so special about the nobel gases?
 
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Hint: what is so special about the nobel gases?
l = n-1, always, for noble gases. Where does m[tex]_{s}[/tex] come into play?
 
Ben Niehoff
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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
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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
Science Advisor
Homework Helper
4,782
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l = n-1, always, for noble gases. Where does m[tex]_{s}[/tex] come into play?

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

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