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Suppose Electron spin != 1/2

  1. Apr 16, 2008 #1
    1. The problem statement, all variables and given/known data

    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. Relevant 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.
  2. jcsd
  3. Apr 16, 2008 #2


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    Hint: what is so special about the nobel gases?
  4. Apr 16, 2008 #3
    l = n-1, always, for noble gases. Where does m[tex]_{s}[/tex] come into play?
  5. Apr 16, 2008 #4

    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...
  6. Apr 16, 2008 #5


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    ... and instead of TWO electrons filling each N-L orbital, ...
  7. Apr 16, 2008 #6


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