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Forum Rules: Read This Before Posting Topic: Quantum number, *delete me*  (Read 3049 times)

0 Members and 1 Guest are viewing this topic. mche

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« on: October 05, 2006, 02:42:00 AM »
Quote
eSuppose that you discoverd from anaother universe that obeyed the following restrictions on
quantum numbers:

n>0
l+1<n
ml= +1 or -1
ms=+1/2

Assume that Hund's rule still applies, What would be the numbers of the first 3 noble gases in that universe

I've difficulty in figuring out the answers.
I guess the principal quantum number n is 2, because n has to be larger than the angular momentum number l plus one , even l is zero, the left side has same value as n.

And then I guess
n=2, l = 0
n=3, l = 0, 1
n=4, l = 0, 1, 2
n=5, l = 0, 1, 2, 3
and so on

Then I don't get the ml part.
Is it like that? [2, 0, +1, +1/2], [2, 0, -1, -1/2]
So the n=2 level has one orbital and 2 subshell, the total number of electron in level n=2 is to equal 2
Then n =3 [3, 0, +1, +1/2], [3, 0, -1, +1/2], [3, 1, +1, +1/2], [3, 1, -1, +1/2]
has 4 orbitals and 4 subshells and the total lof 4 electrons.

The first 2 noble gases atomic # is 2 and 6?

thx! FeLiXe

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« Reply #1 on: October 09, 2006, 03:04:18 PM »
2 and 6 seem pretty reasonable. I don't understand what the deal with Hund's rule is. because there is nothing to talk about if there is only one spin. it would rather be some kind of Pauli exclusion.

[2, 0, -1, -1/2] is supposed to be [2, 0, -1, +1/2] I guess

and for the next n you'd have three l levels with two ml's each
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