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Unpaired electronS

by quasar987
Tags: electrons, unpaired
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quasar987
#1
Nov10-06, 12:08 AM
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In Reif's thermo book, one can read

"Consider a substance which contains n magnetics atoms per unit volumes and which is placed in an external magnetic hield B. Assume that each atom has spin 1/2 (corresponding to one unpaired electron) and an intrinsic magnetic moment of [itex]\mu[/itex]."

He makes it sound like it's possible to have more than one unpaired electron. Is this so? I interpret "unpaired" as "there is one electron in a state r and spin up (resp. down) such that there are no electrons in state r with spin down (resp. up). But as soon as you add one more electrons, it will get in state r with spin down (resp. up) so there are no more unpaired electrons. Hence it's impossible to have more than 1 unpaired. Is this how it work?
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inha
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Nov10-06, 03:22 AM
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I'm not sure if I understand you completely but... look up Hund's rules. An atom can have more than one unpaired electron. They're in different angular momentum states though.
dextercioby
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Nov10-06, 03:29 AM
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Surely if you've ever studied chemistry you might know that "unpaired electrons" occur when writing the electron configurations for various atoms. For instance, the Carbon atom has 2 unpaired electrons, N has 3, etc.

Daniel.

quasar987
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Nov10-06, 09:56 AM
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Unpaired electronS

What is an unpaired electron then?

I've said what I remember from chemistry. You fill up the states two by two since an up and a down can occupy the same state. If there is an even number of electrons, there there is no unpaired electrons; if the number is odd there is one.
dextercioby
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Nov13-06, 02:27 AM
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Nope, Hund rules make it that the Carbon atom which had 6 electrons altogether have the configuration

1s2 2s2 2p2

, but the 2 electrons in the "p" orbital are uncoupled (i.e. in different total angular momentul states), meaning that the spins are not antiparallel in the same orbital (p_{x}), but are alligned & parallel in the 2 orbitals:2p_{x} & 2p_{y}, rendering the total spin 1, instead of 0.

Daniel.
quasar987
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Nov13-06, 10:57 AM
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Ah, I see, thank you.


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