Excited states


by zezima1
Tags: states
zezima1
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#1
Mar8-12, 11:54 AM
P: 123
My book calculates the ratio of probability to find an atom in an excited state vs finding it in the ground state in the sun and gets approx 1/109.
Essentially this must mean that the ratio of the multiplicities of the system must also be equal to this, i.e.:

[itex]\Omega[/itex]2/[itex]\Omega[/itex]1 = 1/109

How can this be possible? Say you excite the atom. The energy that the surroundings in the sun loses is tiny compared to its total energy. So shouldnt the two multiplicities be approximately the same?
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mathman
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#2
Mar8-12, 03:38 PM
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There is something very confusing about your question. There are few (if any) atoms in the sun. Just about everything is fully ionized.
zezima1
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#3
Mar8-12, 04:31 PM
P: 123
It's in the sun's atmosphere.

mathman
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#4
Mar9-12, 06:21 PM
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Excited states


Quote Quote by zezima1 View Post
It's in the sun's atmosphere.
I am not sure what you mean by the sun's atmosphere, since it is entirely in a gaseous or plasma state. The surface is close to 6000 K. I presume everything is highly ionized, but I am no expert on the subject.
zezima1
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#5
Mar10-12, 04:45 AM
P: 123
Then say its somewhere, where everything is not ionized. This has no relevance for my question, which is about the conceptual understanding of the boltzmann distribution!
Modulated
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#6
Mar10-12, 12:08 PM
P: 44
Quote Quote by zezima1 View Post
Essentially this must mean that the ratio of the multiplicities of the system must also be equal to this
Really? This would imply that every energy level is equally likely to be filled, which I hope seems unlikely. So what difference between the ground and the excited state havenít you taken into account?


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