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Excited statesby zezima1
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#1
Mar812, 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 1/10^{9}.
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 = 1/10^{9} 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? 


#2
Mar812, 03:38 PM

Sci Advisor
P: 6,104

There is something very confusing about your question. There are few (if any) atoms in the sun. Just about everything is fully ionized.



#3
Mar812, 04:31 PM

P: 123

It's in the sun's atmosphere.



#4
Mar912, 06:21 PM

Sci Advisor
P: 6,104

Excited states



#5
Mar1012, 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!



#6
Mar1012, 12:08 PM

P: 44




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