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Deriving the boltzmann factors 
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#1
Mar912, 12:20 PM

P: 1,005

My classmate has already asked about this I think, but since I can't find the post I'll ask again..
Take a single atom with degenerate energy levels E1, E2 etc. We place the atom in a resevoir and want to find the probability of probability for finding the atom in the different energy states. So we note that the ratio of probabilities must be the ratio of the multiplicities for the heat resevoir corresponding to the energy levels. Thus we have: p2/p1 = Ω2/Ω1 which using S = kln(Ω) can be written as: p2/p1 = exp((S2S1)k) Invoking the thermodynamic identity using dS = (U2U1)/T we get: p2/p1 = exp((U2U1)kT) and we have arrived at the boltzmann factors, which only depend on temperature. For me this is a fantastic result, but I just wish, that I understood it. In my view, it should somehow depend on the characteristica of the resevoir such as how many particles and energy it stores.  these do after all determine the multiplicity!!! Let's take an example: Let's say you have an resevoir with an energy of 10^{10}J. If our atom is excited to a state with higher energy it "steals" some energy from the resevoir. But this energy portion is tiny compared to the total energy, and thus the multiplicity of the resevoir shouldn't reduce significantly. But if you calculate the ratio of the boltzmann factors at 300K for the ground state in hydrogen and the first state, you get that the probability of finding the atom in the ground state is overwhelmingly more probable  another way of saying that the multiplicity for the ground state system is far bigger. How is this possible when the multiplicities are almost the same from my logic?!?!?!?! 


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