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Old Nov 10th 2008, 11:29 AM   #1
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Join Date: Oct 2008
Posts: 12
Maxwell Boltzmann distribution.

1. The problem statement, all variables and given/known data
The energy difference between the first excited state of mercury and the ground state is 4.86 eV.
(a) If a sample of mercury vaporized in a flame contains 10^20
atoms in thermal equilibrium at 1600K, calculate the number of atoms in the n=1 (ground) and n=2 (first-excited) states. (Assume the Maxwell-Boltzmann distribution applies and that the n=1 and n=2 states have equal statistical weights.)
2. Relevant equations
Maxwell Boltzmann Distribution
3. The attempt at a solution
I thought that since they have the same statistical weight, there must be 5*10^19 particles in each state. But I don't think it is the good answer since we use this number for another exercice and it doesn't yield the good answer.
I don't know how to figure out these number using Maxwell Boltzmann distribution.
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