(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Maxwell Boltzmann distribution

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