The energy difference between the first excited state of mercury and the ground

Question

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 Maxwell- Boltzmann distribution applies and that the n = 1 and n = 2 states have equal statistical weights).

I know the MB distribution is f(E) = Ae^(-E/KbT), and since these are equal in weight:
n1 = Ae^(-E1/KbT) and n2 = Ae^(-E2/KbT). Therefore, n2/n1 = e^(-deltaE/kbT). Also, n1+n2 = N. Using these two equations, I solved and got n2 = 49100 atoms, which is insanely small compared to the 10^20 atoms, which can't be right.

Explanation / Answer

Well it seams you do the right work The excitation potential for mercury is 5.0 ev. So it is not a surprise that at such a high energy you have vary small number of particles in the ground state.

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