How Does Maxwell-Boltzmann Distribution Calculate Excited Mercury Atoms?

[inferno298]

Homework Statement

You will recall from our discussion of the Franck-Hertz experiment that the energy difference between the first excited state of mercury and the ground state is 4.86 eV. If a sample of mercury vaporized in a flame contains 1.02×10^20 atoms in thermal equilibrium at 1613 K, calculate the number of atoms in the first excited state. Assume that the Maxwell-Boltzmann distribution applies and that the n = 1 (ground) and n = 2 (first excited) states have equal statistical weights.

Homework Equations

n2/n1=g(E2)/g(E1)*e^((E1-E2)/(k T))
k=Boltzmann constant = (1.3807*10^-23), divide by 1.602*10^-19 to get in eV

The Attempt at a Solution

So I felt like it was almost a plug and chug. The distinct states at n=2, first excited state, is 8. For n=1 g(E1)= 2. Sooo :
n2/n1=(8/2)*e^((-4.86 eV)/(k*1613))
I got 2.626*10^-15, so that's the ratio of atoms in the first excited state compared to the ground state? So I just multiply it by the number of atoms 1.02*10^20 and I get 267884 atoms. Answer isn't correct though and I can't figure out what I may be missing.

 

[DrClaude]

Assume that the Maxwell-Boltzmann distribution applies and that the n = 1 (ground) and n = 2 (first excited) states have equal statistical weights.

Have you made use of the last part of this statement?

 

[inferno298]

I guess I am not too sure of what that means.

 

[inferno298]

nmv I got it Thanks!