Boltzmann Distribution Relative Population of 2 atomic states

[dozappp]

Homework Statement

The relative population of two atomic population states in equilibrium is given by Boltzmann Distribution:

n1/n0 (proportional to) e^(-ε/(κT)) , where ε is the energy difference between the two states, T is the temperature and κ is the Boltzmann constant = (1.38 x 10^(-23) J/K). For the transition from n = 2 to n = 1, the energy difference is 10.2 eV. Calculate the population of the n = 1 state compared to the ground state for a temperature of T = 6000K.

Homework Equations

n1/n0 (proportional to) e^(-ε/(κT))
I don't know if I need any other equations, but I tried and tried so perhaps I do.

The Attempt at a Solution

What I want is n1/n0. So I use the information given by the problem to solve for the constant of proportionality.

n2/n1 = C * e^(-10.2ev/(1.38*10^-23 J/K * 6000 K )
n2/n1 = C * 3.75567 × 10^-9
C = (n2/n1)/(3.75567 × 10^-9)

so n1/n0 = C * e^(-ε/KT)

but the problem is I don't have the epsilon for this energy difference, and I don't have n2/n1 to calculate C. halps me please.

 

[TSny]

The wording of the problem statement is confusing. It states that the energy difference for $n = 2$ and $n = 1$ is 10.2 eV. This corresponds to atomic hydrogen where $n = 1$ is the principle quantum number for the ground state. Yet the problem asks about the transition from the $n = 1$ state to the ground state!

Also, there are several degenerate states corresponding to the first excited energy level, $n = 2$. I guess we ignore fine structure, etc. It's not clear to me if degeneracy is to be taken into account.

Perhaps someone else can clarify this.