[Capital]

I'm struggling with part A. The question says energy is released as the electron goes from high quantum numbers to quantum number, n = 2 (the first excited state)

Since wavelength increases along the x-axis, energy would decrease along the x axis. So I'm looking for the transition state with the third (A) and fourth (B) lowest energy transitions. How do I find this? I thought it matters where the electron is originally. (i.e: what n is) Because as the electron goes from n = 5 to n = 4, the energy would be less than the energy needed to go from n = 4 to n = 3 and so on.

Once I know the energy transition, I can finish part b by using that information and solving for the atomic number (z) using E = -2.18*10^-18(z^2/n^2).

Thanks for any help.

[baloot]

The trick is that every single line is a transition to the first excited state, n=2. And since wavelength increases as we move right, as you said, energy decreases, thus, the line farthest to the right, is the lowest energy transition to n=2. This lowest energy transition is n=3 to n=2. The line left of that is n=4 to n=2, and the line left of that (A) is n=5 to n=2. B is thus n=6 to n=2.