Cool question. I had to think about it for a bit. Although it turns out that my first instinct was correct, I had to verify the math.

In this question, we're dealing with one-electron species, which keeps it relatively simple. You don't have to worry about subshells, so the calculation is actually pretty easy.

First, realize that the energy of each shell for any one-electron species = -A/n^{2}. A can be measured by adding one electron to a zero-electron species and determining how much energy is given off when the electron ends up in the n=1 shell from the n = ∞ shell (i.e., from infinitely far away, where we define E = 0 (A/∞ = 0)).

In going from n = ∞ to n = 1, E = -A/1^{2}, or E = -A.

In going from n = ∞ to n = 2, E = -A/2^{2}, or E = -A/4.

The difference between the two will be the n=1 to n=2 energy gap.

ΔE = 3/4 A

Now, the value of A will depend on the charge of the nucleus. Higher charge means stronger pull on the electron, and a higher value for A, meaning a higher value for ΔE.

The wavelength of the light is inversely related to ΔE.

Hope this helps, and isn't doing your homework for you.

Dick

P.S. This question might be applicable to a general chemistry class, although it also might be considered a little too advanced and left for physical chemistry.

P.P.S. I finally made it to full member. Woo Hoo!