# Find atomic number from spectral lines

## Homework Statement

An atom or ion with one electron has energy levels $E_n=-A/n^2$. Tw neighboring lines in its spectrum at room temperature have wavelengths $\lambda_1=97.5$ nm and $\lambda_2=102.8$ nm. (Note that $hc=1.240\times 10^{-6}$ eV m).

(a)What is the constant A?
(b) Identify the atom.

## The Attempt at a Solution

I know that it must be either Hydrogen (line n=3 and n=4 of the Lyman series) or Li++ (which should have frequencies very close to those of the Hydrogen. However, this is a guess and I don't know how to actually prove that.

Any idea?

## The Attempt at a Solution

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DrClaude
Mentor
I know that it must be either Hydrogen (line n=3 and n=4 of the Lyman series) or Li++ (which should have frequencies very close to those of the Hydrogen.
Which transition in Li2+ are you considering?

Actually I don't know. I suppose the same ones, but I'm not sure at all.

To find A, did you consider using the deBroglie relation?

DrClaude
Mentor
Actually I don't know. I suppose the same ones, but I'm not sure at all.
Do you know the formula that gives the energy levels of hydrogenic atoms?

Do you know the formula that gives the energy levels of hydrogenic atoms?
It is written in the text of the problem. It's the same but with A=13.6 eV.

DrClaude
Mentor
It is written in the text of the problem. It's the same but with A=13.6 eV.
No it's not. ##A = 13.6\ \mathrm{eV}## for hydrogen, but not for Li2+. You have a nucleus with three times the charge: surely the Coulomb attraction must be greater.

You are right. We have an extra factor of $Z^2$ for Hydrogen-like atoms. However, I don't think this solve the problem. What do you suggest?

DrClaude
Mentor
I would write the Rydberg formula for each transition, take the difference between the two, and see if, with the value of ##A## for an atom other than hydrogen, it is possible to get two consecutive levels ##n## and ##n+1## to produce those wavelengths.

That's probably the best thing to do. The point is that you end up with a cubic equation in n, which is quite horrible. However, in principle is exactly solvable so it is probably the right answer. :D