Find atomic number from spectral lines

[Einj]

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.

Homework Equations

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?

 

[DrClaude]

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?

 

[Einj]

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

 

[Gamma]

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

 

[DrClaude]

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?

 

[Einj]

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]

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.

 

[Einj]

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]

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.

 

[Einj]

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