# Find atomic number from spectral lines

1. Jan 21, 2014

### Einj

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 21, 2014

### Staff: Mentor

Which transition in Li2+ are you considering?

3. Jan 21, 2014

### Einj

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

4. Jan 21, 2014

### Gamma

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

5. Jan 22, 2014

### Staff: Mentor

Do you know the formula that gives the energy levels of hydrogenic atoms?

6. Jan 22, 2014

### Einj

It is written in the text of the problem. It's the same but with A=13.6 eV.

7. Jan 22, 2014

### Staff: Mentor

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.

8. Jan 22, 2014

### 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?

9. Jan 23, 2014

### Staff: 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.

10. Jan 23, 2014

### 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