Find atomic number from spectral lines

Click For Summary
SUMMARY

The discussion focuses on determining the constant A and identifying the atom based on spectral lines at wavelengths λ1=97.5 nm and λ2=102.8 nm. Participants confirm that the atom is likely either Hydrogen or Li++, with the energy levels described by the formula E_n=-A/n^2. The correct value of A for Hydrogen is established as 13.6 eV, while for Li++, the value must account for the increased nuclear charge, leading to a different A value. The Rydberg formula is suggested as a method to find the transitions corresponding to the observed wavelengths.

PREREQUISITES
  • Understanding of atomic energy levels and the formula E_n=-A/n^2
  • Familiarity with the Rydberg formula for spectral lines
  • Knowledge of hydrogenic atoms and their energy level calculations
  • Basic concepts of Coulomb's law and its application in atomic physics
NEXT STEPS
  • Research the Rydberg formula and its application to hydrogenic atoms
  • Study the energy level equations for Li++ and other hydrogen-like ions
  • Explore the de Broglie relation and its relevance in atomic transitions
  • Investigate the impact of nuclear charge on spectral lines in multi-electron systems
USEFUL FOR

Students and educators in physics, particularly those studying atomic structure and spectroscopy, as well as researchers interested in hydrogenic atoms and their spectral properties.

Einj
Messages
464
Reaction score
59

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?
 
Physics news on Phys.org
Einj said:
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?
 
Einj said:
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?
 
DrClaude said:
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.
 
Einj said:
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?
 
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
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
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
Replies
16
Views
3K
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K