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Find atomic number from spectral lines

  • Thread starter Einj
  • Start date
  • #1
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Homework Statement


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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
DrClaude
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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?
 
  • #3
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Actually I don't know. I suppose the same ones, but I'm not sure at all.
 
  • #4
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To find A, did you consider using the deBroglie relation?
 
  • #5
DrClaude
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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?
 
  • #6
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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.
 
  • #7
DrClaude
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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.
 
  • #8
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You are right. We have an extra factor of [itex]Z^2[/itex] for Hydrogen-like atoms. However, I don't think this solve the problem. What do you suggest?
 
  • #9
DrClaude
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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
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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
 

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