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Atomic Physics - Xrays and energy

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    5 The energy levels of a molybdenum atom are approximately:
    K shell ….. – 20.00 KeV
    L shell …... – 2.52 KeV
    M shell …... – 0.23 KeV

    i) Estimate the energy of the Kα, K[tex]\beta[/tex] and Lα x-ray emissions.
    ii) What are the wavelengths of these emissions?

    2. Relevant equations

    3. The attempt at a solution

    I think K[tex]\alpha[/tex] is from L shell to k shell??
    so EK[tex]\alpha[/tex]=-2.52KeV--20KeV

    and the K[tex]\beta[/tex] is from M shell to K shell?
    then find the difference again for the energy

    I don't know what the L[tex]\alpha[/tex] is but possibly L shell to ground state or M shell to L shell??

    not sure if anything im doing there is right?

    shouldn't be hard once I have energy's
    E=hf so E=hc/[tex]\lambda[/tex]

    not sure if anything im doing here is right???

    Thanks for any help!
  2. jcsd
  3. Sep 27, 2010 #2


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  4. Sep 27, 2010 #3
    Thanks ehild and wiki,
    so the Lα transition is from M to L.
    I'm a bit confused as to whether or not the element is relevant to any of the calculations, from what I read on wiki it seems related by the amount of energy gained form each transition but since I have those values I shouldn't even need to know that the element is molybdendum?
    also I did the wave length calculation for the first transition and it came out to 1.24*10^-9m, this seems about right for an xray? (changed energy from KeV to J)

  5. Sep 27, 2010 #4


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    The x-ray lines are named by the letter of the final state. The element is not relevant as you were given the level energies. Check the magnitude of the wavelength, I got a different result.

  6. Sep 28, 2010 #5
    Yeah, I converted 17.48eV instead of 17.48KeV
    now I get 7.1*10^-11 m?????
    seems like a tiny wavelength.
  7. Sep 28, 2010 #6


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    X rays are of very short wavelength. Read the first sentences of the article.

  8. Sep 28, 2010 #7
    Thanks for helping ehild!
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