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Atomic absorption spectrum

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data

    The following photon wavelengths are observed in absorption at room temperature from an ionized atomic gas with a single electron orbiting the nucleus: λ=
    13:5 nm, 11:4 nm, 10:8 nm. Use this data to determine the effective Rydberg constant
    and the nuclear charge.

    2. Relevant equations



    3. The attempt at a solution

    I know that energy is emitted/absorbed when electron transits between different energy levels. 2hxc0ag.png

    Thing is, I don't even know what the n=1 energy is, to find the rydberg constant. I can't tell what the energy levels are just from the photons absorbed, as they could be between any two levels.
     
  2. jcsd
  3. Jun 9, 2013 #2

    mfb

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    Staff: Mentor

    Don't give up before you even started.
    Did you convert the wavelength values to photon energies?
    Which patterns do you expect in the energy spectrum?
     
  4. Jun 9, 2013 #3
    I expect, the gaps between energy levels to become smaller as you go higher up..
     
  5. Jun 9, 2013 #4
    13.5nm -> 92.1 eV
    11.4nm -> 109.0 eV
    10.8nm -> 115.1 eV

    So

    n=1 is 92.1 eV
    n=2 is 109.0 eV
    n=3 is 115.5 eV

    Then what is the point of giving us the other 2 absorption wavelengths?
     
  6. Jun 9, 2013 #5

    mfb

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    Staff: Mentor

    Photon energies are not the energies of states!

    Which photon energies (not electron energy states) do you get for a hydrogen atom? Which ratios do you have between those values? Do you see a similarity to your problem?
     
  7. Jun 9, 2013 #6
    For Hydrogen:

    E1 = -13.6 eV
    E2 = -3.4 eV
    E3 = -1.51 eV
    E4 = -0.850 eV

    Based on the wavelengths given, 92.1eV, 109eV, 115.5eV are differences in energy between En and E1.

    The ratio of first energy level between the gas and hydrogen = Z2, where Z is the proton number of the gas.
     
  8. Jun 9, 2013 #7
    Ok using the relation ΔE = (1- 1/n2)

    First emission:
    92.1 = E1(3/4)

    Second emission:
    109 = E1(8/9)

    Third Emission:

    115.5 = E1(15/16)

    These ratios match, so somehow these are the emissions from the second, third and fourth energy levels.

    E1 = (13.6)Z2

    Solving, Z = 3 (Lithium) and E1 = 122.4eV, R = (13.6eV)/hc = 1.09*107
     
    Last edited: Jun 9, 2013
  9. Jun 10, 2013 #8

    mfb

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    Staff: Mentor

    That is correct.

    I think the factor of 9 is missing here, and the last value should have units.
     
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