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Atomic Energy Levels - Calculating the Minimum Energy of Light Emitted

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=51788&stc=1&d=1349951739.jpg
    Determine the longest wavelength of light emitted from this atom.

    2. Relevant equations
    [itex]\lambda=\frac{hc}{E}[/itex]


    3. The attempt at a solution
    I calculated the lowest energy of the light emitted: [itex]E_{min}=2.1(eV)[/itex]
    Then the longest wavelength would be: [itex]\lambda_{max}=\frac{hc}{E_{min}}=\frac{4.14×10^{-15}×3×10^{8}}{2.1}≈5.91×10^{-7}(m)[/itex]

    What confused me is the energy difference between the ionization level and n=4 is even smaller than 2.1 eV: [itex]1.6-0=1.6(eV)[/itex]. Am I going to take this energy difference as the minimum energy emitted? Also, can you please point out if the way I take the absolute value of the energy levels and subtract them is correct?

    Thank you very much.
    LovePhys
     

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  3. Oct 11, 2012 #2
    A transition is necessary for a photon to be emitted. If the energy of an electron is raised to the 0 eV level it will be free of the atom and will not be available for further transitions. The energy of the photon as it transist from the upper energy level to the lower is Eupper - Elower so the two negatives makes the lower energy value positive.
     
    Last edited: Oct 11, 2012
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