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Wavelength - Yellow line in the spectrum of a sodium lamp

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data
    The outermost electron in a sodium atom can be in these energy states: (1)-0.82 aJ, (2) -0,48 aJ, (3) -0.31 aJ [itex]1 aJ= 10^{-18} J[/itex].

    When we study the spectrum of a sodium lamp, we see a yellow line.
    Use the above given energy states to determine the wavelength of this yellow light. Visual light: 400 nm to 750 nm


    2. Relevant equations

    [itex]E_{photon}=hf [/itex] where h is planck's constant: [itex]h=6.63 \cdot 10^{-34}[/itex]

    [itex]c=f\lambda[/itex]


    3. The attempt at a solution

    I can only find one ultraviolet and one infrared wavelength from the given energy states.

    frequency: [itex]f=\frac {E_f}{h} = \frac {-.31 \cdot 10^{-18}J -(-.82 \cdot 10^{-18} J))}{6.63 \cdot 10^{-34} Js} \approx 7.69 \cdot 10^{14}[/itex]

    Wavelength: [itex]\lambda = \frac{c}{f}=\frac {3 \cdot 10^8 m/s}{7.69 \cdot 10^{14}}=3.90 \cdot 10^{-7}= 390 nm[/itex]

    This is in the UV-part of the spectrum. Similar problem arises with the other energy transiteration.

    However the answer key says 585 nm.

    Please help!
     
  2. jcsd
  3. Aug 24, 2011 #2

    kuruman

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    Homework Helper
    Gold Member

    You must have done something wrong. One of the differences does give 585 nm. Check your work. Also, you can use E = hc/λ to relate directly the energy to the wavelength.
     
  4. Aug 25, 2011 #3
    That's true, yesterday I only thought of two possible movements between the energy states, but there were three. And the third one was the one I should use. Thank you for taking time to help...
     
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