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Converting Energy Lines from Electron Volts

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Henergy.GIF

    "Using the conversion factor from eV (electron volts) to Joules, determine which energy line for an electron dropping from one energy level to another in the energy diagram for hydrogen above would show a line at 435 nm in an emission spectrum."


    2. Relevant equations
    1 eV = 1.6 x 10-19 J
    h = 6.63 x 10-34 J•s
    c = 2.99 x 108 m/s
    En = nhf
    λ = c/f
    λ = E/h
    f = E/h


    3. The attempt at a solution

    First, I converted all the eV;
    -.04 • 1.6 x 10-19 = -6.4 x 10-21
    -.05 • 1.6 x 10-19 = -8 x 10-21
    -.08 • 1.6 x 10-19 = -1.28 x 10-20
    -1.5 • 1.6 x 10-19 = -2.4 x 10-19
    -3.4 • 1.6 x 10-19 = -5.44 x 10-19
    -13.6 • 1.6 x 10-19 = -2.18 x 10-18

    Next, I think I'm supposed to find all the frequencies using;
    f = E/h

    Or maybe I'm supposed to find the wavelengths with;
    λ = c/f
    λ = E/h (is this formula even correct?)

    Could I just put it the desired answer and see which matches?;
    4.35 x 10-7 = c/f
    4.35 x 10-7 = E/h
    f = E/h

    c/ 4.35 x 10-7 = f
    2.99 x 108 / 4.35 x 10-7 = 6.87 x 1014

    6.87 x 1014 •*h = E
    6.87 x 1014 • 6.63 x 10-34 = 4.55 x 10-19

    It doesn't seem to match anything...

    Retry previous steps adding n into the equation;

    6.87 x 1014 • h • n = E
    6.87 x 1014 • 6.63 x 10-34 = E/n
    4.55 x 10-19 = E/n
    4.55 x 10-19 • n = E

    Go back to all n from original graph;

    4.55 x 10-19 • 6 = 2.73 x 10-18
    4.55 x 10-19 • 5 = 2.28 x 10-18
    4.55 x 10-19 • 4 = 1.82 x 10-18
    4.55 x 10-19 • 3 = 1.37 x 10-18
    4.55 x 10-19 • 2 = 9.1 x 10-19
    4.55 x 10-19 • 1 = 4.55 x 10-19

    I don't exactly understand what I'm supposed to get for 435 nm. The "nm" means nanometers, right? Is that the wavelength? Is it the frequency? Is it something else...? What is "a line at 435 nm in an emission spectrum"?
     
  2. jcsd
  3. May 8, 2009 #2

    berkeman

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    Staff: Mentor

    Remember that the transition of the electron is *between* those energy levels in the diagram, so you should be calculating the deltas, not using the energies themselves. And E=hf is the correct equation to use after that...
     
  4. May 8, 2009 #3
    So in the graph, the -.04 eV is the ∆eV? eV(before dropping energy level) - eV(after falling energy level) = ∆eV?
     
  5. May 8, 2009 #4

    berkeman

    User Avatar

    Staff: Mentor

    The energy delta for an n = 6-->5 transition is 0.1eV, and so on. You can also try working backwards -- that may be quicker. Calculate the photon energy in eV for the wavelength of light that you are given, and look for a delta-eV transision that matches on the graph...
     
  6. May 8, 2009 #5
    Whew! Thanks so much, I think that'll help a LOT! :)
     
  7. May 8, 2009 #6
    4.35 x 10-7 = c/f
    c / 4.35 x 10-7 = f
    2.99 x 108 / 4.35 x 10-7 = f
    f = 6.87 x 1014

    f = E/h
    f • h = E
    6.87 x 1014 • 6.63 x 10-34 = E
    E = 1.04 x 1048

    1.6 x 10-19 / 1.04 x 1048 = ∆eV

    Oops... I'm sure I did something wrong... Do I at least have the right sort of idea? Or is my approach completely off?
     
  8. May 8, 2009 #7
    ∆eV = -.4 - (-.5) = .1 eV
    f = E/h
    f = (.1)(1.6 x 10-19) / 6.63 x 10-34
    f = 2.4 x 1013
    λ = c/f
    λ = 2.99 x 108 / 2.4 x 1013
    λ = 1.24 x 10-5

    Working backwards;
    λ = 4.35 x 10-7
    4.35 x 10-7 = 2.99 x 108 / f
    2.99 x 108 / 4.35 x 10-7 = f
    f = 6.89 x 1014
    f • h = E
    f • 6.63 x 10-34 = ∆eV • 1.6 x 10-19

    6.89 x 1014 • 6.63 x 10-34 / 1.6 x 10-19 = ∆eV
    ∆eV = 2.85 ≈ 2.9

    So the energy level change is from E2 to E5 because -.5 - (-3.4) = 2.9 eV. Is that right?
     
  9. May 8, 2009 #8

    berkeman

    User Avatar

    Staff: Mentor

    I didn't check it in detail, but it looks correct. One thing that would help would be to include units on each line of your calculation, and carry them along with the quantities. That makes it a lot easier to follow the calculation, and helps you to catch errors in terms as you go.

    Good job.
     
  10. May 8, 2009 #9
    Good tip, and thanks for the help!

    Also, I noticed some of my mistakes from the original equations. The change in eV, as you said, was .1, whereas I had the numbers displaced by a factor of 10 with an extra zero to the right of the decimal. It kind of screwed up my powers...

    I think I finally have it, though, hehehe.
     
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