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Finding wavelength of visible light in a well

  1. Oct 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine what colors of visible light would be absorbed by electrons in an infinite well, N = 3.1 nm. The effective mass for an electron is one-fifteenth of the standard electron mass.

    2. Relevant equations

    En = n2h2/(8mL2)

    E = hf
    f = c/λ

    3. The attempt at a solution
    E1 = 9.4039*10-17 J
    E2 = 3.7616*10-16 J
    E3 = 8.4635*10-16 J
    E4 = 1.5046*10-15 J

    ΔE1-2 = 70.5 nm
    ΔE2-3 = 42.3 nm

    Nothing in the visible.
     
    Last edited: Oct 29, 2016
  2. jcsd
  3. Oct 29, 2016 #2

    Simon Bridge

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    Please show your reasoning - all you've given me is a bunch of letters and numbers.
    For instance - what energies of photon can be absorbed by the electron in the ground state?
     
  4. Oct 30, 2016 #3
    h = 6.626*10^-34
    m2-kg/s
    An electron can only absorb decrite (quantized) energies of a photon.
    E2 − E1 = E ⇒ 70.5 nm
    After using f = E/h, then λ = c/f.
    Or λ = c/(E/h) = c*h/ΔE = 70.5 nm.

    Since this is the highest wavelength that can be absorbed by the electron, and visible light is a higher frequency (a minimum of 400 nm), no colors are absorbed.
     

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  5. Oct 30, 2016 #4

    ehild

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    Your energy values are too high. Check the calculations.
     
  6. Oct 30, 2016 #5
    En = n2h2/(8mL2)
    En = n2(6.626e-34)2/(8(9.11e-31/15)(3.1e-9)2)
    En = 9.4029*10-20n2 J

    And then for the rest:
    E = E2-1 = 2.82087e-19
    λ = ch/E = 3e8*6.626e-34/2.82087e-19
    λ = 7.0468e-7 = 704.68 nm

    So red.
     
    Last edited: Oct 30, 2016
  7. Oct 30, 2016 #6

    ehild

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  8. Oct 30, 2016 #7
    Is it not just taking the effective mass, me/15, as given in the question statement?
     
  9. Oct 30, 2016 #8

    ehild

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    I do not see 15 in your formula.
     
  10. Oct 30, 2016 #9
    Sorry, it was in the Wolfram Alpha link, but I forgot to change it in the writing here. (I'm trying to get this done. I've spent all weekend on this one assignment...reading...learning...but not getting much gainful progress on the actual questions...So ya, I'm trying to go not as slow as I've been doing.)
     
  11. Oct 30, 2016 #10

    ehild

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    OK, so are there any visible wavelengths?
     
  12. Oct 30, 2016 #11
    700ish = Dark red.
    upload_2016-10-30_15-5-53.png
     
  13. Oct 30, 2016 #12

    ehild

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    All right. Any more?
     
  14. Oct 30, 2016 #13
    Light? Oh, yes let me check...
     
  15. Oct 30, 2016 #14

    ehild

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    The electron can be on the second level with some probability...
     
  16. Oct 30, 2016 #15
    ΔE2-1 = 2.82087e-19
    ΔE3-2 = 4.70146e-19
    ΔE4-3 = 6.58204e-19

    h = m2-kg/s
    6.626e-34

    λ = c*h/ΔE

    ΔE2-1 ⇒ λ2-1 = 705 nm (dark red)
    ΔE3-2 ⇒ λ3-2 = 423 nm (purple)
    ΔE4-3 ⇒ λ4-3 = 302 nm (UV ~ not visible light)

    Thank you very much ehild for helping!
     
  17. Oct 30, 2016 #16

    ehild

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    You are welcome :oldsmile:
     
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