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Electron in one dimensional infinite square well

  1. Apr 27, 2013 #1
    An electron in the ground state of a one-dimensional infinite square well of width 1.10 nm is
    illuminated with light of wavelength 600 nm. Into which quantum state is the electron excited?


    ok so I first calculated the engery of the electron in the first ground state of the square well using
    [itex]\frac{1240^{2}}{8*511000*1.10^{2}}[/itex]
    I get .310 eV

    Then i calculate the energy of the photon
    [itex]\frac{1240}{600}[/itex]

    and i get 2.066 eV

    Then i do
    2.066=[itex]\frac{1240^{2}}{8*511000*1.10^{2}}[/itex]*[itex]n^{2}[/itex]-.310
    and solve for n i get the 3rd state. that answer says it should be in the same state n=1, I am not sure what I am doing wrong.
     
  2. jcsd
  3. Apr 27, 2013 #2

    mfb

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    Staff: Mentor

    If that fraction is 0.31 eV, how did you get n=3 as result? The solution differs significantly from 3 - indicating that the electron does not absorb this wavelength at all.
     
  4. Apr 27, 2013 #3
    The .310 I calculated using the energy for a particle in a box E=[itex]\frac{h^{2} c^{2}}{8*m*c^{2}*L^{2} }[/itex]*[itex]n^{2}[/itex], for n=1 for a the electron in ground state. I got n=3 by using E(photon)= (energy in a box) [itex]\frac{h^{2} c^{2}}{8*m*c^{2}*L^{2} }[/itex]*[itex]n^{2}[/itex]-Energy of the electron in ground state (.310). I just then solved for n.





    so how does the ratio compare?
     
  5. Apr 27, 2013 #4
    Does the photon need to have the exact energy level to excite the electron into a different state? since the ground state is .311 eV and the 1 st excited state is 1.244 eV and the 2nd excited state is 2.799 for this length of box. And because a single photon of the light has 2.066 eV and does not match any of the other excited states it does not get absorbed?
     
  6. Apr 28, 2013 #5

    mfb

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    Staff: Mentor

    Pretty much.
    Exactly.
     
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