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Transitional state of electrons.

  1. Apr 3, 2005 #1
    This is how I went about it but I am unsure if it is correct..
    An n=3 to n=1 transition from an electron trapped in a rigid box produces a 240 nm photon. What is the width of the box?


    λ = hc / E:

    E = hc / λ

    E = (1.06E-34 J*s * 3E8m/s) / 240nm

    = 1.33E-19 J

    E = h2 / 8mL2

    L2 = h2 / 8mE
    L = √ h2 / 8mE

    L = √ ((1.06E-34 J*s) 2 / (8*9.11E-31kg*1.33E-19 J))

    = 1.08E-10 m
     
  2. jcsd
  3. Apr 3, 2005 #2

    Andrew Mason

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    Homework Helper

    You are missing a factor of 8:

    Allowed values of electron momentum are:

    [tex]p = \frac{hn}{2L}[/tex]

    Since [itex]E = p^2/2m[/itex], the allowed energy values are:

    [tex]E = \frac{n^2h^2}{4L^2}\frac{1}{2m} =n^2\frac{h^2}{8mL^2}[/tex]

    The change in energy between n=3 and n=1 results in the photon (E=hv):

    [tex]\Delta E = (9-1)\frac{h^2}{8mL^2} = \frac{hc}{\lambda}[/tex]

    [tex]L^2 = 8\frac{h\lambda}{8cm}[/tex]

    [tex]L = \sqrt{\frac{h\lambda}{cm}}[/tex]

    AM
     
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