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Using e = hc/λ to find wavelength of light emitted

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello there, a question that I have been given is to "find the wavelenght of light emitted from an electron jumping from the 2nd to the 6th orbital (or vise versa if that matters). The atom is hydrogen"

    2. Relevant equations
    [itex]E= \frac{hc}{λ}[/itex]


    3. The attempt at a solution
    I've already worked out this question using the Rydberg Formula where I got 410 nm (which I know is correct because I crossed checked it using wikipedia and my text book). However, this is part of a quatumn mechanics assignment, and I was suppose to use the forumla [itex]E= \frac{hc}{λ}[/itex]

    But I have no idea about how to apply the forumla to get the wavelenght. But if I made [itex]λ[/itex] the subject, it would read to be

    [itex]\frac{hc}{E} = λ[/itex]

    I think what I need to find out is the value of E (well, obviously thats what I need to find). But how do I relate the different energy levels with the orbials in which the electron travels?

    Surely its not just 6 -2 = 4 :tongue:

    thanks - miniradman
     
  2. jcsd
  3. May 17, 2012 #2
    The Rydberg formula is based on that formula for the energy of a photon; giving an answer that way ought to be sufficient. The way it is derived is to say that the energy levels are given by [tex]E_n = -\frac{13.6\ \text{eV}}{n^2}[/tex] and then to subtract [itex]E_{n_i}[/itex] from [itex]E_{n_f}[/itex].
     
  4. May 17, 2012 #3

    turin

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    Homework Helper

    No, it is not. It is based on Rydberg's direct empirical observation that wavenumbers come in series. It was basically a matter of (discrete) curve fitting (and almost two decades before Einstein's relation).

    @miniradman,

    Your instructor probably wants you to use the quantum-mechanical fact that the energy levels in the Hydrogen atom are discrete, the Einstein relation that you give, and the conservation of energy (where the electron's energy is lost as a photon). So:

    Ei = (-13.6 eV)/ni2

    Ef = (-13.6 eV)/nf2

    Eγ = (h⋅c)/λ

    Ei = Ef + Eγ

    (Of course, you need the hc to be in eV⋅m, not J⋅m.)
     
    Last edited: May 17, 2012
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