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Diameter of Hydrogen

  1. Dec 11, 2006 #1
    1. The problem statement, all variables and given/known data
    The Lyman alpha radiation for hydrogen (nf=2, ni=1) has a wavelength of 1216 angstroms. Using the simplified model of an infinite 1D potential well, derive an estimate of the diameter of the hydrogen atom. How does this value compare with twice the Bohr radius 2r=1.06anstroms?


    2. Relevant equations
    I imagine maybe De Broglie's relation, or the energy of a potential well.


    3. The attempt at a solution
    I have no idea how I can relate a potential well to the radiation given off by hydrogen. I realize that I'm supposed to work this out myself, but my exam is tomorrow morning and would appreciate if I could get the solution, or at least a hefty hint.
     
  2. jcsd
  3. Dec 11, 2006 #2

    Physics Monkey

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    Hi ultimateguy, you know the energy levels of an infinite well, right? What frequency of light would be emitted if an electron made a transition from n=2 to n=1 in an infinite well of size L?
     
  4. Dec 11, 2006 #3
    The energy levels for an infinite well are [tex]E_{n}=\frac{n^2h^2}{8mL^2}[/tex] and the frequency is [tex]\omega=\frac{E}{\hbar}[/tex].

    EDIT: Is the length of the potential well the same as the diameter of the atom?
     
    Last edited: Dec 11, 2006
  5. Dec 11, 2006 #4

    Physics Monkey

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    First, the frequency for n=2 to n=1 is [tex] \omega = (E_2 - E_1)/\hbar [/tex], not just E.

    Second, they clearly want you to approximate the atom with the potential well, so what do you think will correspond to the atom diameter?
     
  6. Dec 11, 2006 #5
    I think that the length of the well approximates the atom. For [tex]E_1[/tex] the wavelength [tex]\lambda=2L[/tex] and for [tex]E_2[/tex] that [tex]\lambda=L[/tex]. Although this is the wavelength of the wavefunction and not the photon.
     
    Last edited: Dec 11, 2006
  7. Dec 11, 2006 #6

    Physics Monkey

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    If the length of the well approximates the diameter of the atom then you should figure out what the length of the well is.
     
  8. Dec 11, 2006 #7
    I tried doing that already. I had [tex]E_2-E_1=\frac{3h^2}{8mL^2}[/tex]. I then substituted into [tex]\omega=\frac{E_2-E_1}{\hbar}[/tex] to get [tex]\omega=\frac{3h\pi}{4mL^2}[/tex] and solving for L I get [tex]L=\sqrt{\frac{3h\lambda}{4m}}[/tex]. I got 7.77x10^-21m using this.
     
  9. Dec 11, 2006 #8

    Physics Monkey

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    You did your algebra wrong. For one thing you forget a factor of c, the speed of light.
     
  10. Dec 11, 2006 #9
    Don't know where c comes in.
     
  11. Dec 11, 2006 #10

    Physics Monkey

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    What are the units of frequency? How did you relate angular frequency to wavelength? What are the units of wavelength? Notice anything? Your current expression does not have the right units, so there is no way you could get some number of meters from it.
     
  12. Dec 11, 2006 #11
    [tex]\omega=\frac{2c\pi}{\lambda}[/tex]. Damn it I'm an idiot.
     
  13. Dec 11, 2006 #12
    My equation for length is [tex]L=\sqrt{\frac{3h\lambda}{8mc}}[/tex]. If I use the proton mass for m I get 7.76x10^-12m. This isn't right, is it?
     
  14. Dec 11, 2006 #13

    Physics Monkey

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    Nope. Why are you using the proton mass?
     
  15. Dec 11, 2006 #14
    I figured it was a hydrogen atom, has one proton. Now I see that it is an electron mass I need to use, and I get 3.325x10^-10m. I guess the electron mass is used because that is what's in a potential well. That must be right :P.
     
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