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De Broglie Wavelength of an electron

  1. Oct 15, 2014 #1
    1. The velocity of the electron in the ground state of the hydrogen atom is 2.6 x 10^8 m/s. What is the wavelength of this electron in meters?


    2. De Broglie's equation: lamda = h/p
    p=mv



    3. The attempt at a solution...

    (6.626 x 10^-34) / (2.6 x 10^8 x 9.11 x 10^-31)

    = 2.798 x 10^-12 meters

    This isn't the right answer...obviously...
     
  2. jcsd
  3. Oct 15, 2014 #2

    e.bar.goum

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    Quickly (sorry, posting and running), and with wolfram alpha, I get 1.7 * 10^-12 m. Which doesn't sound too far off to me (and neither does your answer), considering 2.6 * 10^8 m/s is about 500 keV of kinetic energy for an electron.

    Why do you suspect your answer wrong?
     
  4. Oct 15, 2014 #3

    Simon Bridge

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    Is the electron relativistic enough to make a difference?
    Is the wavelength consistent with DeBroglie's hypothesis about the allowed energies of the electron?
     
  5. Oct 16, 2014 #4

    ehild

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    Check the problem text. The speed of the electron is much less than the speed of light in the ground state of the H atom.

    ehild
     
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