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De Broglie Wavelength of Electron

  1. Mar 22, 2007 #1
    1. The problem statement, all variables and given/known data
    Ok, question is: " Calculate the de Broglie wavelength for an electron that has kinetic energy a)50.0 eV b) 50.0 keV and c) 3.00 eV d) What If? A photon has energy 3.00 eV. Find its wavelength.

    2. Relevant equations


    1/2 m(v^2)

    *lambda* = planck's constant / momentum

    3. The attempt at a solution

    Was just going to use E = hf then find wavelength from f, but realized that v is unknown unless I can use classical equation K=1/2m(v^2) but I was not sure I can do this??? If I can, problem is easy, if I can't... will need some help... Thanks!
  2. jcsd
  3. Mar 22, 2007 #2
    You have the energy, what do you want to find v for? Use the planck's constant with eV in it, and your units should work out.
  4. Mar 23, 2007 #3
    Are you familiar with the De Broglie relationship, or are you studying in advance? ;-)

    You need to use the final 2 equations you stated.

    From E=1/2 * m * v^2

    2E = m v^2
    2Em = (m v)^2
    2Em = p^2 p=momentum
    p = (2Em) ^ 0.5
    h/lambda = (2Em) ^0.5

    Should be easy to find wavelength with the above equation.
  5. Mar 23, 2007 #4
    mindscrape - don't need v, need v to find p, or at leaast i thought i needed it until I see that I can do what QuantumCrash suggests...

    Was not sure if I could use de Broglie wavelength equation WITH Kinetic energy of a particle equation. i.e. Wasn't sure if i could consider electron particle AND wave in the same situationi. I thought maybe had to consider electron only as wave or only as particle depending on situation. But now I see can do both, Thanks for the help!!
  6. Mar 23, 2007 #5
    hahaha just realized that this is what de Broglie wavelength is all about anyways! wave-particle duality. wow, funny
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