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De Broglie Wavelength at Relativistic velocity

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle has charge e and masse m0. It is accelerated by a charge V to a relativistic velocity. Show that its de Broglie wavelength is:
    [tex] \lambda = \frac{h}{\sqrt{2m_0eV}} (1+\frac{eV}{2m_0c^2})^{-\frac{1}{2}}[/tex]

    2. Relevant equations
    [tex] \lambda = \frac{h}{p} [/tex]
    Conservation of energy can be used. Our potential energy is:
    [tex] PE = eV [/tex]
    Kinetic energy is:
    [tex] KE = \frac{1}{2}m_0v^2 [/tex]

    3. The attempt at a solution
    I tried using:
    [tex] eV = \frac{1}{2}m_0v^2 [/tex]
    That gave me:
    [tex] \sqrt{\frac{2eV}{m_0}} = v [/tex]

    This matches what is in the denominator below Planck's constant (when multiplied with m0). However, the expression in the parenthesis is what doesn't make sense to me. My answer is:

    [tex] \lambda = \frac{h}{\sqrt{2m_0eV}} (1-\frac{2eV}{m_0c^2})^{\frac{1}{2}}[/tex]
    Why is this wrong?
     
  2. jcsd
  3. Feb 10, 2015 #2

    lightgrav

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    is only correct for NON-relativistic speeds ...
     
  4. Feb 10, 2015 #3

    BvU

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    Here you find the right expression for p to use. It works. I think the difference is a factor gamma somewhere. I can't make out how you go from your expression for v to your last answer for ##\lambda##.
     
  5. Feb 11, 2015 #4
    I found a way that gave me the correct solution. However, I do not understand it.

    [tex] eV = \sqrt{p^2c^2 + m_0^2c^4} - m_0c^2 [/tex]
    This gave me the correct answer for p:
    [tex] p = ((\frac{eV}{c})^2 + 2em_0V)^{\frac{1}{2}} = \sqrt{2m_0eV}\sqrt{1 + \frac{eV}{2m_0c^2}}[/tex]

    However, since the particle is at rest at first, why shouldn't the left hand side of the expression above be eV + m0c2?
     
  6. Feb 11, 2015 #5

    lightgrav

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    ?? where do you think the "- mc2" on the right-hand side came from?
     
  7. Feb 11, 2015 #6
  8. Feb 11, 2015 #7

    BvU

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    There was a Show button on the page where the link in post #2 pointed. Sorry I didn't point that out (or gave that as a link directly)
     
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