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De Broglie wavelength in terms of accelerated potential difference

  1. Mar 9, 2013 #1
    Problem:

    Derive a formula expressing the de Broglie wavelength (in Å) of an electron in terms of the potential difference V (in volts) through which it is accelerated.

    Solution (so far):

    The textbook's answer is the following,

    [itex]\lambda=12.27[V(\frac{eV}{2m_{0}c^{2}}+1)]^{-\frac{1}{2}}[/itex]

    I'm having some trouble getting there. I started by noting that if an electron is accelerated from rest through a potential difference V, it gains a kinetic energy

    [itex]\frac{1}{2}mv^{2}=eV[/itex]

    So therefore,

    [itex]v=\sqrt{\frac{2eV}{m}}[/itex]

    Since [itex]\lambda=\frac{h}{mv}[/itex]

    Therefore, [itex]\lambda=\frac{h}{\sqrt{2meV}}[/itex]

    Also, [itex]\frac{h}{\sqrt{2me}}\approx 12.27 Å[/itex]

    So I get, [itex]\lambda=\frac{12.27Å}{\sqrt{V}}[/itex]

    I don't know where I went wrong. Any suggestions?
     
    Last edited: Mar 9, 2013
  2. jcsd
  3. Mar 9, 2013 #2
    Relativity?
     
  4. Mar 9, 2013 #3
    If I use the formula [itex]m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex] in [itex]\lambda=\frac{h}{\sqrt{2meV}}[/itex]

    everything gets extremely complex and looks nothing like what the answer is supposed to look like. Can you be more specfic?
     
  5. Mar 9, 2013 #4
    I think you should start by writing down the energy relativisticly.
     
  6. Mar 9, 2013 #5
    [itex]\frac{m_{0}v^{2}}{2\sqrt{1-\frac{v^{2}}{c^{2}}}}=eV[/itex]

    Is that what you mean?
     
  7. Mar 9, 2013 #6

    collinsmark

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    In relativity, kinetic energy is [itex] T = \gamma mc^2 - mc^2 [/itex]. Momentum is [itex] \vec p = \gamma m \vec v [/itex], where [itex] \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} [/itex].
     
  8. Mar 11, 2013 #7

    collinsmark

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    By the way, when I say [itex] m [/itex] here, I mean the same thing as [itex] m_0. [/itex] I don't distinguish "mass" from "relativistic mass". Most physicists don't these days. Mass is the rest mass, period. If you prefer, kinetic energy is [itex] T = \gamma m_0 c^2 - m_0 c^2 .[/itex]

    [Edit: by the way, if you're curious, if you take the Taylor series expansion of [itex] \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right) m_0 c^2 - m_0 c^2 [/itex] around [itex] v \approx 0[/itex] you'll find that it reduces to the Newtonian [itex] \frac{1}{2} m_0 v^2 [/itex] for [itex] v [/itex] near 0. Neato.]
     
    Last edited: Mar 11, 2013
  9. Mar 13, 2013 #8
    I have no issue with you referring to rest mass as just mass - as long as it's clear to everyone to what you are referring. Why cause unneeded ambiguity (especially in a section of the forum aimed towards students seeking help, and looking to develop good physics practices)?

    As for the problem I originally asked, if anyone is interested, I obtained the required solution by the following:

    Begin with (pc)^2+(E_0)^2=E^2

    Solve for p and substitute the result into the de Broglie wavelength equation λ=h/p.

    Next, eliminate E by introducing kinetic energy, then use the fact that K=eV.

    Simplify and be sure to express the constant quantity h/sqrt(2em_0) in angstroms (≈12.27Å).

    No thanks to voko or collinsmark! - Just kidding, at least you tried to help. :biggrin:
     
  10. Mar 13, 2013 #9
    Also, it's interesting to note the result reduces to the equation I derived first incorrectly in the event V << c^2.
     
  11. Mar 13, 2013 #10

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    My point (and I assume the same point voko was hinting toward), is "relativistic" mass is not used anymore. Instead, "invariant" mass is used, which is the same thing as the rest mass.

    Kinetic energy:
    [tex] T = (\gamma -1) mc^2 [/tex]

    Total energy of the system:
    [tex] E = \gamma mc^2 [/tex]

    3-Momentum:
    [tex] \vec p = \gamma m \vec v [/tex]

    Relationship between 3-momentum and total energy of the system:
    [tex] p^2 c^2 + m^2c^4 = E^2 [/tex]

    And the mass [itex] m [/itex] is the invariant mass. If you wish to directly substitute [itex] m_0 [/itex] in for [itex] m [/itex], feel free to do so.

    http://en.wikipedia.org/wiki/Mass_in_special_relativity
     
  12. Mar 14, 2013 #11
    Thanks for rewording what I already had stated!
     
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