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Calculation of wavelength of EMW produced due to an accelerated charged particle

  1. Aug 5, 2009 #1
    if an electron is accelerated by 2eV energy. then what will be the wavelength of emw produced? can we use E=hc/wavelength here? should we use always de broglie hypothesis for this? what will be the difference between wavelength of electron and wavelength of emw? are they same or not?
     
  2. jcsd
  3. Aug 5, 2009 #2
    This is diving into particle/wave duality. Yes, a moving electron has a wavelength. Yes, a photon has a wavelength. Yes, if a moving electron suddenly comes to a halt it will release a photon. The connection between all of these concepts is through the de Broglie relations.

    The two universal laws that are the same for both massive particles (electrons, He atoms, etc) and for EM waves are:

    [tex]E = h f \quad \&\quad p = \frac{h}{\lambda}.[/tex]

    For EM waves, the relation between energy and momentum goes as:

    [tex]E = h f = \frac{h c}{\lambda} = c\, p.[/tex]

    Where for momentum the energy and momentum relation is the more familiar:

    [tex]KE = \frac{p^2}{2 m}[/tex]

    So to answer your question:
    If an electron is traveling with an *Energy* of 2eV and releases all of it into 1 photon with energy 2eV, then they will have the same wavelength.
     
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