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Wavelength - differences in equations

  1. Feb 19, 2008 #1

    If I wanted to find a wavelength of electrons having an average speed of 1.7e+8, would use lamdba = h/ mv or lamdba = hc/ E? I noticed that the 2nd eq is used mostly for transition states when an electron either falls or jumps from its n shell. What's the major difference?

    Thanks in advance
  2. jcsd
  3. Feb 19, 2008 #2
    In the beginning of 20th century the interplay of special relativity of Einstein and the photon quantization hypothesis of Planck (that light is emitted or absorbed in parcels of energy E = h nu, called photons) lead to the formula

    lambda = h / p

    applicable to photons. Around 1922 de Broglie guessed that material particles like electrons may also have a wave associated with them so he postulated the same exact formula applies to electrons and other material particles.

    The moment of the mass zero relativistic photons is p = E/c which leads to lamdba = hc/ E.
    The moment of non-relativistic electrons is p = mv, leading to lamdba = h/mv.

    Your two formulas are just two particular cases of the same master formula.
    Last edited: Feb 19, 2008
  4. Feb 19, 2008 #3


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    Staff: Mentor

    I assume the speed is in m/sec.

    Neither one. The correct starting point for the wavelength is [itex]\lambda = h / p[/itex] where p is the momentum of the particle.

    [itex]\lambda = h/mv[/itex] uses the non-relativistic momentum [itex]p=mv[/itex] instead of the relativistic momentum [itex]p = mv / \sqrt {1 - v^2/c^2}[/itex]. Your speed is more than half the speed of light, so it makes a significant difference.

    [tex]\lambda = hc/E[/itex] applies only to massless particles like photons, for which [itex]E = pc[/itex], that is, [itex]p = E/c[/itex].
  5. Feb 20, 2008 #4
    Thanks guys

    I'm also wondering about what electron transitions correspond to a UV light coming out from a hydrogen atom. How do I calculate that? And how do I differentiate that from the visible light?
  6. Feb 20, 2008 #5
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