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Calculating energy from de Broglie wavelength

  1. Feb 23, 2015 #1
    Given the relationships: [itex]\lambda = \frac{h}{p} = \frac{h}{mv}[/itex] and [itex]E = hf[/itex] for wavelike non-relativistic matter, and [itex]v = \lambda f[/itex] for a general wave, one can obtain the result:
    [itex]E = \frac{h^2}{m \lambda^2}[/itex].

    Whilst for particulate matter, we have [itex]E = \frac{1}{2}mv^2[/itex], which when combined with the assumptions above gives:
    [itex]E = \frac{h^2}{2m \lambda^2}[/itex] which is the generally accepted answer.

    Does anyone know why these two results differ by a factor of 2 and why the first is incorrect?
     
  2. jcsd
  3. Feb 23, 2015 #2
    I think its because in the first one we also consider the reflected wave other than the absorbed wave!
     
  4. Feb 23, 2015 #3

    mfb

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    Staff: Mentor

    You are mixing formulas for phase velocity and group velocity here in an incorrect way.
    ##v_f = \lambda f## uses the phase velocity, which is always faster than the speed of light for massive particles. The other formulas are for the group velocity, which corresponds to the "motion" of the particle.

    @Faris Shajahan: This has nothing to do with reflection and absorption, all formulas are valid in vacuum.
     
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