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Broglie wavelength

  1. Aug 2, 2008 #1
    Helium becomes a super°uid at temperatures T <
    2:18 K. A super°uid °ows with no viscosity. This behaviour
    can only be explained using quantum physics and it can
    only happen if the de Broglie wavelength of a helium atom,
    of mass m, is comparable to the inter-atomic spacing of the
    °uid. What could be an expression for
    the de Broglie wavelength?

    Can anyone tell me where to learn this?

    I'm currently working on Fundamentals of Physics, would I understand this when I finish the book? What level is this?

    Thanks in advance
  2. jcsd
  3. Aug 2, 2008 #2


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    The debroglie wavelength is a concept in modern/ quantum physics. If you have an edition of Fundamentals that includes a section on modern physics, then the book should cover the DeBroglie wavelength.

    In short, the Debroglie wavelength expresses the wave-particle nature of matter by assigning a wavelength to matter particles. A particle's deBroglie wavelength is given by:

    [tex]\lambda_{deBroglie} = \frac{h}{p}[/tex]

    where p is the particle's momentum, and h is Planck's constant.

    Also, for an introductory summary of the topic:

  4. Aug 3, 2008 #3
    I hadn't thought of super-fluids in that way..

    The OP may also want to google for thermal velocity distributions (and Bose-Einstein statistics).
  5. Aug 3, 2008 #4
    oh I know that equation and stuff but I can't get an answer. Its a mutiple choice and I think the (A) Based on it looks like the mv is just changed {answers shown below on the pic I think}.

    Is K a constant?
    Last edited: Mar 5, 2009
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