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Wavelength of the Earth.

  1. Mar 2, 2012 #1
    So I was reading about De Broglie's theory of particle-wave duality the other day and I came across the equation: λ=h/p. I expect most of you are familiar with this equation but if you're not, it is: wavelength = Planck's length over momentum (mass x velocity).

    So I thought i'd try and find the Earth's wavelength.

    So first I found out what I would have to do to find p, which is (5.9742x1024kg * (approx) 3x104 m/s).

    So... 6.626x10-34/(5.9742x1024kg * (approx) 3x104 m/s)

    This is equal to... 3.697×10^-63 meters.

    When I first saw that, I was mind blown. Will someone explain why it's wavelength is that small?

    Thanks, Ben - i'm unsure of this because well, i'm quite young (15).
  2. jcsd
  3. Mar 2, 2012 #2
    The question should not be why the wavelength is small (the numbers simply make it so), the question should be what it means.
    And such small a wavelength means that the Earth (if we consider it as a monolithic ball) is highly unlikely to exhibit any wave-like properties (to interfere, diffract, etc.) and is much more particle-like than wave-like.

    By the way, it is nice to hear of someone who managed to find out about and become interested in De Broiglie's waves at the age of 15. Cheers!

    Oh, and h is not Planck's length, it is Planck's constant. You got the number correct, though. Planck's length is a related but different thing.
  4. Mar 2, 2012 #3


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    No need to worry too much about this. The de Broglie wavelength only starts to show itself for small objects (particles), which really do tend to get bent round corners when they go through a narrow slot. You can spend a lot of time watching people walk through a doorway, waiting for them to be diffracted. It ain't going to happen to any measurable degree :wink:
  5. Mar 3, 2012 #4
    Thanks guys! Slight confusion on Plancks constant, so sorry for that!
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