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Well, as I am learning about quantum physics, one of the first topics I came across was De Broglie's wave equation. $$\frac{h}{mc} = \lambda$$ As is obvious, it relates the wavelength to the mass of an object. However, what came to my mind is the photon. Doesn't the photon have zero mass? Therefore, won't the wavelength be infinity and the particle nature of the particle non existent? Pretty sure there is a flaw in my thinking, please point it out to me!

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I've linked an existing question that explains the de Broglie wavelength for photons. Basically $\lambda = h/p$ and photons have a non-zero momentum. –  John Rennie Jun 18 '14 at 5:44
Yeah... I read the post, but I don't think it answers my question. –  Gummy bears Jun 18 '14 at 5:45
@JohnRennie I don't think it's a duplicate either, since the root of the confusion here is the use of $h/mc$ rather than $h/p$. –  David Z Jun 18 '14 at 5:46
@DavidZ I knew that it is momentum that is supposed to be in the equation, however I thought that momentum can be replaced with $mv$ for every object. However, I didn't know that there is a different equation for momentum of a photon. –  Gummy bears Jun 18 '14 at 5:59

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What you have there isn't actually de Broglie's equation for wavelength. The equation you should be using is

$$\lambda = \frac{h}{p}$$

And although photons have zero mass, they do have nonzero momentum $p = E/c$. So the wavelength relation works for photons too, you just have to use their momentum. As a side effect you can derive that $\lambda = hc/E$ for photons.

The equation you included in your question is something different: it gives the Compton wavelength of a particle, which is the wavelength of a photon that has the same electromagnetic energy as the particle's mass energy. In other words, a particle of mass $m$ has mass energy $mc^2$, and according to the formulas in my first paragraph, a photon of energy $mc^2$ will have a wavelength $\lambda = hc/mc^2 = h/mc$. The Compton wavelength is not the actual wavelength of the particle; it just shows up in the math of scattering calculations.

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I see, thank you for pointing that out. Another small question I have is if the Heisenberg Uncertainty principle states that it is impossible to find the exact momentum and velocity of a particle or it is extremely difficult? –  Gummy bears Jun 18 '14 at 5:47
No, the uncertainty principle is a statement about waves. It only addresses the correlated uncertainties of position and momentum (in a particular technical sense), but it says nothing about the uncertainty in momentum by itself. –  David Z Jun 18 '14 at 6:33

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