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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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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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Yes, photons have a de Broglie wavelength, because photons have momentum associated with them when they are in motion even though they don't have a rest mass.

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$c=\lambda\nu$, so $\lambda=\frac{c}{\nu}$ and $\Delta E=\frac{hc}{\lambda}$.

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  • $\begingroup$ Please edit using MathJax. This is a useful skill for this website and the norm. $\endgroup$ – StephenG Feb 12 '17 at 7:12
  • $\begingroup$ I edited your post to use mathjax only to show you, how simple is it. Although I think, it is still not really okay. $\endgroup$ – user259412 Feb 12 '17 at 7:49

protected by rob Dec 11 '17 at 20:54

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