I have seen the equation for de Broglie wavelength derived through equating Einstein's $E=mc^{2}$, and Planck's $E=hf$, using a substitution from $c=f\lambda$ to make things in terms of wavelength. From this the following result is derived: $$ \lambda = \frac{h}{mc} $$ This makes sense to me under the assumption that such a wavelength exists. However, at this point a substitution is made for $c$ changing it to $v$. How is this valid? The speed of light is a specific velocity and in the case of Einstein's mass energy equivalence is important that it is such, and not just any velocity. Is there a good reason for this? And if not and this is just meant to be a dumbed down version to give to high school students, does anyone have a correct way of arriving at the desired result?
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1Does this help? Proof of the de Broglie wavelength for electron. – Philip Jun 27 '20 at 05:32
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You could read it from the man himself, to really go deeper, amazing how he brought such insight just with his PhD thesis: http://aflb.ensmp.fr/LDB-oeuvres/De_Broglie_Kracklauer.pdf – UrasGungorPhys Jun 27 '20 at 05:47
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Related : About de Broglie relations, what exactly is E? It's energy of what?. – Frobenius Jun 27 '20 at 08:09