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  What should be the de Broglie's wave-length?

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I have learnt that the de Broglie's as λ=h/p (where h is Planck's constant,p is the momentum of the particle), we can derive it from equating Einstein's mass-energy equivalence and the energy of a photon, E=hν, then replacing 'c' with the velocity of any particle.

Waves are associated with moving particle, so as the wave length. The equation $E=mc^2$ is applicable for particles at rest; I recently found that the general eqn is $E^2=(mc^2)^2+(pc)^2$  Why do we use $E=mc^2$, when $E^2=(mc^2)^2+(pc)^2$is the eqn for moving particles? A particle at rest cannot have a wave length - isn't that so?

Closed as per community consensus as the post is undergraduate-level
asked Oct 13, 2014 in Closed Questions by RogUE (0 points) [ revision history ]
recategorized Oct 14, 2014 by Dilaton

@RogUE Welcome to PO. Try to use LaTeX in your formulae. I have done so in part of your question correcting a missing power in the formula. Yes, $E^2=(mc^2)^2+(pc)^2$ is indeed the correct relativistic energy-momentum (aka dispersion) relation. I like to joke that this should be formula that appears in the T-shirt that has three lines which read: E=ma^2E=mb^2; E=mc^2. Of course, it is obvious that $E=mc^2$ holds for a particle of mass $m$ at rest and is a special case of $E^2=(mc^2)^2+(pc)^2$.  For massless particles like the photon, one obtains $E=pc$ on setting $m=0$. I will let someone else answer the de Broglie wavelength bit.

This is an undergraduate level question, more suitable for physics stackexchange. No downvote, but vote to close.

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