# De broglie equation

What is the de Broglie wavelength? Also, does the $\lambda$ sign in the de Broglie equation stand for the normal wavelength or the de Broglie wavelength? If $\lambda$ is the normal wavelength of a photon or particle, is $\lambda \propto \frac{1}{m}$ true?

Can a wave to decrease its amplitude or energy with the increase of mass?

(My chemistry teacher told me that all matter moves in the structure of a wave and because of de Broglie's equation matter with less mass shows high amplitude and wavelength while matter with great mass shows less amplitude and wavelength. So as a result of that we see that objects like rubber balls, which have a great mass relative to the electron, move in a straight line because of its mass. However, I haven't found any relationship between wavelength and amplitude.)

• Hi Shawn, and welcome to Physics Stack Exchange! It seems that you're asking a few questions in one here. Could you try to focus your question to ask one specific thing? – David Z Feb 25 '13 at 18:47

The sign $\lambda$ in de Broglie's equation, $$\lambda=\frac h p,$$ is indeed the de Broglie wavelength of the object involved. It is the only wavelength one can meaningfully give a material particle, and "normal wavelength" is meaningless in that context. de Broglie's relation is also true for a photon (though it amounts to a calculation of a photon's momentum), where the wavelength is the usual wavelength of light.
Though this is a flawed picture, you can see de Broglie's relation as describing a "wave of probability" associated with the material particle. This wave must have total "weight" equal to $1$, which means that the particle must be somewhere. (The relative "weights" of the wave in different volumes give the relative probabilities of us finding the particle there.) Particles with higher mass have smaller de Broglie wavelengths, so they can be localized better. If they are located in smaller volumes their amplitudes (specifically: their probability densities) must be bigger, to keep the total "weight" equal to $1$.