Electron frequency and wavelength

[new_id_7]

Hi, I'm studying the wave nature of particles.

My book says that the de broglie wavelength of an electron is h / momentum, or h / (mv).
I also have that E = hf = momentum²/(2m) = 1/2 mv².
I know that for a photon the frequency is speed of light / wavelength.


I'm wondering why the frequency for an electron is not 1/ wavelength. Is there any equation relating f and lambda for a particle (electron)?

Thanks for any help.

 

[Naty1]

DeBroglie predicted the wavelength of matter waves would be the same as for the photon.
The releatonship you want is the one you stated: f = c/lambda

Experimental confirmation is discussed here:
http://en.wikipedia.org/wiki/DeBroglie_wavelength

 

[conway]

Naty's answer seems very wrong to me. The wavelength of a particle is tied to its momentum. The frequency is related to its energy. When you combine the wavelength and frequency to get the apparent velocity of the electron, the answer comes out wrong. You don't get the correct velocity.

The resolution of this paradox comes from a careful analysis of the roles of phase velocity vs. group velocity.

 

[jtbell]

The releatonship you want is the one you stated: f = c/lambda

If c is the speed of light, that works only for photons.

For particles with mass, $f = v_{phase} / \lambda$ which gives

$$v_{phase} = f \lambda = \left( \frac{E}{h} \right) \left( \frac{h}{p} \right)<br /> = \frac{E}{p} = \frac {\sqrt{(pc)^2 + (m_0 c^2)^2}}{p}$$

 

[Naty1]

Duh, seems wrong to me, too; right your guys are!..I should have posted f = E/h
the equations are right there in the wiki reference, too.