# Electron matter wave

1. Feb 4, 2006

### Psi-String

Photon's energy $$E=\frac{hc}{\lambda}$$

and photon's momentum $$p =\frac{E}{c} = \frac{h}{\lambda}$$

The textbook say electron's momentum is $$p=\frac{h}{\lambda}$$

I wonder that can the electron's energy be calculated by

$$E=\frac{hc}{\lambda}$$ ?

If it can, what kinds of energy does E involve? It's not only kinetic energy.
But it's interesting that when the electron has a high kinetic energy, say, 1 GeV, I can calculate it's wavelength by

$$\lambda = \frac{hc}{K}$$
though the answer is very very very little different from that calculated by another way:

By $$(pc)^2 = K^2 + 2Kmc^2$$ we know $$p$$
then substitude this p into $$\lambda =\frac{h}{p}$$

What reason cause this difference?

And when the electron's kinetic energy is low, say, 1eV, I got his wavelength by:

$$p= \sqrt{2mK}$$ and substitude this p into $$\lambda =\frac{h}{p}$$

which answer is very different from the answer calculated by$$\lambda = \frac{hc}{K}$$

Why??????????? Is it because when electron has very fast speed, kinetic energy takes most of the part of its total enertgy?

(It's a big mess, if someone can't understand what I'm trying to express, plz tell me)

Last edited: Feb 4, 2006
2. Feb 4, 2006

### Staff: Mentor

No.

The electron has rest mass, the photon does not.

The electron's velocity (speed) is dependent on it's energy, the photons speed is always c, and is completely independent of it's energy.

The total energy of a particle involves kinetic energy and rest energy. Perhaps it is worthwhile plotting total energy, kinetic energy and rest energy to compare their relative values.

3. Feb 4, 2006

### Staff: Mentor

[added] Argh, Astronuc slipped in ahead of me! That's what I get for going out for a cup of coffee while I was typing this. :grumpy:

No, that equation works only for photons and other massless particles. In general, you have to use

$$E=\sqrt{\left(\frac{hc}{\lambda}\right)^2 + (mc^2)^2}$$

See what happens when m = 0?

This equation comes from $E = \sqrt{(pc)^2 + (mc^2)^2}$ upon substituting $p = h / \lambda$.

For a particle with mass, E includes kinetic energy and "rest energy" (the energy associated with the particle's mass).

$$E = K + E_0 = K + mc^2$$

So the relationship between p and K is

$$pc = \sqrt{E^2 - (mc^2)^2}$$

$$pc = \sqrt{(K + mc^2)^2 - (mc^2)^2}$$

$$pc = \sqrt{K^2 - 2Kmc^2}$$

Last edited: Feb 4, 2006
4. Feb 4, 2006

### Psi-String

I got it. THANKS guys!