Photon's energy [tex] E=\frac{hc}{\lambda} [/tex](adsbygoogle = window.adsbygoogle || []).push({});

and photon's momentum [tex] p =\frac{E}{c} = \frac{h}{\lambda} [/tex]

The textbook say electron's momentum is [tex] p=\frac{h}{\lambda} [/tex]

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

[tex] E=\frac{hc}{\lambda} [/tex] ?

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

[tex] \lambda = \frac{hc}{K} [/tex]

though the answer isvery very very littledifferent from that calculated by another way:

By [tex] (pc)^2 = K^2 + 2Kmc^2 [/tex] we know [tex]p[/tex]

then substitude this p into [tex] \lambda =\frac{h}{p} [/tex]

What reason cause this difference?

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

[tex] p= \sqrt{2mK} [/tex] and substitude this p into [tex] \lambda =\frac{h}{p} [/tex]

which answer is very different from the answer calculated by[tex] \lambda = \frac{hc}{K} [/tex]

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)

Thanks in advance

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# Electron matter wave

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