# Kinetic energy of Electron [closed]

I had to find the kinetic energy of electron with wavelength $2$ pm. I used the formula $$KE = \frac{p^2}{2m} = \frac{h^2}{\lambda^2 2m}$$

which gave me result, $KE = 376.9$ KeV. But the answer given in the book is $292$ KeV. Am I making a mistake in calculating it?

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## closed as too localized by David Z♦Jan 2 '12 at 8:35

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Hi orion - In general, if there's some conceptual issue that you think is causing you to get the wrong answer, you can ask about that, but this site is not meant to be a place to check your work. (That being said, I get the same result as you...) –  David Z Jan 2 '12 at 8:36
I get 292.9keV. Note that this is relativistic case. –  Adam Zalcman Jan 2 '12 at 8:48
@ Adam Zalcman: Thanks. Can you tell me how did you knew that this is relativistic case? Sorry for the ignorance, but I really don't know how to know from the wavelength that an electron is relativistic or not? Please guide me. –  orion Jan 2 '12 at 8:52
The kinetic energy you got using non-relativistic formula was of the same order of magnitude as electron's rest energy (511keV). If that happens, it is worth calculating both and checking how much they agree. –  Adam Zalcman Jan 2 '12 at 8:57
@ Adam Zalcman: Thanks for this important point. –  orion Jan 2 '12 at 9:14