According to the de Broglie relation, $\lambda=h/p$ where $h$ is the (not reduced) Planck Constant and $p$ is the magnitude of the relativistic 3-momentum. So, it should be that $\lambda$ should be the order of $10^{-15}\,\mathrm{m}$. Thus, the electron's 3 momentum must have the magnitude of about $4.135\,\mathrm{eV\cdot s \cdot m^{-1}}$. Using the relation $E^2=(mc^2)^2+(pc)^2$, I obtain $E=1240.5\,\mathrm{MeV}$. Since the rest energy of an electron is about $0.511\,\mathrm{MeV}$. I conclude that the electron must have the kinetic energy of $1240\,\mathrm{MeV}$. Is my calculation correct?
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$\begingroup$ what is $\hbar c$? All experimentalist have this memorized. $\endgroup$– JEBCommented Oct 6, 2018 at 15:06
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$\begingroup$ Why should I use the reduced Planck Constant? Isn't $h$ what should be used? $\endgroup$– KeithCommented Oct 6, 2018 at 15:07
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$\begingroup$ $E = \hbar \omega $. Scratch that: $p = \hbar k$, and $p$ is the conjugate variable to position. $\endgroup$– JEBCommented Oct 6, 2018 at 15:08
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$\begingroup$ BTW particle physicist have long since figured out how to make life easy on themselves. According to my copy of the Particle Physics Booklet, $\hbar c = 197 \,\mathrm{MeV\,fm}$, which (after you stick that pesky factor of $2\pi$ back in leads to ... $\endgroup$– dmckee --- ex-moderator kittenCommented Oct 6, 2018 at 16:55
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Yes.
We can also use the more direct method of $E = \frac{hc}{\lambda}$ to obtain the answer.
answered Oct 6, 2018 at 16:02
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$\begingroup$ I think that formula applies to only photons. My calculation is for electrons. $\endgroup$– KeithCommented Oct 6, 2018 at 23:39
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$\begingroup$ @Keith: That formula applies to electrons as well. The only condition is that it is a free particle. $\endgroup$ Commented Oct 7, 2018 at 3:25
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$\begingroup$ Why? The formula implies that the particle has speed of the light $\endgroup$– KeithCommented Oct 8, 2018 at 0:40