# Energy and momentum of a relativistic electron

The question is to find the magnitude, $p$ of the electron's momentum in the unit of MeV/$c$, given that the kinetic energy of the electron is 2.53 MeV. The answer provided by the book says, \begin{align} p&=c^{-1}\sqrt{E^2-m^2c^4} \\ &=c^{-1}\sqrt{3.04^2-0.511^2}=3.00\,{\rm MeV}/c \end{align}

But I don't understand why I can't do it another way. That is, divide the total energy by the speed of light and get the momentum: \begin{align} p&=\frac{E}{c}\\ &=3.04\,{\rm MeV}/c \end{align}

$E = pc$ is only true for massless particles. For massive particles you have the mass-shell relation:
$E^2 = m^2c^4+p^2c^2$
After you use $E=T+mc^2$ and you can find $p$
• Incidentally, in high-energy particle physics we do often use $p\approx E/c$ even for massive particles, when the particle's mass is so small as to be negligible compared to its kinetic energy. – David Z Jul 1 '15 at 10:32