# Relativistic electrons

A. An electron with a rest energy of 0.511 MeV is accelerated to a kinetic energy of 4.50 keV. What is the electron’s speed? Compare the kinetic energy to the rest energy to see if you need relativity.

B. For the kinetic energy of the previous problem, what is the electron’s momentum in units of MeV/c (do not enter any units)?

A) Kinetic energy $(E c )<< Rest-energy (m_0 c^2)$. It means you can safely apply the non-relativistic approximation.

(A relativistic computation is needed only for cases when $E c > (m0*C^2)/10$

This is because $E tot = E c +E0$ or $m c^2 = Ec + m0*c^2$.

In cases when $E c/m0*c^2 <<1$ one can write $m c^2 = approx. m0*c^2$

$E c = m0*v^2/2$

$m0 = (rest-energy)/c^2 =0.511*10^6*1.6*10^{-19}/9*10^{16} =9.08*10^{-31} =9.1*10^{-31} kg$

$v = \sqrt{2Ec/m0} = \sqrt{(2*4.5*10^3*1.6*10^{-16}/9.1*10^{-31})} =$

$=39779612,6 m/s =3.978*10^7 m/s$

B) Total energy E = Kinetic energy + rest energy $= 280 +0.511 = 280.511$ MeV

$E = \sqrt{(P^2*C^2 + m0^2*C^4)}$

$E^2 = P^2*C^2 +m0^2*c^4$

$P^2*C^2 = 280.511^2 – 0.511^2 =78686.16 Mev^2$

$P*C = 280.510 Mev$

$P = 9.35*10^{-7} (MeV/C)$