# Electron Scattering Velocities

So, I'm being taught about the many different types of scattering; one of these is electron scattering. In my notes, it says that the electrons are accelerated to high energies, of around 6 GeV.

After being curious of what velocities these electrons are travelling at, I decided to try and work it out:

• 6 GeV = 9.6x10-10 J
• Assuming the kinetic energy of the electrons, $9.6*10^{-10} = \frac{mv^{2}} {2}$
• After making $v$ the subject, $v = \sqrt{\frac{2E_{k}} {m}} = \sqrt{\frac{2 * 9.6*10^{-10}} {9.11*10^{-31}}}$ = 4.59x1010 ms-1

This sounds absurd, since the calculated velocity is much higher than the speed of light; as far as I'm aware, no object can have a velocity greater than this.

Can anyone spot any errors in my calculations/thinking etc? It's making me slightly confused.

That the velocity comes out close to (or in your case even greater than) the speed of light is an indicator that you have to use the relativistic formulas. As it is explained there in the limit of low energies, when the particles move slowly compared to $c$ the usual formula $T= mv^2/2 = p^2/2m$ is a very good approximation.

To make it short, the right formula to use is

$$E = \gamma m c^2$$

where $\gamma$ is the Lorentz factor

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

• Ah, fair enough. I suppose my error comes from my lack of knowledge; I haven't learnt about relativity yet. Thanks for the answer though. – user193170 Apr 21 '18 at 21:05
• It's worth noting that the relativistic expression is still correct at low speeds. – rob Apr 21 '18 at 22:38

Your mistake is the assumption of total energy is equal to kinetic energy. In this case,the electron has a kinetic energy and a rest energy. Remember $$E^{2}=(pc)^{2}+(mc^{2})^{2}$$