1. ## Electron velocity

The Fermi Gamma-ray Space Telescope can detect gamma rays with energies ranging from 10 keV to 300 MeV. For each of those energy extremes, find the resulting kinetic energy and speed of an electron created by the gamma ray as part of an electron-positron pair. Assume that the electron has half of the gamma ray's energy.

I did the next thing:

$E=E_{electron}+K_{electron}=E/2+K_{electron}$

From this I got that the kinetic energy and speed of the electron are 5 keV and 4.2x107 m/s, respectively when E=10 keV.

But when E=300 MeV, K=150 MeV, which leads me to get a speed greater than c. Where's the mistake?

2. ## Re: Electron velocity

Originally Posted by Cesc1
The Fermi Gamma-ray Space Telescope can detect gamma rays with energies ranging from 10 keV to 300 MeV. For each of those energy extremes, find the resulting kinetic energy and speed of an electron created by the gamma ray as part of an electron-positron pair. Assume that the electron has half of the gamma ray's energy.

I did the next thing:

$E=E_{electron}+K_{electron}=E/2+K_{electron}$

From this I got that the kinetic energy and speed of the electron are 5 keV and 4.2x107 m/s, respectively when E=10 keV.

But when E=300 MeV, K=150 MeV, which leads me to get a speed greater than c. Where's the mistake?
You probably want $K = ( \gamma - 1) mc^2$. (The relativistic kinetic energy.)

-Dan

3. ## Re: Electron velocity

Originally Posted by topsquark
You probably want $K = ( \gamma - 1) mc^2$. (The relativistic kinetic energy.)

-Dan
Yeah, that's what I had to do. Thank you!