# Electron energy from beta decay

I read in my IB-physics book that the average energy for an electron in the beta decay of Potassium-40 is 0.44 MeV. However this would imply the electron have a velocity of 3.9E8 m/s, i.e. faster than light. So I figured this 0.44 MeV must be the sum of its mass-energy and kinetic energy. But then I calculated the mass-energy of the electron and that equals to 0.51 MeV.

What am I missing here? What does the statement 'average energy for an electron' mean in this instance?

• How are you calculating the speed of the electron? Are you using this: en.wikipedia.org/wiki/Electron#Motion_and_energy ?
– Matt
Jan 14, 2016 at 9:15
• Ah, no, I used the classical formula. Silly me. Jan 14, 2016 at 9:34

The question has been answered in a comment, but for completeness the problem with your calculation is that you need to use the relativistic expression for the kinetic energy. The non-relativistic expression:

$$T = \tfrac{1}{2}mv^2 \tag{1}$$

is an approximation that only works well for velocities well below the speed of light, or put another way it only works well for kinetic energies well below the rest mass energy $$mc^2$$. For these higher velocities we need to use the relativistic expression for the total energy:

$$E^2 = p^2c^2 + m^2c^4 \tag{2}$$

where $$p$$ is the relativistic momentum:

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

and the kinetic energy is then:

$$T = E - mc^2 \tag{3}$$

Alternatively equations (2) and (3) can be combined and rearranged to yield:

$$T = (\gamma - 1)mc^2$$