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?


1 Answer 1


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:

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


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