Consider the inelastic collision e−+e−→e−+e−+e−+e+ in which an electron-positron pair is produced in a head-on collision between two electrons moving in opposite directions at the same speed.
What is the speed of an electron with the threshold kinetic energy?
To calculate the speed of an electron with the threshold kinetic energy in the given inelastic collision, we need to consider the conservation of momentum and energy.
Let's denote the initial electron velocity as "v" and the final electron velocity as "v'". The mass of an electron is "m", and the speed of light is "c".
Conservation of momentum: In this head-on collision, the initial momentum of the system is zero, and the final momentum is four times the momentum of an electron:
0 = 4mv' - mv
Simplifying the equation gives:
4mv' = mv
v' = v/4
Conservation of energy: The threshold kinetic energy corresponds to the minimum energy needed to produce the electron-positron pair. This energy is equal to twice the rest energy of an electron (2mc^2):
KE = 2mc^2
The kinetic energy of an electron can be calculated using the relativistic kinetic energy formula:
KE = (γ - 1)mc^2
where γ is the Lorentz factor defined as γ = (1 - v^2/c^2)^(-0.5)
Setting this equation equal to 2mc^2 and solving for v gives:
(γ - 1)mc^2 = 2mc^2
γ - 1 = 2
γ = 3
Now we can calculate the speed of an electron with the threshold kinetic energy:
γ = (1 - v^2/c^2)^(-0.5)
3 = (1 - v^2/c^2)^(-0.5)
Squaring both sides of the equation gives:
9 = (1 - v^2/c^2)^(-1)
1 - v^2/c^2 = 1/9
v^2/c^2 = 8/9
v^2 = (8/9) * c^2
v = (8/9)^(1/2) * c
Therefore, the speed of an electron with the threshold kinetic energy is v = (8/9)^(1/2) * c.