When an energetic incoming electron hits another electron at rest (the target),

Question

When an energetic incoming electron hits another electron at rest (the target), some of the energy can be converted into mass in the form of an electron-positron pair. So the reaction is e-+e-_ e-+e-+e-+e+ . The minimum energy to do this is not simply 2mec2, because momentum conservation does not allow all of the kinetic energy to be converted into mass. It turns out that the minimum kinetic energy for which the reaction works is in the case where all four particles have the same velocity and are moving in the same direction. Using both energy and momentum conversation, find the kinetic energy for the incoming electron for this minimum-energy case. Express as a multiple of the electron’s rest energy, i.e. KEmin=(?#)mec2.

Explanation / Answer

from the given data
the reaction is

e- + e- --> e- + e- + e- + e+

now, the minimum kinetic energy required for this reaction to proceeed be KE
then let rest mass of positron and electron be m
then

given that speeds of all the four particles inthe products is v and in the same direction
then
from conservation of energy

2mc^2 + KE = 4mc^2 + 4*KE'
where KE' is KE of product particles

also, if the initial electorn momentum is p
KE = E - mo*c^2
E = sqrt(p^2c^2 + mo^2c^4)
(KE + mc^2)^2 = p^2c^2 + m^2c^2 = KE^2 + m^2c^4 + 2mc^2*KE
p = sqrt(KE^2 + 2mc^2*KE)/c
from conservation of momentum
p = 4p'
where p' is momentum of each product particle

sqrt(KE^2 + 2mc^2*KE)/c = 4*sqrt(KE'^2 + 2mc^2*KE')/c
(KE^2 + 2mc^2*KE) = 16*(KE'^2 + 2mc^2*KE') = (4mc^2 + 4KE')^2 - 16m^2c^4
KE(KE + 2mc^2) = (2mc^2 + KE)^2 - 16n^2c^4
(KE + 2mc^2)(KE - 2mc^2 - KE) = -16m^2c^4
KE + 2mc^2 = 16m^2c^4/2mc^2 = 8mc^2
KE = 6mc^2

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