Relativity?

A few important points to be noted to approach the problem :

• We want the collision to be as $inelastic$ as possible, so that the energy released can be used to produce the electron -positron pair , thus we can evaluate the $minimum$ kinetic energy

• All the four particles produced move with the same velocity in the minimum kinetic energy case

• The Lorentz factor is given by $\gamma = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$ , where $v$ is velocity of the particle

• The momentum and $total$ energy of a particle moving at a speed $v$ comparable with $c$ is given by $\gamma mv ~~ and ~~\gamma mc^2$

• Apply momentum conservation, keeping in mind the fact that you need to include the factor $\gamma$ in your equation, and that this factor will not be the same for the new set of particles released .

• Apply energy conservation keeping in mind that the $total$ energy is conserved,that is you need to include rest mass energy in your equation.

• For a particle of mass $m$ , the rest mass energy is given by $E = mc^2$ . Thus if the minimum $energy$ that the electron should possess comes out to be $\gamma mc^2$ ,the minimum $kinetic$ energy would be $( \gamma -1 )mc^2$

In the end, you will end up with equations like this :

$\gamma mc^2 + mc^2 = 4m\beta c^2$ ......... ( where $\beta$ is the Lorentz factor for the new set of particles released,it being the same for all of them because of the "minimum" condition given )

$\gamma mu = 4 \beta mv$ ....... (where $u$ and $v$ are the velocities of the incident electrons and the emergent particles respectively )

On solving, $\gamma = 7$

$\therefore \color{#3D99F6}{ k = 7-1 = 6}$

Relevant wiki: Special Relativity

It is important to note the fact that in the case of minimum kinetic energy we may expect the system to have the same momentum for all the particles.means that their velocities must be the same.(This could be proved mathematically as well but since the system has an apparent symmetry in terms of masses, means that the particles are identical in mass, this observation follows).

Next, we go for momentum and later energy conservation.Note that the rest energy is given by E(o)=mc^2.where m is the rest mass and the total energy by $ymc^2$ (y denotes the Lorentz factor). So, applying we have $4y*v*=yv$ (where v* is the common velocity and y* the subsequent Lorentz factor).and we use energy next $4y*mc^2=mc^2+ymc^2$ .

Now, that we have 2 equations and 2 unknown (it may seem 4 but the Lorentz factor is related to velocity as $y=√1-(v/c)^2$ .) that are $v and v*$ (or someone can say $y and y*$ )we can solve for them to obtain $y=7$ . After obtaining $y=7$ the kinetic energy of the initial electron is hence KE= $(y-1)mc^2.$ = $6mc^2$ ..............(Ans)