Pair production: photon collides with an electron During pair production, a high energy photon collides with an electron which then creates an electron–positron pair, but what happens to the electron that the photon collides with?
I am asking because I am trying to set up the laws of conversation of energy but I am not sure what happens to the electron after the collision.

Suppose the high energy photon collides with the electron which then produces the electron-positron pair, will the conversation of energy look like the following:
$$E_{photon} + E_{initial \ electron} = E_{electron \ pair} + E_{positron \ pair} + \frac{1}{2}KE_{electron \ pair} + \frac{1}{2}KE_{positron \ pair}$$
$$\therefore E_{\lambda} + mc^2 = 3E$$
 A: Nothing happens to it.
The electron - or the nucleus or whatever other object that interacts with the photon to make pair production possible - just participates for momentum conservation, since a single photon producing two massive particles is kinematically forbidden. The usual pair production reaction is 
$$ \gamma + Z \to f^+ + f^- + Z,$$
where $Z$ is e.g. an electron or a nucleus and $f$ is some charged particle being produced.
A: An electron does not create an electron-positron pair. A photon creates an electron-positron pair. For a photon to do this it must be "off mass shell". It must be a high energy product of an earlier collision.
A: A photon does not need to collide with an electron to have pair production.
During pair production, a high energy photon (near a nucleus) ceases to exist, and the photon's energy is transformed into a electron-positron pair. Energy needs to be conserved, thus, the energy of the incoming photon needs to be at least the energy (rest masses) of the electron-positron pair.

For pair production to occur, the incoming energy of the interaction must be above a threshold of at least the total rest mass energy of the two particles, and the situation must conserve both energy and momentum.[1]

https://en.wikipedia.org/wiki/Pair_production
Now momentum needs to be conserved too, that is why the incoming photon needs to be near a nucleus, which receives a recoil.
