Ejected Electrons with 0 KE? So I was taught that:
Kinetic Energy (of electron) = Energy (of photon) - Ionization Energy
If E(photon) = IE, then KE=0 of the electron. 
What does this physically/theoretically mean? 
My current thoughts/interpretation is that enough energy/force is applied to ionize the electron so it is 'sufficiently far' from the atom, and then I guess it just moves with whatever speed it is moving at with natural kinetic laws, since no more energy/force is being applied....?
Any clarifications would be greatly appreciated, thanks in advance.
 A: This is the analogue of a projectile getting launched at exactly the escape velocity, something you may remember from studying gravity in freshman physics.
Here we're talking about the photoelectric effect. The electron jumps out of the material into air or vacuum, overcoming the force of attraction that tries to keep it bound inside the material (the force of attraction comes from the surface field, the image force effect, etc.).
If the electron jumps out with too little energy, it cannot escape, but gets pulled right back in. On the opposite extreme, if the electron jumps out with much more than enough energy to escape, it will not only break free of the material but also still energy left over, i.e. it will travel away from the material with a large kinetic energy.
You are asking about the borderline case. Here the electron has just barely enough energy to escape the material, with no energy left over. So it will slow down as it moves away from the material, and get slowed down more and more as it gets farther and farther away. It will never quite come to a stop, but its velocity will approach zero.
A: Just read over this, think the escape velocity makes sense when you look re hash the equation as varying in either time or distance from the gravitational body and think of a single Hydrogen atom with one proton and one electron.
KE(t) + PE(t) = 0
KE(R) + PE(R) = 0
For a constant electron mass m, constant proton mass M, variable distance R, and distance varying velocity v(R).
KE(R) + PR(R) = (1/2)mv(R)^2 -GmM/R = (1/2)v(R)^2 - GM/R
The escape velocity would be the velocity that satisfies the above condition net zero energy condition.
So, v_esc(R) = sqrt(2GM/R)
If we consider the situation where the proton and electron theoretically occupy the same space (R approaches 0) we get.
v_esc(R=0) = inf
Which would mean that no amount of energy would ever separate them.
But another solution exists, when evaluated from some finite offset from the origin of the "gravitational" mass.
KE(R_i) + PE(R_i) = KE(R_f) + PE(R_f)
(1/2)(v(R_i)^2 - v(R_f)^2) = -GM/(R_f - R_i)
v(R_i)^2 - v(R_f)^2 = 2*M/(R_i - R_f)
Now we assume some constant fixed offset greater than zero (surface of proton/earth is different than center of proton/earth), and we want to know what the required initial velocity is to achieve a final velocity of zero as time or displacement approach infinity.
v(R_i)^2 - 0 = 2M/R_i - (2M/R_f -> inf) = v(R_i)^2 = 2M/R_i
v_esc(R_i) = sqrt(2M/R_i)
