Why is momentum not quantised in the photoelectric effect? I know that the momentum transfer mechanism in photoelectric effect is pretty complex etc, but why do we quantise the Energy of the photon but not the momentum ?
Edit: had written conserved instead of quantised
 A: Just because we don't use the equation $$\vec{p}_2=\vec{p}_1+\int_{t_1}^{t_2}\vec{F}\mathrm d t \tag{1}$$
doesn't mean that we ignore momentum. It's simply that there is a large mass (compared to the photoelectron and the mass-energy added by the photon, and that mass absorbs a lot of the momentum of the photon. That integral at the end of (1) tells us how much momentum is transfered into or out of the system. If the mass of the photosensitive substance is included in your system, then the momentum of the system will remain constant. The problem is that the force is very complicated, and we can't measure the momentum of the big mass after the interaction occurs. So, we don't deal with it. Plus, it wouldn't tell us anything useful for the problem.
We can determine the energy of the incoming photon (by its wavelength), the energy of the photoelectron (indirectly by measuring a stopping potential), and the amount of work needed to free the electron (by looking at a variety of wavelengths of photons). And we assume that the kinetic energy or temperature rise due to the interaction is very small compared to the work function.
Basically, we can account for all the energy changes. We can't account for all the momentum changes. But we trust that both energy and momentum (along with their "currents" of work and impulse, respectively) are conserved.
By the way, (1) is the statement of conservation of momentum. And conservation doesn't require constancy within the defined system. Impulse (that integral) tells us how the system momentum changes.
