Why electron-positron annihilation releases two gamma photons? I have heard that a gamma photon can convert into an electrons positron pair( I don't know how that works), but see when the reverse thing happens shouldn't a single photon be release but in books it is given that during the annihilation two gamma photons are generated,  does it have to do something with momentum conservation?
 A: You should not say "gamma photon". It's either "gamma particle" or, better, just 'photon'.
A single photon cannot decay into a pair of electron and positron both of which would then "fly away". This is indeed a consequence of momentum conservation. Using the more appropriate language of quantum electrodynamics (QED), one can say that in such a vertex:

(which is the only one in QED — here it's just drawn in different ways) at least one of particles has to be off-shell. In other words, it has to be a virtual particle, which does not exist for any measurable time and corresponds to the higher orders of the perturbation theory.
As a consequence, in QED you don't have processes with just one vertex. However, you have few reactions of the second order.
All those can be read off from the following picture (assuming various directions of the time axis):



*

*Left to right: two photons into electron and positron (the latter corresponds to the opposite direction of the arrow)

*Right to left: electron and positron annihilate into two photons

*Up to down: photon-electron scattering

*Down to up: photon-positron scattering


As an exercise, you can think what are the other possible reactions of the second order. Also, you can try to draw a picture corresponding to reaction "$\gamma+\gamma\to\gamma+\gamma$".
The impossibility of a reaction involving three real particles and just a single vertex due to the momentum conservation is a purely classical result. And quantum corrections cannot overcome this anyhow. Having an electron-positron pair, you can always switch to their center of momentum frame, in which $\vec{p}_{\text{total}}$ vanishes identically. However, since $\vec{p}_{\gamma}=\vec{p}_{\text{total}}=0$, one has $E_{\gamma}=\vec{p}_{\gamma}^2=0$ — basically, no photon, the reaction does not exist.
