I have read that photons can be absorbed by the protons inside the atom and this would result in a fission, nucleon ejection, or photon emission. Assuming the photon has sufficient amount of energy, does the nucleus have to undergo a fission? Is it possible that the photon is emitted back just like nothing happened?
Bonus: If the reaction / result is random, what is the statistical data on it?
Start by considering an isolated hydrogen atom absorbing a photon with an energy of 10.2 eV i.e. undergoing a transition from the 1s to 2p level. The 2p state decays back to the 1s state by emitting a photon because it cannot do anything else. There is simply no other way to release the energy.
But now consider a proton absorbing a photon with an energy of 294 MeV and transitioning to a delta particle (specifically to the $\Delta^+$). It will eventually decay back to a proton and release the 294 MeV, but this is a huge amount of energy. For comparison medical X-ray photons have an energy of a few tens of keV, so the energy released in a delta particle decay is ten thousand times greater.
And in particle reactions energy can be converted to matter in accordance with Einstein's famous equation $E=mc^2$, so while the decay could release the 294MeV as a photon it could also convert some of that energy to particles and release the particles instead. An electron has a mass of only 0.511 MeV, a muon is 106 MeV and a pion is 135 MeV so the energy could be converted to any of these particles as well as a photon. The result is that there are lots of different ways a delta particle can decay, and it is a general rule that in particle physics if a process can happen then it does happen. It's just a matter of what the probability of the process is.
The various ways particles can decay are tabulated in the particle data group handbook. For the delta that we get by exciting a proton (1232 MeV mass) the most probably decay is to release a pion. The 294 MeV decay energy goes into the mass of the pion and the kinetic energy of the pion and proton after the decay. The next most probably decay is to release a photon, but this happens only 1% of the time.
The PDG doesn't list any other decay options for the 1232 MeV delta. There may be other decay modes that are so unlikely they are never observed. The delta is only the lowest excited state of a proton and there are many other higher energy states. See the question Can neutrons and protons have excited states? for more on this. The higher energy states have even more energy and therefore even more different ways to decay.
It is a fairly general result that any process that can absorb quanta (such as photons) inelastically can also scatter the same quanta, and the scattering cross section $\sigma_{sc}$ is always at least as great as the inelastic cross section $\sigma_{in}$.
For a nucleus, this means that any nucleus that can absorb a photon of energy $E=\hbar\omega$ will also scatter photons of energy $E$. Scattering by something like a nucleus can be viewed as a coherent absorption and reemission process. An incoming electromagnetic wave sets the charged protons in the nucleus oscillating, transferring energy to the nucleus. This energy can be further transferred to other nuclear modes, which may lead to fission or other further nuclear processes. However, the most likely outcome, after the nuclear protons are set oscillating at the photon frequency $\omega$, is that they will reemit a photon with that same frequency. And this is exactly what constitutes a scattering processes: an incoming photon is replaced by an outgoing photon of the same frequency.
