Why do electrons jump between orbitals? When an electron is excited to higher energy levels, it will jump back to the same level from which it was excited. Why does it develop "sentiment" with that level?
 A: I don't know if I'm understanding your question right, but I think you are trying to pose a deeper question than it might seem at first sight... In ordinary quantum mechanics, when you study the hydrogen atom, you derive a set of solutions for the electron wavefunction using the Schrödinger equation (with different values of the energy). These are the different orbitals in which the electron can live.
Now, in principle, these states are stationary, i.e., once you place an electron in an orbital, say the one defined by the quantum numbers $n=2$, $l=1$, $m=0$; the wavefunction does not change with time (technically, only its modulus squared is constant, but that's only a small detail since the energy remains the same), so basically the electron never returns to the ground state according to this theory. Obviously this must be wrong since we know that the electron does return to the ground state (or at least to a state with lower energy), and quite fast. The key point here is that the hydrogen atom is not an isolated system: even in vacuum, when you consider the electromagnetic field from a quantum field theory point of view you get that this vacuum field in some sense interacts with the hydrogen atom, causing the electron to decay to a lower energy state. I'm not an expert in these matters, so I won't try to explain in detail what this means basically because I have never worked out exactly how the process goes (maybe somebody else could develop this point?), but you can read something more here:
https://en.wikipedia.org/wiki/Spontaneous_emission#Theory
A: This is the hydrogen atom energy level solutions, as an easy example.

The electron sits at the ground state, and can be kicked up to an excited state by the appropriate photon i.e. given that the photon has the quantized energy  needed. For each energy level one can calculate using the solutions of the Schrodinger equation, the probability for the electron to fall down to one of the lower energy levels, releasing a photon with the energy difference. It does not have to be the ground state, there could be cascades. This happens because the potential is attractive. If it were not for the quantization of the energy levels, the electron would radiate continuously and would fall into the proton and become a neutron. Quantization assures the stability, but still the attractive force ensures that there is a high probability of filling lower energy states, and rest at the ground state.
In complicated atoms and molecules the quantum numbers of the states become important and the probabilities of decay may depend on angular momentum states and spin states etc ( including the Pauli exclusion principle, quantum mechanics needs to be studied in depth to really understand the subject)
A: The statement in the question is not fully correct. When an electron in a system is excited from a lower to a higher energy level the system becomes excited. That is to say, system reaches to a state which is not favourable. This is an unfavourable state for the system because the excited electron leaves a hole behind.
There are several different mechanisms for the system to relax. The "drop" of the initially excited electron back to its original level is just one of the options. Some other electron can actually fill the hole by emitting a photon with suitable energy. The system can emit an Auger electron. If the system is a molecule it can even dissociate. The probability for these different relaxation "channels" can be calculated.
A: An electron in a higher energy orbital does not necessarily decay directly into the ground state. If an electron in a hydrogen atom is excited into, say, the $n=4$ energy level, all of the following are possible decay paths:


*

*$4 \rightarrow 3 \rightarrow 2 \rightarrow 1$

*$4 \rightarrow 3 \rightarrow 1$

*$4 \rightarrow 2 \rightarrow 1$

*$4 \rightarrow 1$


Each of these decay paths release a number of photons equal to the number of arrows. In fact, the helium-neon (HeNe) laser uses the $n=3 \rightarrow n=2$ transistion of neon to emit light.
