What causes radioactivity? Is it a quantum mechanical effect? I'm just curious what causes radioactivity. I've been told that in the case of alpha decay, since the nucleus is quantum mechanical, there is a probability that the configuration of protons and neutrons is in such a way as to have an alpha particle outside the nucleus, and then the electric repulsion pushes it away. Is this true? Does quantum mechanics give a probability of this sort of thing happening which would be connected to the half life? 
Also the second part of my question: why don't we have many other kinds of decay, why alpha particles or beta only (and fission etc. artificially)? Thanks a lot in advance, if you can point to articles or books I can read on the subject it would also be a plus.
Note: for the configuration, I mean some kind of quantum entanglement or uncertainty principle.
 A: Radioactivity is the result from a confluence of  special relativity  and quantum mechanics.
Special relativity introduces the generalized energy, $E=mc^2$ , which allows the energy conservation to count in the sum the rest masses of the particles which comprise a nucleus. In this relativistic energy conservation we find some nuclear isotopes which are at a higher energy level than a possible reorganization of the constituents. 
In classical physics also, if there exists a lower energy state for a system, it is metastable,  the system  will end up at the lowest energy in the end. When working at dimensions of the nucleus, quantum mechanics has to be used to estimate the probabilities of transitioning to the lower energy state. In addition the lower energy channels opened have to conserve the quantum numbers of the system.
The decays can be alpha, beta, or gamma (photons) depending on the type of nucleus and the type of binding and quantum numbers of the energy states.  
Alpha is possible because it is a tightly bound nucleus  and the split into an alpha and the rest has a good probability as long as the energy level is higher than the energy level of the final-nuclei states. The excess goes into kinetic energy.
Beta comes from  nuclei that have either a high neutron count or a high proton count . The repulsion of too many protons makes the lower energy state more favorable, and the decay possibilities of too many neutrons allow for a probability of neutron decay, to bring up the proton numbers. The lifetimes in general are connected to the probability calculated from the quantum mechanical models.
Gamma is more straightforward and comes because a nucleus has been found in an excited state( usually because it is a fragment from another decay) and it goes to a lower energy state emitting a gamma.
Most of the isotopes have been studied and their lifetimes are consistent with quantum mechanical nuclear models
A: You make many questions in one, all of them have their own answer.


*

*Just to clarify, nuclear decay and nuclear reaction are two totally separated and different things. Radioactivity occurs naturally, spontaneously. You have to sit and wait for the nucleus to decay. A nuclear reaction is forced, is something you obtain by, for example, shooting a particle beam into a target. I am going to explain radioactivity, because nuclear reactions is something wide and huge.


Yes, everything gets "quantum" when you go to very small contexts (microscopic contexts, like atoms, nuclei, particles, etc). 
Nucleons inside the nucleus (i.e., protons and neutrons) are bounded particles (bounded by the nuclear potential) organized in quantum shells, in a similar (but not equal) way as the electrons do organize inside an atom. Nucleons are well described by quantum wavefunctions and their associated quantum numbers (energy, spin, nucleus and particle angular moments, etc.). The energies separating those shells are quantized, which means, you can only move through shells (or energy levels) by absorbing or radiating energy (or particles) in discrete values. In some contexts, you can see a particle as a form of discrete energy, the most common example is the photon (particle) = gamma ray (energy). In nuclear decays, there is always some gamma emission happening (unless you decay directly to the lower state of the daughter nucleus). These energies are generally solutions of the Hamiltonian of the system, if you are able to solve the Hamiltonian in the first place, which is not an easy problem, especially for exotic and unknown nuclei.
These energy values separating shells are not the same for the same shells in different nuclei, because when you start adding and removing nucleons, the interactions between them change, and that affects the nuclear potential, which in turn affects the energy separation of the levels (the position where the shells lie, to say it in a "human" language, which get pushed upwards or downwards). When the energy separation between shells increases, that may be a sign of higher stability (because you need more energy to move from one shell to the next), while with short energy separations it's so easy to jump to the next energy level.
Radioactivity is caused because the nucleus gets unstable. There are many reasons why the nucleus can get unstable, and it will try to get rid of that instability and reach the lowest energy state of the system (in this case the system is the nucleus), like everything in Nature.
The common scenario is: you have an unstable parent nucleus that decays. Depending on the case, it emits some particles.
$$
\text{Parent nucleus }\to\text{ daughter nucleus }+\text{ particles or tiny nuclei }+\text{ energy}.
$$
Some of the energy released in the decay is used as kinetic energy by the rests of the decay, but some energy may be saved as excitation energy in the daughter nucleus. Since the nucleons in the daughter nucleus are organized in a certain way according to shells, quantum numbers, energies, etc., that nucleus will have a set of energy levels that is like a footprint, and the nucleons of that nucleus will be promoted to higher energy levels using the excitation energy we mentioned above.
An image is worth a thousand words, so please check this diagram of a well known radioactive reference source, 60Co, where you can see how the parent (60Co) decays into Ni, and Ni has a set of excitation levels where you can end up when coming from 60Co (its complete level scheme is much more complex, but if you are coming from a decay, you can only access or "populate" certain energy levels, while others are forbidden by quantum physics laws).

Image Source: National Nuclear Data Centre
When the daughter nucleus finds itself in an excited state, (one is at 1332 keV and the other is at 2158 keV) it will try to get rid of the instability by emitting particles, in this case, gamma rays of precise (quantized) energies (826 keV, 2158 keV and 1332 keV). This is the footprint of Ni for that decay.
Instabilities in the nucleus can be produced by the imbalance between the number of protons and the number of neutrons. Because the protons repulse each other since they have the same charge, when you have huge nuclei with a huge amount of protons, you need to increase the amount of neutrons to add more nuclear potential to the system, and keep it bounded. That's why light nuclei have the same amount of neutrons as of protons, but heavy nuclei have some protons and a bunch of neutrons. Some of the experiments researchers do, consist of changing that balance by adding or removing particles. If you start adding protons or removing neutrons from a nucleus, the Coulomb repulsion will make it unstable and it will start decaying, spitting protons, neutrons, electrons, positrons, whatever it needs to get rid of. But the interesting fact is that if you add neutrons to the system, it will also become unstable! The limit to which you can add protons or neutrons to a nucleus until it stops being stable and starts to decay emiting nucleons of the same type you are adding, are called the "proton and neutron drip lines".
In the previous example, 60Co has 27 protons and 33 neutrons, those 33 neutrons make the nucleus unstable, so it gets rid of one of them by means of the beta minus decay process, where a neutron decays into a proton, an electron and an electron antineutrino:
\begin{align}
n&\to p+e+\nu
\\
\text{(i.e. }^{60}\mathrm{Co}&\to{}^{60}\mathrm{Ni}+e+\nu\text{)}
\end{align}
The resulting daughter nucleus that has 28 protons and 32 neutrons is 60Ni. Part of the available final energy is shared as kinetic energy by the electron and antineutrino, and part of the energy will be used to excite the Ni, which will eventually gamma-decay. If that configuration of 28 protons and 32 neutrons was still unstable, the Ni will find a way to decay, and you may have a decaying chain, like that of Thorium, Radium, etc.
Alpha, beta and gamma decays are just a way to get rid of that instability, and they are by no way the only decay modes in the universe. There are lots of exotic decay modes when you produce exotic nuclei in an accelerator (or in a star, where they are produced daily). Proton emission, neutron emission, fission, double beta decay, two proton decay, clusterization, halo nuclei, are just some of the ways a nucleus handle these unstable situations. The decay mode that will take place depends on a number of things, energy, number of nucleons, spins, etc. Even temperature may have a role (for example, in stars, I can think of the CNO cycle) The different decaying modes can also compete with each other (for example, alpha and beta, beta and particle emission, etc.).
The nuclear potential describes these behaviors. However, we are very far from having a unified nuclear model or theory. What we have today is a model that describes stable nuclei and their isotopes (the ones you can find in the periodic table of the elements). They have long been studied and are well understood. The liquid drop model and the shell model are two of the first attempts to explain what happens in the nucleus. For example, check the semi-empirical mass formula, which takes into account not only quantum effects, but Coulomb effects originated by the fact that protons are charged particles.


An example of some nuclear levels with spins and parities assigned. The difficult thing is to calculate where those levels are positioned in terms of energy, and what are the values for their spins and parities, etc. This is the problem quantum calculations try to solve when trying to find a unified nuclear potential model that explains what we see in the data. (Image source: Evidence for a spin-aligned neutron–proton paired phase from the level structure of 92Pd, B. Cederwall et al., Nature 469, 68 (2011), arXiv:1101.2187.)

But what happens when you search outside those stable nuclei region? We don't know. The shell model doesn't apply anymore, new shells appear, new configurations. All of them are governed by quantum mechanics and spins, and energies, and wave functions and the like. We have not found a nuclear potential model that explains every radioactive decay in any region of the nuclear chart. You can find all the exotic nuclei known to date in this updated nuclear chart: 

http://www.nndc.bnl.gov/chart/ 

People have even tried to explain the nuclear potential in terms of quarks and have failed. The nucleus being a many body system, is a complicated problem to solve. In particle physics, the theory tells you where to go measure to find a new particle. In nuclear physics, you have to measure first, and then try to find a theory that explains the data. It's a phenomenological field. 
The alpha emission
As I said above, the position of the energy levels in the nucleus and the shells are also affected by another force: the Coulomb force, because you have a charged particle: the proton, and you will have a repulsive force acting against the nuclear force that wants to get the protons together. The story of the nucleus is the story of these two forces acting against each other and finding an equilibrium at the end. The heavier the nucleus is, the more protons it has, and the bigger the Coulomb force is, so the bigger the repulsion is as well. If $Z$ is the number of protons, Coulomb increases as $Z^2$, while the nuclear potential increases as $Z$ (check the Weizsäcker mass formula, the explanation for each term in there will give you an insight of what goes on inside the nucleus).
Since an alpha particle is made of two protons and two neutrons, the nuclear force is stronger than the Coulomb repulsion in order for those two protons to get together and form an alpha nucleus, so it is a very tightly and bound structure. Its emission is spontaneous, since it is the disintegration product that has the lightness mass to released energy ratio (its mass is very small compared to the sum of the mass of its constituents). If you look at a plot of the number of nucleons vs the bounding energy, the helium nucleus has one of the highest ratios, and is also an extremely stable atom in the atomic world.
Coulomb forces an energy barrier, an energy wall that the alpha particle has to go through in order to escape the nucleus. In the alpha decay, this happens thanks to the tunnel effect (search for quantum tunneling in the Wikipedia). The alpha particle has then a quantum probability of crossing the tunnel and being emitted. In my opinion, the book that best explains this decay is "Introductory nuclear physics" by Kenneth S. Krane, around page 251. The discussion there is a quite interesting read.
A: You are not required to regard decay as quantum mechanical (QM) phenomenon. Regard it as a nuclear reaction, which occurs spontaneously. There are similar reactions in chemistry, like ozone decay.
Modern QM is unable to calculate decay probabilities and half-lives, they are only measured experimentally. In principle it is possible to model nucleus in the computer in QM based virtual reality, but today's computers are not powerful enough for this. This is because nucleus is glued bu strong force, which consist of a lot of virtual gluons, which is hard to model precisely.
The nuclear reactions are guided by binding energy, and the law of entropy (the law of minimum of energy).
Exactly like chemical reactions. 
The binding energy is the energy, required to spend to completely destruct a bound object, i.e. to put all it's parts to infinity. 
The entropy law says, that energy wants to dissipate. 
Since many nuclei have some binding energy, this means that nucleons want to bind. Because being bound, they will allow more energy to dissipate.
Below is the plot of binding energy of stable nuclei

Nobody knows why does this plot look like this. If one knows he/she will get a Nobel.
The higher the plot, the more "profitable" nucleus it describe.
The most binding energy is the energy of Iron. 
Nobody knows why. If one knows he/she will get a Nobel :)
This means that all nucleus want to be Iron. This is the reason why all pre-iron nucleus tend to fusion and all post-iron nucleus tend to fission. 
Also note the peak of alpha particle (He4). This is why emitting alpha particle is the most "popular" reaction. Emitting of C12 or O16 apparently not possible.
Nobody knows why alpha particle has such a high binding energy. If one knows he/she will get a Nobel :)
