The strong nuclear force that holds together an atomic nucleus is very strong indeed, but it is of extremely short range, so the total nuclear binding energy is roughly proportional to its mass number. The repulsive electric forces among the protons in the atomic nucleus, however, are of unlimited range, so the total disruptive electromagnetic force within a nucleus is roughly proportional to the square of its atomic number. A nucleus with 210 or more nucleons is so large that the strong nuclear force holding it together can just barely counterbalance the electromagnetic repulsion between the protons it contains. Alpha decay occurs in such nuclei as a means of increasing stability by reducing size. The weak nuclear force, which is responsible for other types of radioactive decay is not directly involved in this process.

But why should alpha particles, helium nuclei, be emitted in preference to, say, single protons or some other type of particle? Part of the answer comes from the fact that wave function symmetry or antisymmetry must be conserved by any spontaneous event. That is, an atomic nucleus cannot by itself change from exhibiting Fermi-Dirac statistics (if it had an odd number of nucleons) to exhibiting Bose-Einstein statistics (if it had an even number of nucleons). The fact that an alpha particle has an even number (4) of nucleons ensures that an alpha-emitting nucleus will remain with an odd or an even number of nucleons after the event, which would not be the case if a proton or some other particle with an odd number of nucleons were involved. The rest of the answer comes from the high binding energy of the alpha particle. A simple computation of the total disintegration energy, which would be given by the equation:

E = (mi - mf - mp)c2

where mi is the mass of the initial nucleus, mf is the mass of the final nucleus after particle emission, and mp is the mass of the particle that was emitted. This equation shows that the disintegration energy for an alpha particle would usually be possible from energy within the nucleus itself, while other decay modes would require extra energy from somewhere else. For example, performing the calculations for Uranium-232, an alpha particle emission would release 5.4 MeV, while a single proton would require 6.1 MeV. Thus an alpha particle emission would be far more likely. This disintegration energy is very nearly the energy of the emitted alpha particle, except that because conservation of momentum must be satisfied, part of this energy becomes kinetic energy of the atomic nucleus that emitted the particle (in layman's terms, the nucleus will recoil). However, because the mass numbers of most alpha emitters exceed 210, much heavier than the alpha particle (with a mass number of 4) this difference is in general quite small.

A problem comes up from these calculations as the typical potential energy barrier produced by the strong nuclear force, keeping an "escapee" alpha particle bound to the nucleus is very high, typically around 25 MeV, roughly the amount of work that must be done against the electromagnetic force to bring an alpha particle from a point at infinity to a position right next to a nucleus but just outside the range of the strong force's influence. An alpha particle can thus be imagined as being inside a box whose walls require at least 25 MeV of energy to penetrate. As we have seen, typical alpha particle energies are only in the range of 4 to 9 MeV, leaving us plenty short of energy.

Quantum mechanics on the other hand, provides us with a ready explanation. The theory of alpha decay, developed in 1928 by George Gamow and others, was hailed as a very striking confirmation of quantum theory. The answer is that the alpha particle manages to escape from the potential field of the strong force by quantum tunneling its way out. The probabilities are extremely small, especially for low-energy alpha particles, but they are not zero. Working out the details of the theory enables one to compute the half-life of the alpha-emitting radioisotope. One of the consequences of the theory is that long-lived radioactive isotopes tend to emit lower energy alpha particles than those that have a short lifespan. For instance, thorium-232, which has a half-life of 13 billion years, emits alpha particles with an energy of 4.05 MeV, while polonium-212 (half-life of 3 × 10-7 second) emits 8.95 MeV alphas.

Alpha particles, due to their great mass and because of their high electric charge (two elementary charges), are not very harmful on external exposure. A piece of paper, clothes, even the layer of dead skin cells on a person are sufficient to stop these particles (they ionize these materials and turn into ordinary helium atoms). They also are of comparatively short range as collisions with air molecules would turn them into regular helium. A typical alpha particle will travel about 10 cm in air before it becomes harmless.

The problems come if an alpha emitter is absorbed internally: they then don't have to travel very far to reach living tissue and can cause serious damage that way, as alpha particles are very strongly ionizing radiation. Radon, a radioactive gas commonly present in uranium deposits, is a particular hazard, as inhaling this gas causes alpha particles to be emitted straight into lung tissue, ionizing the cells and changing their structure. Next to smoking, it is the most common single cause of lung cancer. Depleted uranium is also a big emitter of alpha particles, and dust from depleted uranium armor and munitions is becoming a serious health risk in places like Iraq, Kosovo, and Afghanistan where the United States used these weapons heavily (the US denies this is a health risk of course). Inhaling this kind of dust is even worse than inhaling radon, as radon leaves the respiratory tract when you exhale, but uranium dust stays there. It is also widely believed to be one of the major causes behind the so-called Gulf War syndrome. Let's not even think about what might happen if you ingest an alpha emitter...

Sources:

Arthur Beiser, Concepts of Modern Physics