There are many ways unstable nuclei may decay. There are conservation laws which nature obeys in all decay schemes. These are:[list][*][u]Conservation of Energy[/u]: The decay will only occur spontaneously in nature if the rest energies of the products of the reaction are less than the rest energy of the reactant(s). Any missing energy after the reaction is realized by kinetic energy of the products and perhaps photons, which I hope you recall have no rest energy or associated rest mass.[/*][*][u]Conservation of Linear Momentum[/u]: In the rest frame of the initial nucleus the momentum is zero. This being the case, the sum of the momenta (they are vectors) of the products must be zero as well. The particle we call a neutrino was discovered by an apparent violation of energy and momentum conservation in [math]\beta[/math]-decay which we will discuss below.[/*][*][u]Conservation of Angular Momentum[/u]: I mentioned back in mechanics that it's a good thing that we don't call angular momentum something like [math]\vec{p}_{rotational}[/math], because that would imply that it's another form of linear momentum while it's not. So it got a different symbol [math]\vec{L}[/math] and is separately conserved in nature. In the context of particle decay, angular momentum is accounted for by the intrinsic angular momentum or spin of the particles as well as the orbital angular momenta in composite particles. For example, decay of a neutron into an electron and a proton is forbidden since they are all spin 1/2 particles. The spins may add or subtract, but that only leads to spin 1 or 0 for the products.[/*][*][u]Conservation of Electric Charge[/u]: The total electric charge of products must equal that of the reactants.[/*][*][u]Conservation of Nucleon Number[/u]: When nuclear reactions take place the total number of nucleons remains constant. Protons may convert to neutrons or vice versa, but the number A is constant.[/*][/list]We will spend some time in this section discussing some of the decay schemes that obey these conservation laws.

In accordance with the binding energy per nucleon (B/A) discussed earlier, recall that the maximum B/A occurred around Fe-56. Higher values of this ratio suggest more stable arrangement of nucleons. At higher Z values the B/A dropped. So if we start with a large nucleus, and it can break into two smaller nuclei such that that average B/A rises at the expense of lost rest energy, such a reaction will take place spontaneously in nature. A very common case is that a small, stable piece of a nucleus breaks off. The most stable small pieces are really helium nuclei, which in this context are called alpha particles. When such pieces break off, it is called [math]\alpha[/math]-decay (alpha decay). The reaction may be written as follows:[br][br][center][math]^A_ZX_N\rightarrow ^{A-4}_{Z-2}X'_{N-2}+\alpha[/math], in which [math]\alpha = ^4_2He_2.[/math][/center]The energy of such a reaction (and any reaction) may be calculated by the difference in rest energies of reactants minus products. We call this energy [math]Q=\Delta mc^2[/math]. The change in mass is often called the [b]mass defect[/b].[br][br]The process of [math]\alpha[/math]-decay is an example in nature of barrier tunneling. The alpha particles inside the nucleus do not have sufficient energy to escape the strong nuclear attraction of the rest of the nucleus. Therefore decay is a classically forbidden process. However, in line with earlier discussions of barrier tunneling, there is a very low, but non-zero tunneling probability associated with this decay. The higher the associated probability, the shorter the corresponding half-life of the nucleus.

[i]Consider as an example the alpha decay of an unstable nucleus. A way to think of this decay is to think of the alpha particle bouncing off the walls of a box, where the walls are a confinement due to a strong nuclear attractive force between the alpha particle and the rest of the nucleus.[br][br]The solution goes like this: If we treat the alpha particle as a particle in a box of width L = 2r, where r is nuclear radius, then we can find the alpha particle's energy which will be all kinetic. This could be done using either WS or Schrödinger's equation. Knowing the energy will allow us to find the velocity of the alpha particle. Knowing that velocity, we can calculate the time it takes to travel the distance across the nucleus, and therefore very simply we can arrive at the time between its escape attempts. [br][br]Let's do the math now. The energy in the ground state of a particle in a box is [math]E_1=\frac{h^2}{8mL^2}[/math]. Using classical kinetic energy since the alpha particle will not be traveling much more than 10% the speed of light, we can find the speed by setting [math]E=1/2mv^2[/math]. That gives [math]v=\sqrt{2E/m}[/math]. Putting these two pieces togther, we get [math]v=\frac{h}{2mL}[/math]. Lastly, the time between collisions is just the time it takes an alpha particle traveling at that speed to cross the nucleus (a distance L). That time is [math]\Delta t=L/v = \frac{2mL^2}{h}[/math]. Knowing this and the tunneling probability, we can calculate the decay constant of a nucleus which is just [math]\lambda = \frac{\text{tunneling probability}}{\text{time between attemps}}=\frac{Prob}{\Delta t}[/math]. Knowing the decay constant also means we can calculate the half-life. Naturally to find a numerical result, we need to calculate the tunneling probability, which can be done if we know the barrier height and width as seen in the quantum mechanics chapter.[/i]

Left on their own, neutrons will decay with a half-life of around 10.3 minutes. The reason for this is that the mass of a neutron is higher than the mass of a proton by a tiny bit. Nature would prefer to change the neutron into a proton along with some other required pieces to conserve charge and other quantities listed above. The reaction looks like this: [math]n\rightarrow p+e+\overline{\nu}_e[/math], which means that the neutron decays into a proton, an electron and an electron antineutrino. (Can you determine the neutrino spin?) In general, [math]\beta[/math]-decay may be written: [br][br][center][math]^A_ZX_N\rightarrow ^A_{Z+1}X'_{N-1}+\beta+\bar{\nu}[/math] where [math]\beta=^{0}_{-1}e^-_0[/math][/center][br]Inside a nucleus there are energy levels for protons and neutrons much like there are for electrons. If there are too many neutrons in a nucleus, some will have to be at very high energy levels as compared with the highest energy protons. If this is the case, then if the neutron can decay into a proton (plus electron and electron antineutrino) and drop down in energy level such that energy is still overall being released in the process, it will occur spontaneously in nature. This is called[math]\beta[/math][b]-decay[/b] (beta decay). The electron and antineutrino do not stay in the nucleus when this occurs, but rather form the associated radioactivity. Electrons like this are easily stopped by a few millimeters of metal, or larger distances of other materials. The high energy electrons released in such processes often exceed the speed of light within materials (not the vacuum speed, but v=c/n) as they pass through materials such as water, and quickly radiate energy away as Cherenkov radiation which glows bright blue.

As mentioned earlier, the nucleus has an energy level hierarchy not unlike electrons orbitals which we have briefly discussed and which are elaborated upon in chemistry classes. Such quantized levels suggest the ability to absorb or emit photons just as electrons do. The gaps between nuclear energy levels are in the range of 100 keV to a few MeV. Recall that electron energy levels are around 1eV-10eV. [br][br]Such excited states typically occur in one of two ways: The absorption of a high energy photon of the right energy may excite a nuclear state, or some of the left-over energy from alpha or beta decay may be momentarily captured as a nuclear excited state of a product nucleus. Such excited states typically have very short half-lives of around 1 ns to 1 ps, but may occasionally survive for days or years.[br][br]Energy and momentum must be conserved when the high energy gamma ray photon is emitted, so the nucleus will recoil a little just as a BB gun will upon firing a BB. I didn't want to use a hand gun analogy since the recoil is very small but worth noting. What it means is that the emitted photon does not get all of the energy associated with the difference in the nuclear energy levels, but just a small fraction less since the recoiling nucleus takes a tiny bit of the energy away from the reaction in the form of kinetic energy. Gamma decay is written: [br][br][center][math]X^*\rightarrow X+\gamma[/math][/center][br]where the asterisk denotes a nuclear excited state.