Binding energy!

First of all, apologies if this page/post takes a while to load. I've chosen to use more than my usual number of equations because I think the math is moderately enlightening in this case.

Atomic nuclei are an incredibly interesting topic. While they seem straightforward (they just have mass and charge, right?), they are in fact very difficult to describe from pure theory, because quantum field theory is hard. A particularly interesting thing to examine is the curve of binding energy per nucleon. This plots the binding energy per nucleon compared with the total number of nucleons (protons and neutrons) in each stable nucleus. What this essentially describes is how energetically favorable it is for that many nucleons to form a nucleus as compared with just existing independently. A positive value for the binding energy means that it takes less energy to stick all of the protons and neutrons together than to keep them separate. What that translates to is the bizarre fact that a nucleus weighs less than the total mass of its constituent parts!

(Thank you to Mathematica's IsotopeData for the help in plotting all this)

The elements with very high binding energies per nucleon are the stablest. The king of these is iron-56, which is located conveniently at the top (the red dot in the graph above). To the left of that, fusion is profitable from an energy perspective. To the left, fission is energetically favorable. The binding energy curve is really important for nuclear physics.

There's this great thing called the semi-empirical mass formula, which is physicists' attempt to fit the curve of binding energy to a nice function. Then they go back after the fact and explain each term in terms of the physics it represents. It's a little hokey, to be sure, but who am I to talk? The formula is this:
$B.E.=a_1A-a_2A^{2/3}-a_3\frac{Z^2}{A^{1/3}}-a_4\frac{\left(Z-\frac{A}{2}\right)^2}{A}+\frac{a_5}{\sqrt{A}}\left(\begin{array}{c}1\\0\\-1\end{array}\right)$
Before you panic, that fifth term isn't really a vector. It's a sort of piecewise function that I'll explain in due course. $Z$ is the total number of protons in the nucleus, and $A$ is the total number of nucleons (protons and neutrons - an excellent example of isospin, which I'll have to save for another post).

The first term, $a_1 A$, is the strong force volume term. Basically nuclei are held together by the strong nuclear force, without which they seriously wouldn't exist. It has a positive contribution to the binding energy, because the greater the volume, the more the strong force pulls inwards. The volume is proportional to $A$, because of course the more nucleons you have, the greater the volume, right? In this case, that model (called the liquid drop model) seems to work quite well, but with quantum mechanics and field theory in play, it's important to remember that nothing is ever exactly what you think it should be.

The second term, $a_2A^{2/3}$, is the strong force surface term. In essence, it's a correction to the first term, which assumed that all nucleons were in the middle of the nucleus and experienced forces all around them. Instead the nucleons on the outside (think surface area here) experience less strong force, since it's adjacent to fewer other nucleons, so this term has a negative contribution to the binding energy.

The third term, $a_3\frac{Z^2}{A^{1/3}}$, is called the electrostatic term. It has a negative contribution from packing all that positive charge into a small space (seriously, nuclei are tiny!). The form of the term (proportional to charge squared and inversely proportional to the radius of the sphere) is derived from basic electricity and magnetism.

Once you get to the fourth term, quantum mechanics kicks in. What this term expresses is the desire of a stable nucleus to have around the same number of protons and neutrons. This is as a result of the way in which protons and neutrons fit into "orbitals," just like electrons. They're fermions, so two protons can fit in a single energy level, and two neutrons can fit in a single energy level. Because protons and neutrons are identical to the strong nuclear force (again, an isospin discussion will have to happen at some point), their energy levels are approximately the same, so it is energetically favorable to have close to the same number of protons and neutrons, or the unhappy neutrons (if there are more of them) are forced into higher energy levels. What the fourth term describes is a negative contribution to the binding energy as the balance between protons and neutrons is skewed.

Now we get to that crazy fifth term, the one that looks like a vector. As promised, it's a piecewise function. The column-vector like part is one if there are an even number of protons and neutrons in the nucleus (an even-even nucleus). It is zero for an even-odd or odd-even nucleus, and negative one for an odd-odd nucleus. This one seems like a real mystery at first; why should the nucleus care whether the number of protons or neutrons is divisible by two? The answer, as before, comes from the spin of the proton and neutron. Because there are two spin states (up and down), two protons fit into each energy level. The same goes for neutrons. And it turns out it's a lot more energetically favorable to have full energy levels than half-full ones. So when there are an even number of both protons and neutrons in the nucleus, the fifth term increases the binding energy (thus decreasing the mass of the nucleus). The opposite occurs for odd-odd nuclei.

There's a sixth term that accounts for gravity. Its coefficient is so tiny that it can be quite safely ignored when dealing with chemical elements, but it is really handy for things like neutron stars, which are essentially enormous nuclei held together by the strong force and gravity.