At the risk of running afoul of [sup]1[/sup]copyright issues, I’ll submit a response a similar question about Iron from 2001. Here is the response by our own mod bibliophage The following words are all his:
One model of the nucleus that is pretty easy to envision is called “the liquid drop model.” In an actual drop of water, water molecules in the middle of the drop are attracted to all the neighboring molecules. Molecules near the surface have no neighbors on the outside, so the net force on surface molecules is toward the center. This is surface tension, which tends to keep the water drop spherical. Something similar keeps the nucleus together, but it gets a big complicated.
In the nucleus, there are attractive forces (the strong nuclear force) between all the nucleons (protons and neutrons), but also repulsive forces (the electric force) between the protons. (Protons repel each other because they are all positively charged.)
The stability of the nucleus depends on a lot of different things, which are encapsulated in a formula called Weizäcker’s semiempirical formula [sup]2[/sup], which has four terms:
- The first term depends only on the number of nucleons. It takes into account the attractive force between all the nucleons, the strong nuclear force. This term is positive because it tends to make the nucleus stable. This term tends to make heavier nuclei more stable than light ones.
- The second term depends on the number of nucleons (to the 2/3 power. This term is negative.
- The third term depends on the number of protons (squared) and the number of nucleons (to the 1/3 power). It takes into account the electrical repulsion between the protons. (The reason it doesn’t depend solely on the number of protons is that with more neutrons, the protons are less closely packed and repel each other less.) This term is negative because it tends to make the nucleus less stable. This term tends to make nuclei with many neutrons more stable than those with few neutrons.
- The fourth and final term is the one most difficult to understand. It takes into account the “asymmetry energy.” Isotopes with equal numbers of neutrons and protons tend to be especially stable. For example, Carbon-12 is composed of exactly 6 protons and 6 neutrons, and is much more stable than any other nucleus composed of 12 nucleons.
The reason is that protons can’t all fit into the same energy state, and the neutrons can’t all fiti into the same neutron state. (Neutron energy states are separate from proton energy states.) When you add an extra proton (or neutron), it goes into a higher energy state than the rest. Imagine a hypothetical molecule with six protons and no neutrons (Carbon-6). The sixth proton goes into a very high energy state, but there are a lot of empty neutron energy states that have gone unfilled, an unstable combination.
So the fourth term tends to make nuclei more stable that have an equal number of protons and neutrons. This term is zero for “balanced,” negative for all others. The bigger the difference between the number of protons and neutrons, the more negative the number will be.
Iron-56 (with 26 protons and 30 neutrons) is as stable as it is because it represents a compromise between these four considerations. Carbon-12, for example, loses few points for #3, but also gets more points for #1 and #2. Uranium-238 gets more points for #1 and #2, but loses more points for #3 and #4. It’s true that Iron-56 is the most stable isotope, but there are a lot of isotopes with about 40-70 nucleons that are almost as stable.
[sup]1[/sup]The original thread seems to have been deleted. I printed a copy a long time ago, because the answer was too good to risk losing.
[sup]2[/sup] B = 15.753 A – 17.804 A[sup]2/3[/sup] – 0.7103 Z[sup]2[/sup]/A[sup]1/3[/sup] – 94.77(½A – Z)[sup]2[/sup]/A
Where B is the binding energy in MeV, Z is the number of protons, and A is the number of nucleons.
If my calculations are right, the formula would predict that the binding energy per nucleon. . .
For Carbon-12: 7.0 MeV
For Iron-56: 9.9 MeV
For U-238: 7.7 MeV