text: The protons and neutrons that comprise an atomic nucleus behave almost identically within the nucleus. The approximate symmetry of isospin treats these particles as identical, but in a different quantum state. This symmetry is only approximate, however, and the nuclear force that binds nucleons together is a complicated function depending on nucleon type, spin state, electric charge, momentum, etc. and with contributions from non-central forces. The nuclear force is not a fundamental force of nature, but a consequence of the residual effects of the strong force that surround the nucleons. One consequence of these complications is that although deuterium, a bound state of a proton (p) and a neutron (n) is stable, exotic nuclides such as diproton or dineutron are unbound. The nuclear force is not sufficiently strong to form either p-p or n-n bound states, or equivalently, the nuclear force does not form a potential well deep enough to bind these identical nucleons.
  Stable nuclides require approximately equal numbers of protons and neutrons. The stable nuclide carbon-12 (C) is composed of six neutrons and six protons, for example. Protons have a positive charge, hence within a nuclide with many protons there are large repulsive forces between protons arising from the Coulomb force. By acting to separate protons from one another, the neutrons within a nuclide play an essential role in stabilizing nuclides. With increasing atomic number, even greater numbers of neutrons are required to obtain stability. The heaviest stable element, lead (Pb), has many more neutrons than protons. The stable nuclide Pb has Z = 82 and N = 124, for example. For this reason, the valley of stability does not follow the line Z = N for A larger than 40 (Z = 20 is the element calcium). Neutron number increases along the line of beta stability at a faster rate than atomic number.
  The line of beta stability follows a particular curve of neutron–proton ratio, corresponding to the most stable nuclides. On one side of the valley of stability, this ratio
  is small, corresponding to an excess of protons over neutrons in the nuclides. These nuclides tend to be unstable to β decay or electron capture, since such decay converts a proton to a neutron. The decay serves to move the nuclides toward a more stable neutron-proton ratio. On the other side of the valley of stability, this ratio is large, corresponding to an excess of neutrons over protons in the nuclides. These nuclides tend to be unstable to β decay, since such decay converts neutrons to protons. On this side of the valley of stability, β decay also serves to move nuclides toward a more stable neutron-proton ratio.