The atomic nucleus

Nearly all of an atom’s mass is concentrated in a nucleus only a few femtometres across. The nucleus contains protons and neutrons, collectively called nucleons, and its structure reflects a competition between the short-range nuclear interaction, electric repulsion and quantum effects.

Nuclide notation keeps the counts explicit

^A_ZX where Z = protons, A = total nucleons, and N = A - Z neutrons.

Changing Z changes the element. Changing N while keeping Z fixed gives an isotope of the same element. Nuclear charge is +Ze, while most atomic mass comes from nucleons.

A and Z are counts, not masses measured in kilograms. They identify the nuclear composition.

The nuclear size scale

A useful empirical relation is R ≈ r₀A^1/3 with r₀ about 1.2 fm. Because volume scales with R³, nuclear density is therefore roughly constant across many nuclei. Doubling A does not double the radius.

Nuclear stability is a balance. The strong interaction binds nearby nucleons, while proton–proton electric repulsion grows in importance for large Z.

Binding and mass defect

A bound nucleus has less rest mass than the same protons and neutrons separated to infinity. The difference Δm corresponds to binding energy B = Δmc². This is an energy of the whole system; it is not evidence that individual nucleons lose their identity.

Stability is not determined by one number

Neutron-to-proton ratio, shell structure, pairing and available decay channels all matter. A nucleus can have large binding energy per nucleon and still be radioactive if a lower-energy state is accessible. “Strongly bound” and “absolutely stable” are not equivalent statements.

A useful scale comparison: a nucleus with A = 125 has a radius of only about 6 fm from the A^1/3 rule, whereas the surrounding atom is typically tens of thousands of femtometres across. The nucleus therefore occupies an extremely small fraction of the atomic volume. This is why scattering experiments can probe a compact positive centre even though chemistry is governed mainly by the much larger electron cloud.

Worked examples

1. Neutrons in uranium-238

Solution

Z=92 and A=238, so N=A-Z=146.

2. Radius scaling

Solution

If A increases by a factor of 8, A^(1/3) doubles, so the nuclear radius is roughly twice as large.

3. Mass defect energy

Solution

A mass defect of 0.010 u corresponds to about 0.010×931.5=9.32 MeV of binding energy.