What a nuclear mass tells us about binding

Choose the separated particles as the reference

Write $Z$ for proton number and $N$ for neutron number. Their sum $A=N+Z$ is the mass number. Imagine the free nucleons infinitely far apart, with no mutual interaction. Their total rest energy exceeds the nuclear ground-state energy by the positive binding energy $B$.

Nuclear masses are conveniently expressed in unified atomic mass units, $\mathrm u$. The conversion $1\,\mathrm u\,c^2\simeq931.494\,\mathrm{MeV}$ turns a mass difference into an energy; a megaelectronvolt is a million electronvolts. The speed of light $c$ supplies the conversion between rest mass and rest energy.

A table of atomic masses includes electrons. Using hydrogen atoms on the separated side and a neutral atom on the bound side cancels the electron rest masses. The deuteron calculation below neglects the small isotope difference between hydrogen and deuterium electronic binding. For heavier nuclei, electronic binding corrections can reach kiloelectronvolts or more and must be assessed against the required precision.

Source: Wang et al.

A worked calculation from mass excesses

AME2020 stores a convenient small quantity called mass excess: $\Delta=(M-A\,\mathrm u)c^2$. Its sign refers to an integer multiple of the atomic mass unit. A positive mass excess does not by itself mean that a nucleus is unbound relative to free nucleons.

The frozen table gives $8071.31806\,\mathrm{keV}$ for the neutron, $7288.971064\,\mathrm{keV}$ for hydrogen and $13135.722895\,\mathrm{keV}$ for deuterium. Hydrogen plus a neutron has the same total mass number as deuterium, so the integer mass-unit contributions cancel. Add the first two excesses and subtract the third.

The result is about $2.224566\,\mathrm{MeV}$. This is the energy required to dissociate a deuteron into a proton and a neutron. In the reverse capture process, radiation and recoil carry away the released energy. As a separate check, twice the tabulated binding per nucleon, $1112.2831\,\mathrm{keV}$, agrees after rounding.

Source: Wang et al.

Ask separately about the last neutron

Dividing binding energy by mass number makes nuclei of different sizes easier to compare. That average hides the situation of the last neutron. To remove one neutron, compare the original nucleus with its neighbor having the same proton number and one fewer neutron. The difference in their binding energies is the one-neutron separation energy.

Consider an illustrative nucleus bound by $160\,\mathrm{MeV}$ and a neighbor bound by $159.5\,\mathrm{MeV}$. Removing the last neutron costs only $0.5\,\mathrm{MeV}$. A strongly bound core can coexist with a weakly bound outer neutron. Chapter 8 turns that small separation energy into a long spatial decay length.

Separation energies often alternate between neighboring even and odd neutron numbers because pairing correlations favor pairs. A two-neutron difference reduces that alternation and can reveal a smoother trend. A sudden decrease beyond a closed shell suggests that newly added neutrons must occupy a more costly orbital.

Source: Wang et al.

Masses are the first clue

A total binding energy does not specify a unique internal arrangement. Motion in an average field, pairing, vibration and rotation can all contribute. Spectral lines and scattering add information about the motion and spatial structure. We next give one neutron the simplest useful environment: a spherical well of finite depth.

Before reading a numerical trend, identify the isotope, energy reference and uncertainty. An evaluation combines evidence into recommended values. AME marks extrapolated masses with a hash sign. Treating those entries as independent measurements would overstate how firmly the trend is known.

Source: NNDC

A different probe reads a different structure

Masses reveal energy differences. Electron scattering adds electromagnetic information. The existing form-factor experiment uses a proton target and varies electron energy and detector angle at fixed momentum transfer to separate electric and magnetic responses. It studies the proton itself; its graph is not a three-dimensional density reconstruction of a whole nucleus.

The neutron-shell experiment works at a different level. It fills states in a spherical mean field and calculates density from their shared wavefunctions. In Chapter 5 an incident neutron instead probes a calcium nucleus through elastic scattering. Identify the probe, target and observable before deciding what a picture tells you.

Source: M. N. Rosenbluth