Mass Defect and Binding Energy

A bound nucleus has less rest mass than the total rest mass of its separated nucleons. The difference is the mass equivalent of energy released when the bound system forms.

Learning objectives

You will define mass defect, calculate total and per-nucleon binding energy, use atomic or nuclear masses consistently, interpret the binding-energy curve, and distinguish stronger binding from greater total binding energy.

Mass bookkeeping

Using nuclear masses,

$$\Delta m=Zm_p+Nm_n-m_{\mathrm{nucleus}}.$$

The binding energy is

$$B=\Delta m c^2.$$

With masses in unified atomic mass units,

$$1\,\mathrm{u}\,c^2 \approx931.5\,\mathrm{MeV}.$$

$\Delta m$ is positive for an ordinarily bound nucleus under this definition.

Begin with separated nucleons at rest and far enough apart to neglect their mutual interaction energy. If energy $B$ leaves when the bound nucleus forms, conservation gives

$$(Zm_p+Nm_n)c^2 =m_{\mathrm{nucleus}}c^2+B.$$

Rearranging gives $B=\Delta m c^2$. The mass is not missing; the bound system has lower rest energy.

Atomic masses versus nuclear masses

Tabulated nuclide masses are often neutral-atom masses. A convenient atomic-mass form uses hydrogen-atom mass $m(\mathrm{^1H})$:

$$\Delta m =Zm(\mathrm{^1H})+Nm_n-m_{\mathrm{atom}}.$$

Electron masses then cancel consistently to a good approximation. Mixing proton masses with neutral-atom product masses creates an electron-bookkeeping error.

Electron binding energies are small compared with typical nuclear binding energies but not zero. Precision work must use a convention and data evaluation appropriate to the required uncertainty.

Binding energy per nucleon

$\frac{B}{A}$ is a useful average measure for comparing nuclear binding across different $A$. It rises rapidly for light nuclides, peaks broadly near iron and nickel, and declines slowly for very heavy nuclides. This pattern explains why fusion of light nuclei and fission of very heavy nuclei can both release energy.

It does not by itself predict reaction rate or whether a particular pathway is accessible.

For a proposed transformation, compare complete initial and final systems. The smooth curve hides shell, pairing, and deformation effects found in evaluated masses.

Assumptions and precision

Keep more digits during intermediate work than in the final result. Mass differences can be small compared with the masses being subtracted, so early rounding can distort $\Delta m$. State whether masses are atomic or nuclear, use one convention throughout, and report units.

Common mistakes

• Mixing atomic and nuclear mass conventions.
• Reporting mass defect without converting to energy.
• Comparing total $B$ when the intended comparison is $\frac{B}{A}$.
• Interpreting lost rest mass as nonconservation rather than energy accounting.

Test Your Knowledge

1. Convert $0.00500\,\mathrm{u}$ to energy using $931.5\,\frac{\mathrm{MeV}}{\mathrm{u}}$.
2. Why may atomic masses be used with hydrogen-atom mass?
3. What broad binding-energy trend permits both fusion and fission to release energy?
4. A nucleus has $B=56.0\,\mathrm{MeV}$ and $A=7$. Find $\frac{B}{A}$ with units.
5. Diagnose the error in using proton masses initially and neutral-atom masses finally.

Solutions

1. $4.66\,\mathrm{MeV}$.
2. Electron masses cancel consistently between the neutral atoms in the bookkeeping expression.
3. Products can move toward the broad maximum in binding energy per nucleon.
4. $\frac{B}{A}=8.00\,\frac{\mathrm{MeV}}{\mathrm{nucleon}}$.
5. The calculation mixes nuclear and atomic mass conventions, so electron rest masses do not cancel consistently.

Connection forward

Mass-energy equivalence generalizes this bookkeeping from bound nuclei to complete nuclear transformations and emitted radiation.

Sources

• NIST Reference on Constants, Units, and Uncertainty for mass-energy conversion constants.
• Atomic Mass Evaluation resources at the IAEA for evaluated mass data.