AcademyNuclear Physics
Academy
Binding Energy
Level 1 - Physics topic page in Nuclear Physics.
Principle
Binding energy is the energy equivalent of the mass missing from separated nucleons.
Notation
\(\Delta m\)
mass defect
kg, u
\(B\)
nuclear binding energy
J, MeV
\(B/A\)
binding energy per nucleon
MeV
\(m_{\mathrm p}\)
proton mass
kg, u
\(m_{\mathrm n}\)
neutron mass
kg, u
\(M\)
nuclear or atomic mass
kg, u
Method
Derivation 1: Mass defect
A bound nucleus has less mass than its separated nucleons.
Separated nucleons
\[m_{\mathrm{sep}}=Zm_{\mathrm p}+Nm_{\mathrm n}\]
Mass defect
\[\Delta m=m_{\mathrm{sep}}-M_{\mathrm{nucleus}}\]
Derivation 2: Energy equivalent
Mass difference converts to binding energy by mass-energy equivalence.
Binding energy
\[B=\Delta mc^2\]
Per nucleon
\[\frac{B}{A}=\frac{\Delta mc^2}{A}\]
Derivation 3: Atomic masses
Using neutral atomic masses cancels electron masses when the electron count balances.
Hydrogen replacement
\[m_{\mathrm p}+m_e\approx m_{\mathrm H}\]
Atomic-mass form
\[\Delta m=Zm_{\mathrm H}+Nm_{\mathrm n}-M_{\mathrm{atom}}\]
Rules
Mass defect
\[\Delta m=Zm_{\mathrm p}+Nm_{\mathrm n}-M_{\mathrm{nucleus}}\]
Binding energy
\[B=\Delta mc^2\]
Atomic-mass form
\[\Delta m=Zm_{\mathrm H}+Nm_{\mathrm n}-M_{\mathrm{atom}}\]
Examples
Question
A nucleus has
\[\Delta m=0.030\,\mathrm u\]
Find \(B\) in MeV.Answer
Use
\[1\,\mathrm u\,c^2=931.5\,\mathrm{MeV}\]
\[B=0.030(931.5)=28\,\mathrm{MeV}\]
Checks
- A positive mass defect gives a positive binding energy.
- Use nuclear masses or atomic masses consistently.
- Larger \(B/A\) means the nucleons are more tightly bound on average.
- Energy released in reactions comes from increased total binding energy.