Nuclear Binding Energy Calculator
Enter proton count, neutron count and nuclear mass to compute the mass defect Δm and binding energy E_B = Δm·c².
Input Data
Results
At a glance:Nuclear binding energy is the energy that holds a nucleus together. The nucleus is lighter than the sum of its free nucleons; the mass defect Δm = Z·m_p + N·m_n − m_nucleus corresponds to binding energy E_B = Δm·c² (m_p = 1.007276 u, m_n = 1.008665 u). The binding energy per nucleon E_B/A (A = Z+N) peaks near iron-56 at ≈8.8 MeV/nucleon, the most stable nucleus. Light nuclei release energy by fusion and heavy nuclei by fission, both moving toward iron. He-4: Δm ≈ 0.0304 u, E_B ≈ 28.3 MeV (7.07 MeV/nucleon).
Formula
Mass defect: Δm = Z·m_p + N·m_n − m_nucleus
Binding energy: E_B = Δm·c²
Per nucleon: E_B/A
1 u = 1.6605391e-27 kg = 931.494 MeV/c²
$$\Delta m = Z m_p + N m_n - m_{nucleus}$$$$E_B = \Delta m \, c^2$$How to Use
- Enter proton count Z, neutron count N and nuclear mass (u).
- The calculator returns Δm (u and kg), E_B (MeV and J) and binding energy per nucleon.
Case Studies
Mass-energy and the Sun
The Sun's p-p chain: 4¹H → ⁴He + energy.
4 protons = 4.029104 u, He-4 nucleus = 4.001506 u.
Δm ≈ 0.0276 u → ~25.7 MeV (≈6e11 J per gram of H).
U-235 fission energy
U-235 + neutron → two medium nuclei + neutrons.
Products have ~8.5 MeV/nucleon vs U-235's 7.6.
≈0.2 u defect → ~200 MeV per fission — basis of nuclear power and weapons.
FAQ
Why is iron-56's binding energy per nucleon maximal?
Iron-56 sits at the balance between the short-range strong nuclear force and the long-range Coulomb repulsion among protons. Light nuclei have a high surface-to-volume ratio (lower E_B/A); heavy nuclei suffer growing Coulomb repulsion. Iron-56 is the optimum, the endpoint of fusion and fission.
Where does the mass defect go?
It is not lost but converted to binding energy E_B = Δmc², released as photons/kinetic energy in fusion/fission, or stored as reduced rest mass in the nucleus — mass-energy conservation, not mass conservation.
How many MeV per u?
1 u = 1.6605391e-27 kg; ×c² gives 1.4924e-10 J = 931.494 MeV. This conversion lets you read MeV directly from a mass difference.
Why nuclear mass, not atomic mass?
Atomic mass includes electrons; nuclear mass = atomic mass − Z·m_e. This tool uses nuclear mass to simplify. If you only have atomic mass, subtract Z×0.0005486 u (error <0.1%).
Which releases more: fission or fusion?
Per reaction, U-235 fission ≈ 200 MeV, 4H→He fusion ≈ 26.7 MeV. Per nucleon, fusion is far more efficient (~6.7 MeV/nucleon vs ~0.85), which is why stars burn hydrogen.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.