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Nuclear Binding Energy Calculator

Physics
Pick a nuclide and its tabulated AME2020 atomic mass fills in Z, A and the measured mass. Everything downstream — mass defect, binding energy, per-nucleon energy, mass excess, packing fraction — follows from that one number.
Each example replaces every input, so you can load one on top of anything.

Formula in use

E_B = [Z·m_H + N·m_n − M_atom] · c²

Inputs

Picking one fills in Z, A and its AME2020 atomic mass.
For the headline conversion row.
Significant figures, 0 to 10.
Binding energy per nucleon
8.79034 MeV

492.259 MeV over 56 nucleons

Near the peak — very stableAt the peak — neither releases much

Primary results — Fe-56

CompositionZ = 26, N = 30, A = 56
Free constituent mass56.4634 u
Measured mass55.9349 u
Mass defect Δm0.528462 u
Total binding energy E_B492.259 MeV
Binding energy per nucleon8.79034 MeV
In joules7.88686e-11 J
Per mole4.74958e+10 kJ/mol

Secondary metrics

Mass excess Δ-60.6064 MeV
Packing fraction-0.00116185
Fractional mass defect0.935937 %
Mass basis usedAtomic mass (includes Z electrons)
Full disassembly of 1 kg202.71 kt TNT

Where the mass went

26 × m_H + 30 × m_n = 56.4634 u

Bound nuclide = 55.9349 u

The highlighted sliver is the mass defect — 0.528462 u, or 0.9359 % of the total, drawn wider than life so it is visible at all. That sliver is worth 492.259 MeV.

Semi-empirical mass formula

Even Z, even N — pairing adds

+882.0Volume−260.5Surface−120.8Coulomb−6.8Asymmetry+1.5Pairing495.4TotalBinding energy contribution (MeV)
TermMeV
Volume+882
Surface−260.543
Coulomb−120.796
Asymmetry−6.77143
Pairing+1.49399
Predicted E_B495.384
Predicted per nucleon8.84614
Model − experiment+3.12448
Deviation+0.634723 %

Coefficients (MeV)

Binding energy per nucleon

024682126250Mass number AFe-56 peak, 8.7903 MeV← fusionfission →Fe-56MeV / nucleon
Minimum mass number plotted.
Maximum mass number plotted.

The line is the liquid-drop model walked along its own valley of stability; the faint dots are the 76 measured nuclides in the built-in table. Both turn over near A = 56, which is why fusion releases energy below the peak and fission releases it above.

Physical constants

Leave these blank to use CODATA values. Overriding one lets you reproduce a textbook answer that used rounded constants.

Default 1.007276467
Default 1.008664916
Default 1.007825032

Step by step

Neutron number
  N = A − Z = 56 − 26 = 30

Free constituents (Z·m_H + N·m_n)
  26 × 1.007825032 + 30 × 1.008664916 = 56.463398309 u

Mass defect Δm
  56.463398309 − 55.93493633 = 0.528461979 u

Binding energy E_B = Δm · c²
  0.528461979 u × 931.49410242 MeV/u = 492.259 MeV

Per nucleon
  492.259 / 56 = 8.79034 MeV/nucleon

In joules
  8.775318e-28 kg × 8.98755179e+16 m²/s² = 7.88686e-11 J

Per mole
  492.259 MeV × 9.6485332e+7 kJ/mol per MeV = 4.74958e+10 kJ/mol

Mass excess Δ = (M − A·u)·c²
  (55.93493633 − 56) × 931.49410242 = -60.6064 MeV

Packing fraction (M − A)/A
  (55.93493633 − 56) / 56 = -0.00116185

SEMF − experiment
  495.384 − 492.259 = +3.12448 MeV (+0.634723 %)

About This Tool

Nuclear Binding Energy Calculator — Mass Defect, the Iron Peak and Reaction Q-Values

A nucleus weighs less than the parts it is made of. That is not a measurement error; it is the whole of nuclear energy in one sentence. This nuclear binding energy calculator takes the difference — the mass defect — and turns it into energy through E = Δm·c², reporting the total, the per-nucleon figure that decides whether a nuclide would rather fuse or split, and the mass excess and packing fraction that nuclear data tables quote alongside them.

How the mass defect becomes energy

Work with atomic masses, as every published table does, and the constituents are Z hydrogen atoms plus N neutrons, so the Z electrons cancel between the two sides:

Δm = Z·m_H + N·m_n − M_atom, then E_B = Δm × 931.49410242 MeV/u

For iron-56, 26 hydrogen atoms and 30 neutrons come to 56.463398 u, against a measured 55.934936 u. The missing 0.528462 u is worth 492.259 MeV, which across 56 nucleons is 8.7903 MeV per nucleon. Less than one per cent of the mass has gone, and it is worth roughly fifty million times the energy of a chemical bond.

Why the curve peaks near iron

Plot binding energy per nucleon against mass number and you get the most important graph in nuclear physics. It climbs steeply through the light elements, flattens into a broad plateau around A = 50–68, and declines slowly through the heavy ones. Two effects are competing: the strong force is short-ranged and saturates, so attraction grows roughly with A, while electrostatic repulsion between protons is long-ranged and grows with Z². Nickel-62 at 8.7945 MeV is the champion, with iron-56 a hair behind — iron wins the astrophysical argument only because supernova nucleosynthesis produces more of it.

Everything else can move downhill toward that plateau. Light nuclei get there by fusion; heavy nuclei get there by fission. Uranium-235 sits at 7.591 MeV per nucleon, about 1.2 MeV below the peak, and multiplying that shortfall across 235 nucleons is where the roughly 173 MeV of a fission event comes from.

The semi-empirical mass formula

When no measured mass exists, the Weizsäcker formula models the nucleus as a charged liquid drop and estimates the binding energy from Z and A alone:

E_B = a_V·A − a_S·A^(2/3) − a_C·Z(Z−1)/A^(1/3) − a_A·(A−2Z)²/A + δ

With the standard coefficients, iron-56 breaks down as +882.000 volume, −260.543 surface, −120.796 Coulomb, −6.771 asymmetry and +1.494 pairing, totalling 495.384 MeV. That is +3.124 MeV above the measured value, a deviation of +0.635 % — the model slightly over-binds nuclei near the peak.

A widely reprinted breakdown is wrong
Several sources give the middle three terms for iron-56 as −296.8, −77.2 and −16.9 MeV, and a model deviation of −2.4 MeV. Substitute into the formula above and none of those appear: the surface term is −260.543, the Coulomb term is −120.796, the asymmetry term is −6.771, and the deviation is positive. This calculator computes every term from the formula rather than carrying a printed number forward, so you can check it yourself with the step-by-step panel.

Q-values, and the units that trip people up

A reaction's Q-value is the mass that disappears across it: Q = [Σm(reactants) − Σm(products)]·c². Positive means exothermic. For U-235 + n → Ba-141 + Kr-92 + 3n the answer is +173.28 MeV, which over 235 g/mol works out at 7.11 × 10¹³ J/kg, or about 17.0 kilotons of TNT per kilogram of uranium. Deuterium–tritium fusion yields only 17.59 MeV per event, but spread over five nucleons instead of 236 it is several times more energy per kilogram of fuel.

Per mole, and the factor of a thousand
One MeV per nucleus is 9.648533 × 10⁷ kJ/mol. Iron-56's 492.26 MeV is therefore 4.7496 × 10¹⁰ kJ/mol — equivalently 4.7496 × 10¹³ J/mol. The same digits appear in print under both labels, and only one of them is kilojoules.

Atomic mass or nuclear mass?

Almost always atomic. Tables list atomic masses, and the atomic convention cancels the electrons to within a few hundred eV. If you feed a genuine bare nuclear mass in while the atomic basis is selected, the missing electrons inflate the binding energy by about 0.511 MeV per proton — 13 MeV for iron. Because nothing in nature exceeds 8.7945 MeV per nucleon, the calculator flags any result above 8.9 as an input mistake rather than a discovery.

Mass excess and packing fraction

Two older quantities describe the same data. The mass excess Δ = (M − A·u)·c² measures how far a mass sits from a whole number of mass units; for iron-56 it is about −60.61 MeV. Aston's packing fraction (M − A)/A is the same idea normalised per nucleon, −1.162 × 10⁻³ for iron, and plotting it was how the shape of the binding curve was first discovered in the 1920s — before anyone knew what held a nucleus together at all.

Frequently Asked Questions

Is the Nuclear Binding Energy Calculator free?

Yes, Nuclear Binding Energy Calculator is totally free :)

Can I use the Nuclear Binding Energy Calculator offline?

Yes, you can install the webapp as PWA.

Is it safe to use Nuclear Binding Energy Calculator?

Yes, any data related to Nuclear Binding Energy Calculator only stored in your browser (if storage required). You can simply clear browser cache to clear all the stored data. We do not store any data on server.

How does this nuclear binding energy calculator work?

It weighs the parts against the whole. Add up Z hydrogen atoms and N neutrons, subtract the measured mass of the nuclide, and what is missing is the mass defect — for iron-56, 0.528462 u out of 56.463398 u. Multiplying by 931.49410242 MeV/u turns that missing mass into 492.259 MeV of binding energy, and dividing by the 56 nucleons gives the 8.7903 MeV per nucleon that puts iron at the top of the stability curve. Everything else on the page — mass excess, packing fraction, the joule and per-mole figures, the TNT equivalent — is that same number in different clothes.

Why does the semi-empirical mass formula give 495.384 MeV for iron-56 when the measured value is 492.260?

Because the liquid-drop model over-binds nuclei near the iron peak by a few MeV, and this is what its own arithmetic produces. With the standard coefficients the five terms are +882.000 volume, −260.543 surface, −120.796 Coulomb, −6.771 asymmetry and +1.494 pairing, summing to 495.384 MeV. That is 3.124 MeV above experiment, a deviation of +0.635 %. Several sources quote a negative deviation of about −2.4 MeV here; work the terms through and the sign comes out positive. The model is doing well to land within 0.7 % of a 492 MeV quantity using nothing but Z and A.

Should I enter an atomic mass or a nuclear mass?

Atomic, almost always — that is what every published table gives, including the built-in one here. Under the atomic convention the constituents are Z hydrogen atoms rather than Z bare protons, so the Z electrons appear on both sides of the subtraction and cancel to within their own binding energy, a few hundred eV against hundreds of MeV. If you enter a genuine bare nuclear mass while the atomic basis is selected, the electrons go missing from one side and the binding energy comes out roughly 0.511 MeV per proton too high; for iron that is 13 MeV, and the tool flags it because no real nuclide exceeds 8.7945 MeV per nucleon.

Why does the curve peak at iron, and what does that have to do with fusion and fission?

Two effects fight. The nuclear force is short-ranged and saturates, so binding grows roughly in proportion to A, while electrostatic repulsion is long-ranged and grows as Z². Below about A = 50 adding nucleons buys more attraction than repulsion and the per-nucleon binding rises; above about A = 68 the Coulomb term wins and it falls. Anything can move downhill toward the peak: light nuclei by fusing, heavy nuclei by splitting. Uranium-235 sits at 7.591 MeV per nucleon, 1.2 MeV below the peak, and that gap across 235 nucleons is where the roughly 173 MeV of a fission event comes from.

Is the binding energy per mole really 4.75 × 10¹⁰ kJ/mol?

Yes, and the unit matters. One MeV per nucleus is 9.648533 × 10⁷ kJ/mol, so iron-56's 492.26 MeV becomes 4.7496 × 10¹⁰ kJ/mol — or equivalently 4.7496 × 10¹³ J/mol, which is the same figure a thousand times larger and is often misprinted as kJ/mol. For scale, burning a mole of methane releases about 890 kJ. Nuclear binding is roughly fifty million times larger per mole than chemical bonding, which is the entire reason a kilogram of uranium is worth seventeen kilotons of TNT.

How accurate are the results, and what are the limits?

The tabulated masses are AME2020 values carried to their full published precision, so measured-mass results are good to more digits than you will ever need — the binding energy of iron-56 is certain to well under a keV. The semi-empirical mass formula is a different matter: it is a five-parameter fit that typically lands within 1 % for A above about 20, but it ignores shell structure entirely and is badly wrong for light nuclei, missing helium-4 by nearly 20 %. Q-values assume the reaction as written is balanced in both nucleon number and charge, which the tool checks and refuses to compute without.