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Nuclear Binding Energy Calculator

Mass defect, total binding energy and binding energy per nucleon, step by step
Find the binding energy and binding energy per nucleon of helium-4
Find the mass defect and binding energy of deuterium in MeV and joules
Find the binding energy per nucleon of iron-56
Convert a mass defect of 0.030377 u into MeV

Mass Defect and the Binding Energy Formula

A nucleus weighs less than the sum of its parts. That missing mass, the mass defect, is the energy released when the nucleons bound together, and it is also the energy you must supply to pull them apart again.

Δm=ZmH+Nmn−Matom\Delta m = Z m_{\text{H}} + N m_n - M_{\text{atom}}

Eb=Δm c2E_b = \Delta m\, c^2

  • ZZ — proton number; N=A−ZN = A - Z — neutron number
  • mH=1.007825m_{\text{H}} = 1.007825 u — mass of the neutral 1^{1}H atom
  • mn=1.008665m_n = 1.008665 u — neutron mass
  • MatomM_{\text{atom}} — tabulated atomic mass of the nuclide, u
  • Δm\Delta m — mass defect, unified atomic mass units (u)

Why mHm_\text{H} and not the bare proton mass: atomic mass tables include the electrons. Using the hydrogen atom mass puts ZZ electrons on both sides of the subtraction, where they cancel. Mix a bare proton mass with an atomic nuclide mass and you are wrong by Z×0.000549Z \times 0.000549 u.

The assumption people forget: electron binding energies are ignored in this cancellation — an approximation good to a few eV, negligible against MeV.

Converting Mass to Energy

You almost never multiply by c2c^2 in SI. The standard conversion is

1 u=931.494 MeV/c21\ \text{u} = 931.494\ \text{MeV}/c^2

so the working formula is simply

Eb [MeV]=Δm [u]×931.494E_b\ [\text{MeV}] = \Delta m\ [\text{u}] \times 931.494

If a question wants joules, go through the electronvolt: 1 eV=1.602×10−191\ \text{eV} = 1.602 \times 10^{-19} J. The SI route is the same answer the long way — 1 u=1.66054×10−271\ \text{u} = 1.66054 \times 10^{-27} kg, times c2=8.98755×1016c^2 = 8.98755 \times 10^{16} m²/s².

Binding energy per nucleon is the figure that actually tells you about stability:

EbA\frac{E_b}{A}

in MeV per nucleon. It rises steeply through the light nuclei, peaks near 8.798.79 MeV at 56^{56}Fe, and falls slowly thereafter — which is why fusion releases energy below iron and fission releases it above.

When it applies: to nuclear ground states. Excited nuclei and isomers have slightly different masses.

Common Mistakes to Avoid

  • Mixing bare-proton and atomic masses — pick the atomic convention (mHm_\text{H} with MatomM_{\text{atom}}) and stay in it.
  • Using AA instead of the measured mass — the mass number is an integer count of nucleons; the atomic mass is a measured value. Their difference is the effect you are computing.
  • Subtracting in the wrong order — free nucleons minus the nucleus. A negative mass defect means you flipped it.
  • Reporting total binding energy when the question asks per nucleon — divide by AA, not by ZZ.
  • Confusing 931.494931.494 MeV/u with 931.494931.494 MeV — it is a conversion factor per unit mass defect.
  • Rounding the masses too early — the defect is a small difference of large numbers, so keep six decimal places in u until the very last step.

Examples

Step 1: Z=2Z = 2, N=2N = 2
Step 2: Free nucleons: 2(1.007825 u)+2(1.008665 u)=2.015650+2.017330=4.032980 u2(1.007825\ \text{u}) + 2(1.008665\ \text{u}) = 2.015650 + 2.017330 = 4.032980\ \text{u}
Step 3: Δm=4.032980 u−4.002603 u=0.030377 u\Delta m = 4.032980\ \text{u} - 4.002603\ \text{u} = 0.030377\ \text{u}
Step 4: Eb=(0.030377 u)(931.494 MeV/u)=28.296 MeVE_b = (0.030377\ \text{u})(931.494\ \text{MeV/u}) = 28.296\ \text{MeV}
Step 5: Eb/A=(28.296 MeV)÷4=7.074 MeV per nucleonE_b/A = (28.296\ \text{MeV}) \div 4 = 7.074\ \text{MeV per nucleon}
Answer: Δm=0.030377\Delta m = 0.030377 u, Eb≈28.30E_b \approx 28.30 MeV, Eb/A≈7.07E_b/A \approx 7.07 MeV/nucleon

Step 1: Z=1Z = 1, N=1N = 1: free nucleons =1.007825 u+1.008665 u=2.016490 u= 1.007825\ \text{u} + 1.008665\ \text{u} = 2.016490\ \text{u}
Step 2: Δm=2.016490 u−2.014102 u=0.002388 u\Delta m = 2.016490\ \text{u} - 2.014102\ \text{u} = 0.002388\ \text{u}
Step 3: Eb=(0.002388 u)(931.494 MeV/u)=2.2244 MeVE_b = (0.002388\ \text{u})(931.494\ \text{MeV/u}) = 2.2244\ \text{MeV}
Step 4: In joules: (2.2244×106 eV)(1.602×10−19 J/eV)=3.564×10−13 J(2.2244 \times 10^{6}\ \text{eV})(1.602 \times 10^{-19}\ \text{J/eV}) = 3.564 \times 10^{-13}\ \text{J}
Step 5: SI check: Δm=(0.002388)(1.66054×10−27 kg)=3.965×10−30 kg\Delta m = (0.002388)(1.66054 \times 10^{-27}\ \text{kg}) = 3.965 \times 10^{-30}\ \text{kg}; × 8.98755×1016 m2/s2=3.564×10−13 J\times\, 8.98755 \times 10^{16}\ \text{m}^2\text{/s}^2 = 3.564 \times 10^{-13}\ \text{J}
Answer: Eb≈2.224E_b \approx 2.224 MeV =3.56×10−13= 3.56 \times 10^{-13} J, or 1.1121.112 MeV per nucleon

Step 1: Z=26Z = 26, N=30N = 30
Step 2: Protons: 26(1.007825 u)=26.203450 u26(1.007825\ \text{u}) = 26.203450\ \text{u}; neutrons: 30(1.008665 u)=30.259950 u30(1.008665\ \text{u}) = 30.259950\ \text{u}
Step 3: Sum =56.463400 u= 56.463400\ \text{u}
Step 4: Δm=56.463400 u−55.934936 u=0.528464 u\Delta m = 56.463400\ \text{u} - 55.934936\ \text{u} = 0.528464\ \text{u}
Step 5: Eb=(0.528464 u)(931.494 MeV/u)=492.26 MeVE_b = (0.528464\ \text{u})(931.494\ \text{MeV/u}) = 492.26\ \text{MeV}
Step 6: Eb/A=(492.26 MeV)÷56=8.791 MeV per nucleonE_b/A = (492.26\ \text{MeV}) \div 56 = 8.791\ \text{MeV per nucleon}
Answer: Eb≈492.3E_b \approx 492.3 MeV, Eb/A≈8.79E_b/A \approx 8.79 MeV/nucleon — the peak of the curve

Frequently Asked Questions

First the mass defect, Δm = Zm_H + Nm_n − M_atom in unified mass units, then E_b = Δm c². In practice you convert with 1 u = 931.494 MeV/c², so E_b in MeV is just the mass defect in u multiplied by 931.494.

Use the neutral ¹H atomic mass, 1.007825 u, whenever the nuclide mass you are given is an atomic mass — which is how nearly all tables list it. The Z electrons then cancel between the two sides. Using the bare proton mass, 1.007276 u, with an atomic nuclide mass leaves an error of Z electron masses.

Divide the total binding energy by the mass number A, the total count of protons and neutrons. Helium-4 gives 28.30/4 = 7.07 MeV per nucleon; iron-56 gives 492.3/56 = 8.79 MeV per nucleon, the highest of any nuclide.

It measures how tightly bound a nucleus is, so it predicts which reactions release energy. Moving toward the iron-56 peak releases energy — fusion for light nuclei, fission for heavy ones — and moving away from it costs energy.

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