Mass Defect and Nuclear Binding Energy

Introduction

Weigh a helium-4 atom and compare it with the combined mass of two hydrogen atoms and two neutrons. The helium atom is lighter by about 0.8%. That missing mass is not lost: it has been carried away as energy when the nucleons came together. This page makes the idea quantitative using Einstein's relationship E = mc², turning precise mass measurements into the binding energies that govern every nuclear process in this unit.

Core explanation

Binding energy. A nucleus is a bound system, so energy must be supplied to pull it apart into separate protons and neutrons. This energy is the binding energy , B. By mass–energy equivalence, a bound system has less mass than its separated parts by Δm = B ÷ c². This difference is the mass defect .

Mass defect from atomic masses. For a nuclide with Z protons and N neutrons:

Δm = Z × m(¹H) + N × m(n) − M(atom)

Here m(¹H) = 1.007825 u is the mass of a hydrogen atom (a proton plus an electron), m(n) = 1.008665 u is the neutron mass and M(atom) is the tabulated atomic mass. Using hydrogen atoms rather than bare protons is deliberate: Z hydrogen atoms contain Z electrons, which cancel the Z electrons included in the atomic mass. Electron binding energies (a few eV to about 100 keV in total for heavy atoms) are negligible compared with nuclear binding energies of hundreds of MeV.

Converting to energy. The conversion factor between mass and energy is:

1 u × c² = 931.494 MeV

So B (in MeV) = Δm (in u) × 931.494. In SI units, 1 u = 1.660539 × 10⁻²⁷ kg and c = 2.998 × 10⁸ m s⁻¹, giving 1.4924 × 10⁻¹⁰ J, which equals 931.494 MeV because 1 MeV = 1.602 × 10⁻¹³ J.

Typical values.

Nuclide Binding energy B / MeV B per nucleon / MeV --- --- --- ²H (deuterium) 2.22 1.11 ⁴He 28.30 7.07 ¹²C 92.16 7.68 ⁵⁶Fe 492.3 8.79 ²³⁸U 1801.7 7.57

Scale. A chemical bond energy is a few eV per molecule, around 400 kJ mol⁻¹. Nuclear binding energies are millions of eV per nucleus. Forming one mole of ⁴He from free nucleons would release about 2.7 × 10¹² J, some seven million times more than a typical molar bond energy. This ratio explains why nuclear fuels are so energy-dense, and also why chemical mass changes are unmeasurably small: the mass defect of a chemical bond is only about 10⁻¹⁰ to 10⁻⁹ of the molecular mass.

Separation energies. Binding energy concerns the whole nucleus. The energy needed to remove just the last neutron or proton, called the neutron or proton separation energy, is found the same way from the masses of neighbouring nuclides. Separation energies reveal fine structure such as pairing and shell effects that the total binding energy averages out.

Formulae

Δm = Z m(¹H) + N m(n) − M(atom); B = Δm c²; B (MeV) = Δm (u) × 931.494; binding energy per nucleon = B ÷ A.

Step-by-step reasoning

To find the binding energy of any nuclide:

1. Determine Z and N = A − Z. 2. Multiply Z by 1.007825 u and N by 1.008665 u, and add. 3. Subtract the atomic mass of the nuclide to get Δm in u. 4. Multiply by 931.494 MeV u⁻¹ to get B. 5. Divide by A to obtain the binding energy per nucleon if a comparison is required.

Visual explanation

Draw two pans of a balance. On the left put two hydrogen atoms and two neutrons; on the right put one helium-4 atom. The left pan sinks, showing a difference of 0.0304 u. Beside the balance, draw an arrow labelled 28.3 MeV leaving the helium atom — the energy released when it formed.

Real-world analogy

Two magnets that snap together must be pulled apart with effort. If you could weigh them precisely, the joined pair would be minutely lighter than the separated pair, because the work you must do to separate them is stored as energy of the system. Nuclei show this effect clearly because the forces are so strong.

Real-world example

Precision mass measurements in Penning traps, which determine the cyclotron frequency of a single ion in a magnetic field, now reach relative uncertainties below one part in 10¹⁰. These masses feed directly into binding energies, decay energies and astrophysical models of how elements are made in stars.

Why?

Why does a bound nucleus weigh less than its parts? Energy and mass are equivalent. When nucleons bind, the system moves to a lower energy state and the difference leaves as radiation or kinetic energy. The remaining system has less energy and therefore less mass, by exactly B ÷ c².

Common misconception

"Binding energy is energy stored in the nucleus that is released when it breaks apart." The opposite is true: binding energy is the energy you must supply to break a nucleus into free nucleons. Energy is released when nuclei change to arrangements with greater total binding energy.

Worked example

Question: The atomic mass of ⁴He is 4.002603 u. Calculate its mass defect, binding energy and binding energy per nucleon.

Reasoning: Z = 2, N = 2. Mass of parts = 2(1.007825) + 2(1.008665) = 2.015650 + 2.017330 = 4.032980 u. Δm = 4.032980 − 4.002603 = 0.030377 u. B = 0.030377 × 931.494 = 28.30 MeV. Per nucleon: 28.30 ÷ 4 = 7.07 MeV.

Answer: Δm = 0.0304 u; B = 28.3 MeV; 7.07 MeV per nucleon.

Quick check

1. Why are hydrogen atom masses, rather than proton masses, used when the atomic mass of the nuclide is taken from tables? Answer: The electrons in the Z hydrogen atoms cancel the Z electrons included in the tabulated atomic mass.

Exam focus

Show the conversion 931.494 MeV per u explicitly and keep at least six significant figures in masses, because the mass defect is a small difference between large numbers. Rounding atomic masses to three decimal places can destroy the answer entirely.

Advanced insight

Data tables often list the mass excess, Δ = (M − A) × 931.494 MeV, instead of the atomic mass. Mass excesses make calculations easier: the binding energy is B = Z Δ(¹H) + N Δ(n) − Δ(nuclide), with Δ(¹H) = 7.289 MeV and Δ(n) = 8.071 MeV. Because carbon-12 defines the mass scale, its mass excess is exactly zero.

Summary

A nucleus is lighter than its separated nucleons by the mass defect Δm = Z m(¹H) + N m(n) − M(atom). The binding energy B = Δm c², with 1 u equivalent to 931.494 MeV. Binding energies run to hundreds of MeV, millions of times larger than chemical bond energies. B is the energy needed to break a nucleus apart; energy is released when nucleons move into more tightly bound arrangements.

Practice questions

1. Calculate the binding energy of ²H, given its atomic mass 2.014102 u. Answer: Δm = 1.007825 + 1.008665 − 2.014102 = 0.002388 u; B = 0.002388 × 931.494 = 2.22 MeV. 2. The atomic mass of ¹²C is exactly 12 u. Find its binding energy per nucleon. Answer: Δm = 6(1.007825) + 6(1.008665) − 12.000000 = 0.098940 u; B = 92.16 MeV; per nucleon 7.68 MeV. 3. Express a binding energy of 28.3 MeV in joules. Answer: 28.3 × 1.602 × 10⁻¹³ J = 4.53 × 10⁻¹² J. 4. Explain why the mass change in a chemical reaction is never measured on a balance. Answer: Chemical energy changes are only a few eV per molecule, so the mass change is about a billionth of the reactant mass, far below balance sensitivity.