13.2 Mass–Energy Equivalence, Binding Energy and the Nuclear Force
Every nucleus weighs less than the protons and neutrons it is made of. Through $E=mc^2$ that mass defect is the binding energy of the nucleus, and a graph of binding energy per nucleon — peaking near iron — explains why both fission and fusion release energy. A short-range, very strong nuclear force does the binding.
In a hospital PET scanner (positron emission tomography) a patient is given a tracer that emits positrons — the antiparticles of electrons. Each positron travels less than a millimetre before it meets an electron, and the two vanish completely. In their place two gamma-ray photons fly off in opposite directions, each carrying exactly $0.511$ MeV, the energy equivalent of one electron mass. A ring of detectors catches the photon pairs and reconstructs where in the body they were born. Mass has turned into energy, in the precise amount Einstein predicted in 1905. On a vastly larger scale the Sun converts more than four million tonnes of mass into energy every second.
The same principle explains how nuclei hold together. Weigh a helium-4 atom and you find it is about $0.03$ u lighter than two hydrogen atoms and two neutrons. That missing mass is not lost; it is the energy that was released when the nucleons came together, and the energy you would have to supply to pull them apart again — the binding energy. In this section we state the mass–energy relation, define the mass defect and binding energy, read the famous binding-energy-per-nucleon curve, and describe the force that does the binding.
Mass–Energy Equivalence
In Newtonian physics mass and energy are separately conserved. Einstein's special theory of relativity showed that they are not separate: mass is one form of energy.
A body of mass $m$ at rest has an energy (its rest-mass energy) $$E=mc^2,$$ where $c=3.0\times10^8$ m/s is the speed of light in vacuum. If the mass of a system changes by $\Delta m$, an energy $\Delta E=\Delta m\,c^2$ is released (if the mass decreases) or absorbed (if it increases). In every process the total energy, including rest-mass energy, is conserved.
Because $c^2=9\times10^{16}\ \text{m}^2/\text{s}^2$ is so large, a tiny mass corresponds to a huge energy. In chemical reactions the energy released is so small that the change in mass — a few parts in $10^{10}$ — cannot be detected, which is why "conservation of mass" works so well in chemistry. In nuclear reactions the energies are about a million times larger and the mass changes are easily measured.
The energy of one atomic mass unit. Nuclear energies are usually given in MeV ($1\ \text{MeV}=10^6\ \text{eV}=1.602\times10^{-13}$ J) and masses in u, so it is convenient to know the energy equivalent of $1$ u. With precise constants, $$E=(1.66054\times10^{-27}\ \text{kg})(2.9979\times10^8\ \text{m/s})^2$$ $$=1.4924\times10^{-10}\ \text{J}=931.5\ \text{MeV}.$$ So $1\ \text{u}\times c^2=931.5$ MeV, which we write as $1\ \text{u}=931.5\ \text{MeV}/c^2$. In calculations we simply multiply a mass in u by $931.5$ to get an energy in MeV. (Using the rounded values $1.66\times10^{-27}$ kg, $3.0\times10^8$ m/s and $1.6\times10^{-19}$ J gives $934$ MeV — close, but always use $931.5$ MeV for nuclear calculations.)
Mass Defect and Binding Energy
Consider the simplest compound nucleus, the deuteron ${}^{2}_{1}\text{H}$: one proton and one neutron. Its atomic mass is $2.014102$ u. A hydrogen atom and a free neutron together have mass $1.007825+1.008665=2.016490$ u. The deuterium atom is lighter by $0.002388$ u. Every nucleus behaves like this: it is lighter than its separated nucleons.
We will use the following atomic masses throughout this chapter (rounded to six decimal places):
| particle / atom | ${}^{1}\text{H}$ | neutron | ${}^{2}\text{H}$ | ${}^{4}\text{He}$ |
|---|---|---|---|---|
| mass (u) | $1.007825$ | $1.008665$ | $2.014102$ | $4.002603$ |
| atom | ${}^{12}\text{C}$ | ${}^{16}\text{O}$ | ${}^{56}\text{Fe}$ | ${}^{238}\text{U}$ |
|---|---|---|---|---|
| mass (u) | $12.000000$ | $15.994915$ | $55.934936$ | $238.050788$ |
- Read $Z$ and $A$ from the symbol; $N=A-Z$.
- Add up $Zm_\text{H}+Nm_n$ (hydrogen atoms, so the electrons balance).
- Subtract the atomic mass of the nuclide: that is $\Delta M$ in u. Keep all six decimal places — $\Delta M$ is a small difference of large numbers.
- Multiply by $931.5$ to get $E_b$ in MeV; divide by $A$ for $E_{bn}$.
The Binding-Energy-per-Nucleon Curve
Repeat the calculation of Examples 3 and 4 for every known nucleus and plot $E_{bn}=E_b/A$ against the mass number $A$. The result is one of the most important graphs in physics (Interactive 2 draws it).
1. Light nuclei ($A\lt 20$ or so) have small and rapidly rising $E_{bn}$: $1.11$ MeV for ${}^{2}\text{H}$, $5.61$ MeV for ${}^{7}\text{Li}$. Nuclei made of whole "alpha-particle units" — ${}^{4}\text{He}$ ($7.07$ MeV), ${}^{12}\text{C}$ ($7.68$ MeV), ${}^{16}\text{O}$ ($7.98$ MeV) — stand out above their neighbours.
2. Medium nuclei ($30\lt A\lt 170$) have an almost constant $E_{bn}$ of about $8$ to $8.8$ MeV. The curve has a broad maximum of about $8.8$ MeV near $A\approx56$–$62$ (iron and nickel): these are the most tightly bound nuclei.
3. Heavy nuclei have $E_{bn}$ that falls slowly with $A$, to about $7.6$ MeV for uranium.
Why is the middle flat? If every nucleon attracted every other one, the total binding energy would grow like the number of pairs, $\tfrac12A(A-1)\propto A^2$, and $E_{bn}$ would grow in proportion to $A$. It does not: $E_b$ is very nearly proportional to $A$ itself. Each nucleon is bound only to its nearest neighbours — a fixed number, however large the nucleus — so each contributes roughly the same binding energy. This property is called the saturation of the nuclear force, and it follows from its short range (below). It also explains the constant nuclear density of Section 13.1: adding nucleons is like adding drops to a liquid.
Why do light nuclei have lower $E_{bn}$? In a small nucleus a large fraction of the nucleons are at the surface, with fewer neighbours to bind to. Why do heavy nuclei have lower $E_{bn}$? The electric repulsion between protons has a long range, so every proton repels every other one; its effect grows like $Z^2$ and in heavy nuclei it eats noticeably into the binding.
The key consequence — nuclear energy. Any change that moves nucleons from less tightly bound nuclei to more tightly bound ones releases energy, because the products have a larger total binding energy and therefore less mass. There are two ways up the curve:
Fission: split a heavy nucleus ($E_{bn}\approx7.6$ MeV) into two medium ones ($E_{bn}\approx8.5$ MeV). Each of the roughly $236$ nucleons gains about $0.9$ MeV, so about $200$ MeV is released per fission.
Fusion: join two very light nuclei into a heavier one higher up the steep left-hand part of the curve, e.g. two deuterons ($1.11$ MeV each per nucleon) into helium-4 ($7.07$ MeV per nucleon). Section 13.3 develops both.
The Nuclear Force
Inside a nucleus, protons only a couple of femtometres apart repel each other with enormous electric forces (Example 6). Gravity is far too weak to help. A nucleus can hold together only if a third, much stronger, attractive force acts between nucleons: the nuclear force (also called the strong nuclear force).
1. Very strong. At separations of about $1$ fm it is much stronger than the electric repulsion between two protons (roughly a hundred times) — the strongest of the fundamental forces.
2. Short range. It is attractive for separations from about $0.8$ fm to $2$ fm, and falls to almost nothing beyond about $2$–$3$ fm. A nucleon therefore feels only its nearest neighbours (saturation).
3. Repulsive core. At separations below about $0.8$ fm it becomes strongly repulsive, which keeps nucleons from merging and gives nuclear matter its constant density. The potential energy of a pair of nucleons is lowest at a separation $r_0\approx0.8$ fm.
4. Charge independent. The nuclear force between two neutrons, two protons, or a neutron and a proton is essentially the same; it does not depend on electric charge.
Schematic (not to scale): potential energy $U(r)$ of a pair of nucleons — a deep well with a minimum near $r_0\approx0.8$ fm, a steep repulsive core, and almost zero beyond $2$–$3$ fm. The dashed curve is the Coulomb energy $ke^2/r$ of two protons, which is always repulsive and long-ranged but, at these distances, only about $1$ MeV — tiny next to the tens of MeV of the nuclear well.
1. Mixing nuclear and atomic masses. If $M$ is an atomic mass, use $m_\text{H}$ (the hydrogen atom), not $m_p$. Using $m_p$ with atomic masses leaves $Z$ electron masses unbalanced — an error of $0.000549Z$ u, or $0.511Z$ MeV.
2. Rounding too early. $\Delta M$ is a small difference between two large numbers. Rounding the masses to three decimal places can wipe it out. Keep six decimals until the end.
3. Thinking a large binding energy means an unstable nucleus. Binding energy is energy that must be supplied to break the nucleus up. The larger $E_{bn}$, the more stable the nucleus.
4. Comparing total $E_b$ instead of $E_b/A$. Uranium has a far larger total binding energy than iron ($1802$ MeV against $492$ MeV) but is less tightly bound per nucleon. Stability is judged by $E_{bn}$.
5. Thinking fission and fusion contradict each other. Both move nucleons up the $E_{bn}$ curve — from opposite ends, towards the iron peak.
Enrichment only: the semi-empirical mass formula is not in the 2026 CBSE, JEE Main, NEET or JEE Advanced syllabi, which stop at binding energy per nucleon and its variation with mass number.
Treating the nucleus as a drop of charged liquid, C. F. von Weizsäcker (1935) wrote the binding energy as a sum of a few terms, each with a simple physical meaning: $$E_b\approx a_VA-a_SA^{2/3}-a_C\frac{Z(Z-1)}{A^{1/3}}-a_A\frac{(A-2Z)^2}{A}+\delta.$$ The volume term ($a_V\approx15.8$ MeV) is saturation: each nucleon adds the same energy. The surface term ($a_S\approx18.3$ MeV) corrects for surface nucleons with fewer neighbours. The Coulomb term ($a_C\approx0.714$ MeV) is the repulsion of every pair of protons. The asymmetry term ($a_A\approx23.2$ MeV) favours $N\approx Z$. The small pairing term $\delta$ ($+12/\sqrt A$ MeV for even $Z$ and even $N$, $-12/\sqrt A$ for odd–odd, $0$ otherwise) rewards paired nucleons. For ${}^{56}_{26}\text{Fe}$ the terms are $884.8-267.9-121.3-6.6+1.6\approx490.6$ MeV, within $0.4\%$ of the measured $492.3$ MeV. The curve in Interactive 2 is this formula without $\delta$. The formula itself is not examined at board level, but it is the best way to understand the shape of the curve.
Exercises
- 1 Find the energy equivalent of $1.0$ mg of matter, in joules and in kilowatt hours ($1\ \text{kWh}=3.6\times10^6$ J).
- 2 What mass is equivalent to an energy of $1.0$ MeV? Give it in kg ($1\ \text{MeV}=1.6\times10^{-13}$ J) and in u.
- 3 The proton's mass is $1.007276$ u. Find its rest-mass energy in MeV.
- 4 A $100$ W lamp runs for one year ($3.16\times10^7$ s). What mass is equivalent to the energy it uses?
- 5 Burning $1.0$ kg of coal releases about $3.3\times10^7$ J. By how much does the mass of the products fall short of the mass of the reactants? Why is this never noticed in chemistry?
- 6 The neutron's mass is $1.008665$ u. Find its rest-mass energy, and by how many MeV it exceeds that of the proton.
- 7 Using $1\ \text{u}=1.66054\times10^{-27}$ kg, $c=2.9979\times10^8$ m/s and $1\ \text{eV}=1.602\times10^{-19}$ J, show that $1\ \text{u}\,c^2\approx931.5$ MeV.
- 1 Find the mass defect of ${}^{12}_{6}\text{C}$.
- 2 Find the binding energy and binding energy per nucleon of ${}^{12}_{6}\text{C}$.
- 3 Find the binding energy and binding energy per nucleon of ${}^{16}_{8}\text{O}$.
- 4 Tritium, ${}^{3}_{1}\text{H}$, has atomic mass $3.016049$ u. Find its binding energy and binding energy per nucleon.
- 5 Helium-3, ${}^{3}_{2}\text{He}$, has atomic mass $3.016029$ u. Find its binding energy. Suggest why it is less than that of tritium.
- 6 What is the minimum energy needed to separate an ${}^{16}\text{O}$ nucleus completely into protons and neutrons?
- 7 How much energy is needed to split a helium-4 nucleus into two deuterons?
- 8 A nucleus with $A=40$ has a binding energy per nucleon of $8.0$ MeV. What is its mass defect in u?
- 1 Which is more stable, ${}^{12}\text{C}$ ($E_{bn}=7.68$ MeV) or ${}^{7}\text{Li}$ ($E_{bn}=5.61$ MeV)? Explain.
- 2 Sketch the graph of $E_{bn}$ against $A$ and mark its approximate maximum value and position.
- 3 Explain why $E_{bn}$ decreases for heavy nuclei.
- 4 Estimate the total binding energy of a nucleus with $A=120$, taking $E_{bn}\approx8.5$ MeV.
- 5 A nucleus with $A=200$ ($E_{bn}=7.9$ MeV) splits into two nuclei with $A=100$ ($E_{bn}=8.6$ MeV). Estimate the energy released.
- 6 Use the curve to explain why fusing two ${}^{56}\text{Fe}$ nuclei into one nucleus with $A=112$ would absorb energy rather than release it.
- 7 Which property of the nuclear force explains (a) the almost constant $E_{bn}$ for medium nuclei and (b) the constant nuclear density?
- 8 Find the electric potential energy, in MeV, of two protons $1.0$ fm apart ($k=9.0\times10^9$ N m² C⁻²).
- 1 Modelling. How much energy would be released in forming $1.0$ kg of helium-4 from free protons and neutrons? Compare it with the $3.3\times10^7$ J from burning $1.0$ kg of coal. ($1\ \text{u}=1.66\times10^{-27}$ kg, $1\ \text{MeV}=1.6\times10^{-13}$ J.)
- 2 Find the energy needed to remove one proton from helium-3 (${}^{3}\text{He}\to{}^{2}\text{H}+p$), using the masses in this section.
- 3 Fusion of deuterium and tritium releases $17.6$ MeV and involves $5$ nucleons; fission of uranium-235 releases about $200$ MeV and involves $236$ nucleons. Compare the energy released per nucleon.
- 4 Give two separate reasons why the mass of a nucleus in u is not exactly equal to its mass number.
- 5 In stars, three helium-4 nuclei fuse into carbon-12. Find the energy released, (a) from the atomic masses and (b) from the binding energies per nucleon $7.074$ MeV (He-4) and $7.680$ MeV (C-12).
- 6 A nucleus has $Z=3$, $N=4$ and atomic mass $7.016003$ u. A student uses $m_p=1.007276$ u instead of $m_\text{H}$ in the mass-defect formula. By how much is her binding energy wrong, and in which direction?
E1. Board · 3 marks Define mass defect and binding energy of a nucleus. Calculate the binding energy per nucleon of ${}^{16}_{8}\text{O}$, given $m({}^{16}\text{O})=15.994915$ u, $m_\text{H}=1.007825$ u, $m_n=1.008665$ u and $1\ \text{u}=931.5\ \text{MeV}/c^2$.
E2. Board · 2 marks Draw a graph of binding energy per nucleon against mass number. Use it to explain why energy is released both when a heavy nucleus splits and when two light nuclei fuse.
E3. JEE Main The binding energies per nucleon of ${}^{2}_{1}\text{H}$ and ${}^{4}_{2}\text{He}$ are $1.1$ MeV and $7.0$ MeV. The energy released when two deuterons fuse to form one helium-4 nucleus is (A) $23.6$ MeV (B) $26.9$ MeV (C) $13.9$ MeV (D) $19.2$ MeV.
E4. NEET Which of the following is not a property of the nuclear force? (A) It has a short range. (B) It is nearly the same between n–n, p–p and n–p pairs. (C) It obeys an inverse-square law with distance, like gravity. (D) It is the strongest force at separations of about $1$ fm.
E5. JEE Advanced (Integer answer.) A nucleus with mass number $240$ and binding energy per nucleon $7.6$ MeV splits into two nuclei of mass number $120$ each, whose binding energy per nucleon is $8.5$ MeV. Find the energy released, in MeV.
E6. Board · Case study Lithium-7 (${}^{7}_{3}\text{Li}$, atomic mass $7.016003$ u) is used in the cooling systems of some nuclear reactors. (i) (2 marks) Find its mass defect and binding energy, using $m_\text{H}=1.007825$ u and $m_n=1.008665$ u. (ii) (1 mark) Find its binding energy per nucleon. (iii) (1 mark) Helium-4 has $E_{bn}=7.07$ MeV. Which nucleus is more stable?
Research question. A massive star burns carbon, then oxygen, then silicon in its core before it collapses. How much energy does each burning stage release per reaction, and why does the sequence stop near iron?
Why it matters. The elements in your body heavier than hydrogen and helium were made by fusion in stars. Astrophysicists model these burning stages to predict how long each lasts (thousands of years for carbon, about a day for silicon) and which elements a supernova throws into space — models tested against observations by telescopes around the world and against nuclear data measured in accelerator laboratories.
What to do. Use the atomic masses ${}^{12}\text{C}$ $12.000000$ u and ${}^{16}\text{O}$ $15.994915$ u (above), and ${}^{24}\text{Mg}$ $23.985042$ u, ${}^{28}\text{Si}$ $27.976927$ u, ${}^{32}\text{S}$ $31.972071$ u, ${}^{56}\text{Ni}$ $55.942128$ u.
- Treat each stage, in a simplified way, as one reaction: ${}^{12}\text{C}+{}^{12}\text{C}\to{}^{24}\text{Mg}$; ${}^{16}\text{O}+{}^{16}\text{O}\to{}^{32}\text{S}$; ${}^{28}\text{Si}+{}^{28}\text{Si}\to{}^{56}\text{Ni}$. Find the energy released by each, in MeV.
- Calculate $E_{bn}$ for ${}^{24}\text{Mg}$, ${}^{28}\text{Si}$, ${}^{32}\text{S}$ and ${}^{56}\text{Ni}$, and mark all six nuclei on a sketch of the binding-energy curve.
- Compare the energy released per nucleon in each stage with that in hydrogen burning ($26.7$ MeV for $4$ nucleons). What does this suggest about how long each stage can last?
- Use the curve to explain why fusing two nuclei of $A\approx56$ cannot provide energy.
Tools: Calculator or spreadsheet; Interactive 2 for the curve.
Hand in: A table of $Q$ and $Q/A$ for each stage, your annotated curve, and a paragraph on why stars cannot burn beyond the iron peak.
Going further. A star's luminosity during silicon burning is enormous, but most of the energy escapes as neutrinos. Find out why this shortens the stage so dramatically.
🧪 Self-check quiz
Q1. The energy equivalent of $1$ u of mass is about
Q2. The binding energy per nucleon is greatest for nuclei with mass number about
Q3. The mass of a nucleus is
Q4. The nuclear force between two nucleons
Q5. The binding energy per nucleon is almost constant for $30\lt A\lt 170$ because the nuclear force