Physics › Nuclear physics › Mass-energy and binding energy
Mass-energy and binding energy
Weigh a nucleus and its parts separately and the whole comes up lighter than the sum. Einstein's equation turns the missing mass into energy, and one curve built from it, peaking at iron, explains where every joule of nuclear power comes from.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Antimatter and photons and Nuclear radius and density.
IN THIS TOPIC
- Convert between mass and energy using E = mc² and the 931.5 MeV value of the atomic mass unit.
- Calculate a mass difference and a binding energy from nuclear masses.
- Compare nuclei fairly using binding energy per nucleon.
- Sketch the binding energy per nucleon curve, and mark the fusion and fission release regions on it.
COMMON MISCONCEPTION
Mass is always conserved.
Einstein's ledger
The equation held back in the particles unit arrives here with its full meaning.
Mass and energy are one currency, and the exchange rate c2 applies to every energy change, however ordinary. Boil a kettle and the hot water is heavier by a few nanograms; a wound clock spring outweighs a slack one. The changes are immeasurably small because c2 is enormous. Only in the nucleus, where the energies are millions of electronvolts, does the missing mass become large enough to weigh.
Nuclear masses are quoted in the atomic mass unit, u, defined as one twelfth of the mass of a carbon-12 atom and printed in the booklet as 1.661 × 10-27 kg. Run that mass through E = mc2 and out comes the conversion the booklet prints beside it, 1 u is equivalent to 931.5 MeV. Try the multiplication with the booklet's rounded u and c and you land on 934 MeV, which is close enough to show where the number comes from; the printed 931.5 uses unrounded values, so quote the printed one. Mass differences in u then multiply straight into energies in MeV, with no joules involved at all.
The missing mass
Now weigh a helium-4 nucleus against its ingredients, using the booklet's proton and neutron masses. Two protons and two neutrons apart come to 4.0319 u; the assembled nucleus is 4.0015 u. The difference, 0.0304 u, is the mass difference, and it did not vanish. It left as 0.0304 × 931.5 = 28.3 MeV of energy when the nucleus formed.
That energy is the binding energy. It comes out when the nucleus is assembled, and it is exactly the energy you must supply to pull the nucleus completely apart into separate nucleons again. A bound nucleus sits in an energy well, and the depth of that well is written in its missing mass. In any energy change at all, mass changes with it, and nuclear reactions are simply the ones where the change is large enough to weigh.
WORKED EXAMPLE
Helium-4, weighed against its parts
The nuclear mass of helium-4 is 4.00150 u. Using mp = 1.00728 u and mn = 1.00867 u, find its binding energy and binding energy per nucleon.
Parts first. 2 × 1.00728 + 2 × 1.00867 = 4.03190 u, so the mass difference is 4.03190 − 4.00150 = 0.03040 u.
Convert at the booklet rate, 0.03040 × 931.5 = 28.3 MeV of binding energy.
Per nucleon that is 28.3/4 = 7.1 MeV, remarkably high for so light a nucleus. That anomalous firmness is the spike on the curve ahead, and the reason helium-4 is the ash of choice for every fusing star.
The curve that runs the universe
To compare nuclei fairly, divide each binding energy by its nucleon number. The result, binding energy per nucleon, measures how tightly the average nucleon is held, and plotting it against A gives the most consequential graph in nuclear physics.
The curve climbs steeply through the light nuclei, with helium-4 spiking above its neighbours, peaks near iron-56 at about 8.8 MeV per nucleon, then declines gently to uranium at 7.6. Any change that moves nucleons towards the peak leaves them more tightly bound, so the difference is released. Light nuclei climb from the left by fusion; heavy nuclei move in from the right by fission. Iron itself has nowhere better to go, and that is what makes it the ash of the stars. Exam questions ask you to point at the two release regions on the plot, so practise marking them.
Call the peak iron-56 in an exam and you are answering the question asked. The top of the curve is really a flat iron-nickel shelf a few nuclides wide, and nickel-62 edges above iron-56 by about a twentieth of one per cent, a margin too small to draw and too small to change any answer here.
GUIDED PRACTICE
Reading energy off the curve
A nucleus of A = 236 (binding energy per nucleon about 7.6 MeV) splits into fragments near A = 118, where the curve reads about 8.4 MeV. Estimate the energy released, working per nucleon.
Show the working
Each of the 236 nucleons becomes better bound by roughly 8.4 − 7.6 = 0.8 MeV.
Total ≈ 0.8 × 236 = 190 MeV, the textbook fission yield, read straight off the graph. The curve is a working instrument, and this estimate is how to play it.
ASSESSMENT FOCUS
- Keep the units straight. A mass difference in u multiplies by 931.5 to give MeV. Convert to kilograms and joules only when a question demands SI, and say which route you took.
- Check whether you were given nuclear or atomic masses. Atomic masses include the electrons; in most AQA questions the electron masses cancel or the nuclear mass is supplied, but read the data line.
- Define binding energy as the energy needed to separate a nucleus into its individual nucleons, or the energy released on assembly. “The energy holding the nucleus together” names no measurable quantity and does not define it.
- Comparisons between nuclei always use binding energy per nucleon, never the total; a huge nucleus can have a large total yet sit low on the curve.
- Sketching the curve wants four features. A steep rise, the helium-4 spike, a broad peak near A = 56 labelled at about 8.8 MeV, and a gentle fall away to uranium. Mark fusion to the left of the peak and fission to the right, with both arrows pointing towards iron.
CHECK YOURSELF
The nuclear mass of iron-56 is 55.92066 u. Using mp = 1.00728 u and mn = 1.00867 u, find the mass difference, the binding energy, and the binding energy per nucleon.
Show a hint
Iron-56 has 26 protons and 30 neutrons; parts minus whole first.
Show the answer
Separate parts come to 26 × 1.00728 + 30 × 1.00867 = 56.44938 u, so Δm = 56.44938 − 55.92066 = 0.52872 u.
Binding energy = 0.52872 × 931.5 = 492.5 MeV.
Per nucleon that is 492.5 / 56 = 8.8 MeV, right at the top of the curve. Nothing in nature beats that by more than a rounding error.
A bound nucleus weighs less than its parts, and the gap is the binding energy.
Divide by A and iron-56 sits at the top of the curve, near 8.8 MeV per nucleon.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the mass-energy and binding energy questions page.
WHERE TO GO NEXT
- Standard form and prefixes is the maths this lesson leans on, worked through from GCSE.
- Significant figures, units and checking is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Convert between mass and energy using E = mc² and the 931.5 MeV value of the atomic mass unit.
- Calculate a mass difference and a binding energy from nuclear masses.
- Compare nuclei fairly using binding energy per nucleon.
- Sketch the binding energy per nucleon curve, and mark the fusion and fission release regions on it.
Open the full revision checklist to track your progress across the whole unit.