Nuclear Binding Energy and the Mass Defect

Choose a nucleus and the page weighs its protons and neutrons separately, weighs the atom, and turns the missing mass into energy with E = mc². Choose a reaction and the same arithmetic on both sides gives the energy it releases. Every mass comes from the 2020 Atomic Mass Evaluation.

Worked example — Nuclear Binding Energy and the Mass Defect

Neutrons	30
Mass of the parts, weighed separately	56.463398 u
Mass of the atom	55.934936 u
Mass defect	0.528463 u
Binding energy	492.260 MeV
Binding energy per nucleon	8.7904 MeV
Mass of the nucleus per nucleon	0.998583 u
Place among the 98 nuclei on this page	3 by binding, 1 by lightness

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Nuclear Binding Energy and the Mass Defect

A nucleus weighs less than the protons and neutrons it is made of. The difference is the mass defect, and multiplied by c² it is the binding energy B: the energy it would take to pull the nucleus completely apart. Divided by the number of nucleons A it gives B/A, how tightly each nucleon is held on average, and the curve of B/A against A shows which reactions release energy: a reaction that leaves the same protons and neutrons more tightly bound gives out the gain in binding energy, whether it joins light nuclei (fusion) or splits a heavy one (fission). The energy a reaction releases is its Q-value, the mass that goes in less the mass that comes out, times c².

What each symbol means

Z: the number of protons, which fixes the element
N: the number of neutrons, A − Z
Δm: the mass defect: Z hydrogen atoms and N neutrons weighed separately, less the atom. In u, where 1 u is 931.494 MeV
B: binding energy, Δm c². B/A is the figure the curve plots
Q: the energy a reaction releases, the mass lost times c². Positive means energy comes out; it says nothing about how fast the reaction goes, or whether it goes at all

Assumes: Atomic masses, as the evaluation tabulates them, so the electrons are counted on both sides and cancel. Their own binding energy is not separated out, which is also how the evaluation defines B; even for uranium it is under an MeV, a few parts in ten thousand of the nucleus's. The mass per nucleon subtracts Z electron masses to get the nucleus alone, and the electrons' binding, left in, would move it by less than 0.000001 u for iron and nickel, far less than the gap between them. Q is the total release, including what neutrinos carry off, which in the Sun is about 2% and is never seen as heat. Ground states only: the nuclei are taken at rest and not excited.
Breaks when: Nothing in the arithmetic breaks, because it is only subtraction of measured masses. What the page does not tell you is how fast a reaction goes. A positive Q says a reaction can release energy, not that it will: uranium-238 has a positive Q for splitting in two and almost never does, and deuterium and tritium need to be heated to about a hundred million kelvin before they fuse at a useful rate. And the fission row is one pair of fragments; a real reactor produces a spread of them, each with its own Q.

Nickel-62, not iron-56, holds its nucleons tightest 🖖

Press Iron-56 and then Nickel-62. Iron-56 is bound by 8.7904 MeV per nucleon and nickel-62 by 8.7946, with iron-58 between them at 8.7923. Nickel-62 is first not only on this page but over every nucleus in the 2020 Atomic Mass Evaluation, measured or estimated. Iron-56 is the answer to a different question. Divide the mass of the nucleus by its number of nucleons and iron-56 comes out lightest, 0.998583 u against nickel-62's 0.998597, because a proton is lighter than a neutron and iron-56 is 46.4% protons where nickel-62 is 45.2%. Each of the two is third in the other's ranking, and the second panel draws both. M. P. Fewell set out the data in 1995, in a paper titled plainly The atomic nuclide with the highest mean binding energy, and suggested the belief in iron-56 came from the study of how stars make the elements.

Beryllium-8 is bound less than the two helium nuclei it is made of 🖖

Press Beryllium-8. Its binding energy is 56.499 MeV, and two helium-4 nuclei have 2 × 28.296 = 56.591 MeV between them. Eight nucleons are held more loosely as one nucleus than as two alpha particles, so beryllium-8 needs no push: it splits by itself and the 91.84 keV difference is released. Helium-4 is bound so tightly for its size, 7.0739 MeV per nucleon, more than any nucleus up to carbon-12, that nothing of mass 5 or 8 is stable. In the first minutes after the Big Bang that gap stopped the building of nuclei at helium, with traces of deuterium, helium-3 and lithium-7. Stars get across it by joining three helium nuclei at once into carbon-12, which the reaction table puts at 7.275 MeV.

Fission releases more per reaction, fusion more per kilogram 🖖

Press Uranium-235, fission. A neutron and a uranium-235 nucleus become barium-141, krypton-92 and three neutrons, and Q is 173.278 MeV, nearly ten times the 17.589 MeV of deuterium–tritium fusion. Now read the share column instead. The fission turns 0.07881% of its mass into energy, D–T fusion 0.3754% and the Sun's hydrogen burning 0.7119%. Per kilogram of fuel that is 7.08×10¹³ J for fission, 3.37×10¹⁴ J for D–T, almost five times as much, and 6.40×10¹⁴ J for hydrogen to helium. The curve says why. Fission lifts 236 nucleons by 734.23 keV each, from the gentle right-hand slope to the middle; fusion lifts a handful of nucleons up the steep left-hand side, 3.518 MeV each for D–T. The fragments of a fission are radioactive and their later decays add energy that this Q does not count.

Problem solved in full

How much hydrogen does the Sun burn each second? 5 steps
The Sun radiates 3.828×10²⁶ W, the nominal solar luminosity the International Astronomical Union adopted in 2015. How much mass does it lose each second, and how much hydrogen does it turn into helium? Press Helium-4, the Sun for the first two steps.

Open this problem in the tool
1. Four hydrogen atoms weigh 4 × 1.007825 u and one helium-4 atom weighs 4.002603 u. The difference, times c², is the energy of one completed chain:
  
  $Q = (4 \times 1.007825 - 4.002603) u \times 931.494 MeV/u = 0.028697 \times 931.494 = 26.731 MeV$
2. As a share of the mass that went in, which the page prints:
  
  $\frac{Q}{\sum M c ^{2}} = \frac{0 . 028697}{4 . 031300} = 0 . 7119%$
3. The light leaving the Sun is mass leaving it. Divide the power by c²:
  
  $\frac{Δ m}{Δ t} = \frac{L _{⊙}}{c ^{2}} = \frac{3 . 828 \times 1 0 ^{26} W}{( 2 . 998 \times 1 0 ^{8} m/s ) ^{2}} = 4.259 \times 1 0^{9} kg/s$
4. Each kilogram of hydrogen gives up 0.7119% of itself as energy, so the hydrogen that has to be burned is the mass lost divided by that share:
  
  $\frac{4 . 259 \times 1 0 ^{9} kg/s}{0 . 007119} = 5.983 \times 1 0^{11} kg/s$
5. What is left of it becomes helium:
  
  $5.983 \times 1 0^{11} - 4.259 \times 1 0^{9} = 5.941 \times 1 0^{11} kg/s$
Answer

The Sun loses 4.259×10⁹ kg a second, a little over four million tonnes, and to do it turns 5.983×10¹¹ kg of hydrogen, about 600 million tonnes, into 5.941×10¹¹ kg of helium. The luminosity leaves out the neutrinos, which carry off about 2% of each chain's energy, so the true rate of hydrogen burning is about 2% higher than this.

$\overset{m}{˙}_{H} \approx 6 \times 1 0^{11} kg/s \overset{m}{˙}_{γ} = 4.26 \times 1 0^{9} kg/s$

References (3)

Every mass on the page: the 2020 Atomic Mass Evaluation's table of mass excesses and binding energies per nucleon, which this page reproduces to under 1 eV for all 98 bound nuclei it carries: M. Wang, W. J. Huang, F. G. Kondev, G. Audi and S. Naimi, "The AME 2020 atomic mass evaluation (II). Tables, graphs and references." Chinese Physics C 45(3), 2021, 030003. DOI 10.1088/1674-1137/abddaf
The paper that corrected the textbook belief that iron-56 is the most tightly bound nucleus: iron-58 and nickel-62 are both bound more strongly, and nickel-62 most of all. It suggests the belief came from studies of stellar nucleosynthesis: M. P. Fewell, "The atomic nuclide with the highest mean binding energy." American Journal of Physics 63(7), 1995, pp. 653-658. DOI 10.1119/1.17828
The solar luminosity the worked problem starts from, 3.828×10²⁶ W, the nominal value the International Astronomical Union adopted in 2015, rounded from a best estimate of (3.8275 ± 0.0014)×10²⁶ W: A. Prša et al., "Nominal Values for Selected Solar and Planetary Quantities: IAU 2015 Resolution B3." The Astronomical Journal 152(2), 2016, 41. DOI 10.3847/0004-6256/152/2/41

By the 42Math team · Reviewed against published sources. Corrections welcome.