Lesson 1 · 30 min

Nuclear stability and binding energy

Of more than 3500 known nuclides only 253 are stable. The band of stability, the magic numbers and the binding-energy curve tell you why — and why iron and nickel sit at the top of the curve.

In this lesson

Locate the band of stability on the chart of nuclides and explain why the neutron-to-proton ratio rises with mass
Use even–odd patterns and magic numbers to explain which nuclides are especially stable
Calculate the mass defect, binding energy and binding energy per nucleon from atomic masses
Use the five terms of the liquid-drop model to explain the shape of the binding-energy curve and why it peaks near iron and nickel

Put a helium-4 atom on the most precise mass balance: 4.002 603 u. Now weigh its parts separately — 2 hydrogen atoms (each a proton plus an electron) and 2 neutrons — and they add up to 4.032 980 u. The whole is 0.030 377 u, about 0.75 %, lighter than its parts. That mass has not vanished: through $E = mc^2$ it is 28.3 MeV of energy, released when the nucleus formed and needed to take it apart again — the binding energy. At about 7 MeV per nucleon it is roughly a million times a chemical bond (a few eV).

Predict first

Light stable nuclei such as ¹²C, ¹⁶O and ⁴⁰Ca have equal numbers of neutrons and protons. How does the ratio N/Z change for heavier stable nuclei?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

It stays at 1 It climbs gradually, to about 1.5 for lead It falls below 1 There is no pattern

The band of stability

The NUBASE2020 evaluation lists the ground states of 3557 known nuclides, and only 253 of them have never been seen to decay. On the chart they form a narrow band; everything on either side decays, and always in a direction that pushes the nucleus back toward the band:

below the band (too many neutrons): β⁻ decay turns a neutron into a proton, N − 1 and Z + 1;
above the band (too many protons): β⁺ decay or electron capture turns a proton into a neutron;
very heavy nuclei (Z > 82): α decay sheds 2 protons and 2 neutrons at once, and the heaviest also fission spontaneously.

No element beyond lead has a stable isotope, and technetium (Z = 43) and promethium (Z = 61) have none at all. Even ²⁰⁹Bi, long counted as the heaviest stable nucleus, was caught α-decaying in 2003, with a half-life of about 2 × 10¹⁹ years — around a billion times the age of the universe.

Stable nuclides also favour even numbers. Of the 253, 147 have both an even number of protons and an even number of neutrons, while only 4 have both odd: ²H, ⁶Li, ¹⁰B and ¹⁴N. Paired nucleons have lower energy, much like spin-paired electrons in atoms.

When the proton or neutron number equals one of the magic numbers 2, 8, 20, 28, 50, 82, 126, a nuclear shell is full and the nucleus is especially stable: tin (Z = 50) has 10 stable isotopes, more than any other element, and ⁴He, ¹⁶O, ⁴⁰Ca and ²⁰⁸Pb are “doubly magic”. Turn on the magic-number lines in the chart and see how stable nuclides cluster along them.

Binding energy

B = \left[Z\,m(\ce{^1H}) + N\,m_\mathrm{n} - m(\ce{^{A}X})\right]c^2,\qquad 1\ \mathrm{u}\cdot c^2 = 931.494\ \mathrm{MeV}

Use neutral-atom masses: the Z hydrogen atoms bring exactly the Z electrons that atom X has.

$B$: binding energy: what it takes to pull the nucleus apart into free nucleons
$Z,\ N$: proton and neutron numbers, A = Z + N
$m(\ce{^1H}),\ m_\mathrm{n}$: hydrogen-atom mass 1.007 825 u and neutron mass 1.008 665 u
$B/A$: binding energy per nucleon: how tightly the nucleus is held

AME2020 measurements · semi-empirical mass formula

The binding-energy curve

Stable nuclides (AME2020)
Semi-empirical formula (along the valley)

Click any point to choose another nuclide.

⁵⁶Fe Z = 26, N = 30

Measured B/A: 8.790 MeV
Formula B/A: 8.760 MeV
Formula − measured: −0.031 MeV

Term	MeV	per nucleon
Volume term	882.0	15.750
− surface term	−260.5	−4.653
− Coulomb term	−125.6	−2.243
− asymmetry term	−6.8	−0.121
± pairing term	1.5	0.027
Total B	490.6	8.760

Each dot is a stable nuclide. Click ⁴He, ¹²C and ¹⁶O first: they stand above the curve on the left — exceptionally tight light nuclei. Then compare ⁵⁶Fe with ⁶²Ni: which is higher? Finally switch to “Liquid-drop terms” and see which end the surface and Coulomb terms pull down.

The binding-energy curve rises steeply, peaks near A ≈ 60 at about 8.8 MeV, then declines slowly to 7.6 MeV for uranium. It decides where nuclear energy comes from: fusing light nuclei into heavier ones (climbing from the left) releases energy, and so does splitting heavy nuclei into middle-sized ones (climbing back from the right). Both head for iron and nickel at the top.

The liquid-drop model (the semi-empirical mass formula) explains the shape: nucleons are like the molecules of a droplet, attracting only their near neighbours.

B = a_V A - a_S A^{2/3} - a_C\frac{Z^2}{A^{1/3}} - a_A\frac{(A-2Z)^2}{A} \pm \frac{a_P}{A^{1/2}}

Coefficients of Rohlf (1994): aV = 15.75, aS = 17.8, aC = 0.711, aA = 23.7, aP = 11.18 MeV; the pairing term is + for even–even, − for odd–odd and 0 for odd-A nuclei.

$a_V A$: volume term: every nucleon is held by its neighbours
$a_S A^{2/3}$: surface term: surface nucleons lack neighbours — worst for light nuclei
$a_C Z^2/A^{1/3}$: Coulomb term: protons repel — worst for heavy nuclei
$a_A (A-2Z)^2/A$: asymmetry term: the further N is from Z, the less stable
$a_P/A^{1/2}$: pairing term: paired nucleons are more stable

The surface term drags down the left end of the curve and the Coulomb term the right end, so the peak can only lie in between. The formula gives B/A = 8.76 MeV for ⁵⁶Fe (measured 8.790 MeV) and 7.58 MeV for ²³⁸U (measured 7.570 MeV), quite good for A ≥ 40 — but only 5.49 MeV for ⁴He against a measured 7.074 MeV; such tight light nuclei need the shell model.

Worked example

Binding energy per nucleon of iron-56

An atom of ⁵⁶Fe has mass 55.934 936 u. With $m(\ce{^1H}) = 1.007\,825$ u and $m_\mathrm{n} = 1.008\,665$ u, find its binding energy per nucleon.

Concept check

Which of these nuclear changes would absorb energy rather than release it?

²H and ³H fusing into ⁴He and a neutron ²³⁵U capturing a neutron and splitting into ¹⁴¹Ba, ⁹²Kr and 3 neutrons Splitting a ⁵⁶Fe nucleus into two ²⁸Al nuclei Four ¹H ending up as one ⁴He in the Sun

Try the calculation

From $m(\ce{^4He}) = 4.002\,603$ u, $m(\ce{^1H}) = 1.007\,825$ u and $m_\mathrm{n} = 1.008\,665$ u, find the binding energy of the ⁴He nucleus in MeV.

Key takeaways

Only 253 of 3557 known nuclides are stable; they form a band whose N/Z rises from 1 for light nuclei to about 1.5 for lead.
Nuclei below the band undergo β⁻ decay, those above β⁺ decay or electron capture, and the heaviest α decay — all moving back toward the band.
Even–even nuclei and those with magic numbers (2, 8, 20, 28, 50, 82, 126) are especially stable.
Binding energy B = Δm·c²; B per nucleon peaks at 8.795 MeV for ⁶²Ni, so both fusing light nuclei and splitting heavy ones release energy.

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Data and references

F. G. Kondev, M. Wang, W. J. Huang, S. Naimi, G. Audi, “The NUBASE2020 evaluation of nuclear physics properties”, Chinese Physics C 45, 030001 (2021).
M. Wang, W. J. Huang, F. G. Kondev, G. Audi, S. Naimi, “The AME 2020 atomic mass evaluation (II)”, Chinese Physics C 45, 030003 (2021).
J. W. Rohlf, Modern Physics from α to Z⁰, Wiley (1994), §11.3 (semi-empirical mass formula coefficients).
K. S. Krane, Introductory Nuclear Physics, Wiley (1988), chapters 3 and 5.
P. de Marcillac et al., “Experimental detection of α-particles from the radioactive decay of natural bismuth”, Nature 422, 876–878 (2003).