Mass Defect, Binding Energy and Nuclear Stability – Atomic Nuclei and Radioactive Decay (Senior 6 Physics) | Ganzaa

Atomic Nuclei and Radioactive Decay

Lessons

Senior 6 · Physics

Atomic Nuclei and Radioactive Decay

5 lessons

Lesson 2 of 5

Mass Defect, Binding Energy and Nuclear Stability

25 min

A nucleus that weighs less than its parts

A helium-4 nucleus is two protons and two neutrons. Add up their masses from the table in Lesson 1: $2(1.007\,276) + 2(1.008\,665) = 4.031\,882\,\mathrm{u}$ . Now measure a helium-4 nucleus in a mass spectrometer: $4.001\,506\,\mathrm{u}$ . About 0.030 u, roughly 0.75% of the total, has gone missing. It has not been destroyed. When the four nucleons came together they released energy, and by $E = m c^{2}$ that energy carried mass away with it. To pull the nucleus apart again you must put the same energy back. This "missing mass" explains why some nuclei are stable and others decay, and why both splitting and fusing nuclei can release energy.

Mass defect and binding energy

For a nucleus with $Z$ protons and $N$ neutrons, the mass defect is

\Delta m = Zm_p + Nm_n - m_{\text{nucleus}}

and the binding energy is the energy equivalent of the mass defect, $E_B = \Delta m\,c^2$ , or $E_B = \Delta m\,(\text{in u}) \times 931.5\,\mathrm{MeV}$ .

Using atomic masses. Data tables list the masses of neutral ATOMS, which include $Z$ electrons. If you use the mass of a hydrogen-1 atom, $m_{\text{H}} = 1.007\,825\,\mathrm{u}$ (one proton plus one electron), in place of $m_{p}$ , the $Z$ electron masses cancel:

\Delta m = Zm_{\text{H}} + Nm_n - m_{\text{atom}}

(The electrons' own small binding energies, a few eV to keV, are negligible here.)

To compare nuclei of different sizes, divide by the number of nucleons: the binding energy per nucleon $\frac{E_B}{A}$ is the average energy needed to remove one nucleon. The larger it is, the more tightly bound, and so the more stable, the nucleus.

Definition

The mass defect $\Delta m$ of a nucleus is the difference between the total mass of its separate nucleons and the mass of the nucleus. The binding energy $E_B = \Delta m\,c^2$ is the energy needed to separate a nucleus completely into its individual protons and neutrons (equal to the energy released when it forms). Example: helium-4 has $\Delta m = 0.0304\,\mathrm{u}$ and $E_B = 28.3\,\mathrm{MeV}$ , or 7.07 MeV per nucleon.

Worked example 1: helium-4 from atomic masses

Given $m(^{4}_{2}\mathrm{He}\ \text{atom}) = 4.002\,603\,\mathrm{u}$ , $m_{\text{H}} = 1.007\,825\,\mathrm{u}$ and $m_n = 1.008\,665\,\mathrm{u}$ , find the binding energy and the binding energy per nucleon of ${}^{4}_{2}\mathrm{He}$ .

Solution

$\Delta m = 2(1.007\,825) + 2(1.008\,665) - 4.002\,603 = 4.032\,980 - 4.002\,603 = 0.030\,377\,\mathrm{u}$

$E_B = 0.030\,377 \times 931.5 = 28.3\,\mathrm{MeV}$ , and $\frac{E_B}{A} = \frac{28.3}{4} = 7.07\,\mathrm{MeV}$ per nucleon.

The same answer as the opening, where nuclear masses were used: the electrons cancel.

Worked example 2: iron-56 and uranium-238

The atomic masses are $m(^{56}_{26}\mathrm{Fe}) = 55.934\,936\,\mathrm{u}$ and $m(^{238}_{92}\mathrm{U}) = 238.050\,788\,\mathrm{u}$ . Compare their binding energies per nucleon.

Solution

Fe-56 ( $Z = 26$ , $N = 30$ ): $\Delta m = 26(1.007\,825) + 30(1.008\,665) - 55.934\,936 = 0.528\,464\,\mathrm{u}$ ; $E_B = 0.528\,464 \times 931.5 = 492\,\mathrm{MeV}$ ; $\frac{E_B}{A} = \frac{492}{56} = 8.79\,\mathrm{MeV}$ .

U-238 ( $Z = 92$ , $N = 146$ ): $\Delta m = 92(1.007\,825) + 146(1.008\,665) - 238.050\,788 = 1.934\,202\,\mathrm{u}$ ; $E_B = 1802\,\mathrm{MeV}$ ; $\frac{E_B}{A} = \frac{1802}{238} = 7.57\,\mathrm{MeV}$ .

Uranium has almost four times the TOTAL binding energy, but each of its nucleons is held about 1.2 MeV less tightly than in iron. Iron is the more stable nucleus.

The binding energy curve

Figure 2.1: Binding energy per nucleon for a selection of stable and long-lived nuclides (data from the Atomic Mass Evaluation 2020).

Read the graph carefully:

For light nuclei, $\frac{E_B}{A}$ rises steeply. Helium-4 stands out: it is unusually tightly bound, which is why whole helium-4 nuclei (alpha particles) are ejected by heavy nuclei.
It reaches a broad maximum of about 8.8 MeV near iron-56 and nickel-62, the most tightly bound nuclei.
It then falls slowly, to about 7.6 MeV for uranium.

Any change that moves nucleons TOWARDS the peak releases energy. Joining two light nuclei into one (fusion) climbs the left side; splitting a heavy nucleus into two middle-sized ones (fission) climbs the right side. Lesson 5 returns to both.

Why the curve has this shape

Nucleons are held together by the strong nuclear force. It is attractive between any two nucleons, much stronger than the electric force at about 1 fm, but it acts only over about 2 to 3 fm. So each nucleon attracts only its near neighbours. In a small nucleus many nucleons sit on the surface with fewer neighbours, so $\frac{E_B}{A}$ is low. The electric repulsion between protons, however, acts across the whole nucleus: every proton repels every other one. In a heavy nucleus this repulsion grows roughly as $Z^{2}$ and steadily reduces the binding per nucleon. Heavy nuclei therefore need extra neutrons, which add strong-force attraction without adding repulsion.

Worked example 3: estimating the energy of fission from the graph

A uranium-235 nucleus absorbs a neutron and splits into two nuclei of mass number about 117 each. Use Figure 2.1 to estimate the energy released.

Solution

Before: 236 nucleons bound at about 7.6 MeV each. After: 236 nucleons bound at about 8.5 MeV each (read at $A \approx 117$ ). The binding energy increases by about $236 \times (8.5 - 7.6) = 212\,\mathrm{MeV}$ , and this energy is released. So one fission releases roughly 200 MeV, an estimate that agrees with measured values. By comparison, burning one molecule of methane releases about 9 eV: more than twenty million times less.

Nuclear stability: the N–Z plot

Figure 2.2: Stable nuclei lie in a narrow band. Unstable nuclei decay in a way that brings them towards it.

For light nuclei the stable ones have $N \approx Z$ (carbon-12, oxygen-16). Heavier stable nuclei need more neutrons: $\frac{N}{Z} \approx 1.5$ for lead-208, the heaviest stable nuclide. No nucleus with $Z > 82$ is stable. A nucleus away from the band is radioactive: it decays spontaneously, and the type of decay moves it towards the band, as Lesson 3 shows.

Warning

Common mistakes. (1) Thinking binding energy is energy stored IN the nucleus that it can give out. It is energy that was released when the nucleus formed, and that must be SUPPLIED to take it apart. (2) Mixing atomic masses with the proton mass: use $m_{\text{H}}$ with atomic masses, or $m_{p}$ with nuclear masses, never a mixture. (3) Judging stability from the total binding energy: compare the binding energy PER NUCLEON. (4) Rounding too early: the mass defect is a small difference of large numbers, so keep all six decimal places until the end.

Tip

On the binding energy curve, uphill is downhill: every nuclear change that moves nucleons UP the curve, towards iron, releases energy. Fusion climbs from the left, fission from the right.

Binding energy graph (Khan Academy India). Watch how the shape of the binding-energy-per-nucleon curve shows why both fusion of light nuclei and fission of heavy nuclei release energy.

✦

Practice

Use $m_{\text{H}} = 1.007\,825\,\mathrm{u}$ , $m_n = 1.008\,665\,\mathrm{u}$ and $1\,\mathrm{u} = 931.5\,\mathrm{MeV}$ .

Define mass defect and binding energy per nucleon.
The atomic mass of ${}^{16}_{8}\mathrm{O}$ is 15.994 915 u. Find its binding energy and binding energy per nucleon.
The atomic mass of ${}^{7}_{3}\mathrm{Li}$ is 7.016 003 u. Find its binding energy per nucleon. Is lithium-7 more or less tightly bound than helium-4?
The atomic mass of deuterium, ${}^{2}_{1}\mathrm{H}$ , is 2.014 102 u. Find the least energy of a gamma-ray photon that can split a deuteron into a proton and a neutron, and the photon's wavelength ( $hc = 1.24 \times 10^{-6}\,\mathrm{eV\,m}$ ).
Use Figure 2.2 to predict the likely decay of (a) carbon-14, (b) carbon-11, (c) radium-226.
Explain, in terms of the forces between nucleons, why heavy stable nuclei have more neutrons than protons.
Two deuterium nuclei fuse to form helium-4. Using Figure 2.1 and your answer to question 4, estimate the energy released.
Gatete says: "Uranium-238 is the most stable nucleus in the table, because its binding energy of 1802 MeV is by far the largest." Evaluate his statement.

Show answersHide answers

Mass defect: total mass of the separate nucleons minus the mass of the nucleus. Binding energy per nucleon: the total binding energy $\Delta m\,c^2$ divided by the mass number; the average energy needed to remove one nucleon.
$\Delta m = 8(1.007\,825) + 8(1.008\,665) - 15.994\,915 = 0.137\,005\,\mathrm{u}$ ; $E_B = 127.6\,\mathrm{MeV}$ ; 7.98 MeV per nucleon.
$\Delta m = 3(1.007\,825) + 4(1.008\,665) - 7.016\,003 = 0.042\,132\,\mathrm{u}$ ; $E_B = 39.2\,\mathrm{MeV}$ ; 5.61 MeV per nucleon. Less tightly bound than He-4 (7.07 MeV).
$\Delta m = 1.007\,825 + 1.008\,665 - 2.014\,102 = 0.002\,388\,\mathrm{u}$ , so $E_B = 2.22\,\mathrm{MeV}$ . $\lambda = \frac{hc}{E} = \frac{1.24 \times 10^{-6}}{2.22 \times 10^{6}} = 5.6 \times 10^{-13}\,\mathrm{m}$ .
(a) C-14 (Z = 6, N = 8) has too many neutrons: β⁻ decay. (b) C-11 (Z = 6, N = 5) has too many protons: β⁺ decay or electron capture. (c) Ra-226 (Z = 88 > 82): α decay.
The strong force attracts every nucleon to its near neighbours; the electric force makes every proton repel every other proton, over the whole nucleus. As Z grows the repulsion grows faster than the attraction, so extra neutrons are needed: they add strong attraction but no repulsion.
Before: $2 \times 2.22 = 4.4\,\mathrm{MeV}$ . After: $4 \times 7.07 = 28.3\,\mathrm{MeV}$ . Energy released $\approx 28.3 - 4.4 = 23.8\,\mathrm{MeV}$ .
Wrong. A large nucleus has a large total binding energy simply because it has many nucleons. Stability depends on the binding energy PER NUCLEON: 7.57 MeV for U-238 against 8.79 MeV for Fe-56. U-238 is in fact radioactive (it decays by α emission).

Key points

$\Delta m = Zm_{\text{H}} + Nm_n - m_{\text{atom}}$ ; $E_B = \Delta m \times 931.5\,\mathrm{MeV}$ .
$\frac{E_B}{A}$ peaks at about 8.8 MeV near iron and nickel; moving towards the peak (fusion of light nuclei, fission of heavy ones) releases energy.
The short-range strong force binds nucleons; long-range proton repulsion limits the size of stable nuclei and requires $N > Z$ in heavy ones.
Stable nuclei lie in a band on the N–Z plot; none has $Z > 82$ .

Sources and credits

10.3: Nuclear Binding Energy, from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Atomic masses and binding energies: Atomic Mass Evaluation 2020 (AME2020), M. Wang, W. J. Huang, F. G. Kondev, G. Audi and S. Naimi, Chinese Physics C 45 (2021), distributed by the IAEA Atomic Mass Data Center.
Video: Binding energy graph, Khan Academy India - English, embedded from YouTube.

The two graphs, the worked examples and the practice exercises are written by Ganzaa.

1A nucleus that weighs less than its parts

2Mass defect and binding energy

3The binding energy curve

4Why the curve has this shape

5Nuclear stability: the N–Z plot

A nucleus that weighs less than its parts

Mass defect and binding energy

The binding energy curve

Why the curve has this shape

Nuclear stability: the N–Z plot