← Part I: Nuclear Structure
Chapter 2

Binding Energy & Mass Formula

Nuclear Binding Energy

The binding energy of a nucleus is the energy required to disassemble it into free, unbound protons and neutrons. It arises because the mass of a nucleus is less than the sum of the masses of its constituent nucleons:

$$B(A,Z) = \left[Z\,m_p + N\,m_n - M(A,Z)\right]c^2$$

where $m_p = 938.272$ MeV/$c^2$ is the proton mass,$m_n = 939.565$ MeV/$c^2$ is the neutron mass, and$M(A,Z)$ is the nuclear mass. The mass defect $\Delta M = Zm_p + Nm_n - M$ is converted to binding energy via Einstein's $E = mc^2$.

The binding energy per nucleon, $B/A$, is a key quantity. It peaks near$A \approx 56$ (iron/nickel region) at about 8.8 MeV. This peak explains why energy is released by fission of heavy nuclei and fusion of light nuclei.

Semi-Empirical Mass Formula

The Bethe-Weizsacker semi-empirical mass formula (SEMF) parameterizes the nuclear binding energy as a sum of five terms, each with a physical motivation from the liquid drop model of the nucleus:

$$B(A,Z) = a_V A - a_S A^{2/3} - a_C \frac{Z(Z-1)}{A^{1/3}} - a_A \frac{(A - 2Z)^2}{A} + \delta(A,Z)$$

1. Volume Term: $a_V A$

Since each nucleon interacts only with its nearest neighbors (saturation), the total binding energy is proportional to the total number of nucleons. This is the dominant term.

$$B_{\text{vol}} = a_V A, \quad a_V \approx 15.56 \text{ MeV}$$

2. Surface Term: $-a_S A^{2/3}$

Nucleons at the surface have fewer neighbors than those in the interior. The surface area goes as $R^2 \propto A^{2/3}$, reducing the binding:

$$B_{\text{surf}} = -a_S A^{2/3}, \quad a_S \approx 17.23 \text{ MeV}$$

3. Coulomb Term: $-a_C Z(Z-1)/A^{1/3}$

The electrostatic repulsion between protons reduces binding. For a uniform charge sphere of radius $R = r_0 A^{1/3}$:

$$B_{\text{Coul}} = -\frac{3}{5}\frac{e^2}{4\pi\epsilon_0}\frac{Z(Z-1)}{R} = -a_C \frac{Z(Z-1)}{A^{1/3}}, \quad a_C \approx 0.7 \text{ MeV}$$

4. Asymmetry Term: $-a_A(A-2Z)^2/A$

The Pauli exclusion principle favors equal numbers of protons and neutrons. An excess of either type must occupy higher energy levels:

$$B_{\text{asym}} = -a_A \frac{(N-Z)^2}{A}, \quad a_A \approx 23.3 \text{ MeV}$$

This can be derived from the Fermi gas model by computing the kinetic energy difference between symmetric and asymmetric nuclear matter.

5. Pairing Term: $\delta(A,Z)$

Nucleons prefer to pair in spin-zero couples (analogous to Cooper pairs in superconductors):

$$\delta(A,Z) = \begin{cases} +a_P / A^{1/2} & \text{even-even (Z, N both even)} \\ 0 & \text{even-odd or odd-even} \\ -a_P / A^{1/2} & \text{odd-odd (Z, N both odd)} \end{cases}$$

with $a_P \approx 12$ MeV. Even-even nuclei are the most stable, which explains why most stable isotopes have even Z and even N.

Nuclear Stability

For a given mass number A, the most stable nucleus minimizes the mass (maximizes B). Differentiating the SEMF with respect to Z:

$$\frac{\partial B}{\partial Z} = 0 \implies Z_{\text{stable}} = \frac{A}{2} \cdot \frac{1}{1 + \frac{a_C}{4a_A} A^{2/3}}$$

For light nuclei, $Z \approx A/2$ (equal protons and neutrons). For heavy nuclei, the Coulomb term pushes $Z < A/2$, favoring neutron-rich nuclei. This defines the valley of stability in the $(N, Z)$ chart of nuclides.

Nuclei above the valley (neutron-deficient) undergo $\beta^+$ decay or electron capture. Nuclei below (neutron-rich) undergo $\beta^-$ decay. Very heavy nuclei ($Z > 82$) are unstable to alpha decay.

Python Simulation: SEMF Binding Energy Curve

Calculates binding energy per nucleon using the semi-empirical mass formula, showing individual term contributions and comparison with experimental data for key nuclei.

SEMF Binding Energy per Nucleon

Python

Plots B/A vs A with SEMF prediction and individual term contributions

script.py110 lines

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Fortran Implementation

High-performance Fortran implementation that scans all nuclei to find the most stable isotope for each mass number and identifies the binding energy maximum.

SEMF Nuclear Scan

Fortran

Scans all nuclei from A=2 to 270 to find maximum binding energy per nucleon

semf_calculator.f9066 lines

Click Run to execute the Fortran code

Code will be compiled with gfortran and executed on the server

Derivation: SEMF Term by Term

Each term in the semi-empirical mass formula has a physical derivation grounded in the liquid drop model and quantum mechanics. We present the detailed derivation of each contribution.

Volume Term Derivation

In the liquid drop model, the nuclear density is approximately constant:$\rho_0 \approx 0.16$ fm$^{-3}$. Since each nucleon interacts only with its nearest neighbors (saturation), the binding energy is proportional to the number of nucleons:

$$B_{\rm vol} = a_V A$$

From the Fermi gas model, we can estimate $a_V$. The binding energy per nucleon in infinite nuclear matter is:

$$\frac{B}{A}\bigg|_{\rm vol} = -\frac{3}{5}E_F + \frac{1}{2}\bar{V}$$

where $E_F \approx 37$ MeV is the Fermi energy and $\bar{V}$ is the average potential per nucleon. The empirical value $a_V \approx 15.56$ MeV reflects the balance between kinetic energy (positive) and potential energy (negative).

Surface Term Derivation

Nucleons at the surface of the nucleus have fewer neighbors and are therefore less tightly bound. This correction is proportional to the surface area:

$$B_{\rm surf} = -a_S \cdot \frac{\text{Surface area}}{\text{Area per nucleon}} = -a_S \cdot \frac{4\pi R^2}{4\pi r_0^2} = -a_S A^{2/3}$$

since $R = r_0 A^{1/3}$. This is analogous to surface tension in a liquid drop. The nuclear surface tension is:

$$\gamma = \frac{a_S}{4\pi r_0^2} \approx \frac{17.23 \text{ MeV}}{4\pi(1.25)^2 \text{ fm}^2} \approx 0.88 \text{ MeV/fm}^2 \approx 1.1 \text{ MJ/m}^2$$

This is roughly $10^{18}$ times larger than the surface tension of water, reflecting the enormous strength of the nuclear force.

Coulomb Term Derivation

The electrostatic energy of a uniformly charged sphere of charge $Ze$ and radius$R$ is calculated by bringing charge from infinity shell by shell:

$$E_C = \int_0^R \frac{1}{4\pi\epsilon_0}\frac{q(r)}{r}\,dq = \int_0^R \frac{1}{4\pi\epsilon_0}\frac{\frac{4}{3}\pi r^3 \rho_{\rm ch}}{r}\cdot 4\pi r^2 \rho_{\rm ch}\,dr$$

where $\rho_{\rm ch} = 3Ze/(4\pi R^3)$. Evaluating the integral:

$$E_C = \frac{3}{5}\frac{Z^2 e^2}{4\pi\epsilon_0 R} = \frac{3}{5}\frac{e^2}{4\pi\epsilon_0 r_0}\frac{Z^2}{A^{1/3}}$$

Since protons are discrete, we replace $Z^2$ with $Z(Z-1)$ to avoid self-interaction:

$$B_{\rm Coul} = -a_C \frac{Z(Z-1)}{A^{1/3}}, \quad a_C = \frac{3}{5}\frac{e^2}{4\pi\epsilon_0 r_0} = \frac{3}{5}\frac{1.44 \text{ MeV}\cdot\text{fm}}{1.25 \text{ fm}} \approx 0.69 \text{ MeV}$$

Asymmetry Term Derivation from the Fermi Gas Model

Consider a nucleus as two independent Fermi gases (protons and neutrons) confined in the nuclear volume $V = \frac{4}{3}\pi R^3$. The total kinetic energy is:

$$E_{\rm kin} = \frac{3}{5}Z\,E_F^{(p)} + \frac{3}{5}N\,E_F^{(n)}$$

where the Fermi energies are $E_F^{(p)} = \frac{\hbar^2}{2m}(3\pi^2 Z/V)^{2/3}$ and similarly for neutrons. For a symmetric nucleus ($Z = N = A/2$):

$$E_{\rm kin}^{(0)} = \frac{3}{5}\frac{A}{2}\frac{\hbar^2}{2m}\left(\frac{3\pi^2 A}{2V}\right)^{2/3} \times 2 = \frac{3}{5}A\,\bar{E}_F$$

For an asymmetric nucleus with $Z = (A - \delta)/2$, $N = (A + \delta)/2$ where$\delta = N - Z$, expanding to second order in $\delta/A$:

$$E_{\rm kin} = E_{\rm kin}^{(0)} + \frac{\bar{E}_F}{3}\frac{(N-Z)^2}{A} + \mathcal{O}(\delta^4)$$

This identifies the asymmetry coefficient:

$$a_A^{\rm(kin)} = \frac{\bar{E}_F}{3} = \frac{E_F}{3} \approx \frac{37}{3} \approx 12.3 \text{ MeV}$$

The empirical value $a_A \approx 23.3$ MeV is about twice the kinetic energy estimate, because the potential energy also depends on the N-Z asymmetry. The symmetry potential contributes roughly equally to the asymmetry coefficient.

Pairing Term Origin

The pairing term arises from the short-range attractive residual interaction between nucleons in time-reversed orbits $(n\ell j m)$ and $(n\ell j \bar{m})$. The pairing gap in the BCS theory of nuclear pairing is approximately:

$$\Delta \approx \frac{12}{\sqrt{A}} \text{ MeV}$$

Even-even nuclei gain extra binding from both proton and neutron pairing. The total pairing energy is:

$$\delta(A,Z) = \begin{cases} +\Delta_p + \Delta_n \approx +\frac{2 \times 12}{\sqrt{A}} & \text{even-even} \\ \Delta_p \text{ or } \Delta_n & \text{odd-A} \\ 0 & \text{odd-odd (approximate)} \end{cases}$$

The enhanced stability of even-even nuclei is reflected in the observation that 167 of the 253 stable nuclei are even-even, while only 4 stable nuclei are odd-odd ($^2$H, $^6$Li, $^{10}$B, $^{14}$N), all with small A.

Valley of Stability and Drip Lines

The nuclear landscape is bounded by the proton and neutron drip lines, beyond which nucleons are no longer bound. These limits are defined by the separation energies.

Drip Line Conditions

The neutron drip line is reached when the one-neutron separation energy vanishes:

$$S_n(A,Z) = B(A,Z) - B(A-1,Z) = 0 \quad \text{(neutron drip line)}$$

Similarly, the proton drip line is:

$$S_p(A,Z) = B(A,Z) - B(A-1,Z-1) = 0 \quad \text{(proton drip line)}$$

Using the SEMF, the most stable Z for a given A satisfies:

$$Z_0(A) = \frac{A}{2}\cdot\frac{1}{1 + \frac{a_C A^{2/3}}{4a_A}} \approx \frac{A}{2}\cdot\frac{1}{1 + 0.0075\,A^{2/3}}$$

For light nuclei $Z_0 \approx A/2$; for $A = 240$, $Z_0 \approx 93$, giving$N/Z \approx 1.6$. This neutron excess in heavy nuclei is driven by the Coulomb term, which penalizes large Z.

Isobaric Mass Parabolas

For fixed A, the atomic mass as a function of Z forms a parabola (or two parabolas for even A, split by the pairing term):

$$M(A,Z)c^2 = \alpha A - \beta Z + \gamma Z^2 + \text{const.}$$

where $\gamma = 4a_A/A + a_C/A^{1/3}$. For odd-A isobars, there is a single parabola with one stable nucleus at the minimum. For even-A, the even-even and odd-odd parabolas are offset by $2\delta$, potentially allowing multiple stable isobars.

This explains why some even-A isobars have 2 or even 3 stable isotopes (e.g.,$^{124}$Sn, $^{124}$Te, $^{124}$Xe for A=124), while odd-A isobars typically have only one stable member.

Separation Energies and Q-Values

Neutron and Proton Separation Energies

The one-neutron separation energy is the minimum energy needed to remove a neutron:

$$S_n = B(Z,N) - B(Z,N-1) = [M(Z,N-1) + m_n - M(Z,N)]c^2$$

The two-neutron separation energy is often smoother (less odd-even staggering):

$$S_{2n} = B(Z,N) - B(Z,N-2) = [M(Z,N-2) + 2m_n - M(Z,N)]c^2$$

From the SEMF, the neutron separation energy is approximately:

$$S_n \approx a_V - \frac{2}{3}a_S A^{-1/3} - 4a_A\frac{N-Z-1}{A} + \text{pairing}$$

Sudden drops in $S_n$ or $S_{2n}$ at specific N values (N = 8, 20, 28, 50, 82, 126) provide the classic evidence for nuclear shell closures (magic numbers).

Q-Values for Nuclear Reactions

The Q-value of a nuclear reaction $a + A \to b + B$ is the mass-energy difference:

$$Q = (M_a + M_A - M_b - M_B)c^2 = T_b + T_B - T_a - T_A$$

In terms of binding energies:

$$Q = B(b) + B(B) - B(a) - B(A)$$

Important Q-value applications:

  • - Alpha decay: $Q_\alpha = B(\alpha) + B(A-4,Z-2) - B(A,Z)$. Alpha decay is possible when $Q_\alpha > 0$.
  • - Beta decay: $Q_{\beta^-} = [M(Z,A) - M(Z+1,A)]c^2$. In terms of atomic masses, $Q_{\beta^-} = [\mathcal{M}(Z,A) - \mathcal{M}(Z+1,A)]c^2$.
  • - Fusion: $Q_{\rm DT} = B(^4{\rm He}) + B(n) - B(^2{\rm H}) - B(^3{\rm H}) = 17.6$ MeV for the D-T reaction.

Weizsacker Mass Formula Applications

Predicting the Limits of Nuclear Stability

The SEMF predicts that nuclei become unstable to spontaneous fission when the Coulomb energy equals twice the surface energy. The fissility parameter is:

$$x = \frac{E_C^{(0)}}{2E_S^{(0)}} = \frac{a_C Z^2/A^{1/3}}{2a_S A^{2/3}} = \frac{a_C}{2a_S}\frac{Z^2}{A}$$

Spontaneous fission becomes dominant when $x \geq 1$, giving:

$$\left(\frac{Z^2}{A}\right)_{\rm crit} = \frac{2a_S}{a_C} = \frac{2 \times 17.23}{0.70} \approx 49$$

This predicts that nuclei with $Z^2/A > 49$ should be instantly unstable, setting an upper bound on the periodic table around $Z \approx 120$. Shell effects (not included in the SEMF) create islands of stability for superheavy elements near Z = 114, N = 184.

Alpha Decay Q-Value from SEMF

The SEMF predicts the alpha decay Q-value as:

$$Q_\alpha = B(A-4,Z-2) + B(4,2) - B(A,Z)$$

Using $B(^4{\rm He}) = 28.296$ MeV and expanding to leading order:

$$Q_\alpha \approx 28.3 - 4a_V + \frac{8}{3}a_S A^{-1/3} + 4a_C\frac{Z - 1}{A^{1/3}} + 4a_A\frac{A - 2Z}{A} + \cdots$$

This formula correctly predicts that $Q_\alpha > 0$ for heavy nuclei with$A \gtrsim 150$, though shell effects cause significant deviations (e.g., nuclei near $^{208}$Pb have anomalously low $Q_\alpha$ due to the Z=82, N=126 shell closures).

Magic Numbers: Evidence from Binding Energies

Systematic deviations from the smooth SEMF predictions reveal the nuclear magic numbers: 2, 8, 20, 28, 50, 82, 126. These correspond to closed shells in the nuclear potential, analogous to noble gas configurations in atomic physics.

Binding Energy Signatures

Magic nuclei show several distinctive features in their binding energies:

  • - Extra binding: The binding energy of a magic nucleus exceeds the SEMF prediction. The deviation $\Delta B = B_{\rm exp} - B_{\rm SEMF}$ peaks at magic numbers.
  • - Separation energy drops: $S_n$ drops sharply when going from N = magic to N = magic + 1. For example, $S_n(^{208}{\rm Pb}) = 7.37$ MeV but $S_n(^{209}{\rm Pb}) = 3.94$ MeV.
  • - Large first excited state energy: Doubly-magic nuclei like $^{208}$Pb have a first excited state at 2.61 MeV, compared to typical ~0.1 MeV for deformed nuclei.
  • - Isotope/isotone abundance: Elements with Z or N equal to a magic number have more stable isotopes/isotones.

Empirical Mass Parabola Widths

The curvature of the mass parabola for a fixed A is related to the effective asymmetry coefficient. Near shell closures, the effective $a_A$ increases, reflecting the larger energy cost of exciting nucleons across the shell gap:

$$a_A^{\rm eff}(A) = a_A + \text{shell correction terms}$$

The Strutinsky shell correction method quantifies this: the shell energy is defined as the difference between the discrete sum of single-particle energies and its smoothed (Thomas-Fermi) average: $\delta E_{\rm shell} = \sum_i \varepsilon_i - \widetilde{\sum_i \varepsilon_i}$.

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