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Chapter 2 β€’ Theory & Derivations

Unit 2: Atomic Nucleus: Composition, Binding Energy & Structural Models

In-depth treatment of nuclear morphology, spatial density distributions, mass defect energetics, the binding energy per nucleon curve, nuclear stability criteria, the WeizsΓ€cker Semi-Empirical Mass Formula, and fundamental structural frameworks including the Nuclear Shell Model, Magic Numbers, and Collective Deformations.

Β§2.1 Nucleonic Composition & Nuclear Dimensions: Radius Scaling, Density & Charge Distributions

The atomic nucleus is an ultra-dense, quantum many-body system bound by the strong nuclear interaction. It consists of two constituent fermionic species collectively termed nucleons:

  • Protons ($p$): Baryons with positive charge $+e = +1.60217663 \times 10^{-19}\text{ C}$, rest mass $m_p = 1.0072764666\text{ u} = 938.272\text{ MeV}/c^2$, intrinsic spin $s = 1/2$, and isospin projection $T_3 = +1/2$.
  • Neutrons ($n$): Electrically neutral baryons ($q = 0$), rest mass $m_n = 1.0086649158\text{ u} = 939.565\text{ MeV}/c^2$, intrinsic spin $s = 1/2$, and isospin projection $T_3 = -1/2$.

The composition of any nuclide $^{A}_{Z}\text{X}_N$ is defined by:

  • Atomic number $Z$: Number of protons.
  • Neutron number $N$: Number of neutrons.
  • Mass number $A = Z + N$: Total nucleon count.

Nuclear Dimensions and Radius Scaling

Because nucleons are subject to the Pauli exclusion principle and the strong nuclear force exhibits a hard repulsive core at inter-nucleon separations $r < 0.5\text{ fm}$ alongside short-range saturation at $r \approx 1.0 - 1.5\text{ fm}$, nuclear matter is essentially incompressible.

Consequently, the volume of a nucleus $V$ is directly proportional to the total number of nucleons $A$:

$$V = \frac{4}{3}\pi R^3 \propto A \implies R \propto A^{1/3}$$

The nuclear radius $R$ is parameterized by the foundational empirical scaling law:

$$R = R_0 A^{1/3}$$

where $R_0$ is the nuclear radius parameter.

  • High-energy electron scattering measurements (Hofstadter, 1950s) sensitive to the nuclear charge distribution yield:
$$R_0 \approx 1.20\text{ to }1.25\text{ fm} \quad (1\text{ fm} = 10^{-15}\text{ m})$$
  • Nuclear potential measurements (neutron scattering, alpha decay barriers) yield slightly larger values:
$$R_{0,\text{matter}} \approx 1.4\text{ fm}$$

Nuclear Density and Incompressibility

Using $R_0 = 1.25\text{ fm} = 1.25 \times 10^{-15}\text{ m}$ and average nucleon mass $m \approx 1.67 \times 10^{-27}\text{ kg}$:

$$\rho_{\text{nuc}} = \frac{M}{V} = \frac{A \cdot m}{\frac{4}{3}\pi R^3} = \frac{A \cdot m}{\frac{4}{3}\pi R_0^3 A} = \frac{3 m}{4\pi R_0^3}$$

Notice that mass number $A$ cancels out completely!

$$\rho_{\text{nuc}} = \frac{3 (1.67 \times 10^{-27}\text{ kg})}{4\pi (1.25 \times 10^{-15}\text{ m})^3} \approx 2.04 \times 10^{17}\text{ kg/m}^3 \approx 2 \times 10^{14}\text{ g/cm}^3$$

The nuclear matter nucleon number density $\rho_0$ is:

$$\rho_0 = \frac{3}{4\pi R_0^3} \approx 0.17\text{ nucleons/fm}^3$$

This incredible densityβ€”over 200 trillion times denser than liquid waterβ€”is uniform across all nuclei from helium to uranium, confirming that nuclear matter behaves like an incompressible quantum liquid drop.

Woods-Saxon Radial Charge Distribution

High-energy elastic electron scattering demonstrates that nuclei do not possess sharp, hard-sphere boundaries. Instead, the charge density profile $\rho(r)$ is modeled accurately by the Woods-Saxon (two-parameter Fermi) distribution:

$$\rho(r) = \frac{\rho_0}{1 + \exp\left(\frac{r - c}{a}\right)}$$

where:

  • $\rho_0$ is the central interior nuclear density.
  • $c$ is the half-density radius ($c \approx 1.07 A^{1/3}\text{ fm}$), where $\rho(c) = 0.5 \rho_0$.
  • $a$ is the surface diffuseness parameter ($a \approx 0.54\text{ fm}$), governing the rate of density falloff.
  • The skin thickness $t_{90-10}$, defined as the distance over which the density drops from $90\%$ to $10\%$ of $\rho_0$, is universally:
$$t_{90-10} = 4 a \ln(3) \approx 4.394 a \approx 2.4\text{ fm}$$

across virtually all stable nuclei.

Β§2.2 Mass Defect, Einsteinian Equivalence & the Systematics of Binding Energy per Nucleon

One of the most striking findings of precise mass spectrometry (Aston, Bainbridge) is that the precise mass of any bound nucleus $M(A,Z)$ is strictly less than the sum of the rest masses of its constituent free protons and neutrons.

The Mass Defect $\Delta m$

The difference between the total mass of the individual constituent nucleons at infinite separation and the actual bound atomic mass is defined as the mass defect $\Delta m$:

$$\Delta m = \left[ Z \cdot m(^1\text{H}) + (A - Z) \cdot m_n \right] - M(A,Z)$$

where:

  • $m(^1\text{H}) = 1.007825032\text{ u}$ is the atomic mass of neutral hydrogen-1 (accounting for electron mass $m_e$).
  • $m_n = 1.008664916\text{ u}$ is the free neutron mass.
  • $M(A,Z)$ is the neutral atomic mass of the nuclide.
  • $1\text{ u} \equiv \frac{1}{12} m(^{12}\text{C}) = 1.66053906660 \times 10^{-27}\text{ kg} \equiv 931.49410242\text{ MeV}/c^2$.

Total Nuclear Binding Energy ($B$)

According to Einstein's mass-energy equivalence principle ($E = mc^2$), this missing mass was liberated as binding energy during the nucleosynthetic coalescence of the nucleus:

$$B(A,Z) = \Delta m \cdot c^2 = \left[ Z m(^1\text{H}) + (A - Z) m_n - M(A,Z) \right] c^2$$

Binding Energy per Nucleon ($B/A$) Curve Morphology

To compare the relative thermodynamic stability of different nuclear species, we define the binding energy per nucleon $B/A$:

$$\frac{B}{A} = \frac{B(A,Z)}{A}$$

``` B/A (MeV / Nucleon) 9 β–² ⁡⁢Fe (8.79 MeV) ⁢²Ni (8.79 MeV) β”‚ β–² 8 β”‚ ⁴He / \_____________ β”‚ β–² / \_______ 7 β”‚ / \ / \_____ ²³⁸U (7.57 MeV) β”‚ / \____/ \ 6 β”‚ / 5 β”‚ / EXOTHERMIC FUSION EXOTHERMIC FISSION 4 β”‚ / 3 β”‚/ 2 β”‚ Β²H (1.11 MeV) 1 β”‚ 0 └──┴───────────┴─────────────────────────┴────────► Mass Number A 0 56 238 ```

The universal curve of binding energy per nucleon reveals fundamental properties of nuclear forces:

1. Low Mass Region ($A < 20$):

The curve rises steeply from $1.11\text{ MeV/nucleon}$ for deuterium ($^{2}\text{H}$) up to $\approx 8\text{ MeV}$. Prominent periodic spikes appear at $^{4}\text{He}$ ($7.07\text{ MeV}$), $^{8}\text{Be}$, $^{12}\text{C}$ ($7.68\text{ MeV}$), and $^{16}\text{O}$ ($7.98\text{ MeV}$). These peaks reflect the exceptional stability of tightly bound $\alpha$-conjugate nuclei ($Z = N = 2n$).

2. The Global Peak at $A \approx 56 - 62$:

The maximum of the curve occurs in the iron-nickel group:

  • $^{56}_{26}\text{Fe}$: $B/A = 8.790\text{ MeV/nucleon}$
  • $^{62}_{28}\text{Ni}$: $B/A = 8.7946\text{ MeV/nucleon}$ (highest absolute binding energy per nucleon of all known nuclides)
  • $^{58}_{26}\text{Fe}$: $B/A = 8.792\text{ MeV/nucleon}$

These nuclides represent the thermodynamically most stable nuclear states in the cosmos ("iron peak" in stellar nucleosynthesis).

3. High Mass Plateau and Decline ($A > 60$):

Beyond iron, the binding energy per nucleon gently and monotonically declines from $\sim 8.8\text{ MeV}$ down to $7.57\text{ MeV/nucleon}$ for uranium-238 ($^{238}\text{U}$). This decline is driven directly by long-range Coulomb electrostatic repulsion between protons, which scales quadratically with $Z^2$ and overcomes the short-range strong nuclear attraction.

4. Thermodynamic Implications for Nuclear Energy:

  • Nuclear Fusion: Coalescing light nuclei ($A < 56$, e.g., $^2\text{H} + ^3\text{H} \to ^4\text{He} + n$) moves upward along the curve toward higher $B/A$, releasing millions of electron volts per reaction.
  • Nuclear Fission: Splitting heavy nuclei ($A > 200$, e.g., $n + ^{235}\text{U} \to \text{Ba} + \text{Kr} + 3n$) breaks a less tightly bound nucleus ($7.6\text{ MeV/nucleon}$) into two intermediate fragments ($8.5\text{ MeV/nucleon}$), releasing $\approx 0.9\text{ MeV/nucleon} \times 235 \approx 200\text{ MeV}$ of net kinetic energy per fission.

Β§2.3 The Nuclear Valley of Beta-Stability: N/Z Ratios, Neutron Driplines & Proton Driplines

If we plot all known nuclides on a Segrè chart (the chart of nuclides) with proton number $Z$ on the ordinate and neutron number $N$ on the abscissa, stable nuclei form a narrow ribbon termed the valley of beta-stability.

``` Z (Proton Number) 92 β–² / N = Z line β”‚ / 82 β”‚ Pb / β”‚ / Valley of Beta Stability β”‚ / (N > Z for heavy nuclei) β”‚ /β€’ 40 β”‚ Zr /β€’β€’ β”‚ /β€’β€’ 20 β”‚ Ca /β€’β€’β€’ (N β‰ˆ Z for light nuclei) β”‚ /β€’β€’β€’β€’ β”‚ /β€’β€’β€’β€’ └────────────┴──────────────────────────► N (Neutron Number) 0 20 40 82 126 ```

The $N/Z$ Ratio Trajectory

1. Light Stable Nuclei ($Z \le 20$):

For light nuclei, the valley follows the line of symmetry $N = Z$ ($N/Z \approx 1.0$). Examples include $^{4}_{2}\text{He}$, $^{12}_{6}\text{C}$, $^{14}_{7}\text{N}$, $^{16}_{8}\text{O}$, $^{20}_{10}\text{Ne}$, $^{28}_{14}\text{Si}$, and $^{40}_{20}\text{Ca}$. This symmetry arises from the Pauli exclusion principle, which penalizes unequal filling of neutron and proton quantum energy levels.

2. Heavy Stable Nuclei ($Z > 20$):

As $Z$ increases, Coulomb repulsion among protons scales as $Z(Z-1)/R \propto Z^2 / A^{1/3}$. To maintain binding against this disruptive force, the nucleus must incorporate an increasing excess of neutrons, which supply attractive strong nuclear force without adding disruptive electrostatic charge. Consequently, the $N/Z$ ratio increases monotonically:

  • Iron-56: $^{56}_{26}\text{Fe}_{30} \implies N/Z = 30/26 \approx 1.15$
  • Silver-107: $^{107}_{47}\text{Ag}_{60} \implies N/Z = 60/47 \approx 1.28$
  • Lead-208: $^{208}_{82}\text{Pb}_{126} \implies N/Z = 126/82 \approx 1.54$
  • Uranium-238: $^{238}_{92}\text{U}_{146} \implies N/Z = 146/92 \approx 1.59$

Nuclear Decay Modes Relative to the Valley of Stability

Nuclides lying off the central floor of the valley undergo spontaneous radioactive transitions to reach stability:

  • Neutron-Rich Nuclides (Lying below/to the right of the valley):

Excess neutrons result in negative beta decay ($\beta^-$) where a neutron converts to a proton:

$$n \longrightarrow p + e^- + \bar{\nu}_e \quad (\Delta Z = +1, \Delta N = -1, A = \text{constant})$$

For extreme neutron excess, prompt neutron emission ($S_n \le 0$) defines the neutron dripline.

  • Proton-Rich Nuclides (Lying above/to the left of the valley):

Excess protons result in positron emission ($\beta^+$) or electron capture (EC):

$$p \longrightarrow n + e^+ + \nu_e \quad (\Delta Z = -1, \Delta N = +1, A = \text{constant})$$
$$p + e^- \longrightarrow n + \nu_e$$

For extreme proton excess, proton emission ($S_p \le 0$) defines the proton dripline.

  • Superheavy Unstable Nuclides ($Z > 82, A > 209$):

Coulomb repulsion is so massive that beta decay cannot stabilize the nucleus; alpha decay ($\Delta Z = -2, \Delta A = -4$) and spontaneous fission become dominant. The heaviest completely stable nuclide is lead-208 ($^{208}_{82}\text{Pb}$). Bismuth-209 is very weakly alpha active ($T_{1/2} = 2.01 \times 10^{19}\text{ yr}$).

Even-Odd Nuclear Stability Systematics

The pairing interaction in nuclear forces produces dramatic patterns in stable isotope abundances:

| Nucleus Type | $Z$ | $N$ | Number of Stable Nuclides | Average Stable Isotopes per Element | Examples | | :--- | :--- | :--- | :--- | :--- | :--- | | Even-Even | Even | Even | 166 | $\sim 5.5$ | $^{12}\text{C}, ^{16}\text{O}, ^{56}\text{Fe}, ^{208}\text{Pb}$ | | Even-Odd | Even | Odd | 57 | $\sim 1.4$ | $^{13}\text{C}, ^{57}\text{Fe}, ^{207}\text{Pb}$ | | Odd-Even | Odd | Even | 53 | $\sim 1.3$ | $^{19}\text{F}, ^{23}\text{Na}, ^{63}\text{Cu}$ | | Odd-Odd | Odd | Odd | 9 (only 4 stable light) | $< 0.1$ | $^{2}\text{H}, ^{6}\text{Li}, ^{10}\text{B}, ^{14}\text{N}$ ($^{50}\text{V}, ^{138}\text{La}, ^{180m}\text{Ta}$) |

Over $60\%$ of all stable nuclides in the universe are even-even, while only four completely stable odd-odd nuclides exist in nature ($^{2}\text{H}, ^{6}\text{Li}, ^{10}\text{B}, ^{14}\text{N}$). This disparity provides direct evidence that nucleons of identical species couple into anti-parallel pairs ($J^\pi = 0^+$) with enhanced binding energy.

Β§2.4 Nuclear Taxonomy: Isotopes, Isobars, Isotones, Nuclear Isomers & Mirror Nuclei

Nuclear species (nuclides) are classified into specific families based on relationships between their proton count $Z$, neutron count $N$, and total mass number $A$.

1. Isotopes ($\Delta Z = 0$)

Nuclides possessing the same atomic number $Z$ (identical chemical identity) but differing neutron numbers $N$ and mass numbers $A$.

  • Examples:
  • Hydrogen isotopes: $^{1}_{1}\text{H}$ (protium), $^{2}_{1}\text{H}$ (deuterium), $^{3}_{1}\text{H}$ (tritium).
  • Uranium isotopes: $^{234}_{92}\text{U}$, $^{235}_{92}\text{U}$, $^{238}_{92}\text{U}$.
  • Because their electron configurations are identical, isotopes exhibit nearly identical chemical properties (governed by the same electron shell structure), but display distinct nuclear properties (spins, magnetic moments, cross sections, decay half-lives).

2. Isobars ($\Delta A = 0$)

Nuclides possessing the same total mass number $A$ but differing atomic numbers $Z$ and neutron numbers $N$.

  • Examples:
  • $A = 14$: $^{14}_{6}\text{C}$, $^{14}_{7}\text{N}$, $^{14}_{8}\text{O}$.
  • $A = 40$: $^{40}_{18}\text{Ar}$, $^{40}_{19}\text{K}$, $^{40}_{20}\text{Ca}$.
  • Isobars possess completely different chemical identities. In any isobaric chain, Mattauch's Isobar Rule states that two neighboring stable isobars ($\Delta Z = 1$) cannot exist; one must beta decay into the other.

3. Isotones ($\Delta N = 0$)

Nuclides possessing the same neutron number $N$ but differing atomic numbers $Z$ and mass numbers $A$.

  • Examples ($N = 82$, magic neutron shell):
  • $^{138}_{56}\text{Ba}_{82}$, $^{139}_{57}\text{La}_{82}$, $^{140}_{58}\text{Ce}_{82}$, $^{141}_{59}\text{Pr}_{82}$, $^{142}_{60}\text{Nd}_{82}$.
  • Isotones are invaluable in nuclear structure research because variations in binding energy and excitation spectra reflect changes in proton configuration against an identical neutron core.

4. Nuclear Isomers

Nuclides possessing the identical $Z$ and identical $A$ that exist in different, long-lived metastable nuclear energy states (isomeric states, denoted with an "m", e.g., $^{99m}\text{Tc}$, $^{180m}\text{Ta}$).

  • A metastable excited state arises when the angular momentum difference $\Delta I$ between the excited isomer and the lower state is large (high multipolarity, e.g., $E4$ or $M4$) and the transition energy $\Delta E$ is low, severely suppressing the probability of gamma electromagnetic de-excitation (isomeric transition, IT).
  • Example: Technetium-99m ($^{99m}_{43}\text{Tc}$):

The $142.6\text{ keV}$ state has spin-parity $1/2^-$ while the ground state $^{99}\text{Tc}$ has $9/2^+$. The large spin change $\Delta I = 4$ produces an exceptionally long half-life of $T_{1/2} = 6.01\text{ hours}$, making it the premier radiotracer in nuclear medicine.

  • Tantalum-180m ($^{180m}_{73}\text{Ta}$, $I^\pi = 9^-$) has a half-life exceeding $4.5 \times 10^{16}\text{ years}$β€”longer than the age of the universe!

5. Mirror Nuclei

A pair of isobaric nuclei ($A_1 = A_2$) where the proton count of one equals the neutron count of the other:

$$Z_1 = N_2 \quad \text{and} \quad N_1 = Z_2 \implies |Z_1 - Z_2| = 1$$
  • Examples: $(^3_1\text{H}, ^3_2\text{He})$, $(^7_3\text{Li}, ^7_4\text{Be})$, $(^{11}_5\text{B}, ^{11}_6\text{C})$, $(^{15}_7\text{N}, ^{15}_8\text{O})$.
  • Because the strong nuclear interaction is charge-symmetric (the $p-p$, $n-n$, and $p-n$ strong potentials are identical in the same quantum state), the binding energy difference between mirror nuclei is due almost exclusively to the difference in Coulomb electrostatic self-energy:
$$\Delta E_C = B(Z_1, N_1) - B(Z_2, N_2) \approx \frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 R} \left[ Z_2(Z_2 - 1) - Z_1(Z_1 - 1) \right]$$

Measuring this mass difference enables precise experimental determination of the nuclear charge radius $R_0$.

Β§2.5 The Liquid Drop Model & WeizsΓ€cker Semi-Empirical Mass Formula (SEMF)

In 1935, Carl Friedrich von WeizsΓ€cker developed the Liquid Drop Model, analogizing the atomic nucleus to a macroscopic droplet of incompressible charged liquid. Nucleons interact through short-range, saturating attractive strong forces analogous to van der Waals forces in a liquid, counteracted by long-range electrostatic repulsion between protons.

The resulting Semi-Empirical Mass Formula (SEMF) provides an analytical formulation for total binding energy $B(A,Z)$:

$$B(A,Z) = a_v A - a_s A^{2/3} - a_c \frac{Z(Z - 1)}{A^{1/3}} - a_a \frac{(A - 2Z)^2}{A} + \delta(A,Z)$$

``` SEMF Binding Energy Components B(A,Z) = + Volume Term [+ a_v * A]

  • Surface Term [- a_s * A^(2/3)]
  • Coulomb Term [- a_c * Z(Z-1) / A^(1/3)]
  • Asymmetry Term [- a_a * (A - 2Z)^2 / A]

+ Pairing Term [+ Ξ΄(A,Z)] ```

Physical Derivation of the Five Terms

1. Volume Energy Term ($+a_v A$):

Because the strong nuclear force is short-ranged and exhibits saturation, each interior nucleon interacts only with its immediate nearest neighbors ($\sim 12$ nucleons), contributing a constant binding energy independent of total drop size:

$$B_{\text{volume}} = +a_v A \quad (a_v \approx 15.75\text{ MeV})$$

2. Surface Energy Term ($-a_s A^{2/3}$):

Nucleons residing on the nuclear surface have fewer neighboring nucleons than those in the interior, reducing the net binding. By analogy with surface tension in liquids, this deficit is proportional to the surface area of the sphere:

$$\text{Area} = 4\pi R^2 = 4\pi (R_0 A^{1/3})^2 \propto A^{2/3}$$
$$B_{\text{surface}} = -a_s A^{2/3} \quad (a_s \approx 17.80\text{ MeV})$$

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

The $Z$ protons uniformly distributed throughout the nuclear sphere repel each other electrostatically. The classical self-energy of a uniformly charged sphere of radius $R$ is:

$$E_C = \frac{3}{5}\frac{Q^2}{4\pi\varepsilon_0 R} = \frac{3}{5}\frac{(Z e)^2}{4\pi\varepsilon_0 R_0 A^{1/3}}$$

Correcting for quantum self-interaction (a proton does not repel itself, yielding $Z(Z-1)$ pairs):

$$B_{\text{Coulomb}} = -a_c \frac{Z(Z - 1)}{A^{1/3}} \quad \text{where } a_c = \frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 R_0} \approx 0.711\text{ MeV}$$

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

Quantum mechanics (Pauli exclusion principle) dictates that protons and neutrons fill separate fermion energy wells. For a fixed total nucleon count $A$, the state of lowest total kinetic energy occurs when $N = Z = A/2$. Any neutron excess $(N - Z) = (A - 2Z)$ forces nucleons into higher unoccupied quantum states. In a Fermi gas approximation, expanding the energy difference in powers of $(N - Z)$:

$$B_{\text{asymmetry}} = -a_a \frac{(A - 2Z)^2}{A} \quad (a_a \approx 23.70\text{ MeV})$$

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

Due to the spin-orbit pairing force, identical nucleons with antiparallel spins form pairs ($J^\pi = 0^+$) with enhanced binding energy. The pairing term is:

$$\delta(A,Z) = \begin{cases} +\frac{a_p}{A^{1/2}} & \text{for Even-Even nuclei (extra stable)} \\ 0 & \text{for Even-Odd / Odd-Even nuclei} \\ -\frac{a_p}{A^{1/2}} & \text{for Odd-Odd nuclei (least stable)} \end{cases}$$

where $a_p \approx 11.18\text{ MeV}$ (or alternatively parameterized as $a_p / A^{3/4}$).

Prediction of the Most Stable Isobar ($Z_{\text{stable}}$)

For a constant mass number $A$, the binding energy is a quadratic parabola in $Z$:

$$\frac{\partial B(A,Z)}{\partial Z} = 0 \implies \frac{\partial}{\partial Z}\left[ -a_c \frac{Z(Z-1)}{A^{1/3}} - a_a \frac{(A - 2Z)^2}{A} \right] = 0$$

Evaluating the derivative:

$$- \frac{a_c}{A^{1/3}}(2Z - 1) - \frac{a_a}{A}[-4(A - 2Z)] = 0$$
$$\frac{a_c}{A^{1/3}}(2Z) + \frac{8 a_a Z}{A} = 4 a_a + \frac{a_c}{A^{1/3}}$$

Neglecting small terms, we solve for $Z_{\text{stable}}$:

$$Z_{\text{stable}} = \frac{A}{2 + \frac{a_c}{2 a_a} A^{2/3}} \approx \frac{A}{2 + 0.015 A^{2/3}}$$
  • For light nuclei ($A \to 0$): $Z_{\text{stable}} \to A/2$ ($N = Z$).
  • For heavy nuclei ($A = 208$): $Z_{\text{stable}} = \frac{208}{2 + 0.015(208)^{2/3}} = \frac{208}{2 + 0.015(35.1)} = \frac{208}{2.527} \approx 82.3 \implies Z = 82$ (Lead-208), in remarkable agreement with experiment!

Β§2.6 The Nuclear Shell Model: Spin-Orbit Coupling, Magic Numbers & Single-Particle States

While the Liquid Drop Model explains macroscopic binding energies and fission, it completely fails to explain quantum micro-structure:

  1. Exceptional binding energy spikes at specific proton or neutron counts:
$$\mathbf{2, \; 8, \; 20, \; 28, \; 50, \; 82, \; 126} \quad \text{("Magic Numbers")}$$
  1. Nuclei with magic numbers of nucleons (such as $^{4}_{2}\text{He}_2$, $^{16}_{8}\text{O}_8$, $^{40}_{20}\text{Ca}_{20}$, $^{48}_{20}\text{Ca}_{28}$, $^{208}_{82}\text{Pb}_{126}$) are "doubly magic" and possess zero electric quadrupole moments, exceptionally high first-excited state energies, and extraordinarily small neutron capture cross sections.
  2. Ground-state nuclear spins and parities ($J^\pi$) and magnetic dipole moments.

The Shell Model Potential and Mayer-Jensen Spin-Orbit Coupling

In 1949, Maria Goeppert Mayer and J. Hans D. Jensen independently recognized that nucleons move in a mean central potential created by all other nucleons, accompanied by an extraordinarily strong spin-orbit interaction ($\vec{l}\cdot\vec{s}$ coupling):

$$V(r) = V_{\text{central}}(r) + V_{ls}(r) \, (\vec{l} \cdot \vec{s})$$

Mathematical Derivation of Spin-Orbit Level Splitting

For a nucleon with orbital angular momentum $\vec{l}$ and spin $\vec{s}$ ($s = 1/2$), the total single-particle angular momentum is:

$$\vec{j} = \vec{l} + \vec{s} \implies j = l + \frac{1}{2} \quad \text{or} \quad j = l - \frac{1}{2}$$

Squaring $\vec{j}$:

$$\vec{j}^2 = (\vec{l} + \vec{s})^2 = \vec{l}^2 + \vec{s}^2 + 2 (\vec{l} \cdot \vec{s})$$
$$\vec{l} \cdot \vec{s} = \frac{1}{2}\left( \vec{j}^2 - \vec{l}^2 - \vec{s}^2 \right)$$

Evaluating the quantum expectation values:

$$\langle \vec{l} \cdot \vec{s} \rangle = \frac{\hbar^2}{2} [ j(j + 1) - l(l + 1) - s(s + 1) ]$$

With $s = 1/2 \implies s(s+1) = 3/4$:

  1. For state with $j = l + 1/2$:
$$\langle \vec{l} \cdot \vec{s} \rangle = \frac{\hbar^2}{2} \left[ (l + 1/2)(l + 3/2) - l(l + 1) - 3/4 \right] = \frac{\hbar^2}{2} [l^2 + 2l + 3/4 - l^2 - l - 3/4] = +\frac{l}{2}\hbar^2$$
  1. For state with $j = l - 1/2$:
$$\langle \vec{l} \cdot \vec{s} \rangle = \frac{\hbar^2}{2} \left[ (l - 1/2)(l + 1/2) - l(l + 1) - 3/4 \right] = \frac{\hbar^2}{2} [l^2 - 1/4 - l^2 - l - 3/4] = -\frac{l + 1}{2}\hbar^2$$

The energy splitting between the two sub-states is:

$$\Delta E_{ls} = \langle V_{ls} \rangle \left[ \frac{l}{2} - \left(-\frac{l+1}{2}\right) \right] = \langle V_{ls} \rangle \left(l + \frac{1}{2}\right)\hbar^2$$

Crucially, in the nuclear interaction (unlike atomic atomic fine structure), the spin-orbit potential $V_{ls}(r)$ is negative. Therefore, the state with higher total angular momentum $j = l + 1/2$ is pushed downward in energy, while $j = l - 1/2$ is pushed upward!

``` Harmonic Oscillator With Spin-Orbit Splitting Magic Shell Levels (V_ls < 0) Closure ──────────────────────────────────────────────────────────────── 1f (l=3) ────────── 1f_5/2 (6 states) ─────── \ \──────── 1f_7/2 (8 states) ─────── β–Ί [28] MAGIC ──────────────────────────────────────────────────────────────── 1d (l=2) ────────── 1d_3/2 (4 states) ─────── β–Ί [20] MAGIC 2s (l=0) ────────── 2s_1/2 (2 states) 1d (l=2) ────────── 1d_5/2 (6 states) ──────────────────────────────────────────────────────────────── 1p (l=1) ────────── 1p_1/2 (2 states) ─────── β–Ί [8] MAGIC ────────── 1p_3/2 (4 states) ──────────────────────────────────────────────────────────────── 1s (l=0) ────────── 1s_1/2 (2 states) ─────── β–Ί [2] MAGIC ```

For large orbital angular momentum $l$ ($l = 3$ for $f$-states, $l = 4$ for $g$-states), this downward shift is so massive that the $j = l + 1/2$ level drops completely across the oscillator gap into the major shell below. This accounts for every observed magic number:

  • $1s_{1/2}$ (capacity 2) $\implies \mathbf{2}$
  • $1p_{3/2}, 1p_{1/2}$ (capacity $4 + 2 = 6$) $\implies 2 + 6 = \mathbf{8}$
  • $1d_{5/2}, 2s_{1/2}, 1d_{3/2}$ (capacity $6 + 2 + 4 = 12$) $\implies 8 + 12 = \mathbf{20}$
  • Intruder state $1f_{7/2}$ pushed down (capacity 8) $\implies 20 + 8 = \mathbf{28}$
  • $2p_{3/2}, 1f_{5/2}, 2p_{1/2}$ plus intruder $1g_{9/2}$ (capacity 22) $\implies 28 + 22 = \mathbf{50}$
  • $2d_{5/2}, 1g_{7/2}, 1h_{11/2}$, etc. $\implies \mathbf{82}$
  • Intruder $1i_{13/2}$, etc. $\implies \mathbf{126}$

Β§2.7 Collective & Statistical Nuclear Models: Bohr-Mottelson Deformations & the Fermi Gas Model

The spherical Shell Model succeeds near magic numbers, but fails in the mid-shell regions ($150 < A < 190$ and $A > 220$). In these regions, nuclei exhibit:

  • Electric quadrupole moments $Q_0$ up to 30 times larger than the single-particle limit.
  • Rotational band excitation spectra following exact $E_J \propto J(J+1)$ sequences.
  • Strongly enhanced collective electric quadrupole transition rates $B(E2)$.

The Bohr-Mottelson Unified Collective Model (1953)

Aage Bohr, Ben Mottelson, and James Rainwater resolved this disparity by modeling the nucleus as a deformed, non-spherical liquid drop whose collective surface vibrations and rotations couple to individual valence nucleon orbits.

The deformed nuclear surface is parameterized via spherical harmonics $Y_{\lambda \mu}(\theta, \phi)$:

$$R(\theta, \phi) = R_0 \left[ 1 + \sum_{\lambda=2}^{\infty} \sum_{\mu=-\lambda}^{\lambda} \alpha_{\lambda \mu} Y_{\lambda \mu}^*(\theta, \phi) \right]$$

For axially symmetric quadrupole deformations ($\lambda = 2, \mu = 0$), this simplifies in terms of the deformation parameter $\beta$:

$$R(\theta) = R_0 \left[ 1 + \beta \sqrt{\frac{5}{16\pi}} (3\cos^2\theta - 1) \right] = R_0 [ 1 + \beta Y_{20}(\theta) ]$$
  • $\beta > 0$: Prolate spheroid (cigar-shaped, major axis along symmetry axis; predominant in nature).
  • $\beta < 0$: Oblate spheroid (doorknob / disk-shaped).

Rotational Energy Spectra

For an even-even deformed nucleus with ground state $J^\pi = 0^+$, quantum rotation of the collective core about an axis perpendicular to the symmetry axis yields kinetic rotational energy:

$$E_{\text{rot}}(J) = \frac{\hbar^2}{2 \mathcal{I}} J(J + 1), \quad J = 0, 2, 4, 6, 8, \dots$$

where $\mathcal{I}$ is the effective moment of inertia of the deformed nucleus. The excitation energy ratios in a pure rotational band follow:

$$\frac{E(4^+)}{E(2^+)} = \frac{4(5)}{2(3)} = \frac{20}{6} = 3.333$$
$$\frac{E(6^+)}{E(2^+)} = \frac{6(7)}{6} = 7.000, \quad \frac{E(8^+)}{E(2^+)} = \frac{8(9)}{6} = 12.000$$

Experimental spectra for deformed nuclei (e.g., $^{160}\text{Gd}, ^{174}\text{Yb}, ^{238}\text{U}$) match this $3.33$ ratio precisely, confirming collective rotation.

The Fermi Gas Model of Nuclear Matter

For high-energy nuclear reactions and statistical level densities, the nucleus is treated as a degenerate quantum gas of non-interacting fermions (protons and neutrons) confined within a spherical volume $V = \frac{4}{3}\pi R^3$.

The number of spatial momentum states in phase space volume $d^3r \, d^3p$ is:

$$dN = \frac{2}{(2\pi\hbar)^3} d^3r \, d^3p = \frac{2 V}{(2\pi\hbar)^3} 4\pi p^2 dp$$

where the factor 2 accounts for spin degeneracy ($s_z = \pm 1/2$). Integrating from $p = 0$ to the Fermi momentum $p_F$:

$$N = \frac{8\pi V}{(2\pi\hbar)^3} \int_0^{p_F} p^2 dp = \frac{8\pi V}{(2\pi\hbar)^3} \frac{p_F^3}{3} = \frac{V p_F^3}{3\pi^2 \hbar^3}$$

Substituting nuclear matter density $\rho = N / V$:

$$p_F = \hbar (3\pi^2 \rho)^{1/3} \implies k_F = (3\pi^2 \rho)^{1/3}$$

For symmetric nuclear matter ($N = Z = A/2$) with $\rho_0 \approx 0.17\text{ fm}^{-3}$:

$$k_F = \left( 3\pi^2 \cdot \frac{0.17}{2} \right)^{1/3} \approx 1.36\text{ fm}^{-1}$$

The Fermi Energy $E_F$ of a nucleon is:

$$E_F = \frac{p_F^2}{2 m_N} = \frac{\hbar^2 k_F^2}{2 m_N} = \frac{(197.3\text{ MeV}\cdot\text{fm})^2 (1.36\text{ fm}^{-1})^2}{2 (938\text{ MeV})} \approx 38.4\text{ MeV}$$

The average kinetic energy per nucleon in the ground state is:

$$\langle T \rangle = \frac{\int_0^{E_F} E \cdot g(E) dE}{\int_0^{E_F} g(E) dE} = \frac{3}{5} E_F \approx \frac{3}{5}(38.4\text{ MeV}) \approx 23.0\text{ MeV}$$

Adding the average nucleon binding energy ($B/A \approx 8\text{ MeV}$), the total depth of the nuclear potential well is:

$$V_0 = E_F + B/A \approx 38.4 + 8.0 \approx 46 - 50\text{ MeV}$$

Β§2.8 Modern Nuclear Structure & Isospin Formalism: Charge Independence & Wigner Supermultiplets

In 1932, immediately following Chadwick's discovery of the neutron, Werner Heisenberg proposed that the proton and neutron are not fundamentally distinct particles, but two different charge states of a single underlying physical entity: the nucleon.

The Isospin Vector Formalism

Heisenberg introduced the concept of isotopic spin (or isospin, denoted by vector $\vec{T}$ or $\vec{I}$), formulating a mathematical analogy with quantum mechanical spin-1/2: The nucleon possesses total isospin $T = 1/2$. In an abstract three-dimensional "isospace":

  • Proton state: Third component projection $T_3 = +1/2$ (or $-1/2$ by nuclear convention):
$$|p\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
  • Neutron state: Third component projection $T_3 = -1/2$ (or $+1/2$):
$$|n\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

For a nucleus composed of $Z$ protons and $N$ neutrons, the total third isospin component $T_3$ is strictly fixed by its composition:

$$T_3 = \frac{1}{2}(Z - N) = Z - \frac{A}{2}$$

The total isospin quantum number $T$ can take any integer or half-integer value satisfying:

$$|T_3| \le T \le \frac{A}{2}$$

Charge Symmetry and Charge Independence

High-energy scattering experiments demonstrate that the strong nuclear force obeys two profound symmetries:

1. Charge Symmetry: The proton-proton ($p-p$) strong potential equals the neutron-neutron ($n-n$) strong potential in the identical spatial and spin state ($V_{pp} = V_{nn}$).

2. Charge Independence: The proton-neutron ($p-n$) strong interaction equals the $p-p$ and $n-n$ interactions in identical $T = 1$ states ($V_{pp} = V_{nn} = V_{pn}|_{T=1}$).

Mathematically, the nuclear Hamiltonian $\hat{H}_{\text{strong}}$ commutes with the total isospin operator:

$$[\hat{H}_{\text{strong}}, \vec{T}] = 0$$

Consequently, states with identical total isospin $T$ but differing projections $T_3$ form an isobaric multiplet (or isospin analog states) with nearly identical nuclear structure, differing in energy solely due to electromagnetic Coulomb perturbation:

$$\Delta E_{\text{Coulomb}} = \frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 R} \left[ Z_1(Z_1 - 1) - Z_2(Z_2 - 1) \right]$$

Eugene Wigner expanded this into Wigner Supermultiplet Theory, classifying nuclear energy levels according to $SU(4)$ symmetry combining space, spin, and isospin degrees of freedom.

Comparative Systematics of Semi-Empirical Mass Formula (SEMF) Parameter Sets

Different empirical fits to experimental nuclear masses yield slightly varying coefficient sets (all in MeV):

| Parameter Set | Volume $a_v$ | Surface $a_s$ | Coulomb $a_c$ | Asymmetry $a_a$ | Pairing $a_p$ | RMS Mass Error | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | von WeizsΓ€cker (1935) | $15.75$ | $17.80$ | $0.711$ | $23.70$ | $11.18$ | $\sim 2.5\text{ MeV}$ | | Green (1954) | $15.75$ | $17.80$ | $0.710$ | $23.69$ | $12.00$ | $\sim 2.1\text{ MeV}$ | | Myers & Swiatecki (1966) | $15.68$ | $18.56$ | $0.717$ | $28.07$ | $11.00$ | $\sim 1.8\text{ MeV}$ | | MΓΆller et al. (FRDM, 1995) | $16.25$ | $19.00$ | $0.730$ | $32.00$ | $11.50$ | $\sim 0.67\text{ MeV}$ |

Modern Finite-Range Droplet Models (FRDM) incorporate microscopic Strutinsky shell corrections and macroscopic nuclear surface diffuseness, predicting nuclear binding energies for thousands of exotic isotopes out to the neutron and proton driplines.

Easy Example 2.1: Nuclear Matter Density and Radius of Lead-208

Given the nuclear radius constant $R_0 = 1.25\text{ fm}$:

  1. Calculate the charge radius $R$ of the doubly magic nucleus lead-208 ($^{208}_{82}\text{Pb}$).
  2. Determine the nuclear mass density of $^{208}\text{Pb}$ in $\text{kg/m}^3$ assuming a uniform sphere with atomic mass $M = 207.97665\text{ u}$.

Step 1: Nuclear Radius

Using the radius scaling law $R = R_0 A^{1/3}$:

$$A = 208 \implies A^{1/3} = (208)^{1/3} \approx 5.9250$$
$$R = 1.25\text{ fm} \times 5.9250 \approx 7.406\text{ fm} = 7.406 \times 10^{-15}\text{ m}$$

Step 2: Mass Density $\rho$

Volume of the nuclear sphere:

$$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (7.406 \times 10^{-15}\text{ m})^3 = \frac{4}{3}\pi (4.062 \times 10^{-43}) \approx 1.7015 \times 10^{-42}\text{ m}^3$$

Total nuclear mass in kilograms:

$$M = 207.97665\text{ u} \times 1.66054 \times 10^{-27}\text{ kg/u} \approx 3.4535 \times 10^{-25}\text{ kg}$$

Density:

$$\rho = \frac{M}{V} = \frac{3.4535 \times 10^{-25}\text{ kg}}{1.7015 \times 10^{-42}\text{ m}^3} \approx 2.03 \times 10^{17}\text{ kg/m}^3$$

The density is $\approx 2.0 \times 10^{14}\text{ g/cm}^3$ (200 million metric tons per cubic centimeter).

Easy Example 2.2: Exact Binding Energy and Energy Release in Deuteron Fusion

Consider the fusion reaction between two deuterons:

$$^2_1\text{H} + ^2_1\text{H} \longrightarrow ^4_2\text{He} + \Delta E$$

Given the atomic masses:

  • $m(^2_1\text{H}) = 2.01410178\text{ u}$
  • $m(^4_2\text{He}) = 4.00260325\text{ u}$
  • $1\text{ u} = 931.4941\text{ MeV}/c^2$
  1. Calculate the binding energy per nucleon of $^2\text{H}$ and $^4\text{He}$.
  2. Calculate the total energy released $\Delta E$ in $\text{MeV}$ per fusion reaction and per gram of deuterium fuel.

Step 1: Binding Energy per Nucleon

  1. For Deuterium ($^2\text{H}$, $Z = 1, N = 1$):
$$\Delta m(^2\text{H}) = m(^1\text{H}) + m_n - m(^2\text{H}) = 1.00782503 + 1.00866492 - 2.01410178 = 0.00238817\text{ u}$$
$$B(^2\text{H}) = 0.00238817 \times 931.4941\text{ MeV} = 2.2246\text{ MeV}$$
$$\frac{B}{A}(^2\text{H}) = \frac{2.2246\text{ MeV}}{2} = 1.1123\text{ MeV/nucleon}$$
  1. For Helium-4 ($^4\text{He}$, $Z = 2, N = 2$):
$$\Delta m(^4\text{He}) = 2(1.00782503) + 2(1.00866492) - 4.00260325 = 2.01565006 + 2.01732984 - 4.00260325 = 0.03037665\text{ u}$$
$$B(^4\text{He}) = 0.03037665 \times 931.4941\text{ MeV} = 28.2957\text{ MeV}$$
$$\frac{B}{A}(^4\text{He}) = \frac{28.2957\text{ MeV}}{4} = 7.0739\text{ MeV/nucleon}$$

Step 2: Energy Release $\Delta E$

$$\Delta m = 2 m(^2\text{H}) - m(^4\text{He}) = 2(2.01410178) - 4.00260325 = 4.02820356 - 4.00260325 = 0.02560031\text{ u}$$
$$\Delta E = 0.02560031 \times 931.4941\text{ MeV} = 23.8465\text{ MeV}$$

Alternatively: $\Delta E = B(^4\text{He}) - 2 B(^2\text{H}) = 28.2957 - 2(2.2246) = 23.8465\text{ MeV}$.

Energy Per Gram of Deuterium:

Two deuterons ($M \approx 4.028\text{ g/mol}$) release $23.85\text{ MeV}$:

$$\text{Specific Energy} = \frac{23.8465 \times 1.60218 \times 10^{-13}\text{ J}}{2 \times 2.0141 \times 1.66054 \times 10^{-24}\text{ g}} = \frac{3.8206 \times 10^{-12}\text{ J}}{6.688 \times 10^{-24}\text{ g}} \approx 5.71 \times 10^{11}\text{ J/g} = 571\text{ GJ/g}$$

One gram of deuterium yields energy equivalent to combusting over 13 metric tons of oil!

Intermediate Example 2.3: Semi-Empirical Mass Formula Calculation of Iron-56 Binding Energy

Using the WeizsΓ€cker Semi-Empirical Mass Formula with parameters: $a_v = 15.75\text{ MeV}$, $a_s = 17.80\text{ MeV}$, $a_c = 0.711\text{ MeV}$, $a_a = 23.70\text{ MeV}$, and $a_p = 11.18\text{ MeV}$ (with $\delta = +a_p / A^{1/2}$ for even-even):

  1. Compute each of the five individual energetic terms for iron-56 ($^{56}_{26}\text{Fe}$, $Z = 26, N = 30$).
  2. Calculate the total binding energy $B$ and binding energy per nucleon $B/A$.
  3. Compare with the experimental value of $8.790\text{ MeV/nucleon}$ and calculate the percentage discrepancy.

Step 1: Compute Individual SEMF Terms for $^{56}_{26}\text{Fe}$

Here $A = 56$, $Z = 26$, $N = 30$.

  • $A^{1/3} = (56)^{1/3} \approx 3.82586$
  • $A^{2/3} = (3.82586)^2 \approx 14.6372$
  • $A^{1/2} = \sqrt{56} \approx 7.4833$

1. Volume Term:

$$B_{\text{vol}} = a_v A = 15.75 \times 56 = +882.000\text{ MeV}$$

2. Surface Term:

$$B_{\text{surf}} = -a_s A^{2/3} = -17.80 \times 14.6372 = -260.542\text{ MeV}$$

3. Coulomb Term:

$$B_{\text{coul}} = -a_c \frac{Z(Z-1)}{A^{1/3}} = -0.711 \times \frac{26 \times 25}{3.82586} = -0.711 \times \frac{650}{3.82586} = -0.711 \times 169.896 = -120.796\text{ MeV}$$

4. Asymmetry Term:

$$A - 2Z = 56 - 52 = 4$$
$$B_{\text{asym}} = -a_a \frac{(A - 2Z)^2}{A} = -23.70 \times \frac{4^2}{56} = -23.70 \times \frac{16}{56} = -23.70 \times 0.285714 = -6.771\text{ MeV}$$

5. Pairing Term:

Since $Z = 26$ (even) and $N = 30$ (even), the nucleus is even-even ($\delta > 0$):

$$B_{\text{pair}} = +\frac{a_p}{A^{1/2}} = +\frac{11.18}{7.4833} = +1.494\text{ MeV}$$

Step 2: Total Binding Energy and $B/A$

$$B = 882.000 - 260.542 - 120.796 - 6.771 + 1.494 = 495.385\text{ MeV}$$
$$\frac{B}{A} = \frac{495.385\text{ MeV}}{56} \approx 8.846\text{ MeV/nucleon}$$

Step 3: Comparison with Experiment

$$\% \text{ Discrepancy} = \frac{|8.846 - 8.790|}{8.790} \times 100\% = \frac{0.056}{8.790} \times 100\% \approx 0.64\%$$

The SEMF reproduces the binding energy of iron-56 to within $0.64\%$.

Intermediate Example 2.4: Most Stable Isobar for Mass Chain A = 135

Fission of uranium produces radioactive fission products along the isobaric decay chain $A = 135$. Using the SEMF parameters $a_c = 0.711\text{ MeV}$ and $a_a = 23.70\text{ MeV}$:

  1. Derive the theoretical most stable atomic number $Z_{\text{stable}}$ for $A = 135$.
  2. Identify the stable isobar that terminates this decay chain and write down the decay sequence starting from $^{135}_{53}\text{I}$.

Step 1: Calculate $Z_{\text{stable}}$

From the SEMF minimization condition:

$$Z_{\text{stable}} = \frac{A}{2 + \frac{a_c}{2 a_a} A^{2/3}}$$

For $A = 135$:

$$A^{2/3} = (135)^{2/3} \approx 26.326$$
$$\frac{a_c}{2 a_a} = \frac{0.711}{2 \times 23.70} = \frac{0.711}{47.40} \approx 0.01500$$

Denominator:

$$2 + 0.01500 \times 26.326 = 2 + 0.39489 = 2.39489$$
$$Z_{\text{stable}} = \frac{135}{2.39489} \approx 56.37$$

Rounding to the nearest integer yields:

$$Z_{\text{stable}} = 56 \quad (\text{Barium, Ba})$$

Step 2: Isobaric Decay Chain Sequence

The chain terminates at the stable nuclide $^{135}_{56}\text{Ba}$. Starting from iodine-135 (a critical reactor fission product):

$$^{135}_{53}\text{I} \xrightarrow[\beta^-]{6.57\text{ h}} \, ^{135}_{54}\text{Xe} \xrightarrow[\beta^-]{9.14\text{ h}} \, ^{135}_{55}\text{Cs} \xrightarrow[\beta^-]{2.3 \times 10^6\text{ yr}} \, ^{135}_{56}\text{Ba} \text{ (Stable)}$$

Notice that $^{135}\text{Xe}$ is the notorious "reactor poison" with a thermal neutron capture cross section of $2.6 \times 10^6\text{ barns}$.

Intermediate Example 2.5: Nuclear Shell Model Ground-State Spin and Parity Predictions

Using the extreme single-particle Shell Model, predict the ground-state total angular momentum and parity ($J^\pi$) for:

  1. Oxygen-17 ($^{17}_{8}\text{O}$, $Z = 8, N = 9$)
  2. Potassium-39 ($^{39}_{19}\text{K}$, $Z = 19, N = 20$)
  3. Scandium-45 ($^{45}_{21}\text{Sc}$, $Z = 21, N = 24$)

Shell Model Filling Sequence:

Single-particle levels in order of increasing energy: $1s_{1/2}$ (2), $1p_{3/2}$ (4), $1p_{1/2}$ (2) [closure at 8] $1d_{5/2}$ (6), $2s_{1/2}$ (2), $1d_{3/2}$ (4) [closure at 20] $1f_{7/2}$ (8) [closure at 28]

1. Oxygen-17 ($^{17}_{8}\text{O}$):

  • Protons: $Z = 8$ (closed shell, paired, contributes $0^+$).
  • Neutrons: $N = 9$.

The first 8 neutrons fill the closed shell ($1s_{1/2}^2 1p_{3/2}^4 1p_{1/2}^2$). The 9th valence neutron enters the $1d_{5/2}$ orbital ($l = 2, j = 5/2$). Parity: $\pi = (-1)^l = (-1)^2 = +1$. Predicted ground state: $J^\pi = \frac{5}{2}^+$ (Matches experiment exactly).

2. Potassium-39 ($^{39}_{19}\text{K}$):

  • Neutrons: $N = 20$ (magic closed shell, contributes $0^+$).
  • Protons: $Z = 19$.

The proton shell has 19 protons, which is one proton hole short of the $Z = 20$ shell closure. The hole resides in the $1d_{3/2}$ orbital ($l = 2, j = 3/2$). Parity: $\pi = (-1)^2 = +1$. Predicted ground state: $J^\pi = \frac{3}{2}^+$ (Matches experiment exactly).

3. Scandium-45 ($^{45}_{21}\text{Sc}$):

  • Neutrons: $N = 24$ (even, paired, contributes $0^+$).
  • Protons: $Z = 21$.

The first 20 protons fill through $1d_{3/2}$. The 21st proton enters the $1f_{7/2}$ orbital ($l = 3, j = 7/2$). Parity: $\pi = (-1)^3 = -1$. Predicted ground state: $J^\pi = \frac{7}{2}^-$ (Matches experiment exactly).

Advanced Example 2.6: Coulomb Energy Difference of Mirror Nuclei and Nuclear Radius Parameter

The mirror pair Carbon-11 ($^{11}_{6}\text{C}_5$) and Boron-11 ($^{11}_{5}\text{B}_6$) have a measured nuclear mass difference:

$$\Delta M = M(^{11}\text{C}) - M(^{11}\text{B}) = 1.982\text{ MeV}/c^2$$

Assuming the difference in binding energy is entirely due to Coulomb self-energy:

$$\Delta E_C = \frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 R} [Z_1(Z_1 - 1) - Z_2(Z_2 - 1)]$$

and accounting for the neutron-proton mass difference ($m_n - m_H = 0.782\text{ MeV}$):

  1. Determine the experimental Coulomb energy difference $\Delta E_C$.
  2. Calculate the nuclear radius $R$ and the radius parameter $R_0$ for $A = 11$.

Step 1: Relation Between Mass Difference and Coulomb Energy

The mass difference between neutral mirror atoms is:

$$\Delta M c^2 = [M(Z+1, A) - M(Z, A)] c^2 = \Delta E_C - (m_n - m_H) c^2$$
$$\Delta E_C = \Delta M c^2 + (m_n - m_H) c^2$$

Given $\Delta M c^2 = 1.982\text{ MeV}$ and $(m_n - m_H)c^2 = 0.782\text{ MeV}$:

$$\Delta E_C = 1.982 + 0.782 = 2.764\text{ MeV}$$

Step 2: Coulomb Energy Expression

For $^{11}\text{C}$ ($Z_1 = 6$) and $^{11}\text{B}$ ($Z_2 = 5$):

$$Z_1(Z_1 - 1) - Z_2(Z_2 - 1) = 6(5) - 5(4) = 30 - 20 = 10$$

Therefore:

$$\Delta E_C = \frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 R} \cdot 10 = \frac{6 e^2}{4\pi\varepsilon_0 R}$$

Using $\frac{e^2}{4\pi\varepsilon_0} \approx 1.43996\text{ MeV}\cdot\text{fm}$:

$$\Delta E_C = \frac{6 \times 1.43996\text{ MeV}\cdot\text{fm}}{R} = \frac{8.6398\text{ MeV}\cdot\text{fm}}{R}$$

Step 3: Solve for $R$ and $R_0$

$$R = \frac{8.6398\text{ MeV}\cdot\text{fm}}{2.764\text{ MeV}} \approx 3.126\text{ fm}$$

Using $R = R_0 A^{1/3}$ with $A = 11$:

$$A^{1/3} = (11)^{1/3} \approx 2.224$$
$$R_0 = \frac{R}{A^{1/3}} = \frac{3.126\text{ fm}}{2.224} \approx 1.405\text{ fm}$$

This value ($1.40\text{ fm}$) agrees with the accepted strong interaction matter radius parameter.

Intermediate Example 2.7: Collective Rotational Band Energies of Uranium-238

The ground-state rotational band of the deformed even-even nucleus $^{238}_{92}\text{U}$ exhibits its first excited $2^+$ state at an excitation energy of $E(2^+) = 44.91\text{ keV}$.

  1. Assuming a rigid rotor $E(J) = \frac{\hbar^2}{2\mathcal{I}} J(J+1)$, calculate the rotational inertia parameter $\frac{\hbar^2}{2\mathcal{I}}$ in $\text{keV}$.
  2. Predict the excitation energies of the $4^+$, $6^+$, and $8^+$ rotational states.
  3. Compare the predicted values with the experimental energies ($E(4^+) = 148.4\text{ keV}$, $E(6^+) = 307.2\text{ keV}$) and comment on centrifugal stretching.

Step 1: Rotational Inertia Parameter

For $J = 2$:

$$E(2^+) = \frac{\hbar^2}{2\mathcal{I}} 2(2 + 1) = 6 \left(\frac{\hbar^2}{2\mathcal{I}}\right) = 44.91\text{ keV}$$
$$\frac{\hbar^2}{2\mathcal{I}} = \frac{44.91\text{ keV}}{6} = 7.485\text{ keV}$$

Step 2: Predict $4^+$, $6^+$, and $8^+$ Energies

  1. State $4^+$ ($J = 4$):
$$E(4^+) = \left(\frac{\hbar^2}{2\mathcal{I}}\right) 4(5) = 20 \times 7.485\text{ keV} = 149.70\text{ keV}$$
  1. State $6^+$ ($J = 6$):
$$E(6^+) = \left(\frac{\hbar^2}{2\mathcal{I}}\right) 6(7) = 42 \times 7.485\text{ keV} = 314.37\text{ keV}$$
  1. State $8^+$ ($J = 8$):
$$E(8^+) = \left(\frac{\hbar^2}{2\mathcal{I}}\right) 8(9) = 72 \times 7.485\text{ keV} = 538.92\text{ keV}$$

Step 3: Comparison with Experiment and Centrifugal Stretching

  • Experimental $E(4^+) = 148.4\text{ keV}$ vs Predicted $149.7\text{ keV}$ (Discrepancy: $+0.87\%$)
  • Experimental $E(6^+) = 307.2\text{ keV}$ vs Predicted $314.4\text{ keV}$ (Discrepancy: $+2.3\%$)

The experimental energies are slightly lower than the rigid rotor prediction. As the nucleus spins faster at higher $J$, centrifugal forces stretch the deformed nucleus, increasing its moment of inertia $\mathcal{I}$ and lowering the rotational energy spacing (centrifugal stretching correction $-D J^2(J+1)^2$).

Advanced Example 2.8: Schmidt Limits for Single-Particle Nuclear Magnetic Dipole Moments

In the single-particle Shell Model, the magnetic dipole moment $\mu$ of an odd-mass nucleus is determined by the unpaired valence nucleon. The Schmidt limits state that for an odd nucleon with orbital angular momentum $l$ and total angular momentum $j$:

  • Case $j = l + 1/2$: $\mu = j g_l + \frac{1}{2}(g_s - g_l)$
  • Case $j = l - 1/2$: $\mu = \frac{j}{j + 1} \left[ (j + 1) g_l - \frac{1}{2}(g_s - g_l) \right]$

Given:

  • For a proton: $g_l = 1$, $g_s = +5.586$ nuclear magnetons ($\mu_N$).
  • For a neutron: $g_l = 0$, $g_s = -3.826$ nuclear magnetons ($\mu_N$).
  1. Calculate the theoretical Schmidt magnetic moment in $\mu_N$ for oxygen-17 ($^{17}_{8}\text{O}$, valence neutron in $1d_{5/2}$, $l = 2, j = 5/2$).
  2. Calculate the Schmidt magnetic moment for potassium-39 ($^{39}_{19}\text{K}$, proton hole in $1d_{3/2}$, $l = 2, j = 3/2$).
  3. Compare with experimental values ($\mu_{\text{exp}}(^{17}\text{O}) = -1.894\,\mu_N$, $\mu_{\text{exp}}(^{39}\text{K}) = +0.391\,\mu_N$).

Step 1: Schmidt Moment for Oxygen-17 ($^{17}\text{O}$)

Valence neutron: $g_l = 0, g_s = -3.826\,\mu_N$. State is $1d_{5/2} \implies l = 2, j = 5/2 = l + 1/2$. Using the $j = l + 1/2$ Schmidt formula:

$$\mu = j g_l + \frac{1}{2}(g_s - g_l) = \frac{5}{2}(0) + \frac{1}{2}(-3.826 - 0) = -1.913\,\mu_N$$

Comparison: Experimental value is $-1.894\,\mu_N$β€”in exceptional agreement ($<1\%$ difference)!

Step 2: Schmidt Moment for Potassium-39 ($^{39}\text{K}$)

Unpaired proton: $g_l = 1, g_s = +5.586\,\mu_N$. State is $1d_{3/2} \implies l = 2, j = 3/2 = l - 1/2$. Using the $j = l - 1/2$ Schmidt formula:

$$\mu = \frac{j}{j + 1} \left[ (j + 1) g_l - \frac{1}{2}(g_s - g_l) \right]$$

Here $j = 3/2 \implies j + 1 = 5/2$:

$$\mu = \frac{3/2}{5/2} \left[ \frac{5}{2}(1) - \frac{1}{2}(5.586 - 1) \right] = \frac{3}{5} \left[ 2.500 - \frac{1}{2}(4.586) \right] = 0.60 \times [2.500 - 2.293] = 0.60 \times 0.207 = +0.124\,\mu_N$$

Comparison: Experimental value is $+0.391\,\mu_N$. The deviation is due to core polarization of the paired nucleons.

Advanced Example 2.9: Deuteron Non-Spherical Quadrupole Moment and Tensor Force D-State Mixing

The deuteron ($^2_1\text{H}$) is the simplest bound nuclear system ($Z = 1, N = 1$, binding energy $B = 2.2246\text{ MeV}$, spin-parity $J^\pi = 1^+$).

  1. If the nuclear force were purely central (spherical), the deuteron ground state would be a pure orbital S-wave ($l = 0, ^3S_1$). Show that a pure S-state must have an electric quadrupole moment $Q = 0$.
  2. The measured electric quadrupole moment of the deuteron is $Q_{\text{exp}} = +0.00286\text{ barn} = +0.286\text{ fm}^2$.

Explain how this non-zero, positive quadrupole moment proves the existence of a non-central tensor nuclear force and mixing with an orbital D-state ($l = 2, ^3D_1$).

  1. If the wave function is $|\psi\rangle = a_S |^3S_1\rangle + a_D |^3D_1\rangle$, state the approximate percentage of D-state admixture ($a_D^2 \approx 4 - 6\%$).

Step 1: Quadrupole Moment of a Pure S-State

The electric quadrupole moment operator is defined as:

$$\hat{Q} = \frac{1}{e} \int \rho(\vec{r}) (3 z^2 - r^2) d^3r = \sqrt{\frac{16\pi}{5}} \int \rho(\vec{r}) r^2 Y_{20}(\theta, \phi) d^3r$$

For a pure S-state ($l = 0$), the spatial wave function is spherically symmetric:

$$|\psi_S|^2 = \frac{|u(r)|^2}{4\pi}$$

Evaluating the angular integral:

$$\int_{4\pi} Y_{20}(\Omega) d\Omega = 0$$

Due to spherical symmetry, $\langle 3 z^2 - r^2 \rangle = 0$. Therefore, any pure S-state has an electric quadrupole moment $Q \equiv 0$.

Step 2: Proof of Non-Central Tensor Force

Because the experimental quadrupole moment is finite and positive ($Q = +0.286\text{ fm}^2$):

  1. The charge distribution of the deuteron is prolate (cigar-shaped, elongated along the spin axis).
  2. The nuclear force is non-central: it contains a tensor force $\hat{S}_{12}$:
$$\hat{S}_{12} = \frac{3}{r^2} (\vec{\sigma}_1 \cdot \vec{r})(\vec{\sigma}_2 \cdot \vec{r}) - (\vec{\sigma}_1 \cdot \vec{\sigma}_2)$$
  1. The tensor force mixes states with identical total angular momentum $J = 1$ and parity $\pi = (-1)^l = +1$, coupling the dominant $l = 0$ ($^3S_1$) state with the $l = 2$ ($^3D_1$) state:
$$|\psi_D\rangle = a_S |^3S_1\rangle + a_D |^3D_1\rangle$$

Step 3: D-State Admixture

The interference between the S and D wave functions produces the quadrupole moment:

$$Q \approx \frac{\sqrt{2}}{10} a_S a_D \int_0^\infty r^2 u_S(r) u_D(r) dr$$

Modern nucleon-nucleon potential models (Argonne $v_{18}$, CD-Bonn) determine that the deuteron ground state contains approximately $4.5\%$ to $5.8\%$ D-state admixture ($a_D^2 \approx 0.05$). The deuteron is not spherical!

Solved Honors Problems & Derivations

Step-by-step rigorous solutions with full physical, thermodynamic, and nuclear kinematic validation.