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Chapter 1 • Theory & Derivations

Nuclear Structure, Masses, Radii & Nuclear Models

Comprehensive foundation of nuclear physics: subatomic constitution, high-energy electron scattering, charge radius measurement, mass defect, binding energy systematics, semi-empirical mass formula (SEMF), valley of beta stability, electromagnetic moments (Schmidt lines, quadrupole deformation), nuclear force characteristics, and comparative evaluation of the Liquid Drop and Shell Models.

1.1Nuclear Constitution, Quarks, Nucleons & Charge Radii

1. Fundamental Constitution of the Atomic Nucleus

An atomic nucleus is a tightly bound quantum many-body system composed of $A$ nucleons: $Z$ positively charged protons and $N = A - Z$ electrically neutral neutrons. Protons and neutrons are not point particles, but composite hadrons belonging to the baryon family, each formed by three valence quarks bound via the strong interaction mediated by gluons ($SU(3)_C$ color gauge theory):

  • Proton ($p$): Valence quark composition $uud$ (two up quarks, one down quark). Net electric charge $Q = 2(+2/3 e) + (-1/3 e) = +1 e$. Rest mass $m_p = 938.272\text{ MeV}/c^2 = 1.67262 \times 10^{-27}\text{ kg} = 1.007276\text{ u}$. Intrinsic spin $s = 1/2$.
  • Neutron ($n$): Valence quark composition $udd$ (one up quark, two down quarks). Net electric charge $Q = +2/3 e + 2(-1/3 e) = 0$. Rest mass $m_n = 939.565\text{ MeV}/c^2 = 1.67493 \times 10^{-27}\text{ kg} = 1.008665\text{ u}$. Intrinsic spin $s = 1/2$. Free neutrons undergo beta decay ($n \to p + e^- + \bar{\nu}_e$) with a mean lifetime $\tau \approx 879.4\text{ s}$.

2. Nuclear Size & High-Energy Electron Scattering

Because the nuclear strong force saturates, nuclear matter exhibits nearly constant volume density $\rho_0 \approx 0.17\text{ nucleons/fm}^3 = 2.7 \times 10^{17}\text{ kg/m}^3$. Consequently, the mean nuclear charge radius $R$ scales with the cube root of the mass number $A$:

$$R = R_0 A^{1/3} \quad (R_0 \approx 1.20 - 1.25\text{ fm})$$

The definitive experimental measurement of the spatial nuclear charge density $\rho(r)$ was conducted by Robert Hofstadter (1950s) utilizing high-energy elastic electron scattering ($E_e \sim 200 - 1000\text{ MeV}$, where de Broglie wavelength $\lambda = hc/E \sim 0.5 - 2\text{ fm}$ is smaller than the nuclear radius). In the first Born approximation, the differential scattering cross-section is the Rutherford cross-section modulated by the squared nuclear Form Factor $F(\vec{q})$:

$$\left(\frac{d\sigma}{d\Omega}\right) = \left(\frac{d\sigma}{d\Omega}\right)_{\text{Mott}} |F(\vec{q})|^2, \quad F(\vec{q}) = \frac{1}{Z e} \int \rho(\vec{r}) e^{i \vec{q}\cdot\vec{r}/\hbar} d^3r$$

where $\vec{q} = \vec{p} - \vec{p}'$ is the momentum transfer. The experimental form factor maps directly to the Fermi Two-Parameter Charge Distribution:

$$\rho(r) = \frac{\rho_0}{1 + e^{(r - c)/a}}$$

where $c \approx 1.18 A^{1/3}\text{ fm}$ is the half-density radius, and $a \approx 0.54\text{ fm}$ is the surface diffuseness parameter (corresponding to a 90%-to-10% surface thickness $t \approx 4.4 a \approx 2.4\text{ fm}$).

3. Muonic X-Ray Atoms

In a muonic atom, a negative muon ($\mu^-$, mass $m_\mu \approx 206.77 m_e$) replaces an orbital electron. Because the Bohr radius scales inversely with mass ($a_\mu = \frac{m_e}{m_\mu} a_0 \approx \frac{0.529\text{ \AA}}{207} \approx 256\text{ fm}$), the lowest muonic orbits ($1s$) reside largely inside the nuclear volume. The severe perturbation of the Coulomb potential inside the charge sphere shifts the $2p \to 1s$ Lyman muonic X-ray transition energies by hundreds of keV, providing a complementary sub-femtometer measurement of the nuclear charge radius.

1.2Mass Defect, Binding Energy & Mirror Nuclei Coulomb Energy

1. Mass Defect and Nuclear Binding Energy

When $Z$ free protons and $N$ free neutrons coalesce to synthesize a bound nucleus $^A_Z\text{X}$, total energy conservation dictates that a substantial quantity of energy, the nuclear binding energy $B(A, Z)$, is liberated. By Einstein's mass-energy equivalence $E = m c^2$, the rest mass of the bound neutral atom $M(A, Z)$ is strictly less than the sum of its isolated constituents:

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

Using the atomic mass unit definition ($1\text{ u} \equiv \frac{1}{12} M(^{12}\text{C}) = 931.494\text{ MeV}/c^2$), the binding energy per nucleon is defined as:

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

2. The Binding Energy per Nucleon Curve

Plotting $B/A$ against mass number $A$ reveals the fundamental energetic landscape of the universe:

  • Light Nuclei ($A < 20$): Rapid rise with sharp local stability spikes at alpha-conjugate even-even nuclei ($^4\text{He}$, $^8\text{Be}$, $^{12}\text{C}$, $^{16}\text{O}$, $^{20}\text{Ne}$), with $^4\text{He}$ displaying $B/A \approx 7.07\text{ MeV/nucleon}$.
  • Broad Maximum ($A \approx 56 - 62$): The curve attains its global peak at $^{56}\text{Fe}$ ($B/A \approx 8.790\text{ MeV/nucleon}$) and $^{62}\text{Ni}$ ($B/A \approx 8.795\text{ MeV/nucleon}$), representing the most thermodynamically stable nuclear matter.
  • Heavy Nuclei ($A > 100$): Gradual monotonic descent toward $B/A \approx 7.5\text{ MeV/nucleon}$ at $^{238}\text{U}$, driven by the disruptive Coulomb repulsion of $Z^2$ protons scaling faster than short-range nuclear attraction.
  • Nuclear Power Consequences:
    • Nuclear Fusion: Combining light nuclei ($A \ll 56$, e.g., $\text{D} + \text{T} \to {^4\text{He}} + n$) moves up the steep left slope, releasing enormous kinetic energy.
    • Nuclear Fission: Splitting a heavy actinide ($A \sim 235 \to A_1, A_2 \sim 118$) moves up the right slope toward the iron peak, liberating $\approx 200\text{ MeV}$ per fission event.

3. Coulomb Displacement Energy in Mirror Nuclei

Mirror nuclei are pairs of isobars related by exchanging the numbers of protons and neutrons: $Z_1 = N_2$ and $N_1 = Z_2$ (e.g., $^{15}_7\text{N}$ and $^{15}_8\text{O}$, or $^{27}_{13}\text{Al}$ and $^{27}_{14}\text{Si}$). Because the strong nuclear interaction is charge-symmetric, the difference in their binding energies is due solely to the difference in electrostatic Coulomb self-energy $E_C$:

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

For isobars differing by $\Delta Z = 1$ ($Z_1 = Z$ and $Z_2 = Z - 1$):

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

Measuring the beta-decay end-point energy between mirror nuclei provides an independent experimental method to determine the nuclear radius parameter $R_0 \approx 1.20\text{ fm}$.

1.3The Semi-Empirical Mass Formula (SEMF) & Valley of Stability

1. The Weizsäcker Semi-Empirical Mass Formula

Carl Friedrich von Weizsäcker (1935) modeled the nucleus as an incompressible, charged liquid drop of nuclear fluid. The total nuclear binding energy $B(A, Z)$ is parameterized by five physical terms:

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

where the standard empirical coefficients (in MeV) are:

  1. Volume Term ($+ a_v A$, $a_v \approx 15.75\text{ MeV}$): Reflects the short-range, saturating nature of nuclear forces. Each nucleon interacts only with its nearest neighbors, contributing a constant volume energy proportional to $A$.
  2. Surface Term ($- a_s A^{2/3}$, $a_s \approx 17.8\text{ MeV}$): Nucleons at the nuclear surface have fewer neighbors than interior nucleons, reducing binding energy proportional to the nuclear surface area $4\pi R^2 \propto A^{2/3}$ (analogous to surface tension).
  3. Coulomb Term ($- a_c \frac{Z(Z-1)}{A^{1/3}}$, $a_c \approx 0.711\text{ MeV}$): Mutual electrostatic repulsion among all $Z(Z-1)/2$ proton pairs distributed uniformly within a sphere of radius $R = R_0 A^{1/3}$:
    $$E_C = \frac{3}{5} \frac{Z(Z-1) e^2}{4\pi\varepsilon_0 R_0 A^{1/3}} = a_c \frac{Z(Z-1)}{A^{1/3}}$$
  4. Asymmetry Term ($- a_a \frac{(A - 2Z)^2}{A}$, $a_a \approx 23.7\text{ MeV}$): A purely quantum mechanical consequence of the Pauli exclusion principle. Displacing protons into neutron levels increases total Fermi energy proportional to $(N - Z)^2/A = (A - 2Z)^2/A$.
  5. Pairing Term ($\delta(A, Z)$): Arises from the spin-pairing attraction between identical nucleons in time-reversed orbits:
    $$\delta(A, Z) = \begin{cases} + a_p A^{-1/2} \text{ (or } + a_p A^{-3/4}), & \text{even } Z, \text{even } N \text{ (even-even: most stable)} \\ 0, & \text{odd } A \text{ (even-odd or odd-even)} \\ - a_p A^{-1/2} \text{ (or } - a_p A^{-3/4}), & \text{odd } Z, \text{odd } N \text{ (odd-odd: least stable)} \end{cases}$$

    where $a_p \approx 11.2\text{ MeV}$ (or $a_p \approx 34\text{ MeV}$ with $A^{-3/4}$).

2. The Valley of Beta Stability

For a fixed mass number $A$, the binding energy is a quadratic parabola in $Z$. The most stable isobar $Z_0$ corresponds to the maximum binding energy: $\left.\frac{\partial B(A, Z)}{\partial Z}\right|_{Z_0} = 0$:

$$- \frac{a_c (2Z_0 - 1)}{A^{1/3}} + \frac{4 a_a (A - 2Z_0)}{A} = 0 \implies Z_0 = \frac{A}{2 + \frac{a_c}{2 a_a} A^{2/3}} \approx \frac{A}{2 + 0.015 A^{2/3}}$$
  • For light nuclei ($A \ll 40$), the Coulomb term is negligible ($A^{2/3} \ll 1$), giving $Z_0 \approx A/2$ ($N \approx Z$).
  • For heavy nuclei ($A \sim 200$), Coulomb repulsion shifts the valley of stability toward neutron-rich compositions: $Z_0 \approx 82$ for $A = 208$, giving $N/Z \approx 1.54$.

1.4Nuclear Spin, Parity, Magnetic Moments & Quadrupole Moments

1. Nuclear Total Angular Momentum (Nuclear Spin $\vec{I}$) & Parity $\pi$

The total angular momentum of a nucleus in its rest frame, conventionally designated as the nuclear spin $\vec{I}$, is the vector sum of individual nucleon orbital angular momenta $\vec{l}_i$ and intrinsic spin angular momenta $\vec{s}_i$:

$$\vec{I} = \sum_{i=1}^A (\vec{l}_i + \vec{s}_i)$$

Nuclear state parities $\pi = \pm 1$ reflect spatial inversion symmetry: $\psi(-\vec{r}) = \pi \psi(\vec{r})$, where single-particle states have parity $\pi = (-1)^l$. Total nuclear parity is the product over all occupied single-particle orbitals: $\pi = \prod_{i=1}^A (-1)^{l_i}$.

  • Even-Even Nuclei ($Z$ even, $N$ even): In the ground state, all nucleon spins pair off identically in time-reversed orbits ($J^\pi = 0^+$ without exception).
  • Odd-$A$ Nuclei: Spin and parity are determined entirely by the single unpaired valence nucleon: $I = j_{\text{val}}$, $\pi = (-1)^{l_{\text{val}}}$.
  • Odd-Odd Nuclei: Coupling of unpaired proton and neutron: $|j_p - j_n| \le I \le j_p + j_n$ (Nordheim's empirical coupling rules).

2. Nuclear Magnetic Dipole Moments & Schmidt Limits

The nuclear magnetic moment operator is expressed in units of the nuclear magneton $\mu_N = \frac{e\hbar}{2m_p} \approx 5.05078 \times 10^{-27}\text{ J/T}$ (which is $\approx 1/1836$ of the Bohr magneton $\mu_B$):

$$\vec{\mu} = \sum_{i=1}^A \left[ g_l^{(i)} \vec{l}_i + g_s^{(i)} \vec{s}_i \right] \mu_N$$

where the bare nucleon $g$-factors are:

  • Proton: $g_l = 1$, $g_s = +5.5857$
  • Neutron: $g_l = 0$, $g_s = -3.8263$

In the extreme single-particle shell model, the magnetic moment of an odd-$A$ nucleus is generated entirely by the single valence nucleon, yielding the Schmidt limits:

  • For $j = l + 1/2$: $\mu = \left[ (j - 1/2) g_l + \frac{1}{2} g_s \right] \mu_N$
  • For $j = l - 1/2$: $\mu = \frac{j}{j+1} \left[ (j + 3/2) g_l - \frac{1}{2} g_s \right] \mu_N$

3. Electric Quadrupole Moment $Q$ & Nuclear Deformation

The nuclear electric quadrupole moment $Q$ measures the deviation of the nuclear charge distribution $\rho(\vec{r})$ from spherical symmetry:

$$e Q = \int \rho(\vec{r}) (3 z^2 - r^2) d^3r = \int \rho(\vec{r}) r^2 (3\cos^2\theta - 1) d^3r$$
  • Spherical Nucleus ($Q = 0$): $\langle z^2 \rangle = \langle x^2 \rangle = \langle y^2 \rangle = \frac{1}{3}\langle r^2 \rangle$. Occurs for all closed-shell magic nuclei and all $I = 0$ or $I = 1/2$ states.
  • Prolate Ellipsoid ($Q > 0$): Elongated along the spin axis like an American football ($z_{\text{axis}} > x, y$). Common in rare-earth and actinide deformed nuclei.
  • Oblate Ellipsoid ($Q < 0$): Flattened along the spin axis like a discus ($z_{\text{axis}} < x, y$).

1.5Nuclear Forces & Yukawa Meson Exchange Theory

1. Empirical Characteristics of the Strong Nuclear Force

Detailed scattering experiments ($p$-$p$ and $n$-$p$) and deuteron binding properties reveal seven defining characteristics of the nuclear force:

  1. Short Range: Acts strongly across $r \approx 1 - 2\text{ fm}$, vanishing exponentially beyond $r \sim 2.5\text{ fm}$. Possesses a hard repulsive core at $r < 0.5\text{ fm}$ preventing nuclear collapse.
  2. Enormous Strength: At $r \approx 1\text{ fm}$, it is $\sim 100$ times stronger than electromagnetic Coulomb repulsion.
  3. Charge Independence: The nuclear interaction between two nucleons in the same quantum state is identical: $V_{pp} = V_{nn} = V_{np}$ (isospin symmetry $T = 1$).
  4. Charge Symmetry: Invariant under proton-neutron reflection ($V_{pp} = V_{nn}$).
  5. Spin Dependence: The force is significantly stronger when nucleon spins are aligned parallel ($S = 1$, triplet, as in the bound deuteron $^2\text{H}$) than antiparallel ($S = 0$, singlet, where di-proton and di-neutron are unbound).
  6. Non-Central (Tensor) Force: Contains a non-central component $S_{12} = \frac{3}{r^2}(\vec{\sigma}_1\cdot\vec{r})(\vec{\sigma}_2\cdot\vec{r}) - \vec{\sigma}_1\cdot\vec{\sigma}_2$, explaining the non-zero electric quadrupole moment of the deuteron ($Q_d = +0.00286\text{ b}$) and $D$-state orbital mixing.
  7. Saturation: Each nucleon interacts only with immediate nearest neighbors, keeping $B/A$ approximately constant.

2. Yukawa's Meson Exchange Theory (1935)

Hideki Yukawa proposed that the strong nuclear force is mediated by the virtual exchange of massive scalar bosons called pions ($\pi^\pm, \pi^0$). In relativistic quantum field theory, the static wave equation for a scalar field $\phi(r)$ generated by a point nucleon source is the Klein-Gordon equation:

$$\left( \nabla^2 - \frac{m_\pi^2 c^2}{\hbar^2} \right) \phi(r) = - g \delta(\vec{r})$$

The spherically symmetric solution is the Yukawa Potential:

$$V(r) = - g^2 \frac{e^{- r / \lambda_\pi}}{r} = - g^2 \frac{e^{- \mu r}}{r}$$

where $\lambda_\pi = \frac{\hbar}{m_\pi c}$ is the Compton wavelength of the mediating pion. Estimating the force range as $R \approx 1.4\text{ fm}$ using the Heisenberg uncertainty principle ($\Delta E \Delta t \sim (m_\pi c^2) (R/c) \sim \hbar$):

$$m_\pi \approx \frac{\hbar}{R c} = \frac{197.3\text{ MeV}\cdot\text{fm}}{(1.4\text{ fm}) c^2} \approx 140\text{ MeV}/c^2$$

Yukawa's predicted pion was discovered experimentally in cosmic rays in 1947 with rest masses $m_{\pi^\pm} = 139.57\text{ MeV}/c^2$ and $m_{\pi^0} = 134.98\text{ MeV}/c^2$.

1.6Nuclear Models: Liquid Drop vs. The Nuclear Shell Model

1. The Nuclear Liquid Drop Model

The liquid drop model treats the nucleus collectively as a droplet of incompressible quantum fluid. It accurately predicts:

  • Nuclear binding energies across the chart of nuclides via the SEMF.
  • Low-lying collective quadrupole ($\lambda = 2$) and octupole ($\lambda = 3$) vibrational excitations.
  • The mechanism of induced nuclear fission (Bohr-Wheeler theory) through liquid drop ellipsoidal surface distortions.

However, it completely fails to explain the dramatic stability spikes observed at specific nucleon numbers.

2. Experimental Evidence for Magic Numbers

Nuclei possessing specific numbers of protons or neutrons—Magic Numbers: $2, 8, 20, 28, 50, 82, 126$—exhibit extraordinary stability:

  • Abrupt discontinuities in separation energies $S_n$ and $S_p$ (analogous to atomic noble gas ionization potentials).
  • Small neutron capture cross-sections $\sigma_{(n,\gamma)}$.
  • Doubly magic nuclei ($^4_2\text{He}_2$, $^{16}_8\text{O}_8$, $^{40}_{20}\text{Ca}_{20}$, $^{48}_{20}\text{Ca}_{28}$, $^{208}_{82}\text{Pb}_{126}$) possess exceptionally large binding energies and spherical ground states ($Q = 0$).

3. The Nuclear Shell Model & Spin-Orbit Coupling

Maria Goeppert-Mayer and J. Hans D. Jensen (1949, Nobel Prize 1963) resolved the magic number sequence by introducing a strong attractive spin-orbit interaction into the 3D central potential (Woods-Saxon or Harmonic Oscillator):

$$\hat{H} = \hat{H}_0 + V_{ls}(r) \vec{l} \cdot \vec{s}$$

Because $\vec{j} = \vec{l} + \vec{s}$, we have $\vec{j}^2 = \vec{l}^2 + \vec{s}^2 + 2\vec{l}\cdot\vec{s}$, which gives:

$$\langle \vec{l} \cdot \vec{s} \rangle = \frac{1}{2} [j(j+1) - l(l+1) - s(s+1)] = \begin{cases} +\frac{l}{2}, & j = l + 1/2 \\ -\frac{l+1}{2}, & j = l - 1/2 \end{cases}$$

The energy splitting between the two spin-orbit partner states is:

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

Because $V_{ls} < 0$, the $j = l + 1/2$ state is shifted downward in energy. For large orbital angular momentum $l$ (e.g., $1f_{7/2}$, $1g_{9/2}$, $1h_{11/2}$, $1i_{13/2}$), the downward shift is so massive that the $j = l + 1/2$ level drops across the major oscillator shell gap into the shell below. These "intruder states" produce major shell closures at precisely 28, 50, 82, and 126, reproducing the complete sequence of nuclear magic numbers.

EXAM SUCCESS WORKSHOP

Solved University Examination Problems

Step-by-step mathematical solutions to classic university honors examination questions.

SOLVED PROBLEM 1.1

Nuclear Radius and Binding Energy Calculation for Iron-56

Iron-56 ($^{56}_{26}\text{Fe}$) has an atomic mass of $M = 55.9349375 \text{ u}$. The rest masses of a neutral hydrogen atom and a free neutron are $m(^1\text{H}) = 1.007825 \text{ u}$ and $m_n = 1.008665 \text{ u}$. (a) Calculate the nuclear charge radius $R$ assuming $R_0 = 1.22 \text{ fm}$. (b) Determine the mass defect $\Delta m$ in $\text{u}$ and total binding energy $B$ in $\text{MeV}$. (c) Calculate the binding energy per nucleon $B/A$.

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Calculate the Nuclear Radius R
R = R_0 A^{1/3} = (1.22\text{ fm}) (56)^{1/3} = (1.22\text{ fm}) (3.82586) \approx 4.668\text{ fm} = 4.67 \times 10^{-15}\text{ m}

Evaluate R using the standard nuclear scaling law with A = 56.

Step 2: Calculate the Mass Defect Delta m
\Delta m = Z m(^1\text{H}) + (A - Z) m_n - M(^{56}\text{Fe}) = 26(1.007825\text{ u}) + 30(1.008665\text{ u}) - 55.934938\text{ u} = 26.203450\text{ u} + 30.259950\text{ u} - 55.934938\text{ u} = 56.463400\text{ u} - 55.934938\text{ u} = 0.528462\text{ u}

Compute the difference between the constituent masses and the bound atomic mass.

Step 3: Convert Mass Defect to Binding Energy in MeV
B = \Delta m \times 931.494\text{ MeV/u} = (0.528462\text{ u}) \times (931.494\text{ MeV/u}) \approx 492.269\text{ MeV}

Multiply by the atomic mass energy conversion factor.

Step 4: Compute the Binding Energy per Nucleon B / A
\frac{B}{A} = \frac{492.269\text{ MeV}}{56} \approx 8.7905\text{ MeV/nucleon}

Divide total binding energy by mass number A = 56. This matches the peak of the experimental binding energy curve.

Final Answer & Physical Insight

R = 4.67 \text{ fm}, \quad \Delta m = 0.5285 \text{ u}, \quad B = 492.27 \text{ MeV}, \quad B/A = 8.791 \text{ MeV/nucleon}

SOLVED PROBLEM 1.2

Coulomb Displacement Energy and Radius of Mirror Pair Nitrogen-15 and Oxygen-15

The mirror pair $^{15}_7\text{N}$ and $^{15}_8\text{O}$ differ by one proton ($Z_1 = 8, Z_2 = 7$). The atomic mass difference is $\Delta M = M(^{15}\text{O}) - M(^{15}\text{N}) = 2.754 \text{ MeV}/c^2$. The neutron-proton mass difference is $(m_n - m_H)c^2 = 0.782 \text{ MeV}$. (a) Calculate the experimental Coulomb displacement energy $\Delta E_C$. (b) Using the uniform sphere Coulomb model $\Delta E_C = \frac{3e^2}{5(4\pi\varepsilon_0)R}(2Z - 1)$, determine the nuclear radius $R$ and the radius parameter $R_0$.

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Relate Mass Difference to Coulomb Energy Difference
\Delta E_C = [M(^{15}\text{O}) - M(^{15}\text{N})]c^2 + (m_n - m_H)c^2 = 2.754\text{ MeV} + 0.782\text{ MeV} = 3.536\text{ MeV}

The binding energy difference between mirror nuclei equals the mass difference plus the neutron-hydrogen rest mass difference.

Step 2: Formulate the Coulomb Energy Equation for R
\Delta E_C = \frac{3}{5} \frac{e^2}{4\pi\varepsilon_0 R} (2Z - 1) = \frac{3}{5} \frac{1.440\text{ MeV}\cdot\text{fm}}{R} (2 \times 8 - 1) = \frac{0.864\text{ MeV}\cdot\text{fm} \times 15}{R} = \frac{12.96\text{ MeV}\cdot\text{fm}}{R}

Substitute Z = 8 and e^2 / (4 pi epsilon_0) = 1.440 MeV * fm into the formula.

Step 3: Solve for the Nuclear Radius R
R = \frac{12.96\text{ MeV}\cdot\text{fm}}{3.536\text{ MeV}} \approx 3.665\text{ fm}

Divide the Coulomb coefficient by the measured energy shift.

Step 4: Extract the Radius Parameter R_0
R_0 = \frac{R}{A^{1/3}} = \frac{3.665\text{ fm}}{(15)^{1/3}} = \frac{3.665\text{ fm}}{2.4662} \approx 1.243\text{ fm}

Divide R by 15^(1/3) to obtain R_0.

Final Answer & Physical Insight

\Delta E_C = 3.536 \text{ MeV}, \quad R = 3.67 \text{ fm}, \quad R_0 = 1.24 \text{ fm}

SOLVED PROBLEM 1.3

Most Stable Isobar Prediction for A = 125 via Weizsacker Formula

Using the Weizsäcker Semi-Empirical Mass Formula with Coulomb coefficient $a_c = 0.711 \text{ MeV}$ and asymmetry coefficient $a_a = 23.7 \text{ MeV}$: (a) Derive and calculate the theoretical most stable atomic number $Z_0$ for mass number $A = 125$. (b) Determine the stable chemical element corresponding to this isobar and verify whether it matches the known stable nuclide Tellurium-125 ($Z=52$).

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Recall the Formula for the Valley of Beta Stability
Z_0 = \frac{A}{2 + \frac{a_c}{2 a_a} A^{2/3}}

Set the first derivative of the SEMF binding energy with respect to Z to zero.

Step 2: Evaluate the Denominator Terms for A = 125
A^{2/3} = (125)^{2/3} = (5)^2 = 25.0 \implies \frac{a_c}{2 a_a} A^{2/3} = \frac{0.711\text{ MeV}}{2 \times 23.7\text{ MeV}} \times 25.0 = \frac{0.711}{47.4} \times 25.0 \approx 0.0150 \times 25.0 = 0.375

Compute the Coulomb correction to the symmetry denominator.

Step 3: Calculate the Theoretical Value of Z_0
Z_0 = \frac{125}{2 + 0.375} = \frac{125}{2.375} \approx 52.63

Divide A by the total denominator.

Step 4: Identify the Nearest Integer Stable Isobar
Z_0 \approx 52.63 \implies \text{Nearest Integer } Z = 52 \text{ or } 53

Tellurium (Z = 52) has 73 neutrons and is an extraordinarily stable even-Z nuclide (^125_52Te, natural abundance 7.07%). Iodine-125 (Z = 53) decays to Te-125 via electron capture with T_1/2 = 59.4 days. The SEMF prediction matches experiment.

Final Answer & Physical Insight

Z_0 = 52.63 \implies Z = 52 \quad (^{125}_{52}\text{Te}, \text{Tellurium-125})