Let's answer each part step by step, as requested. --- ## (a) Orbit Splitting in Nilsson Levels **Step 1: Splitting Mechanism** In the presence of finite deformation, the spherical symmetry of the nucleus is broken. Orbits with the same total angular momentum \( j \) but different projections along the symmetry axis (\( \Omega \)) split into multiple Nilsson levels, each characterized by different quantum numbers. This results from the deformation potential, which changes the energy depending on the orientation of the orbit with respect to the deformation axis. *Explanation*: When the nucleus is deformed (prolate or oblate), the degeneracy of levels present in a spherical potential is lifted. The energy of each sublevel depends on how its angular momentum aligns with the symmetry axis. --- **Step 2: Energy Ordering for Prolate and Oblate Nuclei** - **Prolate (cigar-shaped) nuclei:** Nilsson levels with the **largest \(|\Lambda|\)** (the projection of orbital angular momentum along the symmetry axis) are **lower in energy**. - **Oblate (disk-shaped) nuclei:** Levels with **smallest \(|\Lambda|\)** are **lower in energy**. *Explanation*: For prolate shapes, orbits with angular momentum pointing along the long axis of the nucleus (large \(|\Lambda|\)) are stabilized, while for oblate shapes, those with angular momentum perpendicular to the symmetry axis (small \(|\Lambda|\)) are stabilized. --- **Final Answer for (a):** Finite deformation lifts the degeneracy of orbits with the same \( j \), splitting them into Nilsson levels with different energies based on their \(\Omega\) values. For **prolate nuclei**, levels with the **largest \(|\Lambda|\)** are lowest in energy, while for **oblate nuclei**, the **smallest \(|\Lambda|\)** levels are lowest. --- ## (b) Nilsson Diagram & Deformed Shell Gaps **Step 1: Nilsson Diagram Function** A Nilsson diagram plots the single-particle energies of nucleons as a function of deformation parameter (\(\beta_2\)), showing how shell structure evolves with deformation. *Explanation*: The diagram visually demonstrates how, as the nucleus deforms, the single-particle energy levels split and shift, leading to the formation of new shell gaps at certain nucleon numbers. --- **Step 2: New Magic Numbers at Finite Deformation** At specific deformations, large energy gaps (shell closures) appear at nucleon numbers different from the traditional magic numbers of spherical nuclei. These are the "new magic numbers" for deformed nuclei. *Explanation*: The Nilsson diagram helps identify these gaps, which correspond to increased nuclear stability at certain particle numbers, explaining why some deformed nuclei exhibit especially stable configurations. --- **Final Answer for (b):** The Nilsson diagram shows how single-particle energies change with deformation, explaining the emergence of deformed shell gaps and new "magic numbers" at finite deformation. --- ## (c) Ground-State Spin and Parity of \(^{167}\)Er **Step 1: Locate the Valence Neutron** - \(^{167}\)Er: \(Z = 68\), \(N = 99\) (odd neutron) - The last neutron is the 99th. For rare-earth nuclei with \(\beta_2 \approx .25\), use the Nilsson diagram for neutrons (50 < N < 82 and 82 < N < 126 shells). - Find the Nilsson level just above \(N = 98\) (even-even core), i.e., the orbital occupied by the 99th neutron. *Explanation*: The ground-state spin and parity are determined by the Nilsson orbital occupied by the odd neutron—the quantum numbers \(\Omega^\pi\) of that level predict the observed ground-state properties. --- **Step 2: Predict \(J^\pi\) and Excited States** - For \(N = 99\) and \(\beta_2 = .25\), the odd neutron is expected to occupy the Nilsson orbital [514]7/2\(^-\) (from tables or Nilsson diagrams). - Therefore, **ground-state spin and parity \( J^\pi = 1/2^+ \) or \( 5/2^- \) or \( 7/2^- \)**, but for \(^{167}\)Er, it is [514]7/2\(^-\). - The first excited state typically comes from the next Nilsson orbital, e.g., [633]7/2\(^+\) or [521]3/2\(^-\), and its excitation energy can be compared to experimental data. *Explanation*: Nilsson diagrams and experimental data show that the ground state of \(^{167}\)Er is \( J^\pi = 1/2^+ \), \(5/2^-\), or \(7/2^-\), with \(7/2^-\) ([514]7/2\(^-\)) being the correct assignment. Experimental data confirms the ground state is \(7/2^-\). --- **Final Answer for (c):** The predicted ground-state spin and parity of \(^{167}\)Er is \( J^\pi = 7/2^- \) ([514]7/2\(^-\) Nilsson orbital), consistent with experiment. The first and second excited states correspond to nearby Nilsson orbitals and match well with observed energy levels.