Unit 5: Electronic Spectroscopy of Atoms: Angular Momentum Coupling & Atomic Term Symbols
Hydrogen atom spectrum, electronic angular momentum, spin-orbit fine structure and Lamb shift, alkali metal spectra and quantum defects, Russell-Saunders L-S vs j-j coupling schemes, atomic term symbols, Hund's rules, Landé interval rule, helium ortho/para states, normal and anomalous Zeeman effect, and Atomic Absorption Spectroscopy (AAS).
§5.1 The Hydrogen Atom Spectrum & One-Electron Energy Levels
The quantum mechanics of the one-electron hydrogenic atom (nuclear charge \(Z\)) is governed by the Coulomb Hamiltonian \(\hat{H} = -\frac{\hbar^2}{2\mu}\nabla^2 - \frac{Z e^2}{4\pi\varepsilon_0 r}\). Solving the radial and angular Schrödinger equations yields quantized energy eigenvalues depending solely on the principal quantum number \(n\):
\[ E_n = -\frac{\mu Z^2 e^4}{32 \pi^2 \varepsilon_0^2 \hbar^2} \frac{1}{n^2} = -\frac{R_H Z^2}{n^2}, \quad n = 1, 2, 3, \dots \]where \(R_H = \frac{\mu e^4}{8 \varepsilon_0^2 h^3 c} \approx 109677.58\text{ cm}^{-1}\) is the Rydberg constant for hydrogen. Spectral transitions obey the Rydberg formula:
\[ \tilde{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]Spectral Series of Hydrogen
- Lyman Series (\(n_1 = 1, n_2 \ge 2\)): Ultraviolet (\(91.2 - 121.6\text{ nm}\)).
- Balmer Series (\(n_1 = 2, n_2 \ge 3\)): Visible (\(364.6 - 656.3\text{ nm}\)), including \(H_\alpha\) at \(656.3\text{ nm}\).
- Paschen Series (\(n_1 = 3, n_2 \ge 4\)): Infrared (\(820.4 - 1875\text{ nm}\)).
- Brackett (\(n_1 = 4\)) & Pfund (\(n_1 = 5\)) Series: Far-infrared.
In the non-relativistic Schrödinger treatment, all orbital states with identical \(n\) (\(s, p, d, f\)) are exactly degenerate, with total degeneracy \(g_n = 2n^2\) (including electron spin).
§5.2 Spin-Orbit Coupling & Fine Structure of Hydrogen
Relativistic corrections to the hydrogen atom Hamiltonian resolve the apparent \(l\)-degeneracy into fine structure. Three relativistic perturbations contribute:
- Relativistic Mass Correction: Kinetic energy expansion \(T = \frac{p^2}{2m} - \frac{p^4}{8 m^3 c^2}\).
- Darwin Term: Contact interaction of s-electrons with the nucleus due to Zitterbewegung.
- Spin-Orbit Coupling: The electron's intrinsic magnetic spin moment \(\vec{\mu}_s = -g_e \frac{e}{2m} \vec{S}\) interacts with the internal magnetic field \(\vec{B}_{\text{int}}\) created by the relative orbital motion of the charged nucleus: \[ \hat{H}_{SO} = \xi(r) \vec{L} \cdot \vec{S} = \frac{1}{2 m^2 c^2} \frac{1}{r}\frac{dV}{dr} \vec{L} \cdot \vec{S} = \frac{Z e^2}{8\pi\varepsilon_0 m^2 c^2 r^3} \vec{L} \cdot \vec{S} \]
Total Angular Momentum J & Energy Splitting
Total angular momentum is \(\vec{J} = \vec{L} + \vec{S}\). Squaring both sides: \(\vec{J}^2 = \vec{L}^2 + \vec{S}^2 + 2\vec{L}\cdot\vec{S}\), which gives:
\[ \vec{L} \cdot \vec{S} = \frac{1}{2} (\vec{J}^2 - \vec{L}^2 - \vec{S}^2) \]The expectation value in state \(|l, s, j\rangle\) is:
\[ \langle \vec{L} \cdot \vec{S} \rangle = \frac{\hbar^2}{2} [j(j+1) - l(l+1) - s(s+1)] \]Combining all three relativistic terms yields the famous Dirac fine structure formula:
\[ E_{n, j} = E_n \left[ 1 + \frac{(Z\alpha)^2}{n} \left( \frac{1}{j + 1/2} - \frac{3}{4n} \right) \right] \]where \(\alpha = \frac{e^2}{4\pi\varepsilon_0 \hbar c} \approx \frac{1}{137.036}\) is the fine-structure constant. States with the same \(j\) (e.g., \(2s_{1/2}\) and \(2p_{1/2}\)) are degenerate in Dirac theory, but quantum electrodynamic (QED) vacuum fluctuations lift this degeneracy by \(1057.8\text{ MHz}\), known as the Lamb shift.
### Advanced Quantum Formalism: Wigner-Eckart Theorem and the Breit-Rabi Equation In atomic spectroscopy, calculating transition intensities and magnetic shifts across arbitrary coupling regimes requires angular momentum tensor algebra. #### 1. The Wigner-Eckart Theorem for Electric Dipole Transitions The electric dipole operator is an irreducible spherical tensor of rank 1: \(\hat{T}_q^{(1)} = \hat{\mu}_q\), where \(q \in \{0, \pm 1\}\). The Wigner-Eckart theorem states that the matrix element of any spherical tensor operator factors cleanly into a geometrical Clebsch-Gordan coefficient (or Wigner 3-\(j\) symbol) and a dynamical **reduced matrix element**: \[ \langle \gamma J M | \hat{T}_q^{(k)} | \gamma' J' M' \rangle = (-1)^{J - M} \begin{pmatrix} J & k & J' \\ -M & q & M' \end{pmatrix} \langle \gamma J || \hat{T}^{(k)} || \gamma' J' \rangle \] Consequences for spectroscopic transitions (\(k = 1\)): - **Triangular condition:** \(|J - J'| \le 1 \le J + J' \implies \Delta J = 0, \pm 1\) (with \(J = 0 \not\rightarrow J' = 0\)). - **Magnetic projection:** \(-M + q + M' = 0 \implies \Delta M = M - M' = q \in \{0, \pm 1\}\). - All relative line intensities within a multiplet or Zeeman pattern are governed entirely by the geometric Wigner 3-\(j\) symbols, leaving the intrinsic electronic radial overlap locked inside \(\langle \gamma J || \hat{\mathbf{r}} || \gamma' J' \rangle\). #### 2. The Breit-Rabi Equation: Intermediate Magnetic Field Splitting When an external magnetic field \(B_0\) is neither strictly weak (Zeeman limit, \(g_F \mu_B B_0 \ll A_{\text{hfs}}\)) nor strictly strong (Paschen-Back limit, \(\mu_B B_0 \gg A_{\text{hfs}}\)), the hyperfine Hamiltonian: \[ \hat{H} = A_{\text{hfs}} \hat{\mathbf{I}} \cdot \hat{\mathbf{J}} + g_J \mu_B B_0 \hat{J}_z + g_I \mu_N B_0 \hat{I}_z \] can be solved analytically for any atom with electronic angular momentum \(J = 1/2\) (such as alkali ground states \(^2S_{1/2}\)) using the **Breit-Rabi formula**: \[ E(F, M_F) = -\frac{\Delta E_{\text{hfs}}}{2(2I + 1)} + g_I \mu_N B_0 M_F \pm \frac{\Delta E_{\text{hfs}}}{2} \sqrt{1 + \frac{4 M_F}{2I + 1} x + x^2} \] where: - \(\Delta E_{\text{hfs}} = A_{\text{hfs}} \left( I + \frac{1}{2} \right)\) is the zero-field hyperfine splitting. - \(x\) is the dimensionless magnetic field parameter: \[ x = \frac{(g_J \mu_B - g_I \mu_N) B_0}{\Delta E_{\text{hfs}}} \approx \frac{g_J \mu_B B_0}{\Delta E_{\text{hfs}}} \] - The plus sign applies to the upper hyperfine manifold \(F = I + 1/2\), and the minus sign applies to the lower manifold \(F = I - 1/2\). **Limiting Behaviors:** 1. **Low-field limit (\(x \ll 1\)):** Expanding the square root via Taylor series: \[ \sqrt{1 + \frac{4 M_F}{2I + 1} x + x^2} \approx 1 + \frac{2 M_F}{2I+1} x + \mathcal{O}(x^2) \] reproduces linear Zeeman splitting with \(g_F \approx \frac{g_J}{2I+1}\). 2. **High-field limit (\(x \gg 1\)):** The square root approaches \(x + \frac{2 M_F}{2I+1}\), decoupling \(\mathbf{I}\) and \(\mathbf{J}\) into linear Paschen-Back states with energies \(g_J \mu_B B_0 M_J + A_{\text{hfs}} M_I M_J\).§5.3 Alkali Metal Spectra, Quantum Defects & Sodium D-Lines
Alkali metal atoms (Li, Na, K, Rb, Cs) possess a single valence electron outside a closed, spherically symmetric noble gas core. Because inner core electrons partially shield the nuclear charge \(Z\), the valence electron experiences an effective nuclear charge \(Z_{\text{eff}}(r)\) that varies with distance:
\[ Z_{\text{eff}} \to 1 \text{ as } r \to \infty, \quad Z_{\text{eff}} \to Z \text{ as } r \to 0 \]Electrons with low orbital angular momentum \(l\) (especially \(s\) and \(p\) orbitals) have non-zero probability density near the nucleus, penetrating the core and experiencing a higher effective charge. This penetration lowers their energy below the hydrogenic level:
\[ E_{n, l} = -\frac{R_y}{(n - \delta_l)^2} = -\frac{R_y}{n_{\text{eff}}^2} \]where \(\delta_l\) is the Rydberg quantum defect, with \(\delta_s > \delta_p > \delta_d > \delta_f \approx 0\).
The Sodium D-Line Doublet
For sodium (\(Z = 11\)), the ground state is \(3s\), represented by term symbol \(^2S_{1/2}\). The lowest excited state is \(3p\), which is split by spin-orbit coupling into two fine-structure levels:
- \(^2P_{3/2}\) (\(j = 1 + 1/2 = 3/2\)): higher energy level
- \(^2P_{1/2}\) (\(j = 1 - 1/2 = 1/2\)): lower energy level
Electric dipole selection rules allow transitions with \(\Delta l = \pm 1, \Delta j = 0, \pm 1\). Radiative decay to the ground state produces the intense yellow doublet:
\[ D_1: \quad 3p\ ^2P_{1/2} \to 3s\ ^2S_{1/2} \quad (\lambda = 589.592\text{ nm}, \tilde{\nu} = 16960.9\text{ cm}^{-1}) \] \[ D_2: \quad 3p\ ^2P_{3/2} \to 3s\ ^2S_{1/2} \quad (\lambda = 588.995\text{ nm}, \tilde{\nu} = 16978.1\text{ cm}^{-1}) \]The fine-structure splitting is \(\Delta \tilde{\nu} = 17.2\text{ cm}^{-1}\) (\(\Delta \lambda = 0.597\text{ nm}\)). The theoretical intensity ratio is \(I(D_2) : I(D_1) = 2 : 1\), matching the statistical degeneracy ratio \((2j+1) = 4 : 2\).
§5.4 Russell-Saunders (L-S) Coupling & Atomic Term Symbols
For light to medium-weight atoms (\(Z \lesssim 30\)), electrostatic Coulomb repulsion between electrons is substantially stronger than relativistic spin-orbit interactions. Under this regime, the system is described by the Russell-Saunders (\(L\text{-}S\)) coupling scheme:
- Individual orbital angular momenta \(\vec{l}_i\) couple via electrostatic forces to form total orbital angular momentum \(\vec{L} = \sum_i \vec{l}_i\), with quantum number \(L = 0, 1, 2, 3, 4, 5 \dots\) designated as letters \(S, P, D, F, G, H \dots\).
- Individual spin angular momenta \(\vec{s}_i\) couple via exchange forces to form total spin angular momentum \(\vec{S} = \sum_i \vec{s}_i\), with total spin multiplicity \(2S + 1\).
- Total orbital \(\vec{L}\) and total spin \(\vec{S}\) couple weakly via spin-orbit interaction to yield total electronic angular momentum \(\vec{J} = \vec{L} + \vec{S}\), where: \[ J = |L - S|, |L - S| + 1, \dots, L + S \]
Standard Term Symbol Notation
Atomic electronic states are designated by the universal term symbol:
\[ ^{2S+1}L_J \]where \(2S+1\) is spin multiplicity, \(L\) is total orbital angular momentum letter, and \(J\) is total angular momentum quantum number. Each term possesses degeneracy \(g_J = 2J + 1\). The total degeneracy of an entire \(L\text{-}S\) multiplet is \((2L+1)(2S+1)\).
§5.5 Hund's Rules & Equivalent Electron Microstate Analysis
When multiple electrons occupy equivalent orbitals (same \(n\) and \(l\)), the Pauli exclusion principle dictates that no two electrons can possess identical sets of all four quantum numbers \((n, l, m_l, m_s)\). Microstate table analysis must be employed to find the permitted terms.
Microstate Analysis for Carbon \(2p^2\)
For the \(p^2\) configuration, there are \(\binom{6}{2} = \frac{6 \times 5}{2} = 15\) allowed microstates. Sorting these by \(M_L = \sum m_l\) and \(M_S = \sum m_s\) decomposes the configuration into three permitted terms:
\[ p^2 \implies {}^1D_2\ (5\text{ states}), \quad {}^3P_{2,1,0}\ (9\text{ states}), \quad {}^1S_0\ (1\text{ state}) \]Total microstates: \(5 + 9 + 1 = 15\).
Hund's Rules for Ground State Terms
Friedrich Hund formulated three empirical rules that uniquely identify the lowest-energy ground term:
- Hund's First Rule: The term with the maximum spin multiplicity (\(S\)) has the lowest energy, because parallel electron spins maximize the exchange stabilization and minimize interelectronic Coulomb repulsion. For \(p^2\), \(^3P\) is lower than \(^1D\) and \(^1S\).
- Hund's Second Rule: For terms with identical multiplicity \(S\), the term with the largest total orbital angular momentum (\(L\)) has the lowest energy.
- Hund's Third Rule: For subshells that are:
- Less than half-full: The level with the minimum \(J = |L - S|\) lies lowest (normal multiplet). For \(p^2\) (2 of 6 electrons), the ground state is \(^3P_0\).
- More than half-full: The level with the maximum \(J = L + S\) lies lowest (inverted multiplet). For \(p^4\), the ground state is \(^3P_2\).
- Exactly half-full: \(L = 0\), so \(J = S\), and there is only a single \(J\) level (e.g., \(p^3 \implies {}^4S_{3/2}\)).
The Landé Interval Rule
The energy separation between consecutive \(J\) levels within an \(L\text{-}S\) term is proportional to the larger \(J\) value:
\[ \Delta E(J, J-1) = E_J - E_{J-1} = A \cdot J \]where \(A\) is the spin-orbit coupling constant of the term.
§5.6 Atomic Selection Rules & The Zeeman Effect
For electric dipole transitions between atomic levels in the \(L\text{-}S\) coupling approximation, conservation of angular momentum and parity dictate the following Laporte selection rules:
\[ \Delta l = \pm 1 \quad (\text{parity must change: } \text{Laporte rule}) \] \[ \Delta L = 0, \pm 1 \quad (\text{except } L = 0 \not\to L = 0) \] \[ \Delta S = 0 \quad (\text{spin multiplicity cannot change}) \] \[ \Delta J = 0, \pm 1 \quad (\text{except } J = 0 \not\to J = 0) \] \[ \Delta M_J = 0\ (\pi\text{-polarization, parallel to field}), \quad \Delta M_J = \pm 1\ (\sigma\text{-polarization, perpendicular}) \]The Zeeman Effect
Applying an external static magnetic field \(\vec{B}_0 = B_0 \hat{z}\) interacts with the total atomic magnetic moment \(\vec{\mu} = -\frac{\mu_B}{\hbar} (\vec{L} + g_e \vec{S})\). In first-order perturbation theory, each level \(J\) splits into \(2J+1\) equally spaced magnetic sublevels:
\[ \Delta E_Z = g_J \mu_B B_0 M_J \]where \(\mu_B = \frac{e\hbar}{2m_e} = 9.27401 \times 10^{-24}\text{ J/T}\) is the Bohr magneton, and \(g_J\) is the Landé \(g\)-factor:
\[ g_J = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \]- Normal Zeeman Effect: Occurs in singlet states (\(S = 0 \implies g_J = 1\)). Every transition splits into a symmetric triplet: an unshifted \(\pi\)-line (\(\Delta M_J = 0\)) and two \(\sigma\)-lines shifted by \(\pm \frac{\mu_B B_0}{h}\).
- Anomalous Zeeman Effect: Occurs whenever \(S \neq 0\), causing different states to possess different \(g_J\) factors, producing complex multiplet splittings (e.g., 4 or 6 lines in the sodium D-lines).
§5.7 Atomic Absorption Spectroscopy (AAS) in Chemical Analysis
Atomic Absorption Spectroscopy (AAS) is an indispensable quantitative analytical technique based on the resonant absorption of optical radiation by free, unexcited ground-state gaseous atoms.
Instrumentation & Operating Principles
- Primary Radiation Source: A Hollow Cathode Lamp (HCL) containing a cathode fabricated from the target element. Sputtered excited atoms emit extraordinarily narrow atomic emission lines (\(\Delta\tilde{\nu} \sim 0.01\text{ cm}^{-1}\)), matching the exact resonant absorption frequencies of the analyte atoms.
- Atomizer:
- Flame AAS (FAAS): Air-acetylene (\(2300^\circ\text{C}\)) or nitrous oxide-acetylene (\(2900^\circ\text{C}\)) flame converts aerosolized liquid samples into ground-state atomic vapor. Detection limits \(\sim 10 - 100\text{ ppb}\).
- Graphite Furnace AAS (GFAAS): Electrothermal heating in a graphite tube through drying, ashing, and rapid atomization steps (\(2000 - 2800^\circ\text{C}\)). Detection limits reach sub-ppb (\(\text{pg}\) masses).
- Monochromator & Detector: Isolates the specific resonant analytical wavelength and rejects broad flame background emission using photomultiplier tubes (PMT) or solid-state detectors.
Quantitative Calibration & Background Correction
Absorption obeys the Beer-Lambert law: \(A = \log_{10}(I_0 / I) = \varepsilon b c\). Background molecular absorption and particulate scattering are corrected using:
- Continuum Source (Deuterium Lamp) Correction: Alternates HCL and broad \(D_2\) pulses.
- Zeeman Background Correction: Uses a strong magnetic field to split the atomic absorption line away from the broad background, providing the gold standard for trace metal analysis in complex biological and environmental matrices.
§5.8 Inductively Coupled Plasma (ICP-OES & ICP-MS) Atomic Emission Spectrometry
While Flame Atomic Absorption Spectroscopy (FAAS) measures atomic absorption at temperatures up to \(2800\text{ K}\), Inductively Coupled Plasma Optical Emission Spectrometry (ICP-OES) harnesses thermal excitation in an atmospheric argon plasma sustained at extreme temperatures of \(6,000 - 10,000\text{ K}\).
The Inductively Coupled Argon Plasma Torch
An ICP torch consists of three concentric quartz tubes surrounded by a water-cooled copper induction coil connected to a radiofrequency (RF) generator (typically \(27.12\text{ MHz}\) or \(40.68\text{ MHz}\) at \(1 - 1.5\text{ kW}\)):
- Argon gas flows tangentially through the outer tube (\(12 - 15\text{ L/min}\)).
- A high-voltage Tesla spark seeds initial seed electrons into the argon stream.
- The oscillating RF magnetic field accelerates the free electrons in closed annular paths, inducing intense ohmic resistance heating that sustains a toroidal, self-perpetuating argon plasma.
- Aerosolized sample solution is injected through the central injector tube directly through the center of the plasma donut.
Plasma Excitation & Spectroscopic Advantages
At \(8,000\text{ K}\), the plasma provides distinct physical advantages over chemical flames:
- Complete Atomization & High Ionization: Refractory metal oxides and carbides (\(\text{Zr}, \text{W}, \text{B}, \text{Al}, \text{Ti}\)) dissociate completely into free atoms and singly charged ions (\(\text{M}^+\)). Chemical matrix interferences are virtually eliminated.
- Simultaneous Multi-Element Detection: Thermally excited atoms and ions emit intense discrete optical lines spanning \(165 - 800\text{ nm}\). Modern Echelle grating polychromators with segmented charge-coupled detectors (SCD) record up to 70 elements simultaneously in a single 30-second measurement.
- Linear Dynamic Range: The thin, optically transparent analytical zone suppresses self-absorption, yielding a linear calibration range exceeding \(5 - 6\) orders of magnitude (\(0.1\text{ ppb} - 100\text{ ppm}\)).
ICP Mass Spectrometry (ICP-MS)
Coupling the ICP torch through water-cooled nickel sampling and skimmer cones into a high-vacuum quadrupole or magnetic sector mass spectrometer (ICP-MS) detects elemental ions directly by their mass-to-charge ratio (\(m/z\)). Detection limits drop by an additional three to four orders of magnitude into the parts-per-trillion (\(\text{ppt}\), \(\text{ng/L}\)) and parts-per-quadrillion (\(\text{ppq}\)) regimes with isotopic precision.
For the ground state electronic configuration of the neutral carbon atom (\(1s^2 2s^2 2p^2\)): (a) Determine all possible Russell-Saunders term symbols \(^{2S+1}L_J\) allowed by the Pauli exclusion principle. (b) Apply Hund's rules to determine the exact term and level of the ground state. (c) Write all the term levels in order of increasing energy.
Step (a): Derivation of allowed terms for p²
For two equivalent p-electrons (\(n=2, l=1\)), the total number of microstates is \(\binom{6}{2} = 15\).
Microstates sorted by \(M_L\) and \(M_S\):
- Maximum \(M_L = 2\) with \(M_S = 0\) (both electrons in \(m_l = +1\) with opposite spins). This belongs to a singlet D term: \(^{1}D\) (\(L=2, S=0\)), consisting of \((2L+1)(2S+1) = 5 \times 1 = 5\) microstates. Since \(J = L = 2\), the term level is \(^{1}D_2\).
- Maximum \(M_S = 1\) with \(M_L = 1\) (electrons in \(m_l = +1, 0\) with parallel spins). This belongs to a triplet P term: \(^{3}P\) (\(L=1, S=1\)), consisting of \((2L+1)(2S+1) = 3 \times 3 = 9\) microstates. Total angular momentum values: \(J = |1-1|, 1, 1+1 = 0, 1, 2\). Term levels: \(^{3}P_0, {}^{3}P_1, {}^{3}P_2\).
- One remaining microstate at \(M_L = 0, M_S = 0\) forms the singlet S term: \(^{1}S\) (\(L=0, S=0\)). Term level: \(^{1}S_0\).
Total microstates: \(5 ({}^{1}D_2) + 9 ({}^{3}P) + 1 ({}^{1}S_0) = 15\).
Step (b): Application of Hund's rules
- Hund's First Rule (Max Multiplicity): The triplet term \(^{3}P\) has \(2S+1 = 3\), which is larger than the singlets (\(2S+1 = 1\)). Therefore, \(^{3}P\) is the lowest energy term.
- Hund's Third Rule (J value): The \(2p\) subshell has 2 electrons out of 6, which is less than half-full. Therefore, the state with the minimum \(J\) value lies lowest: \(J = |L - S| = |1 - 1| = 0\).
Thus, the ground state of carbon is uniquely \(^{3}P_0\).
Step (c): Order of energy levels
Within \(^{3}P\), \(J\) increases with energy (normal multiplet): \(^{3}P_0 < {}^{3}P_1 < {}^{3}P_2\). Between the remaining singlets, Hund's second rule places \(^{1}D_2\) lower than \(^{1}S_0\). The complete order is:
\[ {}^{3}P_0 < {}^{3}P_1 < {}^{3}P_2 < {}^{1}D_2 < {}^{1}S_0 \]The ground configuration of atomic carbon has fine structure levels \(^{3}P_0\), \(^{3}P_1\), and \(^{3}P_2\). Spectroscopic measurements establish that the \(^{3}P_1\) level lies \(16.4\text{ cm}^{-1}\) above \(^{3}P_0\). (a) Using the Landé interval rule, predict the energy separation between \(^{3}P_2\) and \(^{3}P_1\). (b) Determine the spin-orbit coupling constant \(A\) for the \(^{3}P\) term. (c) Calculate the energy of \(^{3}P_2\) relative to the \(^{3}P_0\) ground state.
Step (a): Landé interval rule prediction
The Landé interval rule states that the energy separation between consecutive \(J\) levels is proportional to the larger \(J\):
\[ \Delta E(J, J-1) = A \cdot J \]For the \(^{3}P\) term (\(J = 0, 1, 2\)):
\[ \Delta E(1, 0) = E(^{3}P_1) - E(^{3}P_0) = A \cdot 1 = 16.4\text{ cm}^{-1} \] \[ \Delta E(2, 1) = E(^{3}P_2) - E(^{3}P_1) = A \cdot 2 = 2 \times 16.4\text{ cm}^{-1} = 32.8\text{ cm}^{-1} \]Step (b): Spin-orbit constant A
\[ A = 16.4\text{ cm}^{-1} \]Step (c): Energy of ³P₂ relative to ground state
\[ E(^{3}P_2) - E(^{3}P_0) = \Delta E(1, 0) + \Delta E(2, 1) = 16.4 + 32.8 = 49.2\text{ cm}^{-1} \]Experimental measurement yields \(43.4\text{ cm}^{-1}\), showing excellent agreement with Landé interval rule scaling (deviations \(\sim 10\%\) arise from second-order spin-orbit mixing with the higher \(^{1}D_2\) state).
For the sodium atom transitions corresponding to the yellow D-lines (\(3p\ ^2P_{1/2} \to 3s\ ^2S_{1/2}\) and \(3p\ ^2P_{3/2} \to 3s\ ^2S_{1/2}\)): (a) Calculate the Landé \(g\)-factor \(g_J\) for each of the three states: \(^2S_{1/2}\), \(^2P_{1/2}\), and \(^2P_{3/2}\). (b) In an external magnetic field of \(B_0 = 1.00\text{ Tesla}\), determine the energy shift \(\Delta E\) (in \(\text{cm}^{-1}\)) for each \(M_J\) magnetic sublevel. (c) State the number of Zeeman spectral lines observed for the \(D_1\) and \(D_2\) transitions under selection rules \(\Delta M_J = 0, \pm 1\).
Step (a): Landé g-factor calculations
\[ g_J = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \]- Ground State \(^2S_{1/2}\) (\(L=0, S=1/2, J=1/2\)): \[ g_J = 1 + \frac{\frac{3}{4} + \frac{3}{4} - 0}{2(\frac{3}{4})} = 1 + \frac{1.5}{1.5} = 2 \]
- Excited State \(^2P_{1/2}\) (\(L=1, S=1/2, J=1/2\)): \[ g_J = 1 + \frac{\frac{3}{4} + \frac{3}{4} - 2}{2(\frac{3}{4})} = 1 + \frac{1.5 - 2}{1.5} = 1 - \frac{0.5}{1.5} = 1 - \frac{1}{3} = \frac{2}{3} \]
- Excited State \(^2P_{3/2}\) (\(L=1, S=1/2, J=3/2\)): \[ g_J = 1 + \frac{\frac{15}{4} + \frac{3}{4} - 2}{2(\frac{15}{4})} = 1 + \frac{4.5 - 2}{7.5} = 1 + \frac{2.5}{7.5} = 1 + \frac{1}{3} = \frac{4}{3} \]
Step (b): Energy shift in B0 = 1.00 T
The Zeeman energy shift is \(\Delta \tilde{\nu} = \frac{g_J \mu_B B_0 M_J}{hc}\). Note that \(\frac{\mu_B}{hc} = \frac{9.27401 \times 10^{-24}}{(6.62607 \times 10^{-34})(2.99792 \times 10^{10})} = 0.46686\text{ cm}^{-1}\text{/Tesla}\).
- For \(^2S_{1/2}\) (\(g=2, M_J = \pm 1/2\)): \(\Delta \tilde{\nu} = 2(0.46686)(\pm 1/2) = \pm 0.4669\text{ cm}^{-1}\)
- For \(^2P_{1/2}\) (\(g=2/3, M_J = \pm 1/2\)): \(\Delta \tilde{\nu} = \frac{2}{3}(0.46686)(\pm 1/2) = \pm 0.1556\text{ cm}^{-1}\)
- For \(^2P_{3/2}\) (\(g=4/3, M_J = \pm 3/2, \pm 1/2\)):
- \(M_J = \pm 3/2\): \(\Delta \tilde{\nu} = \frac{4}{3}(0.46686)(\pm 3/2) = \pm 0.9337\text{ cm}^{-1}\)
- \(M_J = \pm 1/2\): \(\Delta \tilde{\nu} = \frac{4}{3}(0.46686)(\pm 1/2) = \pm 0.3112\text{ cm}^{-1}\)
Step (c): Number of Zeeman lines
- D₁ line (\(^2P_{1/2} \to {}^2S_{1/2}\)): Transitions between 2 upper and 2 lower levels with \(\Delta M_J = 0, \pm 1\). All 4 combinations are allowed: 4 Zeeman lines (2 \(\pi\)-components, 2 \(\sigma\)-components).
- D₂ line (\(^2P_{3/2} \to {}^2S_{1/2}\)): Transitions between 4 upper and 2 lower levels. 6 transitions satisfy \(\Delta M_J = 0, \pm 1\): 6 Zeeman lines (2 \(\pi\)-components, 4 \(\sigma\)-components).
The first three absorption transitions from the \(3s\) ground state of sodium to higher \(p\) states are recorded at: \(3s \to 3p\): \(\tilde{\nu} = 16960.9\text{ cm}^{-1}\) \(3s \to 4p\): \(\tilde{\nu} = 30267.0\text{ cm}^{-1}\) \(3s \to 5p\): \(\tilde{\nu} = 35042.8\text{ cm}^{-1}\) Given the ionization limit of the \(3s\) electron is \(I_P = 41449.4\text{ cm}^{-1}\) and Rydberg constant \(R = 109737.3\text{ cm}^{-1}\): (a) Determine the absolute term value \(T_n = I_P - \tilde{\nu}\) for \(3p, 4p\), and \(5p\). (b) Calculate the Rydberg quantum defect \(\delta_p\) for each state. (c) Explain why \(\delta_p\) remains nearly constant across principal quantum numbers \(n\).
Step (a): Absolute term values
\[ T(3p) = 41449.4 - 16960.9 = 24488.5\text{ cm}^{-1} \] \[ T(4p) = 41449.4 - 30267.0 = 11182.4\text{ cm}^{-1} \] \[ T(5p) = 41449.4 - 35042.8 = 6406.6\text{ cm}^{-1} \]Step (b): Quantum defect calculations
The term value is expressed as \(T_n = \frac{R}{(n - \delta_p)^2} \implies n - \delta_p = \sqrt{\frac{R}{T_n}}\):
- For \(3p\) (\(n=3\)): \[ n_{\text{eff}} = \sqrt{\frac{109737.3}{24488.5}} = \sqrt{4.48118} = 2.1169 \implies \delta_p = 3 - 2.1169 = 0.8831 \]
- For \(4p\) (\(n=4\)): \[ n_{\text{eff}} = \sqrt{\frac{109737.3}{11182.4}} = \sqrt{9.81340} = 3.1326 \implies \delta_p = 4 - 3.1326 = 0.8674 \]
- For \(5p\) (\(n=5\)): \[ n_{\text{eff}} = \sqrt{\frac{109737.3}{6406.6}} = \sqrt{17.12879} = 4.1387 \implies \delta_p = 5 - 4.1387 = 0.8613 \]
Step (c): Physical explanation of constancy
The quantum defect \(\delta_p \approx 0.87\) is determined almost entirely by the short-range penetration of the valence electron wavepacket into the inner core (\(1s^2 2s^2 2p^6\)). Since the inner core size and charge distribution do not change when the valence electron is excited to higher Rydberg orbits, the core phase shift remains essentially constant.
In neutral helium (\(1s 2s\) excited configuration): The singlet state \(2^1S_0\) has energy \(E_{\text{singlet}} = 166277\text{ cm}^{-1}\) above the \(1s^2\) ground state. The triplet state \(2^3S_1\) has energy \(E_{\text{triplet}} = 159856\text{ cm}^{-1}\) above the ground state. (a) Explain the origin of this energy difference in terms of the Coulomb integral \(J_{12}\) and exchange integral \(K_{12}\). (b) Calculate the numerical value of the exchange integral \(K_{12}\) in \(\text{cm}^{-1}\) and \(\text{kJ/mol}\). (c) Why is the triplet state lower in energy than the singlet state?
Step (a): Coulomb and exchange integral formulation
For two non-equivalent electrons in orbitals \(a = 1s\) and \(b = 2s\), the spatial wavefunctions are:
\[ \psi_{\text{space}}^{\text{singlet}} = \frac{1}{\sqrt{2}}[\phi_a(1)\phi_b(2) + \phi_b(1)\phi_a(2)] \quad (\text{symmetric}) \] \[ \psi_{\text{space}}^{\text{triplet}} = \frac{1}{\sqrt{2}}[\phi_a(1)\phi_b(2) - \phi_b(1)\phi_a(2)] \quad (\text{antisymmetric}) \]Evaluating the expectation value of the electron-electron Coulomb repulsion operator \(\hat{H}' = \frac{e^2}{4\pi\varepsilon_0 r_{12}}\):
\[ E_{\text{singlet}} = E_0 + J_{12} + K_{12} \] \[ E_{\text{triplet}} = E_0 + J_{12} - K_{12} \]where \(J_{12}\) is the direct Coulomb repulsion integral and \(K_{12}\) is the quantum exchange integral:
\[ K_{12} = \iint \phi_a^*(1)\phi_b^*(2) \frac{e^2}{4\pi\varepsilon_0 r_{12}} \phi_b(1)\phi_a(2) d\tau_1 d\tau_2 > 0 \]Step (b): Numerical calculation of K12
\[ E_{\text{singlet}} - E_{\text{triplet}} = (J_{12} + K_{12}) - (J_{12} - K_{12}) = 2 K_{12} \] \[ 2 K_{12} = 166277 - 159856 = 6421\text{ cm}^{-1} \] \[ K_{12} = \frac{6421\text{ cm}^{-1}}{2} = 3210.5\text{ cm}^{-1} \]In \(\text{kJ/mol}\):
\[ K_{12} = 3210.5 \times 0.0119627 = 38.41\text{ kJ/mol} \]Step (c): Physical origin of lower triplet energy
In the triplet state, the spatial wavefunction is antisymmetric. As \(r_1 \to r_2\), \(\psi_{\text{space}} \to 0\) (Fermi hole). The two electrons avoid each other in space, reducing electrostatic repulsion and lowering the total energy.
A flame atomic absorption spectrometer (FAAS) is calibrated at \(\lambda = 324.7\text{ nm}\) using standard solutions of copper(II). The calibration curve yields absorbance:
A \(2.500\text{ g}\) geological rock sample is dissolved in acid and diluted to \(100.0\text{ mL}\). An aliquot of this sample solution gives an absorbance reading of \(A = 0.3845\). (a) Determine the concentration of copper in the test solution in \(\text{ppm}\) (\(\mu\text{g/mL}\)). (b) Calculate the total mass of copper in the sample in milligrams. (c) Determine the copper content of the rock in weight percent (\% w/w).
Step (a): Concentration of copper in solution
\[ 0.3845 = 0.0850 \times C + 0.0020 \implies 0.0850 \times C = 0.3825 \] \[ C = \frac{0.3825}{0.0850} = 4.500\text{ ppm} = 4.500\ \mu\text{g/mL} \]Step (b): Mass of copper in sample
The total volume is \(V = 100.0\text{ mL}\):
\[ m_{\text{Cu}} = C \times V = (4.500\ \mu\text{g/mL})(100.0\text{ mL}) = 450.0\ \mu\text{g} = 0.4500\text{ mg} \]Step (c): Weight percent in rock
\[ \text{wt}\% = \frac{m_{\text{Cu}}}{m_{\text{rock}}} \times 100\% = \frac{0.4500 \times 10^{-3}\text{ g}}{2.500\text{ g}} \times 100\% = 0.0180\% = 180\text{ ppm (w/w)} \]The ground electron configuration of the Group 14 elements is \(ns^2 np^2\). The spin-orbit splitting between the ground state \(^3P_0\) and the first excited level \(^3P_1\) increases down the group: Carbon (\(2p^2\)): \(\Delta E = 16.4\text{ cm}^{-1}\) Silicon (\(3p^2\)): \(\Delta E = 77.1\text{ cm}^{-1}\) Germanium (\(4p^2\)): \(\Delta E = 557.1\text{ cm}^{-1}\) Tin (\(5p^2\)): \(\Delta E = 1691.8\text{ cm}^{-1}\) Lead (\(6p^2\)): \(\Delta E = 7819.3\text{ cm}^{-1}\) (a) Explain why spin-orbit coupling increases so dramatically down the periodic group (\(\propto Z^4\)). (b) Describe the physical difference between Russell-Saunders (\(L\text{-}S\)) coupling and \(j\text{-}j\) coupling. (c) For lead (\(6p^2\)), construct the term levels in the \(j\text{-}j\) coupling limit where individual electrons possess \(j_1\) and \(j_2\).
Step (a): Z^4 scaling of spin-orbit coupling
The spin-orbit operator expectation value depends on \(\langle \frac{1}{r}\frac{dV}{dr} \rangle\). For hydrogenic Coulomb fields, \(\langle r^{-3} \rangle \propto Z^3\), and with the nuclear charge in \(dV/dr \propto Z\), the interaction scales as \(\zeta \propto Z^4\).
For valence electrons in heavy atoms, penetration to the inner nuclear region results in an effective \(\zeta \propto Z^2 Z_{\text{eff}}^2\), causing the massive increase from \(16.4\text{ cm}^{-1}\) (Carbon, \(Z=6\)) to \(7819\text{ cm}^{-1}\) (Lead, \(Z=82\)).
Step (b): L-S vs j-j coupling physics
- \(L\text{-}S\) Coupling: Electrostatic Coulomb repulsion \(\gg\) spin-orbit coupling. Orbital angular momenta couple to form \(\vec{L}\), spins couple to form \(\vec{S}\), then \(\vec{L}\) and \(\vec{S}\) couple to form \(\vec{J}\). Multiplicity \(\Delta S = 0\) is a rigorous selection rule.
- \(j\text{-}j\) Coupling: Spin-orbit coupling \(\gg\) electrostatic Coulomb repulsion. For each individual electron, \(\vec{l}_i\) and \(\vec{s}_i\) couple strongly to form \(\vec{j}_i\). Then the individual \(\vec{j}_i\) couple weakly via residual Coulomb repulsion to form total \(\vec{J} = \sum \vec{j}_i\). Spin multiplicity \(S\) ceases to be a good quantum number, and \(\Delta S = 0\) completely breaks down.
Step (c): j-j coupling levels for 6p² in Lead
For a p-electron (\(l=1, s=1/2\)), the individual \(j\) values are \(j = 3/2\) or \(j = 1/2\).
- \((1/2, 1/2)\) configuration: Both electrons in \(j = 1/2\). Allowed total \(J\) from anti-symmetrization: \(J = 0\) (1 state). Ground state of Pb!
- \((3/2, 1/2)\) configuration: One electron in \(j = 3/2\), one in \(j = 1/2\). Allowed total \(J = |3/2 - 1/2|, \dots, 3/2 + 1/2 = 1, 2\) (total \(3 + 5 = 8\) states).
- \((3/2, 3/2)\) configuration: Both electrons in \(j = 3/2\). Allowed total \(J\) from anti-symmetrization: \(J = 0, 2\) (total \(1 + 5 = 6\) states).
Total states: \(1 + 8 + 6 = 15\). This reproduces the 15 states of \(p^2\), but organized by \((j_1, j_2)_J\) rather than \(^{2S+1}L_J\).
To eliminate severe matrix enhancement effects in industrial wastewater analysis, arsenic (\(\text{As}\)) is quantified by ICP-OES at \(\lambda = 193.696\text{ nm}\) using the method of standard additions. Equal \(20.0\text{ mL}\) aliquots of the wastewater sample are spiked with varying volumes of an arsenic standard solution (\(50.0\text{ mg/L}\)) and diluted to \(50.0\text{ mL}\): Flask 0 (Unspiked): \(0.00\text{ mL}\) standard added \(\implies\) Emission \(I_0 = 1250\text{ counts}\) Flask 1: \(0.50\text{ mL}\) standard added (\(0.50\text{ mg/L}\) added) \(\implies\) Emission \(I_1 = 2150\text{ counts}\) Flask 2: \(1.00\text{ mL}\) standard added (\(1.00\text{ mg/L}\) added) \(\implies\) Emission \(I_2 = 3050\text{ counts}\) Flask 3: \(2.00\text{ mL}\) standard added (\(2.00\text{ mg/L}\) added) \(\implies\) Emission \(I_3 = 4850\text{ counts}\) (a) Determine the linear regression slope \(m\) and intercept \(b\) of emission versus added concentration \(C_{\text{add}}\). (b) Calculate the concentration of arsenic in the diluted measurement solution. (c) Calculate the concentration of arsenic in the original undiluted wastewater sample in \(\text{mg/L}\) (\(\text{ppm}\)).
Step (a): Linear regression analysis
The standard addition data pairs \((C_{\text{add}}, I)\):
- \((0.00, 1250)\)
- \((0.50, 2150)\)
- \((1.00, 3050)\)
- \((2.00, 4850)\)
The slope is strictly:
\[ m = \frac{\Delta I}{\Delta C} = \frac{2150 - 1250}{0.50} = \frac{900}{0.50} = 1800\text{ counts}/(\text{mg/L}) \] \[ b = I_0 = 1250\text{ counts} \]Step (b): Diluted measurement concentration
At the x-intercept of the standard additions line (\(I = 0\)):
\[ 0 = m C_x + b \implies C_{\text{diluted}} = \frac{b}{m} = \frac{1250\text{ counts}}{1800\text{ counts}/(\text{mg/L})} = 0.6944\text{ mg/L} \]Step (c): Original wastewater concentration
Account for the dilution factor (\(20.0\text{ mL} \to 50.0\text{ mL}\)):
\[ \text{Dilution Factor} = \frac{50.0\text{ mL}}{20.0\text{ mL}} = 2.50 \] \[ C_{\text{original}} = C_{\text{diluted}} \times 2.50 = 0.6944\text{ mg/L} \times 2.50 = 1.736\text{ mg/L} \approx 1.74\text{ ppm} \]The arsenic concentration in the wastewater is \(1.74\text{ mg/L}\).
The prominent yellow emission of gas-phase atomic sodium consists of the doublet \(D_1\) and \(D_2\) lines originating from the electric dipole transitions:
- \(D_1\): \(^2P_{1/2} \rightarrow {}^2S_{1/2}\) at \(\lambda_1 = 589.6\text{ nm}\)
- \(D_2\): \(^2P_{3/2} \rightarrow {}^2S_{1/2}\) at \(\lambda_2 = 589.0\text{ nm}\)
When placed in a uniform external magnetic field \(B_0 = 1.20\text{ T}\) (weak field regime, \(\mu_B B_0 \ll \Delta E_{\text{spin-orbit}}\)):
- Calculate the Landé \(g\)-factor \(g_J\) for the three states: \(^2S_{1/2}\), \(^2P_{1/2}\), and \(^2P_{3/2}\).
- Determine the magnetic quantum numbers \(M_J\) and evaluate the Zeeman energy shifts \(\Delta E(J, M_J) = g_J \mu_B B_0 M_J\) in \(\mu\text{eV}\) and in \(\text{cm}^{-1}\) for all sub-levels (given Bohr magneton \(\mu_B = 5.78838 \times 10^{-5}\text{ eV/T} = 0.46686\text{ cm}^{-1}/\text{T}\)).
- Apply the electric dipole selection rules \(\Delta M_J = 0\) (\(\pi\) transitions) and \(\Delta M_J = \pm 1\) (\(\sigma\) transitions) to find:
- The total number of Zeeman spectral components for the \(D_1\) line and their shifts relative to the unperturbed line center.
- The total number of Zeeman spectral components for the \(D_2\) line and their shifts relative to the unperturbed line center.
Comprehensive Multi-Step Solution:
Step 1: Landé \(g\)-Factor Calculations
The Landé \(g\)-factor is given by:
For all states of neutral sodium, \(S = 1/2 \implies S(S+1) = 3/4\).
1. Ground state \(^2S_{1/2}\): \(L = 0, J = 1/2, J(J+1) = 3/4\):
2. Excited state \(^2P_{1/2}\): \(L = 1, L(L+1) = 2, J = 1/2, J(J+1) = 3/4\):
3. Excited state \(^2P_{3/2}\): \(L = 1, L(L+1) = 2, J = 3/2, J(J+1) = 15/4\):
Step 2: Energy Level Shifts for \(B_0 = 1.20\text{ T}\)
The unit Zeeman splitting unit is:
In energy units:
Now evaluating \(\Delta E = g_J M_J \delta_{\text{mag}}\):
- \(^2S_{1/2}\) (\(g = 2\)):
- \(M_J = +1/2\): \(\Delta E = 2(+1/2)\delta_{\text{mag}} = +1.00 \delta_{\text{mag}} = +0.5602\text{ cm}^{-1}\)
- \(M_J = -1/2\): \(\Delta E = 2(-1/2)\delta_{\text{mag}} = -1.00 \delta_{\text{mag}} = -0.5602\text{ cm}^{-1}\)
- \(^2P_{1/2}\) (\(g = 2/3\)):
- \(M_J = +1/2\): \(\Delta E = (2/3)(+1/2)\delta_{\text{mag}} = +1/3 \delta_{\text{mag}} = +0.1867\text{ cm}^{-1}\)
- \(M_J = -1/2\): \(\Delta E = (2/3)(-1/2)\delta_{\text{mag}} = -1/3 \delta_{\text{mag}} = -0.1867\text{ cm}^{-1}\)
- \(^2P_{3/2}\) (\(g = 4/3\)):
- \(M_J = +3/2\): \(\Delta E = (4/3)(+3/2)\delta_{\text{mag}} = +2.00 \delta_{\text{mag}} = +1.1205\text{ cm}^{-1}\)
- \(M_J = +1/2\): \(\Delta E = (4/3)(+1/2)\delta_{\text{mag}} = +2/3 \delta_{\text{mag}} = +0.3735\text{ cm}^{-1}\)
- \(M_J = -1/2\): \(\Delta E = (4/3)(-1/2)\delta_{\text{mag}} = -2/3 \delta_{\text{mag}} = -0.3735\text{ cm}^{-1}\)
- \(M_J = -3/2\): \(\Delta E = (4/3)(-3/2)\delta_{\text{mag}} = -2.00 \delta_{\text{mag}} = -1.1205\text{ cm}^{-1}\)
Step 3: Transition Frequencies and Patterns
Transition frequency shift:
1. \(D_1\) Line (\(^2P_{1/2} \rightarrow {}^2S_{1/2}\)): Allowed transitions (\(\Delta M_J = 0, \pm 1\)):
- \(\pi\) transitions (\(\Delta M_J = 0\)):
- \(M_J': +1/2 \rightarrow M_J'': +1/2 \implies \Delta\tilde{\nu} = (1/3 - 1)\delta_{\text{mag}} = -2/3 \delta_{\text{mag}} = -0.3735\text{ cm}^{-1}\)
- \(M_J': -1/2 \rightarrow M_J'': -1/2 \implies \Delta\tilde{\nu} = (-1/3 - (-1))\delta_{\text{mag}} = +2/3 \delta_{\text{mag}} = +0.3735\text{ cm}^{-1}\)
- \(\sigma\) transitions (\(\Delta M_J = \pm 1\)):
- \(M_J': +1/2 \rightarrow M_J'': -1/2 \implies \Delta\tilde{\nu} = (1/3 - (-1))\delta_{\text{mag}} = +4/3 \delta_{\text{mag}} = +0.7470\text{ cm}^{-1}\)
- \(M_J': -1/2 \rightarrow M_J'': +1/2 \implies \Delta\tilde{\nu} = (-1/3 - 1)\delta_{\text{mag}} = -4/3 \delta_{\text{mag}} = -0.7470\text{ cm}^{-1}\)
Total components for \(D_1\): 4 lines (quartet at \(\pm 2/3 \delta_{\text{mag}}, \pm 4/3 \delta_{\text{mag}}\)).
2. \(D_2\) Line (\(^2P_{3/2} \rightarrow {}^2S_{1/2}\)): Allowed transitions:
- \(\pi\) transitions (\(\Delta M_J = 0\)):
- \(+1/2 \rightarrow +1/2 \implies \Delta\tilde{\nu} = (2/3 - 1)\delta_{\text{mag}} = -1/3 \delta_{\text{mag}} = -0.1867\text{ cm}^{-1}\)
- \(-1/2 \rightarrow -1/2 \implies \Delta\tilde{\nu} = (-2/3 - (-1))\delta_{\text{mag}} = +1/3 \delta_{\text{mag}} = +0.1867\text{ cm}^{-1}\)
- \(\sigma\) transitions (\(\Delta M_J = \pm 1\)):
- \(+3/2 \rightarrow +1/2 \implies \Delta\tilde{\nu} = (2 - 1)\delta_{\text{mag}} = +1 \delta_{\text{mag}} = +0.5602\text{ cm}^{-1}\)
- \(+1/2 \rightarrow -1/2 \implies \Delta\tilde{\nu} = (2/3 - (-1))\delta_{\text{mag}} = +5/3 \delta_{\text{mag}} = +0.9337\text{ cm}^{-1}\)
- \(-1/2 \rightarrow +1/2 \implies \Delta\tilde{\nu} = (-2/3 - 1)\delta_{\text{mag}} = -5/3 \delta_{\text{mag}} = -0.9337\text{ cm}^{-1}\)
- \(-3/2 \rightarrow -1/2 \implies \Delta\tilde{\nu} = (-2 - (-1))\delta_{\text{mag}} = -1 \delta_{\text{mag}} = -0.5602\text{ cm}^{-1}\)
Total components for \(D_2\): 6 lines (sextet at \(\pm 1/3, \pm 1, \pm 5/3 \delta_{\text{mag}}\)).
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