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Chapter 5 • Theory & Derivations

Quantum Concepts & Detailed Atomic Structure

Rigorous quantum mechanics of the atom: 3D Schrödinger equation in spherical coordinates and hydrogen radial/angular solutions; spatial quantization and the Stern-Gerlach experiment; electron spin and spin-orbit fine structure; Normal and Anomalous Zeeman effects with Landé g-factors; Pauli exclusion principle, Hund's rules, and many-electron term symbols under L-S and j-j coupling.

§5.1 3D Schrödinger Equation for Hydrogenic Atoms: Radial & Angular Solutions

1. Three-Dimensional Schrödinger Equation in Spherical Coordinates

For an electron of mass $m_e$ moving in the spherically symmetric Coulomb potential $V(r) = - \frac{Z e^2}{4\pi\varepsilon_0 r}$ of a nucleus with charge $+Ze$, the time-independent Schrödinger equation is:

$$-\frac{\hbar^2}{2 m_e} \nabla^2 \psi(\vec{r}) + V(r) \psi(\vec{r}) = E \psi(\vec{r})$$

In spherical polar coordinates $(r, \theta, \phi)$, the Laplacian operator $\nabla^2$ decomposes into radial and angular parts:

$$\nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r} \right) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta} \right) + \frac{1}{r^2 \sin^2\theta} \frac{\partial^2}{\partial \phi^2} = \frac{1}{r^2} \frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r} \right) - \frac{\hat{L}^2}{\hbar^2 r^2}$$

where $\hat{L}^2$ is the orbital angular momentum operator.

2. Separation of Variables

We seek separable solutions of the form $\psi(r, \theta, \phi) = R(r) Y(\theta, \phi) = R(r) \Theta(\theta) \Phi(\phi)$. Substituting into the Schrödinger equation yields two uncoupled differential equations:

  1. Angular Equation (Spherical Harmonics):
    $$\hat{L}^2 Y_{lm}(\theta, \phi) = l(l+1) \hbar^2 Y_{lm}(\theta, \phi), \quad \hat{L}_z Y_{lm}(\theta, \phi) = m_l \hbar Y_{lm}(\theta, \phi)$$
    The solutions are the Spherical Harmonics $Y_{lm}(\theta, \phi) = \Theta_{lm}(\theta) \Phi_m(\phi)$, where $\Phi_m(\phi) = \frac{1}{\sqrt{2\pi}} e^{i m_l \phi}$ and $\Theta_{lm}(\theta)$ are the normalized Associated Legendre polynomials $P_l^{|m_l|}(\cos\theta)$.
  2. Radial Wave Equation:
    $$\frac{1}{r^2} \frac{d}{dr}\left( r^2 \frac{dR}{dr} \right) + \left[ \frac{2 m_e}{\hbar^2} \left( E + \frac{Z e^2}{4\pi\varepsilon_0 r} \right) - \frac{l(l+1)}{r^2} \right] R(r) = 0$$
    The term $\frac{l(l+1)\hbar^2}{2 m_e r^2}$ acts as an effective repulsive centrifugal potential barrier preventing electrons with non-zero orbital angular momentum from collapsing into $r = 0$.

3. Bound State Eigenvalues & Associated Laguerre Polynomials

Requiring the radial wavefunction $R(r)$ to remain finite at $r = 0$ and decay exponentially as $r \to \infty$ constrains the energy to discrete quantized values governed by the Principal Quantum Number $n$:

$$E_n = - \frac{m_e Z^2 e^4}{32 \pi^2 \varepsilon_0^2 \hbar^2} \frac{1}{n^2} = - \frac{13.606\,\text{eV} \times Z^2}{n^2} \quad (n = 1, 2, 3, \dots)$$

The normalized radial wavefunctions are expressed in terms of the Associated Laguerre polynomials $L_{n-l-1}^{2l+1}(\rho)$:

$$R_{nl}(r) = - \left( \frac{2 Z}{n a_0} \right)^{3/2} \sqrt{\frac{(n-l-1)!}{2n [(n+l)!]^3}} e^{-\rho/2} \rho^l L_{n+l}^{2l+1}(\rho), \quad \rho = \frac{2 Z r}{n a_0}$$

For the ground state ($n=1, l=0, m_l=0$): $\psi_{100}(r) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}$.

§5.2 The Four Quantum Numbers & Spatial Orbital Probability Densities

1. Hierarchy of the Four Quantum Numbers

Every quantum electronic state in an atom is uniquely specified by a set of four quantum numbers:

  1. Principal Quantum Number ($n = 1, 2, 3, \dots$): Dictates the primary energy shell and the mean distance of the electron from the nucleus ($\langle r \rangle \approx n^2 a_0 / Z$).
  2. Orbital Angular Momentum Quantum Number ($l = 0, 1, 2, \dots, n-1$): Determines the subshell ($s, p, d, f, g$) and the magnitude of the orbital angular momentum:
    $$|\vec{L}| = \sqrt{l(l+1)} \hbar$$
    Notice that for the ground state ($n=1$), $l=0$, so orbital angular momentum is strictly zero ($L = 0$). This directly overturned Bohr's ad-hoc assumption that $L = 1\hbar$ in the ground state!
  3. Magnetic Quantum Number ($m_l = -l, -l+1, \dots, 0, \dots, +l$): Defines the spatial orientation of the angular momentum vector relative to a chosen quantization axis ($z$-axis):
    $$L_z = m_l \hbar \quad (2l + 1 \text{ degenerate orientations})$$
  4. Spin Magnetic Quantum Number ($m_s = \pm 1/2$): Specifies the orientation of the intrinsic electron spin angular momentum along the $z$-axis:
    $$S_z = m_s \hbar = \pm \frac{1}{2} \hbar$$

2. Radial Probability Density Distribution

The probability of finding the electron in a spherical shell of radius $r$ and thickness $dr$ (integrated over all angles) is given by the Radial Probability Density $P(r)$:

$$P(r) dr = \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \, |\psi(r, \theta, \phi)|^2 r^2 dr = r^2 |R_{nl}(r)|^2 dr$$

For the ground state of hydrogen ($n=1, l=0$):

$$P(r) = r^2 \left( \frac{4}{a_0^3} e^{-2r/a_0} \right) = \frac{4 r^2}{a_0^3} e^{-2r/a_0}$$

To find the most probable radius, differentiate $P(r)$ with respect to $r$ and set to zero:

$$\frac{dP}{dr} = \frac{4}{a_0^3} \left( 2r e^{-2r/a_0} - \frac{2r^2}{a_0} e^{-2r/a_0} \right) = \frac{8 r}{a_0^3} \left( 1 - \frac{r}{a_0} \right) e^{-2r/a_0} = 0 \implies r_{mp} = a_0$$

The maximum of the radial probability density curve occurs at exactly $r = a_0 = 0.529\,\text{\AA}$—recovering Bohr's first orbit radius as the peak of a 3D probability cloud.

§5.3 Space Quantization & The Stern-Gerlach Experiment

1. Concept of Spatial Quantization

In classical physics, a magnetic dipole moment $\vec{\mu}$ can orient itself in any arbitrary continuous direction in space relative to an applied magnetic field $\vec{B}$. Quantum mechanics predicts Spatial Quantization: the magnetic dipole can only orient at discrete quantized angles $\theta$ such that its projection along the field axis is quantized:

$$\cos\theta = \frac{L_z}{|\vec{L}|} = \frac{m_l \hbar}{\sqrt{l(l+1)}\hbar} = \frac{m_l}{\sqrt{l(l+1)}}$$

2. The Stern-Gerlach Apparatus & Force Derivation (1922)

Otto Stern and Walther Gerlach tested space quantization by directing a collimated beam of neutral silver atoms ($Z = 47$) through an inhomogeneous magnetic field.

In a uniform magnetic field, a magnetic dipole experiences only a torque ($\vec{\tau} = \vec{\mu} \times \vec{B}$) aligning it, but zero net translatory force ($\vec{F} = \vec{0}$). However, in an inhomogeneous field with gradient $\frac{\partial B_z}{\partial z} \neq 0$, the potential energy is $U = - \vec{\mu} \cdot \vec{B} = - \mu_z B_z$. A net deflecting force acts on the atoms:

$$F_z = - \frac{\partial U}{\partial z} = \mu_z \frac{\partial B_z}{\partial z}$$

Classically, because atomic magnetic moments would be randomly distributed over all directions ($-1 \le \cos\theta \le +1$), the atomic beam should broaden continuously into a single uniform smeared patch on the detector plate.

3. The Experimental Result & Discovery of Electron Spin

Stern and Gerlach observed that the beam did NOT smear continuously. Instead, it split cleanly into two distinct, symmetric, separated traces!

The Theoretical Puzzle & Resolution: Silver has a single valence electron outside closed shells: $[\text{Kr}] 4d^{10} 5s^1$. For an $s$-electron, orbital angular momentum is $l = 0$. If the magnetic moment were purely orbital ($\mu_L$), then $m_l = 0$, so $\mu_z = 0$, predicting zero deflection (a single undeflected spot)! Furthermore, if orbital angular momentum $l$ were responsible, the number of split beams would be $2l + 1$, which is always an odd integer ($1, 3, 5, \dots$). The observed splitting into an even number (2) was impossible under orbital quantization alone!

In 1925, Samuel Goudsmit and George Uhlenbeck solved the enigma by proposing that the electron possesses an intrinsic angular momentum termed Electron Spin $\vec{S}$ with spin quantum number $s = 1/2$. The spin angular momentum magnitude and $z$-component are:

$$|\vec{S}| = \sqrt{s(s+1)}\hbar = \frac{\sqrt{3}}{2} \hbar, \quad S_z = m_s \hbar = \pm \frac{1}{2} \hbar \quad (m_s = +1/2, -1/2)$$

The associated spin magnetic dipole moment is:

$$\vec{\mu}_s = - g_s \left( \frac{e}{2 m_e} \right) \vec{S} = - \frac{g_s \mu_B}{\hbar} \vec{S}$$

where $g_s \approx 2.00232$ is the electron spin $g$-factor, and $\mu_B$ is the Bohr Magneton:

$$\mu_B = \frac{e \hbar}{2 m_e} = \frac{(1.602 \times 10^{-19})(1.0546 \times 10^{-34})}{2(9.109 \times 10^{-31})} \approx 9.274 \times 10^{-24}\,\text{J/T} = 5.788 \times 10^{-5}\,\text{eV/T}$$

The two split beams corresponded to the two discrete spin orientations: "spin-up" ($m_s = +1/2$) and "spin-down" ($m_s = -1/2$).

§5.4 Spin-Orbit Interaction, Thomas Precession & Fine Structure of Alkali Doublets

1. Physical Origin of Spin-Orbit Coupling

In the rest frame of the nucleus, an electron orbits with velocity $\vec{v}$ in an electrostatic field $\vec{E} = - abla V(r)$. Transforming to the instantaneous rest frame of the orbiting electron, the moving nuclear charge creates an effective magnetic field:

$$\vec{B}_{int} = - \frac{\vec{v} \times \vec{E}}{c^2} = \frac{1}{m_e c^2 r} \frac{dV}{dr} (\vec{r} \times m_e \vec{v}) = \frac{1}{m_e c^2 r} \frac{dV}{dr} \vec{L}$$

The intrinsic spin magnetic dipole $\vec{\mu}_s = - \frac{g_s e}{2m_e} \vec{S} \approx - \frac{e}{m_e} \vec{S}$ interacts with this internal magnetic field with energy $U' = - \vec{\mu}_s \cdot \vec{B}_{int}$.

Accounting for relativistic Thomas Precession (a factor of $1/2$ arising from the non-inertial accelerating frame of the rotating electron), the exact Spin-Orbit Hamiltonian $H_{so}$ is:

$$H_{so} = \frac{1}{2 m_e^2 c^2} \left( \frac{1}{r} \frac{dV}{dr} \right) \vec{L} \cdot \vec{S} = \xi(r) \vec{L} \cdot \vec{S}$$

2. Evaluation of the $\vec{L} \cdot \vec{S}$ Scalar Product

The total angular momentum is the vector sum $\vec{J} = \vec{L} + \vec{S}$. Squaring both sides:

$$\vec{J}^2 = (\vec{L} + \vec{S})^2 = \vec{L}^2 + \vec{S}^2 + 2 \vec{L} \cdot \vec{S} \implies \vec{L} \cdot \vec{S} = \frac{1}{2} (\vec{J}^2 - \vec{L}^2 - \vec{S}^2)$$

Evaluating in an eigenstate of total angular momentum $|j, l, s\rangle$:

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

For a single electron ($s = 1/2$), the allowed total angular momentum quantum numbers are $j = l + 1/2$ and $j = l - 1/2$ (for $l > 0$).

  • For $j = l + 1/2$: $\langle \vec{L}\cdot\vec{S} \rangle = +\frac{1}{2} l \hbar^2$ (Energy shifted upward)
  • For $j = l - 1/2$: $\langle \vec{L}\cdot\vec{S} \rangle = -\frac{1}{2} (l + 1) \hbar^2$ (Energy shifted downward)

The spin-orbit interaction splits every state with $l > 0$ into a fine-structure doublet separated by energy:

$$\Delta E_{so} = \frac{2l + 1}{2} \hbar^2 \langle \xi(r) \rangle$$

3. Fine Structure of the Sodium D-Lines

In sodium ($Z = 11$, single valence electron $[\text{Ne}]3s^1$), the first excited $3p$ state ($l = 1, s = 1/2$) splits into two states:

  • ${}^2P_{3/2}$ ($j = 3/2$, higher energy)
  • ${}^2P_{1/2}$ ($j = 1/2$, lower energy)

When the electron decays to the ground state $3s_{1/2}$ ($l = 0, j = 1/2$), it emits two closely spaced yellow spectral lines—the famous Sodium D-Lines:

  1. $D_2$ Line ($3^2P_{3/2} \to 3^2S_{1/2}$): Wavelength $\lambda_2 = 589.0\,\text{nm}$ (higher frequency, larger energy transition)
  2. $D_1$ Line ($3^2P_{1/2} \to 3^2S_{1/2}$): Wavelength $\lambda_1 = 589.6\,\text{nm}$

The splitting $\Delta \lambda = 0.6\,\text{nm}$ ($17.2\,\text{cm}^{-1} \approx 2.13\,\text{meV}$) provides direct macroscopic optical verification of relativistic spin-orbit coupling.

§5.5 Magnetic Field Effects: Normal & Anomalous Zeeman Effect & Paschen-Back Effect

1. The Normal Zeeman Effect (Singlet States, $S = 0$)

Pieter Zeeman (1896) discovered that when a light source is placed in an external magnetic field $\vec{B} = B \hat{z}$, its spectral lines split into closely spaced polarized components.

In atoms where total spin is zero ($S = 0$, such as singlet states in Cadmium, Zinc, or Helium), the interaction Hamiltonian is purely orbital:

$$H_B = - \vec{\mu}_L \cdot \vec{B} = \frac{e}{2 m_e} L_z B = \mu_B B m_l \quad (m_l = -l, \dots, +l)$$

The energy of each level is shifted by:

$$\Delta E = m_l \mu_B B$$

Under the electric dipole selection rule $\Delta m_l = 0, \pm 1$:

$$h\nu = (E_i + m_{li} \mu_B B) - (E_f + m_{lf} \mu_B B) = h\nu_0 + \Delta m_l \mu_B B$$

A single spectral line splits into exactly three components (the Lorentz Triplet):

  1. $\pi$-Component ($\Delta m_l = 0$): Unshifted frequency $\nu = \nu_0$, linearly polarized parallel to $\vec{B}$.
  2. $\sigma$-Components ($\Delta m_l = \pm 1$): Shifted frequencies $\nu = \nu_0 \pm \frac{\mu_B B}{h} = \nu_0 \pm \frac{e B}{4\pi m_e}$, circularly polarized perpendicular to $\vec{B}$.

2. The Anomalous Zeeman Effect & The Landé g-Factor ($S \neq 0$)

For states with non-zero spin ($S \neq 0$), because the spin $g$-factor ($g_s \approx 2$) is double the orbital $g$-factor ($g_l = 1$), the total magnetic moment $\vec{\mu} = - \frac{\mu_B}{\hbar}(\vec{L} + 2\vec{S})$ is not parallel to the total angular momentum $\vec{J} = \vec{L} + \vec{S}$.

By the Wigner-Eckart projection theorem, only the component of $\vec{\mu}$ parallel to $\vec{J}$ survives time-averaging:

$$\vec{\mu}_J = - g_J \frac{\mu_B}{\hbar} \vec{J}$$

where $g_J$ is the Landé $g$-Factor:

$$g_J = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2 J(J+1)}$$

In an external magnetic field $B$, each level splits into $2J + 1$ equidistant sublevels:

$$\Delta E = g_J \mu_B B m_j \quad (m_j = -J, -J+1, \dots, +J)$$

Because the initial and final states have different Landé $g$-factors ($g_i \neq g_f$), the transition frequencies split into complex multiplets (e.g., 4 lines for the Sodium $D_1$ line, 6 lines for the $D_2$ line). This was termed "anomalous" until spin was discovered.

3. The Paschen-Back Effect (Strong Magnetic Field Limit)

When the external magnetic field $B$ is made so powerful that the magnetic interaction energy $\mu_B B$ overwhelms the internal spin-orbit coupling energy ($\Delta E_{so}$), the coupling between $\vec{L}$ and $\vec{S}$ is broken. $\vec{L}$ and $\vec{S}$ precess independently around $\vec{B}$, and the good quantum numbers revert to $(m_l, m_s)$. The energy shift becomes:

$$\Delta E = \mu_B B (m_l + 2 m_s)$$

Applying selection rules $\Delta m_s = 0$ and $\Delta m_l = 0, \pm 1$, the complex anomalous multiplet collapses back into a simple three-line normal Lorentz triplet!

§5.6 Many-Electron Atoms: L-S vs j-j Coupling & Spectroscopic Term Symbols

1. Angular Momentum Coupling Schemes in Multi-Electron Atoms

In an atom containing multiple valence electrons, the electrostatic Coulomb repulsions between electrons compete with internal spin-orbit interactions:

  • $L-S$ (Russell-Saunders) Coupling (Light to Medium Atoms, $Z < 30$): Residual Coulomb interactions dominate over spin-orbit interactions. The individual orbital angular momenta couple strongly into a total orbital angular momentum:
    $$\vec{L} = \sum_i \vec{l}_i$$
    The individual spins couple strongly into a total spin angular momentum:
    $$\vec{S} = \sum_i \vec{s}_i$$
    Finally, $\vec{L}$ and $\vec{S}$ couple weakly via spin-orbit interaction into total angular momentum $\vec{J} = \vec{L} + \vec{S}$, where:
    $$|L - S| \le J \le L + S$$
  • $j-j$ Coupling (Heavy Atoms, $Z > 70$): In heavy atoms with large nuclear charge, spin-orbit coupling is colossal ($\propto Z^4$). For each electron individually, its spin $\vec{s}_i$ couples strongly to its own orbital momentum $\vec{l}_i$ to form $\vec{j}_i = \vec{l}_i + \vec{s}_i$. The individual $\vec{j}_i$ vectors then couple weakly to form total $\vec{J} = \sum \vec{j}_i$.

2. Spectroscopic Term Symbols

Under Russell-Saunders coupling, an atomic energy level is designated by the standard Term Symbol:

$${}^{2S+1}L_J$$

where:

  • $2S + 1$ is the Multiplicity (Singlet for $S=0$, Doublet for $S=1/2$, Triplet for $S=1$, Quartet for $S=3/2$).
  • $L$ is represented by spectroscopic capital letters: $S$ ($L=0$), $P$ ($L=1$), $D$ ($L=2$), $F$ ($L=3$), $G$ ($L=4$).
  • $J$ is the total angular momentum quantum number written as a subscript.

3. Hund's Rules for Ground-State Term Determination

To determine the ground state term of an equivalent electron configuration:

  1. Rule 1: The term with the maximum multiplicity ($2S+1$) has the lowest energy (minimizes electron-electron Coulomb repulsion by keeping spins parallel).
  2. Rule 2: For a given spin multiplicity, the term with the largest total orbital angular momentum $L$ has the lowest energy.
  3. Rule 3: For a given $L$ and $S$:
    • If the subshell is less than half-full, the level with the lowest $J = |L - S|$ has the lowest energy (regular multiplet).
    • If the subshell is more than half-full, the level with the highest $J = L + S$ has the lowest energy (inverted multiplet).

4. Electric Dipole Selection Rules

Allowed radiative transitions are governed by the electric dipole operator selection rules:

$$\Delta S = 0 \quad (\text{Intercombination transitions between different multiplicities are forbidden})$$
$$\Delta L = 0, \pm 1 \quad (\text{with } L = 0 \to L = 0 \text{ strictly forbidden})$$
$$\Delta J = 0, \pm 1 \quad (\text{with } J = 0 \to J = 0 \text{ strictly forbidden})$$
$$\Delta M_J = 0, \pm 1$$

Solved Problem Example 5.1: Spin-Orbit Internal Magnetic Field & Fine Structure Splitting

The Sodium doublet lines have measured wavelengths $\lambda_1 = 589.6\text{ nm}$ ($3^2P_{1/2} \to 3^2S_{1/2}$) and $\lambda_2 = 589.0\text{ nm}$ ($3^2P_{3/2} \to 3^2S_{1/2}$). (a) Calculate the fine structure energy splitting $\Delta E$ between the $3^2P_{3/2}$ and $3^2P_{1/2}$ states in eV. (b) Assuming this splitting is due to the interaction of the electron's spin magnetic moment with an internal magnetic field $B_{\text{int}}$, calculate the effective strength of $B_{\text{int}}$ in Tesla ($\Delta E = 2\mu_B B_{\text{int}}$).

Step 1: Calculate Energy Splitting $\Delta E_{so}$

$$\Delta E_{so} = E_2 - E_1 = h c \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) = h c \left( \frac{\lambda_1 - \lambda_2}{\lambda_1 \lambda_2} \right)$$
$$\Delta \lambda = 589.6\,\text{nm} - 589.0\,\text{nm} = 0.60\,\text{nm} = 6.0 \times 10^{-10}\,\text{m}$$
$$\lambda_1 \lambda_2 \approx (589.3 \times 10^{-9}\,\text{m})^2 \approx 3.473 \times 10^{-13}\,\text{m}^2$$
$$\Delta E_{so} = \frac{(6.626 \times 10^{-34}\,\text{J}\cdot\text{s})(2.998 \times 10^8\,\text{m/s})(6.0 \times 10^{-10}\,\text{m})}{3.473 \times 10^{-13}\,\text{m}^2} = \frac{1.192 \times 10^{-34}}{3.473 \times 10^{-13}} \approx 3.432 \times 10^{-22}\,\text{J}$$

Converting to electron-volts:

$$\Delta E_{so} = \frac{3.432 \times 10^{-22}\,\text{J}}{1.6022 \times 10^{-19}\,\text{J/eV}} \approx 2.142 \times 10^{-3}\,\text{eV} = 2.142\,\text{meV}$$

In wavenumber units:

$$\Delta \bar{\nu} = \frac{\Delta E_{so}}{h c} = \frac{3.432 \times 10^{-22}}{(6.626 \times 10^{-34})(3.0 \times 10^{10}\,\text{cm/s})} \approx 17.27\,\text{cm}^{-1}$$

Step 2: Calculate Effective Internal Magnetic Field $B_{int}$

The energy difference between spin parallel ($j = 3/2$) and spin antiparallel ($j = 1/2$) to the orbital field is:

$$\Delta E_{so} = 2 \mu_B B_{int}$$
$$B_{int} = \frac{\Delta E_{so}}{2 \mu_B} = \frac{3.432 \times 10^{-22}\,\text{J}}{2 \times (9.274 \times 10^{-24}\,\text{J/T})} = \frac{3.432 \times 10^{-22}}{1.8548 \times 10^{-23}} \approx 18.5\,\text{Tesla}$$

The internal magnetic field generated by the electron's orbital motion is colossal—approximately $18.5\,\text{Tesla}$, exceeding typical laboratory benchtop electromagnets!

Solved Problem Example 5.2: Landé g-Factor & Anomalous Zeeman Spectral Splitting

An atom in a weak external magnetic field of $B = 1.50\text{ Tesla}$ undergoes a transition between the ${}^2P_{3/2}$ state and the ${}^2S_{1/2}$ state. (a) Calculate the Landé $g$-factors $g_J$ for both states. (b) Determine the magnetic splitting $\Delta E$ between adjacent $m_j$ levels for each state. (c) Applying the electric dipole selection rules $\Delta m_j = 0, \pm 1$, determine the number of allowed anomalous Zeeman spectral lines and draw an energy-level transition diagram.

Step 1: Calculate Landé $g$-Factors

Using $g_J = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2 J(J+1)}$:

For ${}^2S_{1/2}$ ($L = 0, S = 1/2, J = 1/2$):

$$g_1 = 1 + \frac{\frac{1}{2}(\frac{3}{2}) + \frac{1}{2}(\frac{3}{2}) - 0}{2 \times \frac{1}{2}(\frac{3}{2})} = 1 + \frac{\frac{3}{4} + \frac{3}{4}}{\frac{3}{2}} = 1 + 1 = 2.00$$

For ${}^2P_{3/2}$ ($L = 1, S = 1/2, J = 3/2$):

$$g_2 = 1 + \frac{\frac{3}{2}(\frac{5}{2}) + \frac{1}{2}(\frac{3}{2}) - 1(2)}{2 \times \frac{3}{2}(\frac{5}{2})} = 1 + \frac{\frac{15}{4} + \frac{3}{4} - 2}{\frac{15}{2}} = 1 + \frac{\frac{18}{4} - 2}{\frac{15}{2}} = 1 + \frac{2.5}{7.5} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.333$$

Step 2: Energy Splitting of Sublevels

Bohr magneton energy in $B = 1.50\,\text{T}$:

$$\mu_B B = (5.788 \times 10^{-5}\,\text{eV/T}) \times 1.50\,\text{T} \approx 8.682 \times 10^{-5}\,\text{eV} = 86.82\,\mu\text{eV}$$
  • Ground State ${}^2S_{1/2}$ ($g = 2$): 2 sublevels ($m_j = \pm 1/2$):
    $$\Delta E = g \mu_B B m_j = 2 \times (86.82\,\mu\text{eV}) \times (\pm 1/2) = \pm 86.82\,\mu\text{eV}$$
  • Excited State ${}^2P_{3/2}$ ($g = 4/3$): 4 sublevels ($m_j = +3/2, +1/2, -1/2, -3/2$):
    $$\Delta E = \frac{4}{3} (86.82\,\mu\text{eV}) m_j = (115.76\,\mu\text{eV}) m_j = \pm 173.64\,\mu\text{eV}, \, \pm 57.88\,\mu\text{eV}$$

Step 3: Allowed Optical Transitions

Applying the selection rule $\Delta m_j = m_{j2} - m_{j1} = 0, \pm 1$ yields exactly 6 allowed spectral lines:

  • $\Delta m_j = +1$ (2 lines: $+3/2 \to +1/2$, $+1/2 \to -1/2$) — $\sigma$-polarized
  • $\Delta m_j = 0$ (2 lines: $+1/2 \to +1/2$, $-1/2 \to -1/2$) — $\pi$-polarized
  • $\Delta m_j = -1$ (2 lines: $-1/2 \to +1/2$, $-3/2 \to -1/2$) — $\sigma$-polarized

The single original spectral line splits into a symmetric 6-line anomalous Zeeman sextet.

Solved Problem Example 5.3: Derivation of Allowed Term Symbols for Equivalent p^2 Configuration

For an atom with two equivalent p-electrons ($p^2$ configuration, such as Carbon in its ground state): (a) Determine all possible microstates $(m_{l1}, m_{s1}, m_{l2}, m_{s2})$ satisfying the Pauli exclusion principle. (b) Group these microstates into allowed spectroscopic terms ${}^{2S+1}L_J$. (c) Apply Hund's rules to identify the ground-state term symbol.

Step 1: Total Number of Microstates

For a $p$-subshell ($l = 1$), there are $2(2l+1) = 6$ spin-orbital states. The number of ways to place 2 indistinguishable electrons into 6 states is:

$$N = \binom{6}{2} = \frac{6 \times 5}{2} = 15\,\text{microstates}$$

Step 2: Table of Microstates Classified by $(M_L, M_S)$

Each microstate is $(m_{l1}, m_{s1}; m_{l2}, m_{s2})$ with $M_L = m_{l1} + m_{l2}$ and $M_S = m_{s1} + m_{s2}$, subject to the Pauli exclusion principle: no two electrons can have identical $(m_l, m_s)$.

  • Maximum $M_L = 2$: Requires $m_{l1} = 1, m_{l2} = 1$. Since $m_l$ values are identical, spins must be opposite: $m_{s1} = +1/2, m_{s2} = -1/2 \implies M_S = 0$. This represents 1 state with $(M_L = 2, M_S = 0)$. It must belong to a term with $L = 2, S = 0$, which is ${}^1D$ (${}^1D_2$, containing $2L+1 = 5$ states: $M_L = 2, 1, 0, -1, -2$ all with $M_S = 0$).
  • Maximum $M_S = 1$: Spins must be parallel ($+1/2, +1/2$). The electrons must have different $m_l$. Possible pairs: $(1, 0) \implies M_L = 1$; $(1, -1) \implies M_L = 0$; $(0, -1) \implies M_L = -1$. This gives $M_L = 1, 0, -1$ with $M_S = 1$, which belongs to a term with $L = 1, S = 1$, which is ${}^3P$ ($9$ states: $J = 2, 1, 0$).
  • Remaining State at $(M_L = 0, M_S = 0)$: After subtracting $5$ states for ${}^1D$ and $9$ states for ${}^3P$ from the $15$ total states, exactly $1$ state remains at $(M_L = 0, M_S = 0)$. This belongs to ${}^1S$ (${}^1S_0$, $1$ state).

The allowed terms for a $p^2$ configuration are: ${}^1D_2$, ${}^3P_0, {}^3P_1, {}^3P_2$, and ${}^1S_0$ (Total states = $5 + 9 + 1 = 15$).

Step 3: Determine Ground State using Hund's Rules

  1. Hund's Rule 1 (Maximum Spin Multiplicity): The triplet state ${}^3P$ ($S = 1$, multiplicity 3) has lower energy than the singlets ${}^1D$ and ${}^1S$ ($S = 0$, multiplicity 1).
  2. Hund's Rule 2: There is only one triplet term (${}^3P$), so $L = 1$ is fixed.
  3. Hund's Rule 3 (Subshell Less than Half Full): A $p^2$ subshell has 2 electrons out of 6 possible, so it is less than half full. Therefore, the level with the minimum $J$ value has the lowest energy:
    $$J = |L - S| = |1 - 1| = 0$$

Thus, the ground state term symbol of Carbon is ${}^3P_0$.

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