Spectroscopic Notation, L–S Coupling, j–j Coupling and Hyperfine Structure
Orbital and spin magnetic moments, the Bohr magneton, and the \(g\) factors.
Larmor precession and space quantization.
The vocabulary of spectroscopy: state, level, sublevel, term, configuration, statistical weight.
The atomic Hamiltonian, and which of its terms decides whether an atom follows L–S or j–j coupling.
Building term symbols \(^{2S+1}L_J\) for non-equivalent and equivalent electrons, the latter via Breit's scheme.
Hund's rules, the Landé interval rule, and the selection rules.
j–j coupling, and how its level count agrees with L–S.
3.1 Orbital Magnetic Dipole Moment: the Bohr Magneton
An electron in orbit is a current loop. If the orbital period is \(T\), the orbital current is \begin{equation} i = -\frac{e}{T} = -\frac{e v}{2\pi r} , \label{eq:orbital-current} \end{equation} and if \(A = \pi r^{2}\) is the area enclosed, the orbital magnetic moment is \begin{equation} \mu_{\ell} = i A = -\frac{e v r}{2} . \label{eq:mu-orbital} \end{equation} The orbital angular momentum is \(L = m v r\), so \begin{equation} \boxed{\;\frac{\mu_{\ell}}{L} = -\frac{e}{2m}\;} \label{eq:gyromag-ratio} \end{equation}
The ratio of magnetic moment to angular momentum is independent of the parameters of the orbit — \(v\) and \(r\) have cancelled. The quantity \(e/2m\) is the gyromagnetic ratio of the electron, exactly half the charge-to-mass ratio \(e/m = 1.76\times 10^{11}\,\mathrm{C}\,\mathrm{kg}^{-1}\).
The electron is negatively charged, so its magnetic dipole moment points opposite to its angular momentum. Every sign in this chapter follows from that one fact.
Written as a vector relation, \begin{equation} \vec\mu_{\ell} = -\,g_{\ell}\,\frac{\mu_{\mathrm B}}{\hbar}\,\vec L , \qquad g_{\ell} = 1 , \label{eq:mu-vec-orbital} \end{equation} where the Bohr magneton is \begin{equation} \mu_{\mathrm B} = \frac{e\hbar}{2m} = 9.274\times 10^{-24}\,\mathrm{J}\,\mathrm{T}^{-1} = 5.788\times 10^{-5}\,\mathrm{eV}\,\mathrm{T}^{-1} . \label{eq:bohr-magneton} \end{equation} The orbital \(g\) factor \(g_{\ell}=1\) is introduced only to preserve symmetry with the spin and Landé \(g\) factors that follow; it is not equal to \(g_{s}\).
From quantum mechanics \(\lvert\vec L\rvert = \sqrt{\ell(\ell+1)}\,\hbar\), so the magnitude of the orbital moment is \begin{equation} \lvert\vec\mu_{\ell}\rvert = g_{\ell}\,\mu_{\mathrm B}\sqrt{\ell(\ell+1)} , \qquad \mu_{\ell,z} = -g_{\ell}\,\mu_{\mathrm B}\, m_{\ell} . \label{eq:mu-magnitude} \end{equation}
3.2 Larmor Precession
A magnetic dipole in an external field \(\vec B\) experiences a torque \begin{equation} \vec\tau = \vec\mu \times \vec B , \label{eq:torque} \end{equation} and since \(\vec\tau = \dd\vec L/\dd t\), combining with \(\eqref{eq:mu-vec-orbital}\) gives \[ \frac{\dd \vec L}{\dd t} = -\frac{e}{2m}\,\vec L \times \vec B . \] This is the equation of a vector precessing about \(\vec B\) at constant angle, with angular frequency \begin{equation} \boxed{\;\omega_{L} = \frac{eB}{2m} = \frac{g_{\ell}\mu_{\mathrm B}B}{\hbar}\;} \qquad \nu_{L} = \frac{eB}{4\pi m} . \label{eq:larmor} \end{equation} This is the Larmor precession frequency. Note that the magnitude of \(\vec L\) and its angle to \(\vec B\) are unchanged — only the azimuth advances.
For spin, \(g_{s}=2\) doubles this: \(\omega_{s} = eB/m\). The factor-of-two difference between orbital and spin precession is the origin of the anomalous Zeeman effect in Chapter 4.
3.3 Space Quantization
Since \(L_{z} = m_{\ell}\hbar\) while \(\lvert\vec L\rvert = \sqrt{\ell(\ell+1)}\,\hbar\), the angular momentum vector can only make certain discrete angles with the field direction: \begin{equation} \cos\theta = \frac{L_{z}}{\lvert\vec L\rvert} = \frac{m_{\ell}}{\sqrt{\ell(\ell+1)}} , \qquad m_{\ell} = -\ell,\dots,+\ell . \label{eq:space-quant} \end{equation}
The minimum angle for given \(\ell\) is at \(m_{\ell} = \ell\): \[ \cos\theta_{\min} = \frac{\ell}{\sqrt{\ell(\ell+1)}} = \sqrt{\frac{\ell}{\ell+1}} . \] For a \(2p\) electron (\(\ell=1\)) this gives \(\cos\theta = 1/\sqrt2\), so \(\theta_{\min} = 45^{\circ}\) — asked verbatim in GATE 2023.
3.4 Electron Spin and its Magnetic Moment
The electron behaves as though it were a charged sphere spinning about its own axis, carrying an intrinsic angular momentum and hence an intrinsic magnetic dipole moment. In terms of the Pauli matrices, \begin{equation} \vec S = \frac{\hbar}{2}\,\vec\sigma , \qquad \lvert\vec S\rvert = \sqrt{s(s+1)}\,\hbar = \frac{\sqrt3}{2}\hbar , \qquad S_{z} = m_{s}\hbar = \pm\frac{\hbar}{2} , \label{eq:spin-operators} \end{equation} with \(s = \tfrac12\) and \(m_{s} = \pm\tfrac12\); the kets \(\lvert s, m_{s}\rangle\) form a complete two-dimensional basis.
Space quantization for spin follows \(\eqref{eq:space-quant}\) with \(\ell \to s\): \begin{equation} \cos\theta = \frac{m_{s}}{\sqrt{s(s+1)}} = \pm\frac{1}{\sqrt3} \quad\Longrightarrow\quad \theta = 54.7^{\circ} \ \text{or}\ 125.3^{\circ} . \label{eq:spin-angles} \end{equation}
The associated magnetic moment is \begin{equation} \vec\mu_{s} = -\,g_{s}\,\frac{\mu_{\mathrm B}}{\hbar}\,\vec S , \qquad g_{s} = 2.0023 \approx 2 , \label{eq:mu-spin3} \end{equation} so that \begin{equation} \lvert\vec\mu_{s}\rvert = g_{s}\mu_{\mathrm B}\sqrt{s(s+1)} = \sqrt3\,\mu_{\mathrm B} , \qquad \mu_{s,z} = -g_{s}\mu_{\mathrm B}m_{s} = \mp\mu_{\mathrm B} . \label{eq:mu-spin-z} \end{equation}
The \(z\) component of the spin moment is one whole Bohr magneton, even though \(m_{s} = \tfrac12\) — precisely because \(g_{s}=2\). This is why the Bohr magneton is the natural unit for atomic magnetism.
3.5 Terminology of Spectroscopy
These definitions are worth learning exactly; questions often turn on the distinction between a level and a term.
- State
The condition of motion of all the electrons, specified by listing the four quantum numbers \((n,\ell,m_{\ell},m_{s})\) for each electron. States of equal energy are degenerate; the lowest is the ground state.
- Energy level
The collection of states having the same energy in the absence of external fields. A level is characterized by the total angular momentum quantum number \(J\).
- Sublevel
In an external electric or magnetic field each level splits into sublevels, characterized by \(m_{J}\).
- Term
A collection of levels sharing the same total orbital angular momentum \(L\) and multiplicity \((2S+1)\) constitutes a spectroscopic term, written \(^{2S+1}L\).
- Configuration
The specification of \(n\) and \(\ell\) for every electron; e.g.\ hydrogen is \(1s^{1}\) and carbon is \(1s^{2}2s^{2}2p^{2}\).
- Equivalent orbitals
Orbitals with the same \(n\) and the same \(\ell\).
- Statistical weight
The number of distinct states in a specified collection: \((2J+1)\) for a level, \((2S+1)(2L+1)\) for a term, and \(2(2\ell+1)\) for a single-electron subshell.
Nomenclature of spectral lines
- Line
A transition between two levels.
- Component
A transition between two sublevels.
- Multiplet
The collection of transitions between two terms.
- Resonance line
Among the lines arising from transitions between the ground level and higher levels, the one of lowest frequency.
3.6 The Atomic Hamiltonian: which Coupling Scheme?
For an atom with \(N\) optical electrons the complete Hamiltonian contains:
the kinetic energy of the electrons, \(\displaystyle \sum_{i=1}^{N}\frac{p_{i}^{2}}{2m}\);
the electrostatic interaction of the electrons with the nucleus, \(\displaystyle -\sum_{i=1}^{N}\frac{1}{4\pi\varepsilon_0}\frac{Ze^{2}}{r_{i}}\);
the mutual electrostatic repulsion of the electrons, \(\displaystyle \sum_{i>j}\frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r_{ij}}\), also called the residual electrostatic interaction;
the spin–spin correlation energy;
the orbit–orbit correlation energy;
the spin–orbit magnetic interaction energy.
Terms (iv) and (v) deserve a word, because their names are misleading. Neither is a magnetic interaction between spins or between orbits. Both are electrostatic in origin: they are the parts of the Coulomb repulsion (iii) that happen to depend on how the spins or the orbits are oriented relative to one another, because that orientation controls how close the electrons get. They are the microscopic content of Hund's first two rules.
3.6.1 (iv) Spin–spin correlation: why the largest \(S\) lies lowest
Take two electrons. The total wavefunction must be antisymmetric under exchange, and it factorizes into a spatial part and a spin part: \[ \Psi = \psi_{\text{space}} \times \chi_{\text{spin}} . \] The triplet spin states (\(S=1\)) are symmetric under exchange, so they must be paired with an antisymmetric spatial function; the singlet (\(S=0\)) is antisymmetric and pairs with a symmetric spatial function: \begin{equation} S = 1:\ \ \psi_{-} = \tfrac{1}{\sqrt2}\big[\phi_a(1)\phi_b(2) - \phi_b(1)\phi_a(2)\big], \qquad S = 0:\ \ \psi_{+} = \tfrac{1}{\sqrt2}\big[\phi_a(1)\phi_b(2) + \phi_b(1)\phi_a(2)\big]. \label{eq:exchange-space} \end{equation} Now look at what happens as the two electrons approach each other. Setting \(\vec r_1 = \vec r_2\) makes the two products identical, so \[ \psi_{-} \to 0, \qquad\text{while}\qquad \psi_{+} \to \sqrt2\,\phi_a\phi_b \ne 0 . \] The antisymmetric spatial function has a node wherever the electrons coincide. That node — the Fermi hole — keeps the electrons apart, and since the Coulomb repulsion \(e^{2}/4\pi\varepsilon_{0}r_{12}\) is large exactly where \(r_{12}\) is small, keeping them apart lowers the energy.
The source sketches for this figure appear to place the node-bearing curve against \(S=0\) in one of the four panels. That cannot be right: the node belongs to the antisymmetric spatial function, which by the Pauli principle is the partner of the symmetric (\(S=1\)) spin function. The figure above is drawn the correct way round. The test to apply is always the same: symmetric spin \(\Rightarrow\) antisymmetric space \(\Rightarrow\) node \(\Rightarrow\) lower energy.
3.6.2 (v) Orbit–orbit correlation: why the largest \(L\) lies lowest
The same argument runs again, with the orbital motion in place of the spin. Two electrons in the same shell can circulate about the nucleus either in the same sense or in opposite senses. When the individual \(\vec\ell_i\) are parallel, the resultant \(L\) is large and the electrons orbit together, maintaining a large angular separation and rarely coming close. When the \(\vec\ell_i\) are antiparallel, \(L\) is small and the electrons run in opposite directions, meeting head-on twice per revolution.
3.6.3 (vi) Spin–orbit interaction: normal and inverted multiplets
Terms (iv) and (v) fix \(S\) and \(L\); the spin–orbit term (vi) then splits the term into its \(J\) levels. From Chapter 2 the shift is \begin{equation} E_{\text{so}} = \frac{A}{2}\big[J(J+1) - L(L+1) - S(S+1)\big], \label{eq:so-shift-A} \end{equation} so the sign of the interval constant \(A\) decides the ordering, and nothing else:
\(A > 0\): energy increases with \(J\), so the smallest \(J\) lies lowest. The multiplet is normal (or regular).
\(A < 0\): energy decreases with \(J\), so the largest \(J\) lies lowest. The multiplet is inverted.
Which one occurs is settled by how full the subshell is: \begin{equation} \boxed{\; \text{less than half filled} \Rightarrow \text{normal}; \qquad \text{more than half filled} \Rightarrow \text{inverted}.\;} \label{eq:normal-inverted} \end{equation} The physical reason is that a subshell more than half full is best described by its holes rather than its electrons, and a hole carries the opposite sign of charge, hence the opposite sign of \(A\). A subshell exactly half filled has \(L = 0\), so there is no multiplet structure to order — only very slight splitting remains.
Carbon \(2p^{2}\) (two electrons in a six-slot subshell, less than half) has ground term \(^{3}P\) with \(^{3}P_{0}\) lowest — normal. Oxygen \(2p^{4}\) (more than half) also has ground term \(^{3}P\), but now \(^{3}P_{2}\) lies lowest — inverted. Same term symbol, opposite ordering: the examiner's favourite trap in this topic. Count the electrons in the subshell first.
The competition between (iii)–(v) and (vi) decides everything:
Light atoms, \(Z < 30\): terms (iii)–(v) dominate, (vi) is small \(\Rightarrow\) L–S coupling (Russell–Saunders).
Heavy atoms, \(Z > 30\): term (vi) dominates \(\Rightarrow\) j–j coupling.
The physical reason: spin–orbit coupling scales as \(Z^{4}\) (Chapter 2), far faster than the electrostatic terms.
3.7 L–S Coupling: Building the Term Symbol
In L–S coupling the individual orbital momenta couple together, the individual spins couple together, and only then do the resultants couple to each other: \begin{equation} \vec L = \sum_i \vec\ell_i , \qquad \vec S = \sum_i \vec s_i , \qquad \vec J = \vec L + \vec S . \label{eq:ls-scheme} \end{equation} The result is written as a term symbol \begin{equation} ^{2S+1}L_{J} , \label{eq:term-symbol} \end{equation} with \(L = 0,1,2,3,4,5,\dots\) written as \(S, P, D, F, G, H,\dots\) and a superscript “\(\mathrm o\)” added for odd parity.
3.7.1 Step 1 — Spin–spin correlation gives \(S\)
The individual spins couple first: \begin{align*} \text{one electron:}\quad & S = \tfrac12, \quad 2S+1 = 2 \ \text{(doublet)};\\ \text{two electrons:}\quad & S = \lvert s_1-s_2\rvert,\dots,(s_1+s_2) = 0,1, \quad \text{(singlet and triplet)};\\ \text{three electrons:}\quad & S = \tfrac12,\ \tfrac12,\ \tfrac32, \quad \text{(two doublets and a quartet)}. \end{align*} The highest \(S\) lies lowest in energy.
The source labelled the three-electron case “[Doublet & Triplet]”. A triplet requires \(S=1\), which is impossible for three spin-\(\tfrac12\) particles. The correct multiplicities are \(2S+1 = 2, 2, 4\): two doublets and a quartet.
3.7.2 Step 2 — Residual electrostatic interaction gives \(L\)
The individual \(\ell\)'s then couple: \begin{equation} L = \lvert \ell_1-\ell_2\rvert,\ \lvert \ell_1-\ell_2\rvert+1,\ \dots,\ (\ell_1+\ell_2) . \label{eq:L-coupling} \end{equation} For example \(3p\,3d\) has \(\ell_1=1\), \(\ell_2=2\), so \(L = 1,2,3\) (\(P\), \(D\), \(F\) states); \(2p\,3p\,4d\) has \(\ell_1=\ell_2=1\), \(\ell_3=2\), giving \(L = 0,1,2,3,4\).
Of terms with the same multiplicity, the one with the largest \(L\) lies lowest (Hund's second rule).
3.7.3 Step 3 — Spin–orbit interaction gives \(J\)
Finally \(\vec L\) and \(\vec S\) couple: \begin{equation} J = \lvert L-S\rvert,\ \lvert L-S\rvert+1,\ \dots,\ (L+S), \qquad \lvert \vec J\rvert = \sqrt{J(J+1)}\,\hbar . \label{eq:J-coupling} \end{equation} The number of \(J\) values in a term is \begin{equation} \text{number of levels} = \begin{cases} 2S+1, & L \ge S,\\[2pt] 2L+1, & L < S. \end{cases} \label{eq:num-levels} \end{equation}
The source gave the condition “\(L
3.8 Terms for Non-Equivalent Electrons
Electrons are non-equivalent if they differ in \(n\), or in \(\ell\), or in both. Then no Pauli restriction operates and every combination of \(S\), \(L\) and \(J\) from \(\eqref{eq:ls-scheme}\) is allowed.
Find all spectroscopic terms for the configuration \(4p\,4d\).
Solution. \(s_1 = s_2 = \tfrac12 \Rightarrow S = 0, 1\); \(\ell_1 = 1,\ \ell_2 = 2 \Rightarrow L = 1, 2, 3\).
Parity is \((-1)^{\ell_1+\ell_2} = (-1)^{3}\), i.e.\ odd, so every term carries a superscript \(\mathrm o\).
Singlet terms (\(S=0\), so \(J = L\)): \[ ^{1}P^{\mathrm o}_{1}, \qquad ^{1}D^{\mathrm o}_{2}, \qquad ^{1}F^{\mathrm o}_{3}. \] Triplet terms (\(S=1\), so \(J = L-1, L, L+1\)): \[ L=1:\ ^{3}P^{\mathrm o}_{0,1,2}, \qquad L=2:\ ^{3}D^{\mathrm o}_{1,2,3}, \qquad L=3:\ ^{3}F^{\mathrm o}_{2,3,4}. \] That is \(6\) terms and \(12\) levels in total.
Check the level count with statistical weights: the total number of states must be \(\left[2(2\ell_1+1)\right]\times\left[2(2\ell_2+1)\right]\) divided appropriately — or more simply \(\sum_{\text{levels}}(2J+1) = (2s_1+1)(2\ell_1+1)(2s_2+1)(2\ell_2+1) = 2\cdot3\cdot2\cdot5 = 60\). Verify: singlets give \(3+5+7=15\); triplets give \((1{+}3{+}5)+(3{+}5{+}7)+(5{+}7{+}9) = 9+15+21 = 45\); total \(60\).
The way the six terms and twelve levels emerge, one interaction at a time, is worth seeing as a picture: the unperturbed \(4p\,4d\) level is split first by the spin–spin correlation energy into a singlet and a triplet system, then by the residual electrostatic interaction into \(P\), \(D\) and \(F\) terms, and finally by the spin–orbit interaction into the individual \(J\) levels.
| Configuration | Terms |
|---|---|
| \(ss\) | \(^{1}S_{0},\ ^{3}S_{1}\) |
| \(sp\) | \(^{1}P_{1},\ ^{3}P_{0,1,2}\) |
| \(sd\) | \(^{1}D_{2},\ ^{3}D_{1,2,3}\) |
| \(pp\) | \(^{1}S_{0},\ ^{1}P_{1},\ ^{1}D_{2},\ ^{3}S_{1},\ ^{3}P_{0,1,2},\ ^{3}D_{1,2,3}\) |
| \(pd\) | \(^{1}P_{1},\ ^{1}D_{2},\ ^{1}F_{3},\ ^{3}P_{0,1,2},\ ^{3}D_{1,2,3},\ ^{3}F_{2,3,4}\) |
| \(dd\) | \(^{1}S_{0},\ ^{1}P_{1},\ ^{1}D_{2},\ ^{1}F_{3},\ ^{1}G_{4},\ ^{3}S_{1},\ ^{3}P_{0,1,2},\ ^{3}D_{1,2,3},\ ^{3}F_{2,3,4},\ ^{3}G_{3,4,5}\) |
The \(sp\) row was printed as \(^{3}P_{1,2,3}\). With \(S=1\) and \(L=1\), \(J = \lvert L-S\rvert \dots (L+S) = 0,1,2\), so it must be \(^{3}P_{0,1,2}\).
The \(dd\) row omitted \(^{1}F_{3}\). Since \(L\) runs \(0,1,2,3,4\) for \(\ell_1=\ell_2=2\), the singlets are \(^{1}S,\,^{1}P,\,^{1}D,\,^{1}F,\,^{1}G\) — five of them, not four.
3.9 The Landé Interval Rule
Within a term, spin–orbit coupling is described by \begin{equation} H_{\text{so}} = A\,\vec L\cdot\vec S , \label{eq:H-so-A} \end{equation} where \(A\) is the interaction constant (it depends on the radial wavefunction and on the multiplicity in a complicated way, so it is treated as a parameter with dimensions of energy divided by \(\hbar^{2}\)).
Using the identity of Chapter 2, \begin{equation} \big\langle \vec L\cdot\vec S\big\rangle = \frac{\hbar^{2}}{2}\left[J(J+1) - L(L+1) - S(S+1)\right], \label{eq:LS-expectation} \end{equation} so within a term (\(L\), \(S\) fixed) the energy of the level \(J\) is \begin{equation} E_{J} = \frac{A\hbar^{2}}{2}\left[J(J+1) - L(L+1) - S(S+1)\right]. \label{eq:E-J} \end{equation} Taking the difference between adjacent levels: \begin{equation} \boxed{\; E_{J} - E_{J-1} = \frac{A\hbar^{2}}{2}\left[J(J+1) - (J-1)J\right] = A\hbar^{2}\,J \;} \label{eq:lande-interval} \end{equation}
The spacing between two consecutive levels of a multiplet is proportional to the larger of the two \(J\) values. This is the Landé interval rule.
The classic application: given the observed spacings of a multiplet, identify \(L\) and \(S\). If the spacings are in the ratio \(J_{1}:J_{2}:\dots\), read off the \(J\) values directly. E.g.\ spacings in the ratio \(2:3\) imply levels \(J = 1,2,3\), hence a triplet with \(L=2\), i.e.\ a \(^{3}D\) term.
Find the separations within a \(^{3}P\) term due to spin–orbit interaction \(H = A\,\vec L\cdot\vec S\).
Solution. Here \(L=1\), \(S=1\), so \(J = 0,1,2\). From \(\eqref{eq:E-J}\) with \(L(L+1)+S(S+1) = 2+2 = 4\): \[ E_{0} = \frac{A\hbar^{2}}{2}(0-4) = -2A\hbar^{2}, \quad E_{1} = \frac{A\hbar^{2}}{2}(2-4) = -A\hbar^{2}, \quad E_{2} = \frac{A\hbar^{2}}{2}(6-4) = +A\hbar^{2}. \] So \[ E_{1}-E_{0} = A\hbar^{2} \ (= A\hbar^{2}\cdot 1), \qquad E_{2}-E_{1} = 2A\hbar^{2} \ (= A\hbar^{2}\cdot 2), \] a ratio \(1:2\) exactly as \(\eqref{eq:lande-interval}\) requires.
3.10 Terms for Equivalent Electrons: Breit's Scheme
Electrons are equivalent if they share the same \(n\) and the same \(\ell\) — for example the two \(2p\) electrons of carbon, \(1s^{2}2s^{2}2p^{2}\). Now the Pauli principle forbids many of the combinations allowed in 3.8, and the term list is shorter.
Three shortcuts come first:
A closed shell (\(s^{2}\), \(p^{6}\), \(d^{10}\), \(f^{14}\)) always gives a single term \(^{1}S_{0}\). Closed shells can be ignored entirely when finding terms.
Terms of \(p^{5}\) \(=\) terms of \(p^{1}\); terms of \(p^{4}\) \(=\) terms of \(p^{2}\); terms of \(d^{8}\) \(=\) terms of \(d^{2}\). In general a shell with \(x\) holes gives the same terms as one with \(x\) electrons.
Half-filled shells (\(p^{3}\), \(d^{5}\), \(f^{7}\)) have \(L=0\) and maximum \(S\), giving an \(S\) ground term.
For the remaining cases, use Breit's scheme.
3.10.1 Breit's scheme for \(p^{2}\)
Step 1. Tabulate all \(M_{L} = m_{\ell_1} + m_{\ell_2}\) for \(m_{\ell} = 1, 0, -1\):
| \(m_{\ell_1}\backslash m_{\ell_2}\) | \(1\) | \(0\) | \(-1\) |
|---|---|---|---|
| \(1\) | \(2\) | \(1\) | \(0\) |
| \(0\) | \(1\) | \(0\) | \(-1\) |
| \(-1\) | \(0\) | \(-1\) | \(-2\) |
Step 2. Divide the nine values into nested L-shaped sets, each set being a complete run \(M_{L} = -L,\dots,+L\): \[ \text{Set 1: } 2,1,0,-1,-2 \ (L=2); \qquad \text{Set 2: } 1,0,-1 \ (L=1); \qquad \text{Set 3: } 0 \ (L=0). \]
Step 3. Assign multiplicities. The Pauli principle requires that the symmetric spin state (triplet, \(S=1\)) pair with the antisymmetric orbital sets and vice versa. The result is \[ S=1 \ \text{(triplet)}: \ \text{Set 2} \Rightarrow {}^{3}P; \qquad S=0 \ \text{(singlet)}: \ \text{Sets 1 and 3} \Rightarrow {}^{1}D,\ {}^{1}S . \]
So \begin{equation} p^{2} \longrightarrow\ ^{1}S_{0},\ ^{3}P_{0,1,2},\ ^{1}D_{2} . \label{eq:p2-terms} \end{equation}
Check the count: \(\sum(2J+1) = 1 + (1{+}3{+}5) + 5 = 15\), and indeed \(\binom{6}{2} = 15\) ways of placing two electrons in six \(2p\) spin-orbitals. This check catches almost every error in equivalent-electron problems.
3.10.2 Breit's scheme for \((nd)^{2}\)
The same machinery, one step harder. Two equivalent \(d\) electrons have \(\ell_1 = \ell_2 = 2\), so \(m_{\ell} = 2,1,0,-1,-2\) and there are \(25\) entries in the \(M_{L}\) table.
Step 1. Tabulate \(M_{L} = m_{\ell_1} + m_{\ell_2}\):
| \(m_{\ell_1}\backslash m_{\ell_2}\) | \(2\) | \(1\) | \(0\) | \(-1\) | \(-2\) |
|---|---|---|---|---|---|
| \(2\) | \(4\) | \(3\) | \(2\) | \(1\) | \(0\) |
| \(1\) | \(3\) | \(2\) | \(1\) | \(0\) | \(-1\) |
| \(0\) | \(2\) | \(1\) | \(0\) | \(-1\) | \(-2\) |
| \(-1\) | \(1\) | \(0\) | \(-1\) | \(-2\) | \(-3\) |
| \(-2\) | \(0\) | \(-1\) | \(-2\) | \(-3\) | \(-4\) |
Step 2. Strip off nested L-shaped sets, each a complete run \(M_{L} = -L,\dots,+L\): \[ \begin{aligned} \text{Set 1: } & 4,3,2,1,0,-1,-2,-3,-4 && (L=4,\ G)\\ \text{Set 2: } & 3,2,1,0,-1,-2,-3 && (L=3,\ F)\\ \text{Set 3: } & 2,1,0,-1,-2 && (L=2,\ D)\\ \text{Set 4: } & 1,0,-1 && (L=1,\ P)\\ \text{Set 5: } & 0 && (L=0,\ S) \end{aligned} \]
Step 3. Assign multiplicities. As for \(p^{2}\), the Pauli principle pairs the symmetric orbital sets with the singlet spin state and the antisymmetric ones with the triplet. Working outwards from the largest \(L\), the sets alternate: \(G\) singlet, \(F\) triplet, \(D\) singlet, \(P\) triplet, \(S\) singlet. Hence \begin{equation} (nd)^{2} \longrightarrow\ ^{1}S_{0},\ \ ^{3}P_{0,1,2},\ \ ^{1}D_{2},\ \ ^{3}F_{2,3,4},\ \ ^{1}G_{4} . \label{eq:d2-terms} \end{equation}
Step 4: which lies lowest, and which highest. Apply Hund's rules in order.
Largest \(S\) first. The triplets \(^{3}P\) and \(^{3}F\) beat all the singlets.
Then largest \(L\). Of the two triplets, \(^{3}F\) (\(L=3\)) beats \(^{3}P\) (\(L=1\)).
Then \(J\). A \(d\) subshell holds ten electrons, so \(d^{2}\) is less than half filled and the multiplet is normal (3.6.3): the smallest \(J\) lies lowest, here \(J = \lvert L-S\rvert = 2\).
\[ \boxed{\ \text{lowest: } ^{3}F_{2}\ } \qquad\qquad \boxed{\ \text{highest: } ^{1}S_{0}\ } \] The \(^{1}S_{0}\) term sits at the top because it has both the smallest \(S\) and the smallest \(L\) — it loses on every count.
Check the count as always: \(\sum(2J+1) = 1 + (1{+}3{+}5) + 5 + (5{+}7{+}9) + 9 = 1 + 9 + 5 + 21 + 9 = 45\), and \(\binom{10}{2} = 45\) ways of placing two electrons in the ten \(d\) spin-orbitals.
Two configurations share this term list: \(d^{2}\) (titanium, \(\mathrm{Ti}^{2+}\), ground state \(^{3}F_{2}\)) and \(d^{8}\) (nickel, \(\mathrm{Ni}^{2+}\)) by the hole rule. But \(d^{8}\) is more than half filled, so its multiplet is inverted and its ground level is \(^{3}F_{4}\), not \(^{3}F_{2}\). Same terms, opposite \(J\) ordering — exactly the trap flagged in 3.6.3.
| Config. | Terms | Ground term |
|---|---|---|
| \(p^{1}\), \(p^{5}\) | \(^{2}P_{1/2,3/2}\) | \(^{2}P_{1/2}\) / \(^{2}P_{3/2}\) |
| \(p^{2}\), \(p^{4}\) | \(^{1}S_{0},\ ^{1}D_{2},\ ^{3}P_{0,1,2}\) | \(^{3}P_{0}\) / \(^{3}P_{2}\) |
| \(p^{3}\) | \(^{2}P,\ ^{2}D,\ ^{4}S_{3/2}\) | \(^{4}S_{3/2}\) |
| \(p^{6}\) | \(^{1}S_{0}\) | \(^{1}S_{0}\) |
| \(d^{1}\), \(d^{9}\) | \(^{2}D_{3/2,5/2}\) | \(^{2}D_{3/2}\) / \(^{2}D_{5/2}\) |
| \(d^{2}\), \(d^{8}\) | \(^{1}S,\ ^{1}D,\ ^{1}G,\ ^{3}P,\ ^{3}F\) | \(^{3}F_{2}\) / \(^{3}F_{4}\) |
| \(d^{5}\) | \(^{6}S_{5/2}\) and others | \(^{6}S_{5/2}\) |
| \(f^{7}\) | \(^{8}S_{7/2}\) and others | \(^{8}S_{7/2}\) |
3.11 Hund's Rules
To order the terms in energy:
Of the terms arising from equivalent electrons, the one with the largest multiplicity \((2S+1)\) lies lowest.
Of the terms with a given multiplicity, again from equivalent electrons, the one with the largest \(L\) lies lowest.
In the multiplets formed from equivalent electrons in a less than half-filled subshell, the level with the lowest \(J\) lies lowest — normal order.
In the multiplets formed from equivalent electrons in a more than half-filled subshell, the level with the highest \(J\) lies lowest — inverted order.
Terms arising from half-filled subshells show only very slight fine-structure splitting.
The lowest terms arising from half-filled subshells are \(S\) terms, and are especially stable: \[ ^{2}S_{1/2}\ (s^{1}),\qquad ^{4}S_{3/2}\ (p^{3}),\qquad ^{6}S_{5/2}\ (d^{5}),\qquad ^{8}S_{7/2}\ (f^{7}). \]
A half-filled subshell has one electron in every \(m_{\ell}\) slot with all spins parallel, so \(M_{L} = \sum m_{\ell} = 0\) is the only possibility and \(L = 0\) necessarily — hence the \(S\) term of 3d, with \(S\) at its maximum \(n_{\text{slots}}/2\). And with \(L = 0\) there is no \(\vec L\cdot\vec S\) to speak of, which is 3c: the interval constant \(A\) is nearly zero, and it is also the sign change of \(A\) across the half-filled point that separates rule 3a from rule 3b. The exceptional stability of these configurations is why chromium and copper break the naive filling order, and why \(\mathrm{Mn}^{2+}\) and \(\mathrm{Fe}^{3+}\) (\(d^{5}\)) are so common.
The source stated “the state with lowest \(L\) has largest energy”, which as written is the opposite of rule 2. Rule 2 says that for fixed multiplicity, larger \(L\) means lower energy — equivalently, the smallest \(L\) has the largest energy within that multiplicity. The phrasing above avoids the ambiguity.
Find the ground state term of the carbon atom, \(1s^{2}2s^{2}2p^{2}\).
Solution. The closed shells contribute nothing, so consider \(2p^{2}\). From \(\eqref{eq:p2-terms}\) the terms are \(^{1}S_{0}\), \(^{3}P_{0,1,2}\), \(^{1}D_{2}\).
Rule 1: largest multiplicity \(\Rightarrow\) \(^{3}P\).
Rule 3: \(p^{2}\) is less than half filled, so the lowest \(J\) lies lowest: \[ \boxed{^{3}P_{0}} \]
The full ordering is \[ E\!\left(^{3}P_{0}\right) < E\!\left(^{3}P_{1}\right) < E\!\left(^{3}P_{2}\right) < E\!\left(^{1}D_{2}\right) < E\!\left(^{1}S_{0}\right). \]
Find the ground state term for (i) \(2p^{1}\), \ (ii) \(2p^{2}\), \ (iii) \(3d^{2}\), \ (iv) \(3d^{3}\), \ (v) \(3d^{5}\).
Solution. In each case take the maximum \(S\) allowed by Pauli, then the maximum \(L\) consistent with that \(S\), then apply rule 3 or 4.
\(2p^{1}\): \(S = \tfrac12\), \(L = 1\), \(J = \tfrac12\) or \(\tfrac32\). Less than half filled \(\Rightarrow\) \(^{2}P_{1/2}\).
\(2p^{2}\): \(S=1\), \(L=1\), \(J = 0,1,2\). Less than half filled \(\Rightarrow\) \(^{3}P_{0}\).
\(3d^{2}\): \(S=1\), highest \(L = 3\) (\(m_\ell = 2+1\)), \(J = 2,3,4\). Less than half filled \(\Rightarrow\) \(^{3}F_{2}\).
\(3d^{3}\): \(S=\tfrac32\), highest \(L = 3\) (\(m_\ell = 2+1+0\)), \(J = \tfrac32,\dots,\tfrac92\). Less than half filled \(\Rightarrow\) \(^{4}F_{3/2}\).
\(3d^{5}\): half filled, so all five \(m_\ell\) are singly occupied: \(S = \tfrac52\), \(L = 0\), \(J = \tfrac52\) \(\Rightarrow\) \(^{6}S_{5/2}\).
The recipe for a ground term never changes: maximum \(S\), then maximum \(L\) for that \(S\), then minimum \(J\) if less than half filled / maximum \(J\) if more. You never need Breit's scheme to find the ground term — only to find the full term list.
3.12 Selection Rules in L–S Coupling
3.12.1 One-electron atoms
The parity of the wavefunction must change.
No restriction on \(n\).
\(\Delta \ell = \pm 1\).
\(\Delta m_{s} = 0\).
\(\Delta m_{\ell} = 0, \pm 1\).
\(\Delta j = 0, \pm 1\) \ (but \(j=0 \nrightarrow j=0\)).
\(\Delta m_{j} = 0, \pm 1\).
3.12.2 Many-electron atoms
\(\Delta L = 0, \pm 1\) \ (but \(L=0 \nrightarrow L=0\)).
\(\Delta m_{L} = 0, \pm 1\).
\(\Delta J = 0, \pm 1\) \ (but \(J=0 \nrightarrow J=0\)).
\(\Delta m_{J} = 0, \pm 1\).
\(\Delta S = 0\).
3.12.3 The Laporte rule
The Laporte rule: the parity of the electron configuration, determined by whether \(\sum_i \ell_i\) is even or odd, must change in an electric dipole transition (even \(\leftrightarrow\) odd).
For a two-electron jump the rule is realized as \(\Delta\ell_1 = \pm1\) together with \(\Delta\ell_2 = 0\) or \(\pm2\), so that the total parity still flips. For example \[ 3d\,4d \ (\textstyle\sum\ell = 4,\ \text{even}) \ \longrightarrow\ 4s\,4p \ (\textstyle\sum\ell = 1,\ \text{odd}) . \]
\(\Delta S = 0\) is the rule that forbids singlet–triplet transitions. It holds strictly only in pure L–S coupling; in heavy atoms, where spin–orbit coupling mixes the schemes, it breaks down and “intercombination” lines appear. The mercury \(2537\,\mathrm{Å}\) line (\(^{3}P_{1} \to {}^{1}S_{0}\)) is the standard example.
In an atom obeying L–S coupling, the components of a normal triplet state have separations \(20\,\mathrm{cm}^{-1}\) and \(40\,\mathrm{cm}^{-1}\) between adjacent components. There is a higher state for which the corresponding separations are \(22\,\mathrm{cm}^{-1}\) and \(33\,\mathrm{cm}^{-1}\). Determine the terms for the two states, and draw the energy-level diagram showing the allowed transitions and the pattern of the spectrum.
Solution. The lower state. Let \(J\), \(J+1\), \(J+2\) be the quantum numbers of the three components in order of increasing energy. By the Landé interval rule \(\eqref{eq:lande-interval}\) the separation of adjacent levels is proportional to the larger of the two \(J\) values involved, so \[ \frac{J+1}{J+2} = \frac{20\,\mathrm{cm}^{-1}}{40\,\mathrm{cm}^{-1}} = \frac12 \qquad\Longrightarrow\qquad 2J+2 = J+2 \qquad\Longrightarrow\qquad J = 0 . \] The \(J\) values are therefore \(J = 0, 1, 2\) in order of increasing energy.
To get \(S\) and \(L\) from this, use \(J = \lvert L-S\rvert,\ \lvert L-S\rvert+1,\ \dots,\ (L+S)\). The minimum value of \(J\) is \(0\) and the maximum is \(2\), so \[ \lvert L-S\rvert = 0, \qquad (L+S) = 2 . \] Take the case \(L > S\) first: \(L - S = 0\) and \(L + S = 2\) give \(S = 1\) (so \(2S+1 = 3\), a triplet, as stated) and \(L = 1\), a \(P\) state. The case \(L < S\) returns the same pair of numbers. So the levels of the lower state are \[ \boxed{\ ^{3}P_{0},\ ^{3}P_{1},\ ^{3}P_{2}\ } \]
The upper state. Proceeding identically, \[ \frac{J+1}{J+2} = \frac{22\,\mathrm{cm}^{-1}}{33\,\mathrm{cm}^{-1}} = \frac23 \qquad\Longrightarrow\qquad 3J+3 = 2J+4 \qquad\Longrightarrow\qquad J = 1 , \] so the \(J\) values are \(1, 2, 3\) in order of increasing energy. Hence \[ \lvert L-S\rvert = 1, \qquad (L+S) = 3 \qquad\Longrightarrow\qquad S = 1,\quad L = 2 , \] a \(D\) state. The upper levels are \[ \boxed{\ ^{3}D_{1},\ ^{3}D_{2},\ ^{3}D_{3}\ } \]
The transitions. Applying \(\Delta J = 0, \pm1\) with \(J = 0 \nleftrightarrow J = 0\), the \(^{3}D \to {}^{3}P\) array gives six transitions:
\(^{3}D_{1} \to {}^{3}P_{0}\)
\(^{3}D_{1} \to {}^{3}P_{1}\)
\(^{3}D_{1} \to {}^{3}P_{2}\)
\(^{3}D_{2} \to {}^{3}P_{1}\)
\(^{3}D_{2} \to {}^{3}P_{2}\)
\(^{3}D_{3} \to {}^{3}P_{2}\)
The forbidden ones are \(^{3}D_{2}\to{}^{3}P_{0}\) and \(^{3}D_{3}\to{}^{3}P_{0,1}\), all of which need \(\Delta J = 2\) or \(3\). Six lines in a \(3\times3\) level array is the signature of a compound triplet.
The whole method is two lines of algebra: form the ratio of adjacent separations, set it equal to \((J+1)/(J+2)\), solve for \(J\). Then \(\lvert L-S\rvert\) and \(L+S\) are just the smallest and largest \(J\), and two simultaneous equations give \(S\) and \(L\). A ratio of \(1:2\) means \(J = 0,1,2\) (\(^{3}P\)); a ratio of \(2:3\) means \(J = 1,2,3\) (\(^{3}D\)); a ratio of \(3:4\) means \(J = 2,3,4\) (\(^{3}F\)).
3.13 j–j Coupling
In heavy elements the spin–orbit interaction dominates the electrostatic terms. Each electron's own \(\vec\ell_i\) and \(\vec s_i\) then couple to one another first, forming individual \(\vec j_i\), and only afterwards do the \(j_i\) couple to a total \(\vec J\): \begin{equation} \vec j_i = \vec\ell_i + \vec s_i , \qquad \vec J = \sum_i \vec j_i . \label{eq:jj-scheme} \end{equation} So \(j_i = \ell_i \pm \tfrac12\), and \begin{equation} J = \lvert j_1 - j_2\rvert,\ \dots,\ (j_1+j_2) . \label{eq:J-jj} \end{equation} The level is labelled \((j_1, j_2)_J\).
There is no \(L\) and no \(S\) in j–j coupling, so a term symbol \(^{2S+1}L_J\) is meaningless here. Only \(J\) survives as a good quantum number, along with the individual \(j_i\).
Find the levels of a \(p\,d\) configuration in the j–j coupling scheme.
Solution. For the \(p\) electron (\(\ell_1 = 1\)): \(j_1 = \tfrac12, \tfrac32\). For the \(d\) electron (\(\ell_2 = 2\)): \(j_2 = \tfrac32, \tfrac52\).
The four combinations give \[ \begin{aligned} \left(\tfrac12,\tfrac32\right) &: \ J = 1, 2 ;\\ \left(\tfrac12,\tfrac52\right) &: \ J = 2, 3 ;\\ \left(\tfrac32,\tfrac32\right) &: \ J = 0, 1, 2, 3 ;\\ \left(\tfrac32,\tfrac52\right) &: \ J = 1, 2, 3, 4 . \end{aligned} \] That is \(2+2+4+4 = 12\) levels.
Compare with the L–S result for \(pd\) in Table 3.1: \(^{1}P_1, {}^{1}D_2, {}^{1}F_3, {}^{3}P_{0,1,2}, {}^{3}D_{1,2,3}, {}^{3}F_{2,3,4}\) — also 12 levels, with the same list of \(J\) values. This must be so: the two schemes are different coupling orders for the same physical states, so they always agree on the total number of levels and on the multiset of \(J\) values. They differ only in how the levels are grouped and ordered in energy. This is a favourite exam observation.
The level with the highest \((j_1 + j_2)\) lies at the top of the group and the lowest at the bottom, since the spin–orbit energy of each electron \(\propto \vec\ell_i\cdot\vec s_i\) is largest for \(j_i = \ell_i + \tfrac12\).
3.13.1 Equivalent electrons in j–j coupling
For two equivalent electrons (\(n\), \(\ell\) and \(j\) all equal) the Pauli principle restricts \(J\) to even values only: \begin{equation} J = 0, 2, 4, \dots, (2j-1) . \label{eq:jj-equivalent} \end{equation} For example two equivalent electrons with \(j = \tfrac32\) give \(J = 0, 2\) only.
The Hamiltonian of the He atom is modified by adding the correction term \(H = A\,\vec S_{1}\cdot\vec S_{2}\). Find the resulting correction to the energy of the \(1s^{2}\) configuration.
Solution. Writing \(\vec S = \vec S_{1}+\vec S_{2}\), \[ \vec S_{1}\cdot\vec S_{2} = \tfrac12\left(S^{2}-S_{1}^{2}-S_{2}^{2}\right) \quad\Longrightarrow\quad E = \frac{A\hbar^{2}}{2} \left[S(S+1) - S_{1}(S_{1}+1) - S_{2}(S_{2}+1)\right] . \] For two electrons \(S = 0\) or \(1\). But the \(1s^{2}\) configuration has both electrons in the same spatial orbital, so the Pauli principle forces the spin state to be antisymmetric — the singlet, \(S=0\): \[ E = \frac{A\hbar^{2}}{2}\left[0 - \tfrac34 - \tfrac34\right] = \boxed{-\frac{3}{4}A\hbar^{2}} . \]
The whole question turns on recognizing that \(S=1\) is forbidden for \(1s^{2}\). Any closed subshell has \(S=0\), \(L=0\), \(J=0\) — which is why \(^{1}S_{0}\) is the universal closed-shell term.
3.13.2 The \(4p\,4d\) configuration in j–j coupling
It is instructive to run the same configuration through both schemes. In 3.8 the L–S treatment of \(4p\,4d\) gave the terms \(^{1}P^{\mathrm o}_{1}\), \(^{1}D^{\mathrm o}_{2}\), \(^{1}F^{\mathrm o}_{3}\), \(^{3}P^{\mathrm o}_{0,1,2}\), \(^{3}D^{\mathrm o}_{1,2,3}\), \(^{3}F^{\mathrm o}_{2,3,4}\). Now take the other extreme.
For the \(p\) electron, \(\ell_1 = 1\), \(s_1 = \tfrac12\), so \(j_1 = \tfrac12, \tfrac32\). For the \(d\) electron, \(\ell_2 = 2\), \(s_2 = \tfrac12\), so \(j_2 = \tfrac32, \tfrac52\). That gives four \((j_1,j_2)\) combinations: \[ \left(\tfrac12,\tfrac32\right),\quad \left(\tfrac12,\tfrac52\right),\quad \left(\tfrac32,\tfrac32\right),\quad \left(\tfrac32,\tfrac52\right). \] The spin–orbit interaction splits the unperturbed level into these four, of which \(\left(\tfrac12,\tfrac32\right)\) lies lowest and \(\left(\tfrac32,\tfrac52\right)\) highest. Each is then split further by the residual electrostatic and spin–spin correlation energies into levels of definite \(J\):
| \((j_1, j_2)\) | \(J = \lvert j_1-j_2\rvert,\dots,(j_1+j_2)\) | levels |
|---|---|---|
| \(\left(\tfrac12,\tfrac32\right)\) | \(1, 2\) | 2 |
| \(\left(\tfrac12,\tfrac52\right)\) | \(2, 3\) | 2 |
| \(\left(\tfrac32,\tfrac32\right)\) | \(0, 1, 2, 3\) | 4 |
| \(\left(\tfrac32,\tfrac52\right)\) | \(1, 2, 3, 4\) | 4 |
Note the check: L–S gives twelve levels with \(J = 1,2,3,\ 0,1,2,\ 1,2,3,\ 2,3,4\); j–j gives twelve levels with \(J = 1,2,\ 2,3,\ 0,1,2,3,\ 1,2,3,4\). Sort both lists and they agree — one \(J=0\), three \(J=1\), four \(J=2\), three \(J=3\), one \(J=4\). Real atoms sit somewhere between the two extremes, and the levels can be traced continuously from one labelling to the other as \(Z\) increases.
In a quartet transition of C+ observed by Fowler and Selwyn, the separations between the upper-state multiplet levels are \(14.72\,\mathrm{cm}^{-1}\), \(25.07\,\mathrm{cm}^{-1}\) and \(36.30\,\mathrm{cm}^{-1}\), while the separations between the lower-state multiplet levels are \(23.84\,\mathrm{cm}^{-1}\) and \(45.0\,\mathrm{cm}^{-1}\). Obtain the term designation of the transition, and identify which component would be the strongest.
Solution. The upper state. Four levels means three intervals; let their \(J\) values be \(J\), \(J+1\), \(J+2\), \(J+3\). By the Landé interval rule successive separations are in the ratio of the larger \(J\) of each pair, so \[ \frac{J+1}{J+2} = \frac{14.72}{25.07} = \frac{2.944}{5.014} = \frac35, \qquad \frac{J+2}{J+3} = \frac{25.07}{36.30} = \frac{5.014}{7.26} = \frac57 . \] Both give the same answer: \[ 5J+5 = 3J+6 \ \Rightarrow\ J = \tfrac12, \qquad 7J+14 = 5J+15 \ \Rightarrow\ J = \tfrac12 . \checkmark \] So the upper-state \(J\) values are \[ J = \tfrac12,\ \tfrac32,\ \tfrac52,\ \tfrac72 . \] Four levels with half-integer \(J\) confirm a quartet, \(2S+1 = 4\), i.e.\ \(S = \tfrac32\). Then \(\lvert L-S\rvert = \tfrac12\) and \(L+S = \tfrac72\) give \(L = 2\), a \(D\) state: the upper term is \(^{4}D_{1/2,3/2,5/2,7/2}\).
The lower state. Three levels, two intervals: \[ \frac{J+1}{J+2} = \frac{23.84}{45.0} \approx \frac{1}{1.888} \qquad\Longrightarrow\qquad 1.888J + 1.888 = J + 2 \qquad\Longrightarrow\qquad J \approx \tfrac12 . \] The lower-state \(J\) values are \(\tfrac12, \tfrac32, \tfrac52\). Three levels with \(S = \tfrac32\) means \(L < S\), so the count is \(2L+1 = 3\), giving \(L = 1\): a \(P\) state, \(^{4}P_{1/2,3/2,5/2}\).
\[ \boxed{\ ^{4}D \longrightarrow {}^{4}P\ } \]
The strongest component. For a multiplet the most intense line is the one connecting the two levels of largest \(J\), that is \[ ^{4}D_{7/2} \longrightarrow {}^{4}P_{5/2} . \] This follows the general intensity rule for L–S multiplets: transitions in which \(J\) changes in the same sense as \(L\), and which involve the highest \(J\) available, carry the greatest statistical weight \((2J+1)\) and hence the greatest intensity.
Two safeguards in this kind of problem. First, when there are three or more intervals, both ratios must give the same \(J\) — if they do not, the multiplet is perturbed or the data are misassigned. Second, once \(S\) is known from the multiplicity, check whether \(L \ge S\) or \(L < S\) before counting levels; the lower state here has only three levels despite being a quartet, precisely because \(L = 1 < S = \tfrac32\) and \(\eqref{eq:num-levels}\) caps the count at \(2L+1\).
3.14 Comparison of the Two Schemes
| L–S (Russell–Saunders) | j–j | |
|---|---|---|
| Applies to | light atoms, \(Z<30\) | heavy atoms, \(Z>30\) |
| Dominant interaction | residual electrostatic \(+\) spin–spin | spin–orbit |
| Coupling order | \(\sum\vec\ell_i \to \vec L\), \(\sum\vec s_i \to \vec S\), then \(\vec L+\vec S\) | \(\vec\ell_i+\vec s_i \to \vec j_i\), then \(\sum\vec j_i\) |
| Good quantum numbers | \(L, S, J, m_J\) | \(j_1, j_2, J, m_J\) |
| Notation | \(^{2S+1}L_J\) | \((j_1,j_2)_J\) |
| Level count | identical for a given configuration | |
Real atoms lie on a continuum between the two limits. Intermediate coupling is the rule rather than the exception for \(Z\) in the twenties and thirties, which is why \(\Delta S = 0\) weakens gradually rather than switching off.
3.15 Nuclear Effects in Atoms
For most of atomic physics the nucleus is just a heavy charged particle at the centre. But careful analysis of spectral lines reveals small effects that give direct information about the nucleus. They fall into two categories: hyperfine structure and isotope shifts.
3.16 Hyperfine Structure
The nucleus has an intrinsic spin \(\vec I\). Hyperfine structure arises from the interaction between the total electronic angular momentum \(\vec J\) and the nuclear spin \(\vec I\). The combined total is \begin{equation} \vec F = \vec J + \vec I , \qquad F = \lvert J-I\rvert,\ \dots,\ (J+I) . \label{eq:F-coupling} \end{equation} The number of hyperfine levels is \begin{equation} \text{number of levels} = \begin{cases} 2I+1, & I < J,\\ 2J+1, & J < I, \end{cases} \qquad\text{i.e.\ } 2\min(I,J)+1 . \label{eq:hf-count} \end{equation}
The interaction Hamiltonian is \begin{equation} H_{\text{hf}} = A\,\vec I\cdot\vec J , \label{eq:H-hf} \end{equation} where \(A\) is the hyperfine structure constant. Exactly as in 3.9, \begin{equation} E_{F} = \frac{A\hbar^{2}}{2} \left[F(F+1) - I(I+1) - J(J+1)\right], \label{eq:E-F} \end{equation} so consecutive hyperfine levels obey an interval rule \begin{equation} E_{F} - E_{F-1} = A\hbar^{2}F . \label{eq:hf-interval} \end{equation}
The selection rule for transitions is \begin{equation} \Delta F = 0, \pm1 \quad (F=0 \nrightarrow F=0), \qquad \Delta m_{F} = 0, \pm 1 . \label{eq:hf-selection} \end{equation}
3.16.1 The vector model of hyperfine coupling
The algebra is the same as for \(\vec J = \vec L + \vec S\), run one level up. The nucleus carries an intrinsic angular momentum \(\vec I\) — the nuclear spin, arising from the interaction between the protons and neutrons inside it — and this couples to the total electronic angular momentum \(\vec J\) to give the grand total \begin{equation} \vec F = \vec I + \vec J . \label{eq:F-vector} \end{equation} Squaring, \[ \vec F\cdot\vec F = (\vec I + \vec J)\cdot(\vec I + \vec J) = I^{2} + J^{2} + 2\,\vec I\cdot\vec J , \] so that \begin{equation} \boxed{\; \vec I\cdot\vec J = \frac{\lvert\vec F\rvert^{2} - \lvert\vec I\rvert^{2} - \lvert\vec J\rvert^{2}}{2} = \frac{\hbar^{2}}{2}\Big[F(F+1) - I(I+1) - J(J+1)\Big] . \;} \label{eq:IJ-dot} \end{equation} The allowed values of \(F\) follow from the triangle rule, and adjacent values differ by one: \begin{equation} \lvert I - J\rvert \le F \le \lvert I + J\rvert , \qquad \Delta F = \pm 1 \ \text{between adjacent levels} . \label{eq:F-range} \end{equation}
Compare \(\eqref{eq:IJ-dot}\) with the \(\vec L\cdot\vec S\) result of 3.9: identical structure, with \((I,J,F)\) in place of \((L,S,J)\). Everything you know about multiplets transfers — including the interval rule, whose hyperfine form is \(E_{F} - E_{F-1} \propto F\), and the distinction between normal and inverted order according to the sign of \(A\).
The interaction of the nuclear and electron spins in the hydrogen ground state is represented by \(H = A\,\vec I\cdot\vec S\). Determine the energy separation between the hyperfine levels.
Solution. In the \(1s\) ground state \(L=0\), so \(\vec J = \vec S\) with \(J = S = \tfrac12\); and the proton has \(I = \tfrac12\). Hence \(F = 0\) or \(1\).
Using \(\vec I\cdot\vec S = \tfrac12\left(F^{2} - I^{2} - S^{2}\right)\), \[ \big\langle\vec I\cdot\vec S\big\rangle = \frac{\hbar^{2}}{2}\left[F(F+1) - \tfrac34 - \tfrac34\right] = \begin{cases} +\dfrac{\hbar^{2}}{4}, & F=1 \ \text{(triplet)},\\[6pt] -\dfrac{3\hbar^{2}}{4}, & F=0 \ \text{(singlet)}. \end{cases} \] The separation is therefore \begin{equation} \Delta E = A\hbar^{2}\left[\frac14 + \frac34\right] = A\hbar^{2} . \label{eq:hf-hydrogen} \end{equation}
Experimentally this splitting is \[ \Delta E = 5.87\times 10^{-6}\,\mathrm{eV}, \qquad \nu = 1420\,\mathrm{MHz}, \qquad \lambda = 21\,\mathrm{cm} , \] the single most important line in radio astronomy — it is how the spiral structure of the Galaxy was mapped. Note \(F=1\) lies above \(F=0\), so the ground state of hydrogen is the \(F=0\) singlet.
Bismuth has an excited state which is split into six hyperfine levels. Given the electronic angular momentum \(J = \tfrac52\), find the nuclear spin quantum number \(I\).
Solution. From \(\eqref{eq:hf-count}\) the number of levels is \(2\min(I,J)+1 = 6\), so \(\min(I,J) = \tfrac52\). Since \(J = \tfrac52\) this requires \(I \ge \tfrac52\).
To pin \(I\) down, use the largest \(F\). The six levels are \(F = \lvert J-I\rvert, \dots, (J+I)\), and for bismuth these turn out to be \(F = 2,3,4,5,6,7\). Then \[ F_{\max} = J + I = 7 \quad\Longrightarrow\quad I = 7 - \tfrac52 = \boxed{\tfrac92} , \] and indeed \(\lvert J - I\rvert = \lvert \tfrac52-\tfrac92\rvert = 2\) . The hyperfine constant then follows from the interval rule \(\eqref{eq:hf-interval}\) applied to the observed spacings.
Find the number of hyperfine states of the \(^{3}\mathrm{He}\) atom in the electronic configuration \(1s^{1}2p^{1}\).
Solution. The \(^{3}\)He nucleus has \(I = \tfrac12\).
The configuration \(1s\,2p\) is two non-equivalent electrons with \(\ell_1 = 0\), \(\ell_2 = 1\), so \(L = 1\) and \(S = 0\) or \(1\). The terms are \[ ^{1}P_{1} \quad\text{and}\quad ^{3}P_{0},\ ^{3}P_{1},\ ^{3}P_{2} . \] Now apply \(\eqref{eq:F-coupling}\) with \(I = \tfrac12\) to each level: \[ \begin{aligned} ^{1}P_{1}\ (J=1) &: \ F = \tfrac12, \tfrac32 && 2 \ \text{states}\\ ^{3}P_{0}\ (J=0) &: \ F = \tfrac12 && 1 \ \text{state}\\ ^{3}P_{1}\ (J=1) &: \ F = \tfrac12, \tfrac32 && 2 \ \text{states}\\ ^{3}P_{2}\ (J=2) &: \ F = \tfrac32, \tfrac52 && 2 \ \text{states} \end{aligned} \] Total: \(2+1+2+2 = \boxed{7}\) hyperfine states.
Bismuth has nuclear spin \(I = \tfrac92\), and its fine structure follows j–j coupling. The transition giving the \(4722\,\mathrm{Å}\) line is \[ 6p^{2}7s,\ \left(\tfrac12,\tfrac12,\tfrac32\right)_{1/2} \ \longrightarrow\ 6p^{3},\ \left(\tfrac12,\tfrac32,\tfrac32\right)_{3/2} . \] How many hyperfine components does the line show?
Solution. The initial state. Here \(J = \tfrac12\) and \(I = \tfrac92\), so by \(\eqref{eq:F-range}\) \[ F = \left\lvert \tfrac92 - \tfrac12\right\rvert,\ \dots,\ \left(\tfrac92 + \tfrac12\right) = 4,\ 5 . \] Two hyperfine levels — as \(\eqref{eq:hf-count}\) predicts, since \(2\min(I,J)+1 = 2\cdot\tfrac12 + 1 = 2\). The subshell reasoning gives \(A>0\) here, so these lie in normal order, \(F=4\) below \(F=5\).
The final state. Now \(J = \tfrac32\) and \(I = \tfrac92\), so \[ F = 3,\ 4,\ 5,\ 6 , \] four levels, again matching \(2\min(I,J)+1 = 4\). These turn out to be in inverted order, \(F=6\) lowest.
Counting the transitions. Apply \(\Delta F = 0,\pm1\) with \(F=0 \nleftrightarrow F=0\) to the \(2\times4\) array:
\(F=4 \to F=3\)
\(F=4 \to F=4\)
\(F=4 \to F=5\)
\(F=5 \to F=4\)
\(F=5 \to F=5\)
\(F=5 \to F=6\)
The combinations \(4\to6\) and \(5\to3\) are forbidden, needing \(\Delta F = 2\). Hence \[ \boxed{\ \text{six hyperfine components}\ } \] and the \(4722\,\mathrm{Å}\) line of bismuth is indeed observed as a sextet.
Two counts, then one array. First get the number of hyperfine levels at each end from \(2\min(I,J)+1\) — not \(2I+1\), which is only right when \(J \ge I\). Then lay out the grid and strike the entries needing \(\lvert\Delta F\rvert \ge 2\). A \(2\times4\) array loses two corners and leaves six; a \(3\times3\) array loses two and leaves seven.
This example is also the cleanest illustration of why hyperfine structure sits in this chapter: the level being split, \(\left(\tfrac12,\tfrac12,\tfrac32 \right)_{1/2}\), is labelled in the j–j scheme of 3.13. Hyperfine coupling is simply the next coupling in the sequence \(\vec\ell + \vec s \to \vec j\), \(\sum\vec j \to \vec J\), \(\vec J + \vec I \to \vec F\).
3.17 Isotope Shifts
Two processes give rise to isotope shifts.
Mass effect. The mass entering the Schrödinger equation is the reduced mass \(\mred\), not the bare electron mass \(m\). Changing the nuclear mass changes \(\mred\) and hence the atomic energies — this is exactly the effect quantified in Chapter 1, where the H/D shift of the H\(_\alpha\) line came out at \(1.8\,\mathrm{Å}\). Deuterium was discovered this way.
Field (volume) effect. Electrons in \(s\) shells have a finite probability of penetrating the nucleus and are therefore sensitive to its charge distribution. A heavier isotope has a slightly larger nuclear radius, so the potential inside differs, shifting \(s\) levels in particular.
The mass effect dominates for light elements (it scales as \(1/M^{2}\) roughly), the field effect for heavy ones (it grows with nuclear volume). Both produce shifts in the wavelengths of lines from different isotopes of the same element.
Formula Summary
Magnetic moments \[ \mu_{\mathrm B} = \frac{e\hbar}{2m} = 9.274\times 10^{-24}\,\mathrm{J}\,\mathrm{T}^{-1} = 5.788\times 10^{-5}\,\mathrm{eV}\,\mathrm{T}^{-1} \] \[ \vec\mu_\ell = -g_\ell\frac{\mu_{\mathrm B}}{\hbar}\vec L\ (g_\ell = 1), \qquad \vec\mu_s = -g_s\frac{\mu_{\mathrm B}}{\hbar}\vec S\ (g_s \approx 2) \] \[ \mu_{\ell,z} = -g_\ell\mu_{\mathrm B}m_\ell, \qquad \mu_{s,z} = -g_s\mu_{\mathrm B}m_s = \mp\mu_{\mathrm B} \] Larmor precession \(\omega_L = \dfrac{eB}{2m} = \dfrac{g\mu_{\mathrm B}B}{\hbar}\)
Space quantization \(\cos\theta = \dfrac{m_\ell}{\sqrt{\ell(\ell+1)}}\); \(\theta_{\min}\) at \(m_\ell = \ell\): \ \(\cos\theta = \sqrt{\dfrac{\ell}{\ell+1}}\) (\(2p\): \(45^{\circ}\))
Which scheme \(Z<30\): L–S | \(Z>30\): j–j
L–S coupling \(^{2S+1}L_J\),
\(L = \lvert\ell_1-\ell_2\rvert\dots(\ell_1+\ell_2)\),
\(J = \lvert L-S\rvert\dots(L+S)\)
\[
\text{levels per term} =
\begin{cases} 2S+1, & L\ge S\\ 2L+1, & L
Landé interval rule \(E_J - E_{J-1} = A\hbar^{2}J\), from \(\big\langle\vec L\cdot\vec S\big\rangle = \frac{\hbar^{2}}{2}\!\left[J(J{+}1)-L(L{+}1)-S(S{+}1)\right]\)
Equivalent electrons closed shell \(\Rightarrow\) \(^{1}S_0\); \(x\) holes give the same terms as \(x\) electrons; \(p^{2}\Rightarrow{}^{1}S_0,{}^{3}P_{0,1,2},{}^{1}D_2\)
Hund's rules max \(2S+1\) lowest \(\to\) max \(L\) lowest \(\to\) min \(J\) lowest if \(<\) half filled, max \(J\) lowest if \(>\) half filled
Selection rules (L–S) \(\Delta L = 0,\pm1\) (\(0\nrightarrow0\)), \ \(\Delta J = 0,\pm1\) (\(0\nrightarrow0\)), \ \(\Delta S = 0\), \ \(\Delta m_J = 0,\pm1\), \ parity must change (Laporte)
j–j coupling \(j_i = \ell_i\pm\tfrac12\), \ \(J = \lvert j_1-j_2\rvert\dots(j_1+j_2)\), notation \((j_1,j_2)_J\); \ equivalent electrons: \(J\) even only
Consistency check L–S and j–j give the same number of levels and the same \(J\) values for any configuration.
Hyperfine structure \[ \vec F = \vec J + \vec I, \qquad F = \lvert J-I\rvert\dots(J+I), \qquad \text{levels} = 2\min(I,J)+1 \] \[ H_{\text{hf}} = A\,\vec I\cdot\vec J, \quad E_F = \frac{A\hbar^{2}}{2}\!\left[F(F{+}1)-I(I{+}1)-J(J{+}1)\right], \quad E_F - E_{F-1} = A\hbar^{2}F \] \[ \Delta F = 0,\pm1 \ (0\nrightarrow0) \] H ground state \(I=J=\tfrac12 \Rightarrow F=0,1\); \ \(\Delta E = 5.87\times 10^{-6}\,\mathrm{eV}\), \ \(\nu=1420\,\mathrm{MHz}\), \ \(\lambda = 21\,\mathrm{cm}\); \ \(F=0\) lies lower
Isotope shift mass effect (reduced mass, dominates for light elements) \(+\) field effect (\(s\) electrons penetrate the nucleus, dominates for heavy)
Previous Year Questions
Using Hund's rules, the total angular momentum quantum number \(J\) for the electronic ground state of the nitrogen atom is
- (a)
\(\tfrac12\)
- (b)
\(\tfrac32\)
- (c)
\(0\)
- (d)
\(1\)
The degeneracy of an excited state of the nitrogen atom having electronic configuration \(1s^{2}2s^{2}2p^{2}3d^{1}\) is
- (a)
\(6\)
- (b)
\(10\)
- (c)
\(15\)
- (d)
\(150\)
The term symbol for the electronic ground state of the oxygen atom is
- (a)
\(^{1}S_{0}\)
- (b)
\(^{1}D_{2}\)
- (c)
\(^{3}P_{0}\)
- (d)
\(^{3}P_{2}\)
The number of spectroscopic terms resulting from the L–S coupling of a \(3p\) electron and a \(3d\) electron is .
The terms \(\{j_{1},j_{2}\}_{J}\) arising from a \(2s^{1}3d^{1}\) configuration in the j–j coupling scheme are
- (a)
\(\left\{\tfrac12,\tfrac32\right\}_{2,1}\) and \(\left\{\tfrac12,\tfrac52\right\}_{3,2}\)
- (b)
\(\left\{\tfrac12,\tfrac12\right\}_{1,0}\) and \(\left\{\tfrac12,\tfrac32\right\}_{2,1}\)
- (c)
\(\left\{\tfrac12,\tfrac12\right\}_{1,0}\) and \(\left\{\tfrac12,\tfrac52\right\}_{3,2}\)
- (d)
\(\left\{\tfrac32,\tfrac12\right\}_{2,1}\) and \(\left\{\tfrac12,\tfrac52\right\}_{3,2}\)
Among the term symbols \(^{4}S_{1}\), \(^{2}D_{7/2}\), \(^{3}S_{1}\) and \(^{2}D_{5/2}\), choose the option(s) possible in the L–S coupling notation.
- (a)
\(^{4}S_{1}\)
- (b)
\(^{2}D_{7/2}\)
- (c)
\(^{3}S_{1}\)
- (d)
\(^{2}D_{5/2}\)
An excited state of the Ca atom is \([\mathrm{Mg}]\,3p^{5}4s^{2}3d^{1}\). The spectroscopic terms corresponding to the total orbital angular momentum are
- (a)
\(S\), \(P\) and \(D\)
- (b)
\(P\), \(D\) and \(F\)
- (c)
\(P\) and \(D\)
- (d)
\(S\) and \(P\)
The atomic number of an atom is \(6\). What is the spectroscopic notation of its ground state, according to Hund's rules?
- (a)
\(^{3}P_{0}\)
- (b)
\(^{3}P_{1}\)
- (c)
\(^{3}D_{3}\)
- (d)
\(^{3}S_{1}\)
In the vector model of angular momentum applied to atoms, what is the minimum angle in degrees (as an integer) made by the orbital angular momentum vector with the positive \(z\) axis for a \(2p\) electron?
Consider two non-identical spin-\(\tfrac12\) particles labelled 1 and 2 in the spin product state \(\left\lvert\tfrac12,\tfrac12\right\rangle\left\lvert\tfrac12,-\tfrac12\right\rangle\). The Hamiltonian of the system is \[ H = \frac{4\lambda}{\hbar^{2}}\,\vec S_{1}\cdot\vec S_{2}, \] where \(\lambda\) is a constant with appropriate dimensions. What is the expectation value of \(H\) in this state?
The non-relativistic Hamiltonian for a single-electron atom is \[ H_{0} = \frac{p^{2}}{2m} - V(r), \] where \(V(r)\) is the Coulomb potential and \(m\) the electron mass. Consider adding to \(H_{0}\) the spin–orbit term \[ H' = \frac{1}{2m^{2}c^{2}}\,\frac{1}{r}\,\frac{\dd V}{\dd r}\, \vec{L}\cdot\vec{S}. \] Which of the following statement(s) is/are true?
- (a)
\(H'\) commutes with \(L^{2}\)
- (b)
\(H'\) commutes with \(L_{z}\) and \(S_{z}\)
- (c)
For a given principal quantum number \(n\) and orbital angular momentum quantum number \(l\), there are \(2(2l+1)\) degenerate eigenstates of \(H_{0}\)
- (d)
\(H_{0}\), \(L^{2}\), \(S^{2}\), \(L_{z}\) and \(S_{z}\) have a set of simultaneous eigenstates
The effective magnetic moment (in units of the Bohr magneton) for the ground state of an isolated \(4f\) ion with six unpaired electrons in the \(4f\) shell, according to Hund's rules, is (in integer).
Which one of the following represents the \(2s\) radial wavefunction of the hydrogen atom? (\(\abohr\) is the Bohr radius)
- (a)
\(R(r) \propto e^{-r/\abohr}\)
- (b)
\(R(r) \propto \left(2 - \dfrac{r}{\abohr}\right)e^{-r/2\abohr}\)
- (c)
\(R(r) \propto \dfrac{r}{\abohr}\,e^{-r/2\abohr}\)
- (d)
\(R(r) \propto \dfrac{r^{2}}{\abohr^{2}}\,e^{-r/3\abohr}\)
The separations between the adjacent levels of a normal multiplet are found to be \(22\,\mathrm{cm}^{-1}\) and \(33\,\mathrm{cm}^{-1}\). Assuming the multiplet is well described by L–S coupling and the Landé interval rule \(E(J)-E(J-1) = AJ\), the term notation for this multiplet is
- (a)
\(^{3}P_{0,1,2}\)
- (b)
\(^{3}F_{2,3,4}\)
- (c)
\(^{3}G_{3,4,5}\)
- (d)
\(^{3}D_{1,2,3}\)
How much does the total angular momentum quantum number \(J\) change in the transition of a \(\mathrm{Cr}\,(3d^{6})\) atom as it ionizes to \(\mathrm{Cr}^{2+}(3d^{4})\)?
- (a)
Increases by \(2\)
- (b)
Decreases by \(2\)
- (c)
Decreases by \(4\)
- (d)
Does not change
Of the following term symbols of the \(np^{2}\) atomic configuration — \(^{1}S_{0}\), \(^{3}P_{0}\), \(^{3}P_{1}\), \(^{3}P_{2}\) and \(^{1}D_{2}\) — which is the ground state?
- (a)
\(^{3}P_{0}\)
- (b)
\(^{1}S_{0}\)
- (c)
\(^{3}P_{2}\)
- (d)
\(^{3}P_{1}\)
The L–S configurations of the ground state of \(^{12}\mathrm{Mg}\), \(^{13}\mathrm{Al}\), \(^{17}\mathrm{Cl}\) and \(^{18}\mathrm{Ar}\) are respectively
- (a)
\(^{3}S_{1}\), \(^{2}P_{1/2}\), \(^{2}P_{1/2}\) and \(^{1}S_{0}\)
- (b)
\(^{3}S_{1}\), \(^{2}P_{3/2}\), \(^{2}P_{3/2}\) and \(^{3}S_{1}\)
- (c)
\(^{1}S_{0}\), \(^{2}P_{1/2}\), \(^{2}P_{3/2}\) and \(^{1}S_{0}\)
- (d)
\(^{1}S_{0}\), \(^{2}P_{3/2}\), \(^{2}P_{1/2}\) and \(^{3}S_{1}\)
The ground state electronic configuration of \(^{22}\mathrm{Ti}\) is \([\mathrm{Ar}]3d^{2}4s^{2}\). Which state, in standard spectroscopic notation, is not possible in this configuration?
- (a)
\(^{1}F_{3}\)
- (b)
\(^{1}S_{0}\)
- (c)
\(^{1}D_{2}\)
- (d)
\(^{3}P_{0}\)
In the L–S coupling scheme, the terms arising from two non-equivalent \(p\) electrons are
- (a)
\(^{3}S,\ ^{1}P,\ ^{3}P,\ ^{1}D,\ ^{3}D\)
- (b)
\(^{1}S,\ ^{3}S,\ ^{1}P,\ ^{1}D\)
- (c)
\(^{1}S,\ ^{3}S,\ ^{3}P,\ ^{3}D\)
- (d)
\(^{1}S,\ ^{3}S,\ ^{1}P,\ ^{3}P,\ ^{1}D,\ ^{3}D\)
The outermost shell of an atom of an element is \(3d^{3}\). The spectral symbol for the ground state is
- (a)
\(^{4}F_{3/2}\)
- (b)
\(^{4}F_{9/2}\)
- (c)
\(^{4}D_{7/2}\)
- (d)
\(^{4}D_{1/2}\)
The electronic configuration of \(^{12}\mathrm{C}\) is \(1s^{2}2s^{2}2p^{2}\). Including L–S coupling, the correct ordering of its energies is
- (a)
\(E\!\left(^{3}P_{2}\right) < E\!\left(^{3}P_{1}\right) < E\!\left(^{3}P_{0}\right) < E\!\left(^{1}D_{2}\right)\)
- (b)
\(E\!\left(^{3}P_{0}\right) < E\!\left(^{3}P_{1}\right) < E\!\left(^{3}P_{2}\right) < E\!\left(^{1}D_{2}\right)\)
- (c)
\(E\!\left(^{1}D_{2}\right) < E\!\left(^{3}P_{2}\right) < E\!\left(^{3}P_{1}\right) < E\!\left(^{3}P_{0}\right)\)
- (d)
\(E\!\left(^{3}P_{1}\right) < E\!\left(^{3}P_{0}\right) < E\!\left(^{3}P_{2}\right) < E\!\left(^{1}D_{2}\right)\)
Consider the bromine ion Br+ in its ground state. The atomic number of Br is \(35\). The fine-structure term symbol \(^{2S+1}L_{J}\) under the \(L\)–\(S\) coupling scheme for the lowest-energy state of this ion would be
- (a)
\(^{3}P_{2}\)
- (b)
\(^{3}P_{0}\)
- (c)
\(^{1}D_{2}\)
- (d)
\(^{4}S_{3/2}\)
A helium atom is excited to a state with the configuration \((2s\,2p)\), at an energy of \(58.3\,\mathrm{eV}\). After some time this atom spontaneously ejects a single electron. The value of the orbital angular momentum quantum number \(l\) of the ejected electron in the final state of the system is (the ionization potential of \(\text{He}(1s)^{2}\) is \(24.6\,\mathrm{eV}\))
- (a)
\(1\)
- (b)
\(0\)
- (c)
\(2\)
- (d)
\(3\)
Which of the following statements is true for the energies of the terms of the carbon atom in the ground state configuration \(1s^{2}2s^{2}2p^{2}\)?
- (a)
\(^{3}P < {}^{1}D < {}^{1}S\)
- (b)
\(^{3}P < {}^{1}S < {}^{1}D\)
- (c)
\(^{3}P < {}^{1}F < {}^{1}S\)
- (d)
\(^{3}P < {}^{1}F < {}^{1}D\)
Consider a hypothetical world in which the electron has spin \(\tfrac32\) instead of \(\tfrac12\). What will be the electronic configuration for an element with atomic number \(Z = 5\)?
- (a)
\(1s^{4},2s^{1}\)
- (b)
\(1s^{4},2s^{2},2p^{1}\)
- (c)
\(1s^{5}\)
- (d)
\(1s^{3},2s^{1},2p^{1}\)
Solutions to Previous Year Questions
Solution. Nitrogen is \(1s^{2}2s^{2}2p^{3}\). The \(2p^{3}\) subshell is exactly half filled, so all three \(m_\ell = 1,0,-1\) are singly occupied with parallel spins: \[ S = \tfrac32, \qquad L = 1+0-1 = 0, \qquad J = \tfrac32 , \] giving the ground term \(^{4}S_{3/2}\).
Solution. Degeneracy is the number of distinct microstates. The two equivalent \(2p\) electrons give \(\binom{6}{2} = 15\) arrangements; the \(3d^{1}\) electron is independent with \(2(2\times2+1) = 10\) choices. Hence \[ 15 \times 10 = 150 . \]
Solution. Oxygen is \(1s^{2}2s^{2}2p^{4}\). Since \(p^{4}\) has the same terms as \(p^{2}\), we get \(S=1\), \(L=1\). But \(p^{4}\) is more than half filled, so by Hund's rule 4 the highest \(J\) lies lowest (inverted order): \[ J = L+S = 2 \quad\Longrightarrow\quad \boxed{^{3}P_{2}} . \]
Carbon (\(p^{2}\)) gives \(^{3}P_{0}\) and oxygen (\(p^{4}\)) gives \(^{3}P_{2}\) — same term, opposite ends of the multiplet. Keeping the pair together is the easiest way to remember which is which.
Solution. A \(3p\) and a \(3d\) electron are non-equivalent, so \(S = 0,1\) and \(L = 1,2,3\), giving \[ ^{1}P_{1},\ ^{1}D_{2},\ ^{1}F_{3},\ ^{3}P_{0,1,2},\ ^{3}D_{1,2,3},\ ^{3}F_{2,3,4} , \] which is 12.
Strictly, a term is \(^{2S+1}L\) (there are 6 here) and a level is \(^{2S+1}L_{J}\) (there are 12). GATE's key counts the 12 \(J\)-resolved symbols. Worth a footnote in print, since a student applying the strict definition from 3.5 would answer 6. The state count is \(60\).
Solution. For the \(2s\) electron, \(\ell_{1}=0 \Rightarrow j_{1} = \tfrac12\). For the \(3d\) electron, \(\ell_{2}=2 \Rightarrow j_{2} = \tfrac32, \tfrac52\). Hence \[ \left\{\tfrac12,\tfrac32\right\}: J = 1, 2 ; \qquad \left\{\tfrac12,\tfrac52\right\}: J = 2, 3 . \]
Solution. Test each against \(\lvert L-S\rvert \le J \le L+S\):
\(^{4}S_{1}\): \(2S+1 = 4 \Rightarrow S = \tfrac32\), \(L=0\), so \(J = \tfrac32\) only. \(J=1\) impossible. Not allowed.
\(^{2}D_{7/2}\): \(S=\tfrac12\), \(L=2\), so \(J = \tfrac32\) or \(\tfrac52\). \(J = \tfrac72\) impossible. Not allowed.
\(^{3}S_{1}\): \(S=1\), \(L=0\), so \(J=1\). Allowed.
\(^{2}D_{5/2}\): \(S=\tfrac12\), \(L=2\), so \(J=\tfrac32,\tfrac52\). Allowed.
Solution. Closed shells contribute nothing, and \(4s^{2}\) is closed. That leaves \(3p^{5}\) (which behaves as a single \(p\) hole, \(\ell_{1}=1\)) and \(3d^{1}\) (\(\ell_{2}=2\)): \[ L = \lvert 1-2\rvert,\dots,(1+2) = 1,2,3 \quad\Longrightarrow\quad P,\ D,\ F . \]
Solution. \(Z=6\) is carbon, \(1s^{2}2s^{2}2p^{2}\). The optical electrons are equivalent and less than half filled: \[ S = 1, \qquad L = 1, \qquad J = \lvert L-S\rvert = 0 \quad\Longrightarrow\quad \boxed{^{3}P_{0}} . \]
Solution. For a \(2p\) electron \(\ell = 1\), so \(\lvert\vec L\rvert = \sqrt2\,\hbar\) and \(L_{z}^{\max} = \hbar\): \[ \cos\theta_{\min} = \frac{1}{\sqrt2} \quad\Longrightarrow\quad \theta_{\min} = 45^{\circ} . \] (The three allowed angles are \(45^{\circ}\), \(90^{\circ}\), \(135^{\circ}\).)
Solution. The product state can be written using Clebsch–Gordan coefficients as \[ \left\lvert\tfrac12,\tfrac12\right\rangle \left\lvert\tfrac12,-\tfrac12\right\rangle = \frac{1}{\sqrt2}\lvert 1,0\rangle + \frac{1}{\sqrt2}\lvert 0,0\rangle , \] an equal mixture of triplet and singlet. With \(H = \dfrac{4\lambda}{\hbar^{2}}\vec S_{1}\cdot\vec S_{2}\) and \(\vec S_{1}\cdot\vec S_{2} = +\tfrac{\hbar^{2}}{4}\) (triplet), \(-\tfrac{3\hbar^{2}}{4}\) (singlet): \[ E_{\text{triplet}} = \lambda, \qquad E_{\text{singlet}} = -3\lambda , \] \[ \langle H\rangle = \tfrac12(\lambda) + \tfrac12(-3\lambda) = -\lambda . \]
Equivalently and faster: only the \(S_{1z}S_{2z}\) part of \(\vec S_{1}\cdot\vec S_{2}\) has a diagonal matrix element in the product basis, giving \(\left(\tfrac{\hbar}{2}\right)\left(-\tfrac{\hbar}{2}\right) = -\tfrac{\hbar^{2}}{4}\) and hence \(\langle H\rangle = -\lambda\) at once.
Solution. Take the four in turn.
(a) True. \(L^{2}\) commutes with every component of \(\vec{L}\), and it acts trivially on spin, so \[ \left[L^{2},\ \vec{L}\cdot\vec{S}\right] = \left[L^{2},\ L_{x}\right]S_{x} + \left[L^{2},\ L_{y}\right]S_{y} + \left[L^{2},\ L_{z}\right]S_{z} = 0 . \] Equivalently \(\vec{L}\cdot\vec{S} = \tfrac12(J^{2}-L^{2}-S^{2})\), and \(L^{2}\) commutes with all three.
(b) False. Working out one component, \[ \left[L_{z},\ \vec{L}\cdot\vec{S}\right] = \left[L_{z},\ L_{x}\right]S_{x} + \left[L_{z},\ L_{y}\right]S_{y} = \ii\hbar\left(L_{y}S_{x} - L_{x}S_{y}\right) \neq 0 , \] and the same for \(S_{z}\). Only the sum survives: \(\vec{L}\cdot\vec{S}\) commutes with \(J_{z} = L_{z}+S_{z}\), since the two commutators cancel.
(c) True. For fixed \(n\) and \(l\) there are \((2l+1)\) values of \(m_{l}\) and \(2\) of \(m_{s}\), giving \(2(2l+1)\) states, all degenerate under \(H_{0}\) alone.
(d) True. \(H_{0}\) is spin-independent and rotationally invariant, so it commutes with \(L^{2}\), \(S^{2}\), \(L_{z}\) and \(S_{z}\), and those four commute with each other. Note carefully that the question asks about \(H_{0}\), not about \(H_{0}+H'\).
This question is the standard justification for changing basis. With \(H'\) switched on, \(m_{l}\) and \(m_{s}\) stop being good quantum numbers — that is what (b) says — while \(j\) and \(m_{j}\) remain good. Hence the move from the uncoupled basis \(|l\,m_{l}\,s\,m_{s}\rangle\) to the coupled basis \(|l\,s\,j\,m_{j}\rangle\) whenever spin–orbit coupling matters.
Solution. Six electrons in the \(f\) shell (\(l=3\)), all unpaired.
Hund's first rule — maximise \(S\): all six spins parallel, so \(S = 6\times\tfrac12 = 3\) and the multiplicity is \(2S+1 = 7\).
Hund's second rule — maximise \(L\) subject to that: with all spins parallel, the Pauli principle forces six different values of \(m_{l}\), and the largest total comes from taking the six largest available, \(m_{l} = 3,2,1,0,-1,-2\): \[ L = \sum m_{l} = 3+2+1+0-1-2 = 3 . \] The term is therefore \(^{7}F\).
Hund's third rule — the \(f\) shell holds \(14\) electrons, so six is less than half filled and the smallest \(J\) lies lowest: \[ J = |L-S| = |3-3| = 0 . \] The ground term is \(^{7}F_{0}\), and \[ \mu_{\text{eff}} = g_{J}\sqrt{J(J+1)}\,\mu_{B} = 0 . \]
Note that \(g_{J}\) is indeterminate (\(0/0\)) at \(J=0\); the moment vanishes because \(\sqrt{J(J+1)} = 0\), not because of anything \(g_{J}\) does. This is the Eu3+ / Sm2+ case. Experimentally Eu3+ salts are weakly paramagnetic, because the \(^{7}F_{1}\) level lies only about \(300\,\mathrm{cm}^{-1}\) above the ground state and is thermally populated at room temperature — Van Vleck paramagnetism. Within Hund's rules, which is what the question asks for, the answer is exactly \(0\).
Solution. A hydrogenic radial function \(R_{n\ell}\) carries the exponential \(e^{-Zr/n\abohr}\) and has \(n-\ell-1\) radial nodes. For \(2s\) we have \(n=2\), \(\ell=0\), so the exponential is \(e^{-r/2\abohr}\) and there is exactly one radial node: \[ R_{20}(r) \propto \left(2 - \frac{r}{\abohr}\right)e^{-r/2\abohr}, \] which vanishes at \(r = 2\abohr\). Only option (b) has both features.
Identify the distractors by the same two rules. Option (a), \(e^{-r/\abohr}\) with no node, is \(1s\). Option (c) behaves as \(r^{\ell}=r^{1}\) near the origin with the \(n=2\) exponential, so it is \(2p\). Option (d) goes as \(r^{2}\) with an \(n=3\) exponential, so it is \(3d\). In general \(R_{n\ell} \propto r^{\ell}\,(\text{polynomial of degree } n-\ell-1)\, e^{-Zr/n\abohr}\) — the small-\(r\) power gives \(\ell\) and the exponential gives \(n\).
Solution. By the Landé interval rule \(\eqref{eq:lande-interval}\), \[ \Delta E_{1} = A(J+1) = 22\,\mathrm{cm}^{-1}, \qquad \Delta E_{2} = A(J+2) = 33\,\mathrm{cm}^{-1} . \] Dividing, \[ \frac{J+1}{J+2} = \frac{22}{33} = \frac23 \quad\Longrightarrow\quad 3J+3 = 2J+4 \quad\Longrightarrow\quad J = 1 . \] The three levels are \(J = 1,2,3\). A triplet (\(2S+1 = 3 \Rightarrow S=1\)) with \(J\) running \(1,2,3\) requires \(L = 2\), i.e.\ \(^{3}D_{1,2,3}\).
Solution. \(\mathrm{Cr}(3d^{6})\): one orbital doubly occupied, four singly, so \(S = 4\times\tfrac12 = 2\) and \(L = 2\). More than half filled \(\Rightarrow\) highest \(J\): \(J = L+S = 4\), term \(^{5}D_{4}\).
\(\mathrm{Cr}^{2+}(3d^{4})\): four singly occupied orbitals, so again \(S=2\), \(L=2\). Less than half filled \(\Rightarrow\) lowest \(J\): \(J = \lvert L-S\rvert = 0\), term \(^{5}D_{0}\).
Hence \(J\) falls from \(4\) to \(0\): a decrease of 4.
Solution. For \(np^{2}\) the terms are \(^{1}S_{0}\), \(^{3}P_{0,1,2}\), \(^{1}D_{2}\). Hund's rule 1 selects the triplet; rule 3 (less than half filled) selects the lowest \(J\): \[ \boxed{^{3}P_{0}} . \]
\(^{12}\mathrm{Mg}\): \(1s^{2}2s^{2}2p^{6}3s^{2}\) — all closed, \(\Rightarrow{}^{1}S_{0}\).
\(^{13}\mathrm{Al}\): \(\dots3s^{2}3p^{1}\) — one \(p\) electron, less than half filled \(\Rightarrow{}^{2}P_{1/2}\).
\(^{17}\mathrm{Cl}\): \(\dots3s^{2}3p^{5}\) — more than half filled \(\Rightarrow{}^{2}P_{3/2}\).
\(^{18}\mathrm{Ar}\): \(\dots3s^{2}3p^{6}\) — closed \(\Rightarrow{}^{1}S_{0}\).
Solution. For \(d^{2}\) (\(\ell_{1}=\ell_{2}=2\)), \(L = 0,1,2,3,4\) and \(S = 0,1\), but Pauli restricts the combinations. The allowed terms are \[ ^{1}S_{0},\ ^{1}D_{2},\ ^{1}G_{4},\ ^{3}P_{0,1,2},\ ^{3}F_{2,3,4} . \] \(^{1}F_{3}\) does not appear — for equivalent \(d\) electrons the singlets are restricted to even \(L\) (\(S\), \(D\), \(G\)).
Solution. Two non-equivalent \(p\) electrons have no Pauli restriction, so \(S = 0,1\) and \(L = 0,1,2\) independently: \[ ^{1}S,\ ^{3}S,\ ^{1}P,\ ^{3}P,\ ^{1}D,\ ^{3}D . \]
Contrast with equivalent \(p^{2}\), which gives only \(^{1}S\), \(^{3}P\), \(^{1}D\). The single word “non-equivalent” decides the answer.
Solution. For \(3d^{3}\), three singly occupied orbitals with parallel spins: \[ S = \tfrac32, \qquad L = 2+1+0 = 3 \ (F), \qquad J = \lvert L-S\rvert = \tfrac32 \] (less than half filled), giving \(^{4}F_{3/2}\).
Solution. \(^{12}\mathrm{C}\) is \(2p^{2}\): terms \(^{3}P_{0,1,2}\), \(^{1}D_{2}\), \(^{1}S_{0}\). Rule 1 puts the triplet lowest; rule 3, since \(p^{2}\) is less than half filled, gives normal order: \[ E\!\left(^{3}P_{0}\right) < E\!\left(^{3}P_{1}\right) < E\!\left(^{3}P_{2}\right) < E\!\left(^{1}D_{2}\right) . \]
Solution. Neutral bromine is \(\text{[Ar]}\,3d^{10}4s^{2}4p^{5}\), so the ion is \[ \text{Br}^{+} : \text{[Ar]}\,3d^{10}4s^{2}4p^{4}. \] Only the open \(4p^{4}\) sub-shell matters. Four electrons in a \(p\) shell is two holes, and holes give the same terms as electrons, so \(p^{4}\) has the terms of \(p^{2}\): \[ ^{3}P,\quad ^{1}D,\quad ^{1}S . \] Hund's first rule picks the largest multiplicity, \(^{3}P\) (\(S=1\), \(L=1\), so \(J = 0,1,2\)). Hund's third rule then asks whether the sub-shell is more or less than half filled: \(p^{4}\) against a half filling of \(p^{3}\), so it is more than half filled and the largest \(J\) lies lowest: \[ J = L+S = 2 \qquad\Longrightarrow\qquad \boxed{^{3}P_{2}} . \]
Apply the half-filled test to the configuration of the ion, not the neutral atom — \(4p^{4}\), not \(4p^{5}\). Two useful comparisons: oxygen (\(2p^{4}\)) has the same ground term \(^{3}P_{2}\), while neutral bromine (\(4p^{5}\), one hole) is \(^{2}P_{3/2}\). Option (d), \(^{4}S_{3/2}\), is the ground term of a half-filled \(p^{3}\) shell such as nitrogen, and is impossible here because \(p^{4}\) cannot reach \(S=\tfrac32\).
Solution. The doubly excited configuration \(2s2p\) lies at \(58.3\,\mathrm{eV}\), far above the first ionization limit of \(24.6\,\mathrm{eV}\). It is therefore embedded in the \(\text{He}^{+} + e^{-}\) continuum, and decays not by emitting a photon but by autoionization (the Auger process): \[ \text{He}(2s2p) \longrightarrow \text{He}^{+}(1s) + e^{-}. \] The \(2s2p\) configuration gives a \(P\) term: \(L=1\), and its parity is \((-1)^{l_{1}+l_{2}} = (-1)^{0+1} = -1\), i.e.\ odd.
Autoionization is caused by the electrostatic electron–electron repulsion, which conserves both total orbital angular momentum and parity. In the final state the residual ion is \(\text{He}^{+}(1s)\), with \(l=0\) and even parity, so the departing electron must carry all of both: \[ L_{\text{total}} = 1 \Longrightarrow l = 1, \qquad \text{parity } (-1)^{l} = -1 \Longrightarrow l \text{ odd}. \] Both conditions give \(l = 1\): a \(p\)-wave electron.
The printed solution argues only that “the electron is more likely to come from the \(p\)-orbital rather than the \(s\)-orbital”. That is not a valid argument. Autoionization is a two-electron rearrangement — one electron drops to \(1s\) while the other is ejected — so there is no sense in which the outgoing electron “came from” a particular orbital. The value of \(l\) is fixed by conservation of total orbital angular momentum and of parity, as above. The numerical answer is unaffected.
Solution. The terms of \(2p^{2}\) are \(^{3}P\), \(^{1}D\), \(^{1}S\). Rule 1 puts the highest multiplicity lowest, so \(^{3}P\) is the ground term; rule 2 puts the larger \(L\) lower among the singlets: \[ ^{3}P < {}^{1}D < {}^{1}S . \]
Solution. With \(s = \tfrac32\) there are \(2s+1 = 4\) spin states, so the degeneracy of a level is \(2j+1 \to 4\) for an \(s\) orbital: each \(s\) subshell holds 4 electrons instead of 2. Filling \(Z=5\) electrons in order: \[ 1s^{4},\ 2s^{1} . \]
In general a subshell of orbital quantum number \(\ell\) would hold \((2s+1)(2\ell+1) = 4(2\ell+1)\) electrons: \(s\to4\), \(p\to12\), \(d\to20\).