Which statements about the Raman effect are correct?

At room temperature Stokes lines are more intense than anti-Stokes lines

Rotational Raman transitions obey ΔJ = 0, ±2

A Raman-active mode must change the polarisability

At room temperature Stokes lines are more intense than anti-Stokes lines

Rotational Raman transitions obey ΔJ = 0, ±2

A Raman-active mode must change the polarisability

The symmetric stretch of CO₂ is infrared-active

The Franck–Condon principle states that during an electronic transition in a molecule:

The internuclear distance does not change, so the transition is vertical

The vibrational quantum number cannot change

The rotational quantum number changes by ±2

The Paschen–Back effect is observed when:

The Zeeman energy μ_B B exceeds the fine-structure splitting

The field is too weak to decouple L and S

Atomic and Molecular Physics: Spectra, Coupling Schemes, Fine and Hyperfine Structure, Zeeman and Stark Effects, Molecular Spectra, Raman, NMR, ESR, X-ray and Mössbauer

Section 8 of the GATE Physics paper, the applied end of quantum mechanics. It follows the syllabus: the spectra of one- and many-electron atoms; spin–orbit interaction and the L–S and j–j coupling schemes; fine and hyperfine structure; the Zeeman, Paschen–Back and Stark effects; electric-dipole transitions and their selection rules; the rotational and vibrational spectra of diatomic molecules; electronic transitions and the Franck–Condon principle; the Raman effect; and the NMR, ESR, X-ray and Mössbauer spectroscopies. The constants used are the Bohr magneton μ_B = 9.274 × 10⁻²⁴ J/T = 5.788 × 10⁻⁵ eV/T, h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s, and hc = 1240 eV nm.

1. One- and many-electron spectra, L–S and j–j coupling

In an alkali atom the valence electron sees the nucleus screened by the core, but penetrates the core more for low l. The levels are E = −13.6/(n − δₗ)² eV with a quantum defect δₗ that is largest for s electrons, so the l-degeneracy of hydrogen is lifted: in sodium 3s lies below 3p below 3d. In many-electron atoms the residual electrostatic interaction and the spin–orbit interaction compete, and which is stronger decides the coupling scheme.

L–S (Russell–Saunders) coupling, for light atoms: the electrons’ lᵢ add to L and their sᵢ to S, and then L and S couple to J. Terms are written ^2S+1L_J. Within a term the Landé interval rule holds: the spacing between J and J − 1 is proportional to J.
j–j coupling, for heavy atoms where spin–orbit dominates: each electron’s lᵢ and sᵢ first form jᵢ, and the jᵢ then add to J.
Hund’s rules for the ground term of an equivalent-electron configuration: maximum S; then maximum L; then J = |L − S| if the shell is less than half full and L + S if more. Carbon (2p²) is ³P₀; oxygen (2p⁴) is ³P₂.

2. Fine and hyperfine structure

Fine structure is the splitting of a level by the spin–orbit interaction ξ(r)L·S together with the relativistic correction; in hydrogen both are of order α²Eₙ with α ≈ 1/137, and the result depends only on n and j, so 2s₁/₂ and 2p₁/₂ stay degenerate (until the Lamb shift). In sodium the 3p level splits into ²P₁/₂ and ²P₃/₂, giving the D lines at 589.6 nm and 589.0 nm. Hyperfine structure comes from the nuclear magnetic moment: the nuclear spin I couples with J to F = I + J, and splittings are about a thousand times smaller than fine structure, since the nuclear magneton is 1836 times smaller than μ_B. Hydrogen’s ground state splits into F = 1 and F = 0, 5.9 μeV apart — the 21 cm line at 1420 MHz.

3. Zeeman, Paschen–Back and Stark effects

In a weak field B the level J splits into 2J + 1 sublevels, ΔE = g_J μ_B B m_J, with the Landé factor g_J = 1 + [J(J + 1) + S(S + 1) − L(L + 1)]/[2J(J + 1)]. For ²P₃/₂, g = 4/3; for ²P₁/₂, g = 2/3; for ²S₁/₂, g = 2. When S = 0, g = 1 for every level and each line splits into three equally spaced components, ΔE = μ_B B (the normal Zeeman effect, 57.9 μeV per tesla, 14.0 GHz per tesla); when S ≠ 0 the unequal g-factors give more components (the anomalous Zeeman effect), as for the sodium D lines (4 and 6 components).

Paschen–Back effect: when μ_B B exceeds the fine-structure splitting, L and S decouple and precess separately about B; ΔE = μ_B B(m_L + 2m_S), and the pattern returns to a normal-Zeeman-like triplet.
Stark effect: an electric field mixes states of opposite parity. Hydrogen n = 2, where 2s and 2p are degenerate, shows a linear effect (±3eEa₀); a non-degenerate level such as the ground state shows only a quadratic effect, ΔE = −½αE², through its polarisability α.

4. Selection rules, molecular spectra, Franck–Condon and Raman

Electric-dipole selection rules follow from the vector character of the dipole operator: parity must change, Δl = ±1 for the jumping electron, Δm = 0, ±1, and in L–S coupling ΔS = 0, ΔL = 0, ±1 and ΔJ = 0, ±1 with 0 ↛ 0 forbidden. Transitions violating them are "forbidden" — slower by orders of magnitude, not impossible.

Diatomic spectra
Spectrum	Levels and rule	Pattern
Pure rotational (microwave)	F(J) = BJ(J + 1), B = h/(8π²cI); ΔJ = ±1; needs a permanent dipole	lines at 2B(J + 1), equally spaced by 2B
Vibrational (infrared)	G(v) = ω̄(v + ½) − ω̄x(v + ½)²; Δv = ±1 (harmonic), overtones weak	needs a changing dipole: HCl yes, N₂ no
Rotation–vibration	Δv = +1 with ΔJ = ±1	P and R branches, a gap at the band origin
Rotational Raman	ΔJ = 0, ±2; needs anisotropic polarisability	first lines at ±6B, then spaced 4B

Worked: HCl-like lines spaced 20.7 cm⁻¹ give B = 10.35 cm⁻¹ and I = h/(8π²cB) = 6.626 × 10⁻³⁴/(8π² × 3 × 10¹⁰ cm/s × 10.35 cm⁻¹) = 2.70 × 10⁻⁴⁷ kg m². In an electronic transition the electrons jump far faster than the nuclei move, so the transition is vertical on the potential-energy diagram and its intensity to each upper vibrational level is the overlap of the vibrational wavefunctions — the Franck–Condon principle, which explains the vibrational progressions of electronic bands.

In the Raman effect light scattered inelastically is shifted by a molecular energy: Stokes lines at lower frequency (the molecule gains energy) and anti-Stokes lines at higher frequency, weaker at room temperature because fewer molecules start excited. A mode is Raman-active when it changes the polarisability; in molecules with a centre of symmetry, such as CO₂, the rule of mutual exclusion makes the symmetric stretch Raman-active and IR-inactive, and the antisymmetric stretch the reverse.

5. NMR, ESR, X-ray and Mössbauer spectroscopies

NMR: a nuclear spin in B₀ absorbs at the Larmor frequency ν = γB₀/2π; for protons γ/2π = 42.58 MHz/T, so 63.87 MHz in 1.5 T. The small chemical shift (parts per million) reports the electronic environment, and spin–spin coupling splits lines into multiplets.
ESR (EPR): an unpaired electron absorbs at hν = gμ_B B; with g ≈ 2.0023, that is about 28 GHz/T, so X-band (9.5 GHz) resonance falls near 0.34 T. Hyperfine coupling to nuclei splits the line into 2I + 1 components.
X-rays: an X-ray tube at voltage V gives a continuous spectrum with a sharp Duane–Hunt limit λ_min = hc/(eV) (41.3 pm at 30 kV) and characteristic lines. Moseley’s law for Kα, E ≈ 13.6 × (3/4)(Z − 1)² eV = 10.2(Z − 1)² eV, gives 8.0 keV for copper (Z = 29); the Z − 1 is the screening by the remaining K electron.
Mössbauer: a nucleus bound in a solid emits and absorbs γ-rays without recoil (the lattice takes the momentum), so the line has its natural width. A source is moved at velocity v to Doppler-shift the line by ΔE = (v/c)E; for the 14.4 keV line of ⁵⁷Fe, a shift of 4.8 × 10⁻⁸ eV needs v = 1 mm/s. Isomer shifts, quadrupole splittings and magnetic hyperfine sextets read the nucleus’s chemical and magnetic environment.

Key takeaways

L–S coupling for light atoms, j–j for heavy; Hund: max S, max L, then J = |L − S| below half-filling and L + S above.
g_J = 1 + [J(J + 1) + S(S + 1) − L(L + 1)]/[2J(J + 1)]; normal Zeeman spacing μ_B B; Paschen–Back when μ_B B beats fine structure.
E1: parity changes, Δl = ±1, ΔS = 0, ΔJ = 0, ±1 (0 ↛ 0). Rotational lines 2B apart need a permanent dipole; rotational Raman ΔJ = ±2 starts at 6B.
Franck–Condon: vertical electronic transitions weighted by vibrational overlap; mutual exclusion in centrosymmetric molecules.
NMR 42.58 MHz/T for protons; ESR hν = gμ_B B; Moseley E_Kα ≈ 10.2(Z − 1)² eV; Mössbauer ΔE/E = v/c.

Practice questions (16)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

In the normal Zeeman effect, with μ_B = 5.788 × 10⁻⁵ eV/T, the energy separation between adjacent components in a field of 2.0 T, in μeV, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 115.8
ΔE = μ_B B = 5.788 × 10⁻⁵ × 2.0 = 1.158 × 10⁻⁴ eV = 115.8 μeV. The total spread of the triplet (m = −1 to +1) is twice this, 231.5 μeV.
The Landé g-factor of the ²P₃/₂ level, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.33
L = 1, S = ½, J = 3/2: g = 1 + [15/4 + 3/4 − 2]/(2 × 15/4) = 1 + 2.5/7.5 = 4/3 = 1.33. Swapping the signs of S(S + 1) and L(L + 1) gives 0.67, which is the ²P₁/₂ value.
The Landé g-factor of the ²P₁/₂ level, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.67
L = 1, S = ½, J = ½: g = 1 + [3/4 + 3/4 − 2]/(2 × 3/4) = 1 − 0.5/1.5 = 2/3 = 0.67. The pure-spin value 2 belongs to ²S₁/₂, and 1 to a singlet level.
By Hund’s rules, the ground term of the carbon atom (configuration 2p²) is:
1. ³P₀
2. ³P₂
3. ¹D₂
4. ¹S₀
Show answer
Answer: A — ³P₀
Two p electrons: maximum S = 1, then maximum L consistent with the Pauli principle, L = 1 (a ³P term). The shell is less than half full, so J = |L − S| = 0: ³P₀. ³P₂ is the ground term of oxygen (2p⁴, more than half full); ¹D and ¹S are excited terms.
For an electric-dipole transition of a single electron, the orbital quantum number must change by:
1. ±1
2. 0
3. ±2
4. Any amount
Show answer
Answer: A — ±1
The dipole operator is a vector (l = 1) of odd parity, so the matrix element ⟨l′|r|l⟩ vanishes unless l′ = l ± 1. Δl = ±2 belongs to electric-quadrupole transitions and to rotational Raman scattering.
The pure rotational spectrum of a diatomic molecule shows equally spaced lines 20.7 cm⁻¹ apart. With h = 6.626 × 10⁻³⁴ J s and c = 3.00 × 10¹⁰ cm/s, the moment of inertia in units of 10⁻⁴⁷ kg m², to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.70
The spacing is 2B, so B = 10.35 cm⁻¹, and I = h/(8π²cB) = 6.626 × 10⁻³⁴/(78.957 × 3 × 10¹⁰ × 10.35) = 2.70 × 10⁻⁴⁷ kg m². Taking the spacing itself as B halves the answer to 1.35.
A linear molecule has rotational constant B = 2.0 cm⁻¹. The Raman shift of its first rotational Stokes line, in cm⁻¹, is ____.
Numerical answer — type the value.
Show answer
Answer: 12
Rotational Raman obeys ΔJ = ±2, so the first Stokes line is J = 0 → 2 with shift F(2) − F(0) = 6B = 12 cm⁻¹; later lines are 4B apart. The microwave rule ΔJ = ±1 would give 2B = 4 cm⁻¹.
Which of the following molecules show a pure rotational (microwave) absorption spectrum?
1. HCl
2. CO
3. N₂
4. H₂
Show answer
Answer: A — HCl; B — CO
A rotational absorption spectrum needs a permanent electric dipole moment, which heteronuclear diatomics (HCl, CO) have and homonuclear ones (N₂, H₂) do not. The homonuclear molecules still show rotational Raman spectra, because their polarisability is anisotropic.
Which statements about the Raman effect are correct?
1. At room temperature Stokes lines are more intense than anti-Stokes lines
2. Rotational Raman transitions obey ΔJ = 0, ±2
3. A Raman-active mode must change the polarisability
4. The symmetric stretch of CO₂ is infrared-active
Show answer
Answer: A — At room temperature Stokes lines are more intense than anti-Stokes lines; B — Rotational Raman transitions obey ΔJ = 0, ±2; C — A Raman-active mode must change the polarisability
Anti-Stokes scattering starts from excited levels, which are less populated by the Boltzmann factor. The polarisability is a rank-2 tensor, which allows ΔJ = 0, ±2. The CO₂ symmetric stretch keeps the dipole zero, so it is IR-inactive and Raman-active — mutual exclusion.
The Franck–Condon principle states that during an electronic transition in a molecule:
1. The internuclear distance does not change, so the transition is vertical
2. The vibrational quantum number cannot change
3. The rotational quantum number changes by ±2
4. The molecule must dissociate
Show answer
Answer: A — The internuclear distance does not change, so the transition is vertical
The electrons rearrange far faster than the nuclei move, so the transition is vertical on the potential-energy curves and its strength to each vibrational level is the overlap of vibrational wavefunctions. Δv is not restricted — that is exactly why progressions appear — and dissociation happens only when the vertical line lands above the upper curve’s limit.
The Paschen–Back effect is observed when:
1. The Zeeman energy μ_B B exceeds the fine-structure splitting
2. The field is too weak to decouple L and S
3. An electric field is applied
4. The atom has S = 0
Show answer
Answer: A — The Zeeman energy μ_B B exceeds the fine-structure splitting
In a strong field L and S precess independently about B, and ΔE = μ_B B(m_L + 2m_S). A weak field gives the anomalous Zeeman effect; an electric field gives the Stark effect; S = 0 gives the normal Zeeman effect at any field.
An ESR spectrometer operates at 9.5 GHz. For a free-radical with g = 2.0023, using h = 6.626 × 10⁻³⁴ J s and μ_B = 9.274 × 10⁻²⁴ J/T, the resonance field in mT, to the nearest integer, is ____.
Numerical answer — type the value.
Show answer
Answer: 339
B = hν/(gμ_B) = 6.626 × 10⁻³⁴ × 9.5 × 10⁹/(2.0023 × 9.274 × 10⁻²⁴) = 6.295 × 10⁻²⁴/1.857 × 10⁻²³ = 0.339 T = 339 mT. Leaving out g gives 679 mT.
Using Moseley’s law E(Kα) ≈ 10.2 (Z − 1)² eV, the Kα energy of copper (Z = 29), in keV, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 8.0
E = 10.2 × 28² = 10.2 × 784 = 7997 eV ≈ 8.0 keV. Using Z² instead of (Z − 1)² gives 8.6 keV — the screening by the other K electron is what the "−1" represents.
An X-ray tube is operated at 30 kV. Using hc = 1240 eV nm, the shortest wavelength in its continuous spectrum, in pm, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 41.3
λ_min = hc/(eV) = 1240 eV nm/30 000 eV = 0.04133 nm = 41.3 pm (the Duane–Hunt limit), where an electron gives all its kinetic energy to one photon. It does not depend on the target, unlike the characteristic lines.
In a ⁵⁷Fe Mössbauer experiment (γ-ray energy 14.4 keV), the source must be moved to shift the line by 4.8 × 10⁻⁸ eV. With c = 3.00 × 10⁸ m/s, the required source speed, in mm/s, is ____.
Numerical answer — type the value.
Show answer
Answer: 1
ΔE/E = v/c, so v = c ΔE/E = 3 × 10⁸ × 4.8 × 10⁻⁸/(1.44 × 10⁴) = 1.0 × 10⁻³ m/s = 1 mm/s. The tiny speed is the point: recoil-free lines are so narrow that millimetres per second sweep across them.
Protons have γ/2π = 42.58 MHz/T. Their NMR frequency in a 1.5 T magnet, in MHz, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 63.87
ν = (γ/2π)B = 42.58 × 1.5 = 63.87 MHz. Multiplying γ/2π by 2π again gives the angular frequency in Mrad/s (401.3), not the frequency in MHz.