(a) How do Stokes lines appear in Raman spectrum as per classical and quantum theory of Raman effect ? 20 marks (b) What is Lamb…

(a) Stokes lines in the Raman spectrum

In the classical picture, an incident field E=E0 cos ωt makes the molecular polarizability oscillate as α=α0+(∂α/∂Q)Q, with Q=Q0 cos ω_v t. The induced dipole p=αE contains frequencies ω, ω±ω_v. The ω term gives Rayleigh scattering; ω−ω_v gives Stokes lines, lower in frequency because the molecule gains vibrational energy; ω+ω_v gives anti-Stokes lines. The scattered intensity is proportional to (∂α/∂Q)², so only vibrations that modulate polarizability are seen. The classical result explains the frequency shifts but predicts Stokes and anti-Stokes lines of comparable strength, contrary to experiment; quantum theory supplies the populations and selection rules. C.V. Raman’s 1928 Calcutta experiment used a mercury arc, polarizers, a liquid sample and a spectrograph to observe these weak shifted lines.

Quantum mechanically, a photon hν is scattered through a virtual intermediate state, not by real absorption. If the molecule starts in v = 0 and ends in v = 1, the scattered photon has energy hν − hν_v (Stokes); if it starts in v = 1 and ends in v = 0, the photon has hν + hν_v (anti-Stokes). Rotational Raman lines follow ΔJ = 0, ±2. The virtual states are off-resonant, so energy conservation is satisfied by the molecular vibrational change. The Kramers–Heisenberg dispersion formula gives the amplitude as a sum over states n of products of dipole matrix elements divided by energy denominators, with both emission-then-absorption and absorption-then-emission terms. Placzek’s polarizability theory reduces the intensity to the square of the relevant polarizability tensor component, α_ij (α_xx, α_xy, α_zz, etc.), so a vibration is Raman active only if α_ij changes with the normal coordinate; for a harmonic oscillator Δv=±1. In centrosymmetric molecules, modes that change dipole moment are IR active while modes that change polarizability are Raman active, giving mutual exclusion. The Stokes/anti-Stokes ratio is I_S/I_AS ≈ (ν_S/ν_AS)^4 (N_v+1)/N_v ≈ (ν_S/ν_AS)^4 exp(hν_v/kT), where N_v is the mean vibrational occupation; hence Stokes lines dominate at ordinary temperatures.

(b) The Lamb shift

Dirac theory predicts degeneracy of hydrogen 2S1/2 and 2P1/2, but Lamb and Retherford (1947) found 2S1/2 higher by about 1058 MHz. Bethe's 1947 non-relativistic calculation treated the electron's interaction with the quantized radiation field and introduced mass renormalization to remove the self-energy divergence, obtaining about 1040 MHz. In Welton's picture, zero-point fluctuations of the field jiggle the electron and smear its position; S states, which have finite probability at the nucleus, feel the weakened Coulomb attraction most and are raised. In second quantization, the Dirac and electromagnetic fields are expanded in creation and annihilation operators, and the interaction Hamiltonian is H_int=−e∫ψbar γ^μ A_μ ψ d³x. This allows the electron to emit and reabsorb a virtual photon, represented by the one-loop self-energy Feynman diagram; vacuum fluctuations and radiation reaction shift the energy levels. The one-loop correction is integrated over virtual photon momentum; its divergence is absorbed into the electron mass, while the finite remainder gives the shift. The bare mass and charge are divergent, so renormalization fixes them to observed values, leaving a finite Lamb shift that confirms QED. Vacuum polarization adds a small negative contribution (about −27 MHz).

(c) Electron Paramagnetic Resonance (EPR)

Electron Paramagnetic Resonance detects transitions between electron spin sublevels in a magnetic field. The electron spin magnetic moment is μ=−g μ_B S/ℏ; in B0 the m_s=±1/2 levels split. For an unpaired electron, H=μ_B g S·B/ℏ, giving ΔE=gμ_BB and resonance hν=gμ_BB, or B0=hν/(gμ_B). Hyperfine coupling, H_hf=A I·S, splits the line and identifies nuclear spin and coupling constants; in solids the g-factor is anisotropic, giving orientation-dependent resonance. In powders, anisotropy produces broadened spectra; in single crystals, orientation can be resolved. EPR uses microwave frequencies and lower fields because the electron magnetic moment is much larger than the nuclear magneton.

Differences from NMR:

• Resonating moment: EPR involves the electron magnetic moment (μ_B ≈ 1836 μ_N); NMR involves nuclear moments (of order μ_N).
• Frequency and field: EPR uses microwaves (X-band ≈ 9.5 GHz at ≈ 0.34 T); NMR uses radio frequencies (tens to hundreds of MHz at a few to ≈ 20 T).
• Relaxation: EPR relaxation times are much shorter (strong spin–orbit and dipolar coupling), so lines are broader; NMR relaxation is slower and lines are sharp.
• Samples: EPR needs unpaired electrons (free radicals, transition-metal ions, defects); NMR works on almost all molecules containing magnetic nuclei (¹H, ¹³C).
• Sensitivity and information: EPR is far more sensitive per spin and probes the electronic environment (g-values, hyperfine structure); NMR gives chemical shifts and spin–spin couplings that map molecular structure. bulk nuclei, needs high concentration |

Applications include: detection and identification of free radicals in chemical reactions, polymers and biological systems (for example, radical pairs in photosynthetic reaction centres); study of transition-metal complexes and catalysts; defects and dangling bonds in semiconductors; radiation dosimetry (alanine dosimeters) and ESR dating of tooth enamel, shells and quartz, where trapped electrons record accumulated dose; detection of irradiated food; and site-directed spin labelling of proteins, with pulsed EPR (DEER) measuring nanometre distances. A spectrometer uses a microwave source (klystron or Gunn diode), a resonant cavity and a swept magnetic field, recording the first derivative of absorption.