(a) (i) What are hot bands in vibrational spectroscopy ? Why are they called so ? (8 marks) (ii) The fundamental and 2nd…

(a) (i) Hot bands A hot band is a vibrational transition that starts from a thermally excited level, e.g. v = 1 → 2, instead of from the ground level v = 0. Why "hot": the band's intensity depends on the population of v = 1, N₁/N₀ = e^−hcω̃/k_BT, which is appreciable only at higher temperature. The band grows in strength when the sample is heated and fades on cooling. Position: with anharmonicity, G(v) = ω̃ₑ(v + ½) − ω̃ₑ xₑ(v + ½)², so ν̃(0 → 1) = ω̃ₑ − 2ω̃ₑ xₑ and ν̃(1 → 2) = ω̃ₑ − 4ω̃ₑ xₑ. The hot band therefore lies about 2ω̃ₑ xₑ below the fundamental, as a weak satellite on its low-wavenumber side (for a harmonic oscillator it would coincide with the fundamental). Hot bands are prominent for low-frequency modes (hcω̃ ~ k_BT) and in flames and planetary atmospheres, and their intensity ratio to the fundamental can be used to estimate temperature.

(a) (ii) ¹⁴N¹⁶O: ω̃ₑ, anharmonicity and zero-point energy G(v) = ω̃ₑ(v + ½) − ω̃ₑ xₑ(v + ½)² Fundamental (0 → 1): ω̃ₑ − 2ω̃ₑ xₑ = 1876.06 cm⁻¹ Second harmonic (first overtone, 0 → 2): 2ω̃ₑ − 6ω̃ₑ xₑ = 3724.2 cm⁻¹ Subtracting twice the first from the second: −2ω̃ₑ xₑ = 3724.2 − 3752.12 = −27.92 ω̃ₑ xₑ = 13.96 cm⁻¹ ω̃ₑ = 1876.06 + 2(13.96) = 1903.98 cm⁻¹ (νₑ = cω̃ₑ = 5.71 × 10¹³ Hz) Anharmonicity constant xₑ = 13.96/1903.98 = 7.33 × 10⁻³ Zero-point energy: G(0) = ½ω̃ₑ − ¼ω̃ₑ xₑ = 951.99 − 3.49 = 948.5 cm⁻¹ = 1.88 × 10⁻²⁰ J (0.118 eV)

(b) Probability of the new ground state after symmetric widening The box widens suddenly, so the wavefunction is unchanged just after the change and P = |⟨φ₁|ψ⟩|² (sudden approximation). Take the centre of the box as origin. Old ground state (width a): ψ(x) = √(2/a) cos(πx/a) for |x| ≤ a/2, zero elsewhere. New ground state (width 2a): φ₁(x) = √(2/2a) cos(πx/2a) = √(1/a) cos(πx/2a) for |x| ≤ a. Overlap (only |x| ≤ a/2 contributes): c = (√2/a) ∫_−a/2^a/2 cos(πx/a) cos(πx/2a) dx = (√2/2a) ∫_−a/2^a/2 [cos(3πx/2a) + cos(πx/2a)] dx = (√2/2a) [ (2a/3π) · 2 sin(3π/4) + (2a/π) · 2 sin(π/4) ] = (√2/2a) [ 2√2a/3π + 2√2a/π ] = (√2/2a)(8√2a/3π) = 8/(3π) = 0.849 P = c² = 64/(9π²) ≈ 0.72 So there is a 72% chance of finding the particle in the new ground state. (If instead the box were extended on one side only, from [0, a] to [0, 2a], the result would be 32/(9π²) ≈ 0.36; the symmetric widening keeps the larger overlap.)

(c) (i) Why J = 0 → 1 is not the most intense rotational line The intensity of a line J → J + 1 is proportional (mainly) to the population of the lower level J: N_J/N₀ = (2J + 1) e^−hcBJ(J + 1)/k_BT The degeneracy factor (2J + 1) (the 2J + 1 values of M_J) increases with J, while the Boltzmann factor decreases. For typical molecules at room temperature hcB ≪ k_BT, so the degeneracy wins at first and the population rises through several J values before the exponential takes over. J = 0 has degeneracy 1 and is therefore not the most populated level. The maximum is at dN_J/dJ = 0 ⇒ J_max = √(k_BT/2hcB) − ½ Example: CO (B ≈ 1.93 cm⁻¹) at 300 K gives J_max ≈ 7. Only for very light molecules or very low temperatures (hcB comparable to k_BT) is the J = 0 → 1 line the strongest.

(c) (ii) Rotational Raman lines of H₂ and effect of nuclear spin Reduced mass: μ = m_H/2 = 8.365 × 10⁻²⁸ kg Moment of inertia: I = μr² = 8.365 × 10⁻²⁸ × (0.742 × 10⁻¹⁰)² = 4.61 × 10⁻⁴⁸ kg m² Rotational constant: B = h/(8π²cI) = 6.626 × 10⁻³⁴/(8π² × 3.0 × 10¹⁰ × 4.61 × 10⁻⁴⁸) = 60.8 cm⁻¹ Rotational Raman selection rule ΔJ = ±2; the lines J → J + 2 are displaced from the exciting line by Δν̃ = B[(J + 2)(J + 3) − J(J + 1)] = B(4J + 6)

• J = 0 → 2: 6B = 364.7 cm⁻¹
• J = 1 → 3: 10B = 607.8 cm⁻¹
• J = 2 → 4: 14B = 850.9 cm⁻¹
• J = 3 → 5: 18B = 1094.1 cm⁻¹ These appear as Stokes lines on the low-frequency side of the exciting line, with anti-Stokes lines at the same shifts on the high side; the first line is at 6B and the spacing is 4B = 243 cm⁻¹. Nuclear-spin effect: the protons are fermions (I = ½), so the total wavefunction must be antisymmetric under their exchange. The symmetric nuclear-spin triplet (ortho-H₂, weight 3) goes with odd J, and the antisymmetric singlet (para-H₂, weight 1) with even J. Lines from odd J (607.8, 1094.1 cm⁻¹) are therefore about three times as strong as those from even J (364.7, 850.9 cm⁻¹): an intensity alternation of 1 : 3. No lines are missing.