(a) (i) In a diatomic molecule when one constituent atom is replaced by one of its heavier isotopes, what change takes place in…

(a)(i) When one atom in a diatomic molecule is replaced by a heavier isotope, the reduced mass μ = m₁m₂/(m₁+m₂) increases. The bond length is almost unchanged, so the moment of inertia I = μr² increases. Since the rotational constant is B = h/(8π²Ic), B decreases. For a rigid rotor, rotational lines occur at 2B, 4B, 6B, … so the line spacing 2B decreases and the whole rotational spectrum shifts to lower frequency/wavenumber. If the original molecule is homonuclear and has no pure rotational dipole spectrum, isotopic substitution can break the symmetry and make a weak rotational spectrum appear.

(a)(ii) Taking the usual case of replacing one H atom by D in H₂ (i.e. H₂ → HD), let m_H = m and m_D ≈ 2m. Then μ(H₂) = m/2, μ(HD) = m(2m)/(m+2m) = 2m/3. So μ(HD)/μ(H₂) = (2m/3)/(m/2) = 4/3. Since B ∝ 1/μ, B(HD) = (3/4)B(H₂). Therefore ΔB = B(HD) − B(H₂) = −B(H₂)/4. Using B(H₂) ≈ 60.8 cm⁻¹, ΔB ≈ −15.2 cm⁻¹. If both H atoms are replaced (H₂ → D₂), then B(D₂) = B(H₂)/2 and ΔB ≈ −30.4 cm⁻¹.

(a)(iii) Rigid rotor: E_J = B J(J+1). Non-rigid rotor: E_J = B J(J+1) − D_c J²(J+1)², with D_c > 0 due to centrifugal stretching. Selection rule ΔJ = ±1.

Rigid-rotor levels lie at 0, 2B, 6B, 12B (J = 0 to 3) and give lines at 2B, 4B, 6B, … with equal spacing 2B. For the non-rigid rotor the levels are lowered to 0, 2B − 4D_c, 6B − 36D_c, 12B − 144D_c.

Non-rigid transition wavenumbers: ν = 2B(J+1) − 4D_c(J+1)³. So lines occur at 2B − 4D_c, 4B − 32D_c, 6B − 108D_c, … The spacing decreases as J increases; the lines crowd together at higher frequency. The rigid-rotor spectrum is therefore a series of equally spaced lines, while the non-rigid-rotor spectrum shows unequal spacing due to centrifugal distortion.

(b)(i) Anharmonicity is described by a Morse-type potential rather than a pure harmonic potential. In the harmonic oscillator, levels are equally spaced: E_v = (v+1/2)ħω, and the selection rule is Δv = ±1, giving only one fundamental band. With anharmonicity, E_v = ω_e(v+1/2) − ω_e x_e(v+1/2)² + … The level spacing decreases as v increases, and the levels converge near the dissociation limit. The selection rule relaxes so that weak overtones Δv = ±2, ±3, … also appear. The fundamental band shifts slightly, overtones occur at frequencies lower than exact multiples of the fundamental, and hot bands arise from thermally populated excited vibrational levels. The vibrational spectrum is therefore no longer a single equally spaced set of lines.

(b)(ii) For a rigid rotor, the angular momentum is L = √(J(J+1)) ħ = Iω. Hence ω = √(J(J+1)) ħ / I, and the average period is T = 2π/ω = 2πI/[ħ√(J(J+1))]. For J = 3, √(J(J+1)) = √12 = 2√3. Given I = 0.0264×10⁻⁴⁵ kg.m² = 2.64×10⁻⁴⁷ kg.m² and ħ = 1.0546×10⁻³⁴ J.s, T = 2π(2.64×10⁻⁴⁷)/[(1.0546×10⁻³⁴)(3.464)] T = 4.54×10⁻¹³ s. The average rotational frequency is 1/T = 2.20×10¹² Hz. T ≈ 4.54×10⁻¹³ s.

(c) For σₓ = [[0, 1], [1, 0]], the eigenvalue equation is det(σₓ − λI) = 0: λ² − 1 = 0, so λ = ±1. For λ = +1: b = a, choose a = 1, b = 1. Normalized eigenvector: (1/√2)[[1], [1]]. For λ = −1: b = −a, choose a = 1, b = −1. Normalized eigenvector: (1/√2)[[1], [−1]].

For σᵧ = [[0, −i], [i, 0]], again det(σᵧ − λI) = 0 gives λ² − 1 = 0, so λ = ±1. For λ = +1: −i b = a and i a = b. Choose a = 1, b = i. Normalized eigenvector: (1/√2)[[1], [i]]. For λ = −1: −i b = −a and i a = −b. Choose a = 1, b = −i. Normalized eigenvector: (1/√2)[[1], [−i]]. These eigenvectors are mutually orthogonal and normalized up to an overall phase factor.