Q1(m). Rotational constants of ¹⁴N¹⁶O and ¹⁵N¹⁶O

Answer in brief

(i) B > B′: the heavier ¹⁵N raises the reduced mass μ and the moment of inertia I; the bond length is unchanged, so B = h/8π²cI falls.

Key points

• B = h/8π²cI, I = μr²; isotopic substitution leaves the electronic potential and hence r unchanged (Born–Oppenheimer); only μ changes.
• Reduced mass: μ = 7.4664 u for ¹⁴N¹⁶O and μ′ = 7.7408 u for ¹⁵N¹⁶O.
• B/B′ = μ′/μ = 7.7408/7.4664 = 1.0368 ⇒ B > B′ (option (i)).
• With B = 1.672 cm⁻¹ for ¹⁴NO, B′ = 1.613 cm⁻¹; the line spacing 2B′ is smaller and lines shift to lower wavenumber.
• The same change of μ lowers the vibration wavenumber: ω̃′/ω̃ = √(μ/μ′) = 0.9821.