UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper II
Q15(b). Equilibrium vibrational frequency of an anharmonic oscillator
Question
A diatomic molecule shows an intense absorption (v=0→ v=1) at 2886 cm⁻¹. The anharmonicity constant of this molecule (xₑ) is 0.0174. Calculate the equilibrium vibrational frequency of this molecule.
Answer in brief
νₑ=2990 cm⁻¹.
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
For the anharmonic oscillator G(v)=νₑ(v+12)-νₑxₑ(v+12)².
Δ G(0→1)=G(1)-G(0)=νₑ(1-2xₑ)=2886 cm⁻¹
νₑ=(2886)/(1-2(0.0174))=(2886)/(0.9652)=2990 cm⁻¹
(This is the value for H³⁵Cl, νₑ=2990 cm⁻¹; the overtone v=0→2 lies at 2νₑ(1-3xₑ)=5668 cm⁻¹.)
Coversanharmonic oscillator
All parts of Question 15
- 15(a) Bond length of CN from the rotational constant10 marks
- 15(b) Equilibrium vibrational frequency of an anharmonic oscillator10 marks