UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper II
Q15(a). Bond length of CN from the rotational constant
Question
The rotational constant of ¹²C¹⁴N (rigid rotor) is 1.91 cm⁻¹. Calculate the equilibrium bond length (in pm) of CN.
Answer in brief
rₑ=116.9 pm (≈117 pm).
Key points
- I=2.8×10⁻⁴⁴191=1.466×10⁻⁴⁶ kg m²
- μ=(12×14)/(12+14) amu=6.462 amu=6.462×1.66×10⁻²⁷=1.073×10⁻²⁶ kg
- r=√( Iμ)=1.466×10⁻⁴⁶1.073×10⁻²⁶=1.367×10⁻²⁰ m=1.169×10⁻¹⁰ m=116.9 pm
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
B=(h)/(8π²cI), so I=(h)/(8π²cB) with B=1.91 cm⁻¹=191 m⁻¹ and (h)/(8π²c)=2.8×10⁻⁴⁴ kg m.
- I=2.8×10⁻⁴⁴191=1.466×10⁻⁴⁶ kg m²
- μ=(12×14)/(12+14) amu=6.462 amu=6.462×1.66×10⁻²⁷=1.073×10⁻²⁶ kg
- r=√( Iμ)=1.466×10⁻⁴⁶1.073×10⁻²⁶=1.367×10⁻²⁰ m=1.169×10⁻¹⁰ m=116.9 pm
Coversrotational spectraBond length of CN from the rotational constant
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