In the rotational-vibrational spectrum of an idealized carbon monoxide (CO) molecule, ignoring rotational-vibrational coupling, two transitions between adjacent vibrational levels with wavelength λ₁ and λ₂, correspond to the rotational transition from J' = 0 to J" = 1, and J' = 1 to J" = 0, respectively. Given that the reduced mass of CO is 1.2x^​(-26) kg, equilibrium bond length of CO is 0.12 nm and vibrational frequency is 5×10¹³ Hz, the ratio of λ₁/λ₂ is closest to

Correct option is C

Solution:

​Since there is no coupling between rotational and vibrational spectra, we will ignore that. During the transition, we observe the change in rotational energy.

As we know, the rotational energy in terms of wave number is given by:

ε_J = B × J × (J + 1)

The change in wave number during the transition from J' = 0 to J'' = 1 is:

ν'₁ = ε_J'' - ε_J' = 2B = 1 / λ₁ ....(1)

The change in wave number during the transition from J' = 1 to J'' = 0 is:

​$$\nu'_2 = \varepsilon_{J'} - \varepsilon_{J''} = 2B' = \frac{1}{\lambda_2} \quad \text{....(2)}\\\ \\\text{Divide (2) by (1):} \quad \frac{\lambda_1}{\lambda_2} = \frac{B'}{B''} < 1 \quad \therefore B \propto \frac{1}{r_0^2}$$​​