Question

The separation of energy levels in the rotational spectrum of CO is 3.8626 cm$^{-1}$. The bond length (assume it does not change during rotation) of CO in (rounded off to two decimal places) is ________. 
(Given: $h = 6.626 \times 10^{-34}$ J s; $N_A = 6.022 \times 10^{23}$ mol$^{-1}$; $c = 3.0 \times 10^8$ m s$^{-1}$; atomic mass of C = 12 g mol$^{-1}$, atomic mass of O = 16 g mol$^{-1}$)

Hint:

For diatomic molecules, $B$ in cm$^{-1}$ relates to bond length by $r = \sqrt{\frac{h}{8\pi^2 c \mu B}}$.

Answer 1

Correct Answer: 1.12

Step 1: Relation between rotational constant and moment of inertia. \[ B = \frac{h}{8\pi^2 I c} \] and for diatomic molecules, $I = \mu r^2$ where $\mu$ is reduced mass.
Step 2: Compute reduced mass. \[ \mu = \frac{12 \times 16}{12 + 16} \, \text{amu} = 6.857 \, \text{amu} = 6.857 \times 1.66 \times 10^{-27} = 1.138 \times 10^{-26} \, \text{kg} \] Step 3: Solve for bond length. \[ r = \sqrt{\frac{h}{8\pi^2 \mu c B}} \] Substitute: $B = 3.8626$ cm$^{-1} = 3.8626 \times 100$ m$^{-1}$ \[ r = \sqrt{\frac{6.626\times10^{-34}}{8\pi^2 \times 1.138\times10^{-26} \times 3\times10^8 \times 386.26}} = 1.13 \, \text{Å} \]
Step 4: Conclusion. The bond length of CO = 1.13 Å.