Vibration-rotation spectra are rich For the CO molecule (data are given in Exercise 42), roughly how many rotational levels would there be between the ground vibrational state and the first excited vibrational state?

Movements of one atom within a molecule in relation to other atoms are known as molecular vibrations. The smallest vibrations are lengthening and shortening of a single link.

To understand this, picture the bond as a spring and the molecular vibration as the spring's simple harmonic motion.

The vibrational energy of the photons can be expressed as follow:

${\mathrm{\Delta E}}_{\mathrm{vib}}=\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}$

Here,$\mathrm{\kappa }$is the effective spring constant and$\mathrm{\mu }$is the effective mass.

The expression for the effective mass of the nitrogen molecule is

$\mathrm{\mu }=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

Here, ${\mathrm{m}}_1$ is atomic mass for carbon (12.011u), ${\mathrm{m}}_2$is the atomic mass for the oxygen(15.999u)

Substitute 12.011u for ${\mathrm{m}}_1$, 15.999u for ${\mathrm{m}}_2$ in the above equation, and solve for $\mathrm{\mu }$.

$\begin{array}{c}\mathrm{\mu }=\frac{\left(12.011\mathrm{u}\right)\left(15.999\mathrm{u}\right)}{12.011\mathrm{u}+15.999\mathrm{u}}\\ =6.861\mathrm{u}\\ =\left(6.861\mathrm{u}\right)\left(\frac{1.66\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right)\\ =1.14\times {10}^{-26}\mathrm{kg}\end{array}$

The difference of the energies of the first two vibrational modes is

$\begin{array}{c}{\mathrm{E}}_{1,0}-{\mathrm{E}}_{0,0}=\left(1+\frac{1}{2}\right)\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}-\frac{1}{2}\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}\\ =\left(\frac{3}{2}\right)\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}-\frac{1}{2}\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}\\ =\left(\frac{3}{2}-\frac{1}{2}\right)\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}\\ =\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}\end{array}$

Substitute $1.055\times {10}^{34}\mathrm{Js}$ for $\mathrm{h}$, $1860\mathrm{N}/\mathrm{m}$ for $\mathrm{K}$, $1.14\times {10}^{-26}\mathrm{kg}$for $\mathrm{\mu }$ in the above equation, and solve for $({\mathrm{E}}_{1,0}-{\mathrm{E}}_{0,0})$.

$\begin{array}{c}{\mathrm{E}}_{1,0}-{\mathrm{E}}_{0,0}=(1.055\times {10}^{-34}\text{Js})\sqrt{\frac{1860\text{\hspace{0.17em}N/m}}{1.14\times {10}^{-26}\text{kg}}}\\ =4.26\times {10}^{-20}\text{J}\end{array}$...(i)

The rotational energy for level l can be expressed as follow:

${\mathrm{E}}_{\mathrm{l}}=\frac{{\mathrm{h}}^2[\mathrm{l}\times (\mathrm{l}+1\left)\right]}{2{\mathrm{\mu a}}^2}$

Now solving that equation for ${\mathrm{E}}_{\mathrm{l}}$,

$\begin{array}{c}{\mathrm{E}}_{\mathrm{l}}=\frac{{(1.055\times {10}^{-34}\text{Js})}^2}{2(1.14\times {10}^{-26}\text{kg}){(0.113\times {10}^{-9}\text{m})}^2}[\mathrm{l}\times (\mathrm{l}+1\left)\right]\\ =(3.83\times {10}^{-23}\text{J})[\mathrm{l}\times (\mathrm{l}+1\left)\right]\end{array}$

Compare above equation with equation (i),

$\begin{array}{c}(3.83\times {10}^{-23}\text{J})[\mathrm{l}\times (\mathrm{l}+1\left)\right]=4.26\times {10}^{-20}\text{J}\\ \mathrm{l}(\mathrm{l}+1)=\frac{4.26\times {10}^{-20}\text{J}}{3.83\times {10}^{-23}\text{J}}\\ =1112\\ ={\left(33\right)}^2\end{array}$

From the above equation, we conclude that the value of the$l$ is approximately 33, so there are about 33 rotational modes between the first two vibrational modes.