91Ó°ÊÓ

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:

${\mathrm{Δ·¡}}_{\mathrm{vib}}=\mathrm{h}\sqrt{\frac{\mathrm{κ}}{\mathrm{μ}}}$

Here,$\mathrm{κ}$is the effective spring constant and$\mathrm{μ}$is the effective mass.

The expression for the effective mass of the nitrogen molecule is

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

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 $\mathrm{μ}$.

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

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

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

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

$\begin{array}{c}{\mathrm{E}}_{1,0}−{\mathrm{E}}_{0,0}=(1.055×{10}^{−34}\text{Js})\sqrt{\frac{1860\text{ N/³¾}}{1.14×{10}^{−26}\text{kg}}}\\ =4.26×{10}^{−20}\text{J}\end{array}$...(i)

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

${\mathrm{E}}_{\mathrm{l}}=\frac{{\mathrm{h}}^2[\mathrm{l}×(\mathrm{l}+1\left)\right]}{2{\mathrm{μ²¹}}^2}$

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

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

Compare above equation with equation (i),

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

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.