91Ó°ÊÓ

• A model for molecular vibration is the simple harmonic oscillator. It denotes the relative motion of atoms in a diatomic molecule or
• The simultaneous motion of atoms in a polyatomic molecule along a vibrational "normal mode."

A photon's energy in a vibrational state n is:

${\mathrm{E}}_{\mathrm{Y}}=\mathrm{hÓ\neg }\frac{1}{2}+\mathrm{n}$

The amount of energy required for a photon to go from its initial state $n_i$to its final

state $n_f$is:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{Y}}={\mathrm{E}}_{\mathrm{f}}={\mathrm{E}}_{\mathrm{i}} \\ -\mathrm{hÓ\neg }\frac{1}{2}+{\mathrm{n}}_{\mathrm{f}}-\mathrm{hÓ\neg }\frac{1}{2}+{\mathrm{n}}_{\mathrm{i}} \\ -∆\mathrm{nhÓ\neg } \\ ∆\mathrm{n}-{\mathrm{n}}_{\mathrm{f}}-{\mathrm{n}}_{\mathrm{i}} \end{gathered}$$

The photon's frequency is:

$v=\frac{E_y}{h}=\frac{∆n\mathrm{Ó\neg }}{2\mathrm{Ï€}}$

For oscillation frequency, we can use the harmonic oscillator formula:

$$\begin{gathered} \mathrm{Ó\neg }=\sqrt{\frac{\mathrm{K}}{\mathrm{μ}}} \\ \mathrm{v}=\frac{∆\mathrm{n}}{2\mathrm{Ï€}}\sqrt{\frac{\mathrm{K}}{\mathrm{μ}}} \\ \mathrm{v}=\frac{∆\mathrm{n}\sqrt{\mathrm{K}}}{2\mathrm{Ï€}}{\mathrm{μ}}^{-1/2} \end{gathered}$$

Take absolute value (because frequency must be positive) with respect to reduced

mass $\mathrm{μ}$:

$$\begin{gathered} ∆\mathrm{v}=\left|\frac{∆\mathrm{n}\sqrt{\mathrm{K}}}{2\mathrm{Ï€}}×\frac{-1}{2{\mathrm{μ}}^{3/2}}∆\mathrm{μ}\right| \\ =\frac{∆\mathrm{μ}}{2\mathrm{μ}}\frac{∆\mathrm{n}\sqrt{\mathrm{K}}}{2\mathrm{Ï€}}{\mathrm{μ}}^{-1/2} \\ =\frac{∆\mathrm{μ}}{2\mathrm{μ}}\mathrm{v} \\ \mathrm{μ}=\frac{{\mathrm{m}}_{\mathrm{h}}{\mathrm{m}}_{\mathrm{CI}}}{{\mathrm{m}}_{\mathrm{H}}+{\mathrm{m}}_{\mathrm{CI}}} \\ =\frac{1}{{\displaystyle \frac{1}{{\mathrm{m}}_{\mathrm{CI}}}}+{\displaystyle \frac{1}{{\mathrm{m}}_{\mathrm{LI}}}}} \\ ∆\mathrm{μ}=\frac{-1}{{\displaystyle \frac{1}{{\mathrm{m}}_{\mathrm{CI}}}+\frac{1}{{\mathrm{m}}_{\mathrm{μ\pm õ}}}}} \\ =\frac{\mathrm{μ}}{{\mathrm{m}}_{\mathrm{CI}}^2}∆{\mathrm{m}}_{\mathrm{CI}} \end{gathered}$$

For ${\mathrm{m}}_{\mathrm{CI}}=36$, it use the average value. It goes like this:

$$\begin{gathered} ∆\mathrm{v}=\frac{\mathrm{v}}{2}\frac{\mathrm{μ}∆{\mathrm{m}}_{\mathrm{CI}}}{{\mathrm{m}}_{\mathrm{CI}}^2} \\ =\frac{\mathrm{v}}{2}\frac{∆{\mathrm{m}}_{\mathrm{CI}}/{\mathrm{m}}_{\mathrm{C}}}{1+{\displaystyle \frac{{\mathrm{m}}_{\mathrm{C}}}{{\mathrm{m}}_{\mathrm{U}}}}} \\ \frac{∆{\mathrm{m}}_{\mathrm{CI}}}{{\mathrm{m}}_{\mathrm{CI}}}=\frac{2}{36}-\frac{1}{18} \\ =\frac{{\mathrm{m}}_{\mathrm{CI}}}{{\mathrm{m}}_{\mathrm{H}}}-\frac{36}{1} \\ ∆\mathrm{v}=\frac{1}{2}\frac{1/18}{1+36}\mathrm{v} \\ =\frac{\mathrm{v}}{36×37} \\ ∆\mathrm{v}=7.51×{10}^{-4}\mathrm{v} \end{gathered}$$