Question

Chlorine has two naturally occurring isotopes,${\mathbf{CI}}^{\mathbf{35}}$and ${\mathbf{CI}}^{\mathbf{37}}$. Show that

the vibrational spectrum of HCIshould consist of closely spaced doublets,

with a splitting given by $\mathbf{∆}\mathbf{v}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{51}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{v}$, where v is the frequency of the

emitted photon. Hint: Think of it as a harmonic oscillator, with $\mathbf{\omega }\mathbf{=}\sqrt{\mathbf{k}\mathbf{/}\mathbf{\mu }}$, where

$\mathbf{\mu }$is the reduced mass (Equation 5.8 ) and k is presumably the same for both isotopes.

Short answer

It will identified a difference in HCL's vibrational spectrum using a harmonic oscillator

model.

Show the vibrational spectrum of HCL consist of closely spaced doublets, with a

splitting $∆\mathrm{v}=7.51\times {10}^{-4}\mathrm{v}$,

Step-by-step solution

Definition of harmonic oscillator

• 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."

Determine the photon energy in a vibrational   state

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

${\mathrm{E}}_{\mathrm{Y}}=\mathrm{h\omega }\frac{1}{2}+\mathrm{n}$

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

state $n_f$is:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{Y}}={\mathrm{E}}_{\mathrm{f}}={\mathrm{E}}_{\mathrm{i}} \\ -\mathrm{h\omega }\frac{1}{2}+{\mathrm{n}}_{\mathrm{f}}-\mathrm{h\omega }\frac{1}{2}+{\mathrm{n}}_{\mathrm{i}} \\ -∆\mathrm{nh\omega } \\ ∆\mathrm{n}-{\mathrm{n}}_{\mathrm{f}}-{\mathrm{n}}_{\mathrm{i}} \end{gathered}$$

The photon's frequency is:

$v=\frac{E_y}{h}=\frac{∆n\omega }{2\mathrm{\pi }}$

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

$$\begin{gathered} \mathrm{\omega }=\sqrt{\frac{\mathrm{K}}{\mathrm{\mu }}} \\ \mathrm{v}=\frac{∆\mathrm{n}}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{K}}{\mathrm{\mu }}} \\ \mathrm{v}=\frac{∆\mathrm{n}\sqrt{\mathrm{K}}}{2\mathrm{\pi }}{\mathrm{\mu }}^{-1/2} \end{gathered}$$

Determine the reduced mass

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

mass $\mu$:

$$\begin{gathered} ∆\mathrm{v}=\left|\frac{∆\mathrm{n}\sqrt{\mathrm{K}}}{2\mathrm{\pi }}\times \frac{-1}{2{\mathrm{\mu }}^{3/2}}∆\mathrm{\mu }\right| \\ =\frac{∆\mathrm{\mu }}{2\mathrm{\mu }}\frac{∆\mathrm{n}\sqrt{\mathrm{K}}}{2\mathrm{\pi }}{\mathrm{\mu }}^{-1/2} \\ =\frac{∆\mathrm{\mu }}{2\mathrm{\mu }}\mathrm{v} \\ \mathrm{\mu }=\frac{{\mathrm{m}}_{\mathrm{h}}{\mathrm{m}}_{\mathrm{CI}}}{{\mathrm{m}}_{\mathrm{H}}+{\mathrm{m}}_{\mathrm{CI}}} \\ =\frac{1}{{\displaystyle \frac{1}{{\mathrm{m}}_{\mathrm{CI}}}}+{\displaystyle \frac{1}{{\mathrm{m}}_{\mathrm{LI}}}}} \\ ∆\mathrm{\mu }=\frac{-1}{{\displaystyle \frac{1}{{\mathrm{m}}_{\mathrm{CI}}}+\frac{1}{{\mathrm{m}}_{\mathrm{\mu I}}}}} \\ =\frac{\mathrm{\mu }}{{\mathrm{m}}_{\mathrm{CI}}^2}∆{\mathrm{m}}_{\mathrm{CI}} \end{gathered}$$

Determine the value of∆v

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

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