IChO 2003 · Theory problem 26

Spectrum of CO

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PROBLEM 26 Spectrum of \(\ce{CO}\)

Rotational energy levels of diatomic molecules are well described by the formula \(E_J = B\,J(J+1)\), where \(J\) is the rotational quantum number of the molecule and \(B\) its rotational constant. Constant \(B\) is related to the reduced mass \(\mu\) and the bond length \(R\) of the molecule through the equation

\[ B=\frac{h^2}{8\pi^2\mu R^2}. \]

In general, spectroscopic transitions appear at photon energies which are equal to the energy difference between appropriate states of a molecule (\(h\nu=\Delta E\)). The observed rotational transitions occur between adjacent rotational levels, hence \(\Delta E=E_{J+1}-E_J=2B(J+1)\). Consequently, successive rotational transitions that appear on the spectrum (such as the one shown here) follow the equation \(h(\Delta\nu)=2B\).

By inspecting the spectrum provided, determine the following quantities for \(\ce{^{12}C^{16}O}\) with appropriate units:

a) \(\Delta\nu\)

b) \(B\)

c) \(R\)

Figure 1

Official English text of the problem.

Answers

  • a) \(\boxed{\Delta\nu\approx115\ \mathrm{GHz}=1.15\times10^{11}\ \mathrm{Hz}}\)
  • b) \(\boxed{B\approx3.81\times10^{-23}\ \mathrm{J}}\)
  • c) \(\boxed{R\approx1.13\times10^{-10}\ \mathrm{m}=113\ \mathrm{pm}}\)

Solution

The molecule is carbon monoxide, \(\ce{^{12}C^{16}O}\), with isotopic SMILES [12C-]#[16O+].

a) Frequency spacing, \(\Delta\nu\)

The strong absorption lines appear as minima in transmittance. From the spectrum, the minima near \(460\) and \(1150\ \mathrm{GHz}\) span six successive intervals. Thus,

\[ \Delta\nu\approx\frac{1150-460}{6}\ \mathrm{GHz} =115\ \mathrm{GHz} =1.15\times10^{11}\ \mathrm{s^{-1}}. \]

b) Rotational constant, \(B\)

For the transition \(J\rightarrow J+1\),

\[ h\nu_J =B\big[(J+1)(J+2)-J(J+1)\big] =2B(J+1). \]

Therefore, adjacent spectral lines have spacing

\[ \Delta\nu=\nu_{J+1}-\nu_J=\frac{2B}{h}. \]

Using the standard Planck constant \(h=6.62607015\times10^{-34}\ \mathrm{J\,s}\),

\[ B=\frac{h\Delta\nu}{2} =\frac{(6.62607015\times10^{-34}\ \mathrm{J\,s}) (1.15\times10^{11}\ \mathrm{s^{-1}})}{2} \approx\boxed{3.81\times10^{-23}\ \mathrm{J}}. \]

Here \(B\) has energy units, as defined in the problem.

c) Bond length, \(R\)

Approximating the isotope masses by \(12u\) and \(16u\), the reduced mass is

\[ \mu=\frac{m_{\mathrm C}m_{\mathrm O}}{m_{\mathrm C}+m_{\mathrm O}} =\frac{12\times16}{12+16}\,u =6.85714\,u. \]

Using the standard conversion \(u=1.66054\times10^{-27}\ \mathrm{kg}\),

\[ \mu\approx1.13866\times10^{-26}\ \mathrm{kg}. \]

Rearranging the given expression for \(B\),

\[ R=\sqrt{\frac{h^2}{8\pi^2\mu B}} =\sqrt{ \frac{(6.62607\times10^{-34}\ \mathrm{J\,s})^2} {8\pi^2(1.13866\times10^{-26}\ \mathrm{kg}) (3.80999\times10^{-23}\ \mathrm{J})} }. \]

Hence,

\[ \boxed{R\approx1.13\times10^{-10}\ \mathrm{m}=113\ \mathrm{pm}}. \]

The solution was checked against the official solution.