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I need some help beginning a problem set problem (just want hints, I'd like to figure it out myself).

The problem gives me the IR absorption for a diatomic, as well as the force constant for the bond. It asks me to identify the the average separation (bond length) of the two atoms.

I am stuck because the typical Schrodinger equation for harmonic oscillation takes "displacement from equilibrium length" as its variable: $\hat{H}\psi_{v}(x)=E_{v}\psi_{v}(x)$. A change of variables such as $x = l-l_{0}$ would require I know the average separation, which I am trying to solve for.

Any help or direction to relevant resources is greatly appreciated.

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  • $\begingroup$ You can get the bond lenght via rotational transitions, using the rotational inertia $I=µr^2$, where µ is the reduced mass. From pure vibrational data i don't think it works ... ? $\endgroup$ – Karl Sep 30 '18 at 17:19
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    $\begingroup$ As @Karl mentions, vibrational spectroscopy does not provide any information about bond lengths. Only if a vibrational-rotational (in the infra-red) or pure rotational (microwave) spectrum is measured can bond lengths be obtained. $\endgroup$ – porphyrin Sep 30 '18 at 20:31
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    $\begingroup$ As pointed out in the comments above, one usually needs information on the rotational structure to extract rotational constants and bond lengths. However, if you are willing to make some crude approximations you can estimate the order of magnitude of the bond length by calculating the classical turning points for the harmonic oscillator using the force constant and the reduced mass of your molecule. $\endgroup$ – Paul Sep 30 '18 at 21:27
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    $\begingroup$ See this analysis for HCl. $\endgroup$ – MaxW Sep 30 '18 at 22:51
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    $\begingroup$ Been meaning to get back to y'all -- the professor admitted he should have given the average bond length, and that he intended the exercise to get us to conclude that the average deviation from equilibrium length is $0$.... Thank you guys for an opportunity to really grok the concept, though! I appreciate it. $\endgroup$ – paranomasia Oct 8 '18 at 2:52

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