How Accurate Is the Calculated Spring Constant of a Diatomic Molecule?

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SUMMARY

The calculated spring constant for a diatomic molecule using the provided frequency of 1395 cm-1 yields a value of 9.18 x 10-6 kg/s2. This result is significantly lower than the expected range of 400-800 N/m for bond forces in diatomic molecules. The calculation employs the relation k = μ(2πcv)2, where μ is the reduced mass (8 x 1.66 x 10-27 kg) and v is the frequency in cm-1. Clarification is needed on the conversion from frequency to wavenumber and the experimental methods for measuring frequency.

PREREQUISITES
  • Understanding of angular frequency and its relation to spring constant
  • Knowledge of reduced mass calculation for diatomic molecules
  • Familiarity with the concept of wavenumber in spectroscopy
  • Basic principles of molecular vibrations and bond forces
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  • Research the conversion between frequency and wavenumber in spectroscopy
  • Learn about the experimental techniques for measuring molecular vibrational frequencies
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Chemists, physicists, and students studying molecular vibrations, particularly those involved in spectroscopy and molecular dynamics analysis.

greisen
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Hi,
I have to determine the spring constant for a diatomic molecule. I have the frequency in cm^-1 which is 1395.
I know the relation between the angular frequency and spring constant so I use the following relation:
omega = 2*pi*c*v = sqrt(k/mu)

so I isolate for k = mu*(2*pi*c*v)^2

mu = 8*1.66*10^-27
v = 1395 cm^-1

which gives me k = 9.18*10^-6 kg/s^2

This strikes me as a rather strange result? For a diatomic molecule I would expect something like at least 400-800 N/m for the bond force? What is wrong ? Thanks in advance.
 
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I'm not sure how exactly you go from frequency to wavenumber (or viceversa). It looks like you've used a linear dispersion with some value of c. What value did you use? Where did you get this from? Experimentally, how is the frequency measured (and why is the output in the form of a wavenumber)?
 

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