Frequency of Oscillation of a Molecule

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SUMMARY

The frequency of oscillation, v, of a molecule is directly related to the second derivative of the inter-atomic potential V(r) at the equilibrium separation r = r0, expressed as v = (1/2π)√[d²V/dr²|r=r0/m]. The reduced mass, m, plays a crucial role in this relationship. The discussion emphasizes the importance of understanding the connection between the second derivative of the potential and the spring constant, suggesting the use of a Taylor series expansion around the equilibrium point as a foundational step in solving the problem.

PREREQUISITES
  • Understanding of molecular oscillation frequency
  • Familiarity with inter-atomic potentials, specifically Morse and Lennard-Jones potentials
  • Knowledge of reduced mass in molecular systems
  • Basic calculus, particularly Taylor series expansion
NEXT STEPS
  • Study the derivation of the frequency of oscillation from potential energy functions
  • Learn about the Taylor series expansion and its application in physics
  • Explore the Morse and Lennard-Jones potentials in detail
  • Investigate the concept of reduced mass and its significance in molecular dynamics
USEFUL FOR

Students in physical chemistry, molecular physics, and anyone studying molecular vibrations and inter-atomic forces will benefit from this discussion.

dm9292
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Homework Statement



Show that the frequency of oscillation, v, of a molecule is related to the form of the inter - atomic potential V(r) about the equilibrium separation, r =ro by the expression,

v = (1/2pi).sqrt([d2V/dr2r=ro/(m)])

where m is the reduced mass

Homework Equations



None given

The Attempt at a Solution



I really don't know where to start it was set as a 'challenge question' in a book i have borrowed.

I was thinking the second derivative had something to do with the spring constant. Or it maybe involved the Morse or Leonard Jones Potential. The thing is I don't know how to relate any of these things.

Is the term in square roots the frequency? I.e based on the formula w= 2piv
 
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hmmm...
\nu = \frac{1}{2\pi}\sqrt{\frac{1}{m}\left.\frac{\mathrm{d}^2V}{\mathrm{d}r^2}\right|_{r=r_0}}
yep,
2\pi\nu = \omega
You're on the right track to think about the spring constant. As a first step, try writing a Taylor series for the potential around the point r_0.
 
Thanks
 

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