Estimate Spring Constant of H2 Molecule for Vibrational Frequency

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SUMMARY

The discussion focuses on estimating the spring constant of the hydrogen (H2) molecule for calculating its vibrational frequency. The correct spring constant is approximately 47 eV/A², derived from the potential energy curve near the equilibrium bond length of 0.74 Å. The vibrational frequency is calculated using the formula v = (1/2π) * √(2k/m(proton)), resulting in a frequency of approximately 15,000 Hz. The effective spring constant is equivalent to the second derivative of the potential energy curve evaluated at the minimum point.

PREREQUISITES
  • Understanding of potential energy curves in molecular physics
  • Familiarity with the concept of spring constants in quantum mechanics
  • Knowledge of vibrational frequency calculations
  • Basic understanding of derivatives and Taylor expansion
NEXT STEPS
  • Study the derivation of the spring constant from potential energy curves
  • Learn about the quantum harmonic oscillator model
  • Explore the relationship between vibrational frequency and molecular bond lengths
  • Investigate the application of infrared spectroscopy in molecular vibrations
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Physics students, molecular physicists, and researchers interested in molecular vibrations and potential energy analysis will benefit from this discussion.

GreenLRan
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Estimate the spring constant in units of eV/A^2 for the hydrogen (H2) molecule from the potential energy curve shown below, where r is the distance between protons. From the spring constant and the reduced mass m=1/2m(proton), compute the vibrational frequency. This frequency corresponds to infrared light.

http://geocities.com/greenlran/k.jpg

I tried approximating using V(x)~=V(xe)+1/2k(x-xe)^2 , but i end up with an imaginary term for k. I also tried various other things.. to many to list. but any help would be great!

(correct answer: "approximate V(r) near r=0.74A by V(r)= 1/2k(r-.74)^2 - 4.52eV with k~=47eV/A^2. Freq.(vib)=1/(2pi)*sqrt(2k/m(proton))=1.5e4Hz)
 
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hmm, well I am not a homework helper so take my advice with a grain of salt (I am just a sophmore physics major). I had a question very similar to this recently. The effective spring constant is equivolent to the second deriviative of the potential curve evaluated at the minimum point in the potential curve, presumably where the proton will be oscilating. For this curve, at the .74 angstoms. So, were not talking about the Taylor expansion of the entire curve, just the second derivative term.

Further, w= \sqrt{ \frac{k_{eff}}{m}}, thus v= {2}{pi}^{-1} \sqrt{ \frac{k_{eff}}{m}} where m=1/2m
 
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