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HW3_P30_corrected

HW3_P30_corrected - •...

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Correction to HW 3 Solutions - #30 You may need to know the following info in the future so be sure to examine this. Chapter 6 Problem 30 – corrected C = μ " 2 , where μ is the reduced mass μ = m 1 m 2 m 1 + m 2 = M total 4 , in this problem with equal masses. So, " = C μ = 3.12x10 14 . E = 1 2 h " = 1.64x10 -20 J = 0.102 eV.
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Unformatted text preview: • http://www.cachesoftware.com/mopac/Mopac2002manual/node589.html • http://en.wikipedia.org/wiki/Reduced_mass • http://www.everyscience.com/Chemistry/Physical/Vibrational_Spectroscopy/a.10 19.php...
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  • Spring '09
  • Mass, equal masses, reduced mass, following info, http://www.cachesoftware.com/mopac/Mopac2002manual/node589.html · http://en.wikipedia.org/wiki/Reduced_mass

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