Given wavenumber find spring constant harmonic oscillator

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Homework Help Overview

The discussion revolves around calculating the spring constant for an oxygen molecule modeled as a simple harmonic oscillator, using a given wavenumber of 2061 cm-1. Participants explore the relationship between the spring constant, reduced mass, and energy levels in quantum mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to derive the spring constant using various equations related to energy and frequency, questioning the validity of their calculations and assumptions about units.

Discussion Status

Several participants are exploring different methods to calculate the spring constant, with some providing alternative equations and interpretations. There is no explicit consensus on the correct approach or final value, but productive discussions are ongoing regarding the relationships between the variables involved.

Contextual Notes

Some participants express uncertainty about the derivation of certain equations and the implications of unit conversions. The discussion also highlights the complexity of relating quantum mechanical concepts to classical models.

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Homework Statement


The separation between energies of an oxygen molecule is 2061 cm-1 (wavenumber). Treating the molecule as a simple harmonic oscillator whose fundamental frequency is related to its spring constant and reduced mass, calculate the spring constant for an O2 molecule.

meff = 1.33e-26 kg
h = 4.136e-15 eV-s
1 eV = 8065.45 cm-1

Homework Equations


ω = √(k/meff)

k = ω2eff
ω = 2π(v) = 2π(c/λ)
λ = hc/ΔE

∴ k = 2πc/(hc/ΔE)
= 2πΔE/h

The Attempt at a Solution


ΔE = 2061 cm-1 ⇒ 0.2555 eV
k = 3.88*1014 (which just seems way too big)
unit check: eV/(eV-s) so the above value should be even bigger?? (times the speed of light)
k = 1.1644*1023
Actually, I wrongly assumed that the spring constant is unitless.

N/m (force per unit length) or N-m-1

So instead I found this equation, but I don't know how it was derived:
E0 = (h/2)√(k/meff)
k = meff(2E0/h)
k = 1.33e-26(2*0.2555/4.136e-15)^2
k = 203 (which seems reasonable--but I don't get how the equation comes about even knowing .5mv2 = k)
 
Last edited:
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Hi,

I gather a wavenumber of 2061 cm-1 means a wavelength of 10-2/2061 m (I'm not a spectroscopist :smile:).
With ##\ \lambda f = c \ ## and ##\ \omega = 2\pi f \ ## I then get ##\ \omega = 2\pi {2061\over 10^{-2}}c = 3.88\ 10^{14} \ ## radian/s.
What you call k is actually omega (the dimensions do match). From there to the spring constant gives me quite a different result ...

##E_0 = {1\over 2} \hbar\omega## is the energy of the ground state. But the energy difference between states is a mutltiple of ##\hbar\omega##.
 
reduced mass ~ meff = 1.33e-26 kg
ħ = 1.055*10-34 Js
h = 6.626*10-34 Js
λ = 10-2/2061 m

E = (n+.5)ħw
w = 2πf = 2π(c/λ)

ω = √(k/meff)
k = (ω)2meff
k = (2πc/λ)2*1.33e-26
k = 4*1.33e-26*pi*3e8*2061/10e-2
k = 1.033e-12
 
I get 1.11 x 10-18 kg/s2
 

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