# Given wavenumber find spring constant harmonic oscillator

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1. Oct 30, 2016

1. The problem statement, all variables and given/known data
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

2. Relevant equations
ω = √(k/meff)

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

∴ k = 2πc/(hc/ΔE)
= 2πΔE/h

3. 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: Oct 30, 2016
2. Oct 31, 2016

### BvU

Hi,

I gather a wavenumber of 2061 cm-1 means a wavelength of 10-2/2061 m (I'm not a spectroscopist ).
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 $\hbar\omega$.

3. Oct 31, 2016

### andrevdh

4. Oct 31, 2016

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

5. Nov 1, 2016

### andrevdh

I get 1.11 x 10-18 kg/s2