Given wavenumber find spring constant harmonic oscillator

In summary: O2 molecule. In summary, the separation between energies of an oxygen molecule is 2061 cm-1 and the spring constant for an O2 molecule is calculated to be 1.11 x 10-18 kg/s2 using the formula k = (2πc/λ)2*1.33e-26.
  • #1
adamaero
109
1

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)
 
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  • #2
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##.
 
  • #4
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
I get 1.11 x 10-18 kg/s2
 

FAQ: Given wavenumber find spring constant harmonic oscillator

How is wavenumber related to spring constant in a harmonic oscillator?

The wavenumber, represented by the symbol k, is directly proportional to the square root of the spring constant, represented by the symbol k. This relationship can be expressed mathematically as k = √(mω^2), where m is the mass of the oscillating object and ω is the angular frequency.

Can the spring constant be determined by knowing only the wavenumber?

No, the spring constant cannot be determined solely by knowing the wavenumber. Other factors, such as the mass of the object and the angular frequency, are also needed to calculate the spring constant.

How does changing the wavenumber affect the behavior of a harmonic oscillator?

Changing the wavenumber directly affects the angular frequency of the harmonic oscillator. As the wavenumber increases, the angular frequency and the rate of oscillation also increase. This in turn affects the amplitude and period of the oscillations.

Is there a simple formula for finding the spring constant from the wavenumber?

There is no one simple formula for finding the spring constant from the wavenumber. The formula used will depend on the specific system and variables involved, such as the mass and angular frequency.

How can the wavenumber and spring constant be used to analyze a harmonic oscillator system?

The wavenumber and spring constant can be used to calculate important characteristics of a harmonic oscillator, such as the natural frequency, period, and amplitude. They can also be used to analyze the stability and behavior of the system under different conditions.

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