Vibrational Levels in Molecular Electronic State

[guyvsdcsniper]

Homework Statement

In a diatomic molecule k (N/m) may be estimated from the dissociation energy and bond length by D.A. Let the bond length = 2*10^-10m and the dis. energy = 4.806*10^-19J.

Assuming the vibrational motion can be treated as purely simple harmonic, how many
vibrational levels would you expect to associated with this molecular electronic state if
the molecule were (a) hydrogen, and (b) oxygen ?

Relevant Equations

E=hw(n+1/2)

$K = \frac{N}{m} = \frac{3eV}{bondlength^2} = \frac{4.806*10^-19 J}{(2*10^-10)^2} = 12.015$

Then I know that $H = \frac{1}{2}mωx^2$ where $k = mω^2$ and also $H=ħω(n+\frac{1}{2})$

Therefore, $\frac{1}{2}kx^2 = ħω(n+\frac{1}{2})$

Solving for n, $n = \frac{1/2kx^2}{ħω} - \frac{1}{2}$

for ω I used the fact the $ω = \sqrt(\frac{k}{m})$ where m was the reduced mass of H₂.

So n should tell me the max vibrational levels. When plugging in all my parameters I get 18. I researched online and found this article, https://w.astro.berkeley.edu/~ay216/05/NOTES/Lecture18.pdf , which states
"c. The Electronic Levels - The ground state is X 1Σ+g because it has the quantum numbers: S = 0, I = 0, Λ = 0, & J = 0. It has a manifold of 30 vibrational levels, each with an infinite number of rotational states"

So I am thinking n must be 30 for my problem since the article states H₂ has 30 vibrational levels.

I have found when plugging ~19 for K i get n=30

Any help with this problem?I feel like my k is correct since it checks out with the dimensional analysis, so maybe I am wrong somewhere else?

 

[DrClaude]

Assuming the vibrational motion can be treated as purely simple harmonic, how manySo I am thinking n must be 30 for my problem since the article states H2 has 30 vibrational levels.

The assumption you are asked to make will lead to a strong departure from reality (molecular bonds are anharmonic). Comparison to a real molecule is futile.

I haven't checked your calculation, but getting a smaller number of vibrational levels that the real molecule is a good sign.