vibrational energy-level spacing

[mbradar2]

1. The problem statement, all variables and given/known data
Consider the hydrogen molecule H2 to be a simple harmonic oscilattor with an equilibrium spacing of 0.074 nm, and estimate the vibrational energy-level spacing for H2.
Hint: Estimate the force constant by equating the change in Coulomb repulsion of the protons, when the atoms move slightly closer together than r0, to the "spring" force. That is, assume that the chemical binding force remains aproximately constant as r is decreased slightly from r0


2. Relevant equations
MH = 1.67E-27 kg
Fs = -kr0
Fc = q1q2 / 4π ε r02

3. The attempt at a solution
Fc = Fs gives me
k = -e2 / 4π ε r03
From that, my force constant k = 569.

Side question: what are the units of k? I end up with N/m.. is that right?

However, another classmate said that there is a 2 in front of the e charge, so the numerator would read 2e2. That would make k = 1138, which is the correct answer.

I don't understand where the 2 comes from..?

Thanks!

[kuruman]

The 2 comes from the difference (or Taylor expansion) that you did not consider. Initially, the Coulomb force is

F_0=\frac{e^2}{4\pi \epsilon_0r^{2}_{0}}

When the distance is decreased by an amount ρ, the new Coulomb force is

F=\frac{e^2}{4\pi \epsilon_0(r_{0}-\rho)^2}

You need to set F - F0 equal to kρ and find k for small values of ρ.

[mbradar2]

I don't understand... if I follow your way, what do I use for p? Is 0.01E-9 considered "small" .. it gives me a number for k on the order of 21.

[kuruman]

I don't understand... if I follow your way, what do I use for p?
You don't need a number for ρ. Just do the algebra as I indicated and see what you get.
Is 0.01E-9 considered "small" .. it gives me a number for k on the order of 21.
0.01E-9 what? What does this dimensionless number represent? What we are considering here is displacements from the equilibrium position. That's what ρ is, a number much smaller than whatever the equilibrium position is.

[mbradar2]

I'm very confused.
So your method is:
F - F0 = kp
\frac{e^{2}}{4\pi\epsilon (r_{0}-p)^{2}} - \frac{e^{2}}{4\pi\epsilon (r_{0})^{2}} = kp

\frac{e^{2}}{4\pi\epsilon}
(\frac{1}{(r_{0}-p)^{2}} - \frac{1}{r_{0}^{2}}) = kp


Solving for k, just move the p to 1/p on the other side.
I can try to simply things but it got complicated... I don't get how this helps, or how I can find k without a value of p.

[kuruman]

Can you find what


\frac{1}{(r_{0}-\rho)^{2}} - \frac{1}{r_{0}^{2}}



is as a single fraction? In other words, can you add these two fractions?

[mbradar2]

\frac{r_{0}^{2}-r_{0}+p}{(r_{0}-p)(r_{0}^{2}}

...?

[kuruman]

It seems you need to refresh some basic algebra skills.


\frac{1}{(r_{0}-\rho)^{2}} - \frac{1}{r^{2}_{0}}=\frac{r^{2}{_0}-(r_0-\rho)^2}{r^{2}_{0}(r_0-\rho)^2}=??


*** on edit ***
Formula corrected.

[kuruman]

So...? Can you simplify the numerator?

[mbradar2]

Oh, in my rush I forgot to add the square to (r0-p)2 in the numerator. So..

\frac{2r_{0}p-p^{2}}{(r_{0}-p)^{2}(r_{0})^{2}}

\frac{e^{2}}{4\pi\epsilon}\frac{2r_{0}-p}{(r_{0}-p)^{2}(r_{0})^{2}}

[kuruman]

How did you get the second line? Please write equations, not just pieces of equations.

[mbradar2]

Sorry, I'm not that used to this Latex format so it takes too long to figure out how to type everything properly.

The first line,
\frac{2r_{0}p-p^{2}}{(r_{0}-p)^{2}(r_{0})^{2}}
is what I got when I simplified the part that you asked me to.
\frac{1}{(r_{0}-\rho)^{2}} - \frac{1}{r_{0}^{2}} = \frac{2r_{0}p-p^{2}}{(r_{0}-p)^{2}(r_{0})^{2}}

The second line is substituting that 'simplified' part into the parantheses of my original equation,
\frac{e^{2}}{4\pi\epsilon}(\frac{1}{(r_{0}-p)^{2}} - \frac{1}{r_{0}^{2}}) = kp

So, \frac{e^{2}}{4\pi\epsilon}(\frac{2r_{0}p-p^{2}}{(r_{0}-p)^{2}(r_{0})^{2}}) = kp

[kuruman]

So, \frac{e^{2}}{4\pi\epsilon}(\frac{2r_{0}p-p^{2}}{(r_{0}-p)^{2}(r_{0})^{2}}) = kp
Right. Now observe that for ρ << r0 the numerator in the last fraction becomes what?

[mbradar2]

Okay, so then p ~ 0 and I get what I want:
\frac{2e^{2}p}{4\pi\epsilon r_{0}^{3}}

Now my question is.. where did the negative on your k go? The spring force for this problem is Fc = -kp. But you put a positive k.

Shouldn't I do F0 - F, instead of F - F0? That way, I would get a negative number for Fc, which would cancel out the negative on k, giving me a positive result for k.

..right?

[kuruman]

The negative sign indicates that the vector force is opposite to the vector displacement. Here we are dealing with magnitudes.