What Is the Vibrational Energy-Level Spacing for H2?

[accountkiller]

Homework Statement

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

Homework Equations

M_H = 1.67E-27 kg
F_s = -kr₀
F_c = q₁q₂ / 4π ε r₀²

The Attempt at a Solution

F_c = F_s 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.

 

[accountkiller]

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.

 

[accountkiller]

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}<br /> (\frac{1}{(r_{0}-p)^{2}} - \frac{1}{r_{0}^{2}}) = kp<br />

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

<br /> \frac{1}{(r_{0}-\rho)^{2}} - \frac{1}{r_{0}^{2}}<br /> <br />

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

 

[accountkiller]

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

...?

 

[kuruman]

It seems you need to refresh some basic algebra skills.

<br /> \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}=??<br />

*** on edit ***
Formula corrected.

 

[kuruman]

So...? Can you simplify the numerator?

 

[accountkiller]

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.

 

[accountkiller]

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?

 

[accountkiller]

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 F_c = -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 F_c, 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.