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Estimate the binding energy of the H2 molecule

  1. Aug 21, 2008 #1
    The problem is this:
    Estimate the binding energy of the H2 molecule, assuming the two H nuclei are 0.074 nm apart and the two electrons spend 33% of their time midway between them.

    I assume that this problem uses the equation F=ke2/d.

    So far, I have that E1=F1d, and E2=F2v*t*0.33. Then, we know that the binding energy E should be E1-E2, and therefore E=0.67E1. And E1=ke2/d. Is this correct so far? And where do I go from here? I do know that the correct answer is 4.6 eV, but just can't get there. Any help would be greatly appreciated!
  2. jcsd
  3. Aug 21, 2008 #2


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    You're going to need to explain what E1 and E2 are supposed to be. If you're using F to represent the Coulomb potential energy between two charges (at least that's what you've written), how can Fd be an energy? What exactly is E2? Where would values for 'v' and 't' come from?

    You'll want to go back to the expression for the electric potential energy

    [tex]U = \frac{k \cdot q^2}{d}[/tex]

    Consider that there are four charges here, two protons and two electrons. The electrons act effectively as if 1/3 of each of their charges are at the midpoint of the molecule, attracting the protons. (The particle configuration leads to an asymmetry of electron charge around the nuclei.) What energies do you find between the various pairs of particles?
  4. Aug 21, 2008 #3
    dynamics's tagline says it all. Draw it with those 1/3-charged electrons in there, and then work out all the distances and charges. Luckily potential and energy are scalars.
  5. Aug 22, 2008 #4


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    I don't think we're quite there yet, but here's another thing to consider. As seen from the midpoint or the other nucleus, the charge of each nucleus is partially "screened" (as they say) by the electron charge distribution around it. So the charge on each nucleus must be treated as being effectively something like +(1/3)e.
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