# How does the internuclear repulsion vary in Hydrogenic atom collision?

Hydrogen fusion requires two hydrogen nuclei to get close enough (typically a few fm) to fuse. Much of the problem of creating a fusion reactor is overcoming the Coulomb repulsion between a pair of nuclei - the millions of degrees for Maxwellian distributions, the Bremstrahlung losses for inertial confinement.

If we could align the paths of two neutral Hydrogen atoms (of whichever isotopes), what would the repulsion look like between them as they approach collision? Obviously at long range there is negligable force as both are neutral. But as they approach each other, what happens to the electron distribution?

Intuitively, I expect a bonding cloud to form between the nuclei, and antibonding clouds beyond them. This would presumably attract at first until reaching the usual Hydrogen covalent bond length, after which the internuclear repulsion would increasingly dominate.

But how does that compare to bare ionic collision? How much lower is the potential barrier?

Obviously if it was significantly lower and we could somehow engineer the collision to achieve fusion, the cross section would be larger than ionic fusion, but how much?

Or would the barrier be just as high over the final few femtometers?

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The energy differences associated with the presence, absence, arrangement, etc., of electrons are of order a few to a few tens of electron volts. One way to see this is to note that the size scale of the electron's wavefunction is of order the Bohr radius $a_0$, so the energy levels are of order $e^2/(4\pi\epsilon_0 a_0)$. But to get fusion to happen, as you say, you have to get the two protons to get within a fermi or so of each other. This length scale is something like $10^4$ to $10^5$ times smaller, so it involves energies that are larger by about the same factor -- MeV rather than eV.