Assume that we have a space with just two $\ce{H}$ atoms and their distance to each other is $d$. Let's say they don't have initial velocity. What is the force with which they will act on each other?

This question might have some mistakes in itself, but the main thing I'm trying to learn is how an atom of element $X$ interact with an atom of element $Y$.

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    $\begingroup$ Something else besides F = - dE / dr ? $\endgroup$
    – ssavec
    Commented Nov 22, 2013 at 20:27
  • $\begingroup$ @ssavec: Actually, I'm not familiar with that formula. Can you give a link where I can find somethings about it? $\endgroup$ Commented Nov 22, 2013 at 20:29
  • $\begingroup$ The force is determined by the gradient of the potential energy. See e.g. (en.wikipedia.org/wiki/Scalar_potential) That means, the system tends to adopt the configuration with lowest possible energy. $\endgroup$
    – ssavec
    Commented Nov 22, 2013 at 21:08

2 Answers 2


How one atom interact with another atom is an difficult question and requires quite some calculations, even for a system with only two hydrogen atoms. Basically, this is done by solving the Schrödinger equation for the system. The answer will also depend on the electronics state of the system, especially the spin state.

If the system is in a triplet electronic state, the two atom will experience a repulsive force that approach infinity as you pull the two atoms closer, and that force approaches zero as the distance approaches infinity.

If the system is in a singlet state, the force is zero at the equilibrium bond length of $H_2$ atom(~0.74Å). As the distance gets smaller the atoms experience an increasing repulsive force which approaches infinity at 0. As the two atoms are pulled apart from the equilibrium there is a decreasing attractive force that approaches 0 at infinity.


If you just have two H atoms, nothing else then the only forces acting are gravity and interatomic forces, in particular, the van-der-waals force (note slide 5 which says the van der waals potential is applicable to monatomic gases) . The van der walls force varies and the inverse sixth power, so it will be zero until the hydrogen atoms are quite close, so weak ol' gravity will be doing most of the work until then.

So, you should set up your potential around the center of mass of the system with U including the gravitational and van-der-waals potentials. For all but the closest distances, be prepared for very slow dynamics.

If you don't know how to use potentials, then take the negative of the derivative of the potential for the van der waals and then add in the force of gravity, both acting in the same direction.

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    $\begingroup$ The van der waals energy is inverse sixth power. The force, as a derivative of energy, is inverse seventh power. $\endgroup$
    – user26143
    Commented Nov 23, 2013 at 8:12
  • $\begingroup$ @user26143 my mistake, yes, I was looking at the potential energy function when I was writing this...the force is inverse seventh. $\endgroup$
    – user2603
    Commented Nov 23, 2013 at 11:57

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