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[QUOTE="HAYAO, post: 5940393, member: 558519"] I redrew the figure for you. [ATTACH=full]220243[/ATTACH] First, we need to set some condition and definition. 1) The variable r[SUB]1[/SUB] is the coordinate of the first hydrogen atom. 2) The variable r[SUB]2[/SUB] is the coordinate of the second hydrogen atom. Since this hydrogen molecule is a biatomic molecule, where the only degree of freedom that changes the energy of the molecule itself is its stretching and shrinking. So if we look at this molecule from the perspective of the first hydrogen atom, then the second hydrogen atom is moving toward and away from the first one. With this consideration, we can redefine the coordinate system: 1') The variable r[SUB]1[/SUB] is now set to origin (r[SUB]1[/SUB] = 0). 2') The variable r[SUB]2[/SUB] is now the only independent variable. Now, the bond between two hydrogen atom can be considered as a "spring". If you stretch it, it'll try to contract. If you squeeze it, it'll try to retract. I hope you heard this somewhere or sometime in your education. This is the harmonic oscillator for a spring. Therefore, the atom should follow the Hooke's law. That is, the force is given by F = kx, and the potential energy is given by: [itex]E = \frac{1}{2}kx^{2}[/itex] Now look at this equation. It's a quadratic equation! In our case, we can set x as r[SUB]2[/SUB], so we should get a quadratic potential energy like shown in the grey dotted line in the figure. However, just like how spring breaks when stretched too far, or when the springs are squashed so that the springs break and the two objects fuse, atoms will also break its bond when stretched too far (situation v), and will reach extremely high energy upon extreme contraction of the bond (situation i). The latter enters the realm of high energy physics. Somewhere in between (situation ii, iii, and iv), the atoms oscillate. This is why, the actual potential energy looks like the solid black line in the figure. That is, the potential energy will not infinitely increase when the atoms are stretched. The atoms will reach some point where there is infinitesimal interaction between the two atoms (situation v). [U]That means there is no Coulomb interaction between the two atoms = the potential energy is zero[/U]! The situation iii is called the "equilibrium position". That is, when the atom configuration is in such way so that the potential energy is minimum for this system. In the case of hydrogen molecule, this is 74 pm. The lowering of the energy (approximately 4.52 eV), means that the hydrogen molecule is more stable in this configuration. If it is stretched, it becomes unstable. If it is squeezed, it also becomes unstable. As such, the atoms will attempt going back to the most stable configuration, that is the equilibrium position. This is precisely why [URL='https://www.physicsforums.com/threads/potential-energy-and-internuclear-distance.939381/#post-5939203']blue_leaf77[/URL] said that the gradient of the potential energy curve provides the information of how the force is in this molecule. Gradients are similar to derivatives (which provides the slope of a function), only except in vector form. Say you drop a little marble anywhere on this potential energy curve. Where do you think the marble will go? It will eventually roll to the equilibrium position, right? That is what the potential energy means, and what the figure means. Finally, we need to get deeper into this to understand what [I]really[/I] happens in the hydrogen molecule. You were wondering about the Coulomb potential (the equation you provided). This is not necessarily applicable for this system. The reason is because the bonding between hydrogen atoms is a [B]covalent bond[/B]. While we see classical Coulomb interaction between two nuclei and two electrons (total of four interactions), the actual interaction also involves "[B]exchange interaction[/B]". Exchange interaction is a completely quantum mechanical phenomenon, and there exist no classical analog, but it is what holds the two atoms together. [/QUOTE]
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