# Bond energy - (Force/ potential E vs interatomic separation)

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1. Mar 9, 2015

### vcb003104

Hey guys,

So, I was in my materials lecture today and something the lecturers and the tutors said was slightly confusing.
We were talking about the repulsion and attraction force between atoms. This is all good as I can understand how if two atoms will repel each other if they get really close and attract when close but not as close... and eventually it gets too far for the forces to do anything.

We then went on about how to find E0....
basically just derive a formula, then sub r0 and rearrange it to make it look nice.

However, we came across potential energies vs Interatomic separation.

^ the two graphs
• The reason that I don't understand this graph is that why is the bonding energy the lowest point of this graph?
• When the repulsion force = attraction force (= bonding energy?), net force is 0 but where is there still energy?
• Finally, what does it mean when potential energy = 0 (But then the force is in repulsion according to the two graphs)

I asked my tutor and he said something along the lines of that energy in this case have direction; did not fully understand what he meant.

I know that with gravitational potential energy, the higher you go, the more negative it gets. As gravity acts on things that's even infinitely far. (Therefore there is technically no 0 mgh.) What does it mean when potential energy in this context = 0 though?

Appreciate the help guys!

2. Mar 9, 2015

### Staff: Mentor

The binding energy is slightly higher than the minimum of the potential energy curve because of the zero-point energy of vibration. So $E_0$ is not the binding energy in this graph, it's just the minimum potential energy.
The zero of energy is arbitrary: physics depends only on the relative differences in energy. Here, the zero is chosen to correspond to two non-interacting atoms (infinitely separated), so any $E<0$ corresponds to an attractive interaction.