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Interaction between two atoms

  • Thread starter dlesswater
  • Start date
  • #1
Suppose that the interaction energy between two atoms is given by:

E(R)=-A/(R^2)+B/(R^10)

The atoms form a stable molecule with an inter- nuclear distance of 0.3nm and a dissociation energy of 4 eV.
a) Determine A and B.
b) Calculate the force required to break the molecule. What is the critical distance between the nuclei for which this occurs?

I have two variables which I need to solve for but one equation. My teacher said try taking the derivative and you can then isolate one variable. I don't understand how or why you can so this. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
1,254
3
The for atom to be 'stable' at a distance of 0.3nm, what does that tell you about the Energy at that point (hint: it has to do with your teacher's suggestion).
 
  • #3
Well I know if we take the derivative we will find a slope correct. So do we want the slope to be zero? Meaning the change in energy would be zero. I think that makes since because a stable molecule would presumingly have little change in energy.
 
  • #4
1,254
3
Yeah, you're totally on the right track. The slope does need to be zero, the clearest way of understanding why is that the force is proportional to the gradient of the potential, thus the only way the molecule can be stable (no net force), is if the slope is zero.

So the derivative has to be zero, and you also know the total energy requirement. Thus you have two equations and two unknowns.
 
  • #5
So if I am correct my 2 equations will be:

-A/(R^2)+B/(R^10)=4eV
2A/(R^3)-10B/(R^11)=0

Then solve for A and B right?
 
  • #6
1,254
3
Yeah, that looks good.
Edit: maybe that should be a negative 4.
 

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