How to Solve for Variables Using Derivatives in Atomic Interactions?

In summary, the interaction energy between two atoms is given by the equation E(R)=-A/(R^2)+B/(R^10). The atoms form a stable molecule at a distance of 0.3nm with a dissociation energy of 4 eV. To determine the values of A and B, one can take the derivative of the equation and set it equal to zero. This results in two equations with two unknowns, which can be solved to find the values of A and B. Additionally, the derivative being zero means that the force is proportional to the gradient of the potential, resulting in a stable molecule with minimal change in energy.
  • #1
dlesswater
7
0
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.
 
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  • #2
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
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
Yeah, that looks good.
Edit: maybe that should be a negative 4.
 

Related to How to Solve for Variables Using Derivatives in Atomic Interactions?

1. What is the interaction between two atoms?

The interaction between two atoms is the force that holds them together in a molecule or compound. It is known as a chemical bond and is formed when the atoms share, gain, or lose electrons to achieve a more stable state.

2. What are the different types of interactions between atoms?

The main types of interactions between atoms are ionic, covalent, and metallic bonds. Ionic bonds involve the transfer of electrons between atoms, while covalent bonds involve the sharing of electrons. Metallic bonds occur in metals and involve the sharing of electrons between a lattice of positive ions.

3. How do the properties of atoms affect their interactions?

The properties of atoms, such as their size, charge, and number of valence electrons, determine the type and strength of their interactions. Atoms with similar properties are more likely to form strong bonds, while those with different properties may form weaker bonds or not interact at all.

4. How does the distance between two atoms affect their interaction?

The distance between two atoms can greatly affect their interaction. If the atoms are too far apart, their electrons may not be able to interact and form a bond. If they are too close, the repulsive forces between their electron clouds may prevent them from bonding. A specific distance, known as the bond length, is needed for a stable interaction to occur.

5. Can the interaction between two atoms be broken?

Yes, the interaction between two atoms can be broken. This can happen through various processes such as heating, chemical reactions, or physical force. The strength of the bond between the atoms will determine how easily it can be broken. For example, covalent bonds are generally stronger than ionic bonds, so they require more energy to break.

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