Lennard-Jones Work: Calculating Equilibrium Position

In summary: So, what I mean to equilibrium position is the position of balance between the two atoms, but there is no more information about that.- The force on molecule A when it is at equilibrium or, as you say, in balance with molecule B?At equilibrium, the force on molecule A would be zero.
  • #1
Nacho Verdugo
15
0

Homework Statement


I need to calculate the work donde by the Lennard-Jones Law, considering:

F(r)=F0 [2(σ/r)13-(σ/r)7]

when approximating two atoms from infinity to the equilibrium position between both atoms

Homework Equations



First thing I don't know how to calculate is the equilibrium position (x0) between two arbitrary atoms

The Attempt at a Solution



I just integrated this force, considering

WAB=∫F⋅dr ; r=A to r=B

As I don't know the x0, I just integrate from A to B, to later analyze the result. Assuming A=-∞ and B=x0

W= ∫ F0 [2(σ/r)13-(σ/r)7] dr ; r=A to r=B
= integrating...
= 2F0σ13 ( r-12 /-12) - F0σ7 (r-6/-6) ; still need to evaluate

Considering that A=-∞, I finally obtain:

W= F0σ7/6 (1/x06)-F0σ13/6(1/x012)

Is it right until here?
 
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  • #2
What is the force at the equilibrium position. What does "equilibrium" mean?
What are your limits of integration? Consider particle A already at the origin. Then calculate the work done by the force on particle B as it is brought from infinity to x0.
 
  • #3
kuruman said:
What is the force at the equilibrium position. What does "equilibrium" mean?
What are your limits of integration? Consider particle A already at the origin. Then calculate the work done by the force on particle B as it is brought from infinity to x0.
The statement of this problems is:

In diatomic molecules, the constituent atoms exert attractive forces between themselves at great distances and repulsive forces at short distances. For many molecules the Lennard-Jones law is a good approximation for the modulus of these forces.

$$ F(r)=F_0 \cdot [2(\frac{\sigma}{r})^{13}-(\frac{\sigma}{r})^7] $$

where r is the distance between the center of the nucleus of the atoms, ## \sigma ## a length parameter and ## F_0 ## a constant.

Determine the work to be performed by this force or by an external agent to approximate two atoms from infinity to the position of balance between the two.
- So, what I mean to equilibrium position is the position of balance between the two atoms, but there is no more information about that.

- The limits of integration, as I don't know the position of balance, are ## A=-\infty ## to ## B=x_0 ## (the so called position of balance).

- Didn't catch up the last advice
 
Last edited:
  • #4
Nacho Verdugo said:
So, what I mean to equilibrium position is the position of balance between the two atoms, but there is no more information about that.
How big is the force on molecule A when it is at equilibrium or, as you say, in balance with molecule B?
 
  • #5
kuruman said:
How big is the force on molecule A when it is at equilibrium or, as you say, in balance with molecule B?

I guess it is zero right? I that way it is in equilibrium.
 
  • #6
Exactly. So at what inter-molecular distance r0 is the force zero?
 
  • #7
kuruman said:
Exactly. So at what inter-molecular distance r0 is the force zero?

I don't know :C I took a while to answer cause of that. I think it should be at a huge distance, right?
 
  • #8
It is simpler than you think. The force in general is
$$F(r)=F_0 \cdot [2(\frac{\sigma}{r})^{13}-(\frac{\sigma}{r})^7]$$
At r = r0 you have
$$0=F(r_0)=F_0 \cdot [2(\frac{\sigma}{r_0})^{13}-(\frac{\sigma}{r_0})^7]$$
Do you see what you have to do?
 
  • #9
kuruman said:
It is simpler than you think. The force in general is
$$F(r)=F_0 \cdot [2(\frac{\sigma}{r})^{13}-(\frac{\sigma}{r})^7]$$
At r = r0 you have
$$0=F(r_0)=F_0 \cdot [2(\frac{\sigma}{r_0})^{13}-(\frac{\sigma}{r_0})^7]$$
Do you see what you have to do?
yes! with $$F(r_0)=0$$ I can actually find $$r_0$$ and later evaluate that value in the integral.
 
  • #10
Go for it! :smile:
 
  • Like
Likes Nacho Verdugo
  • #11
kuruman said:
Go for it! :smile:
thank you very much!:biggrin:
 
  • #12
Nacho Verdugo said:
yes! with $$F(r_0)=0$$ I can actually find $$r_0$$ and later evaluate that value in the integral.
 
  • #13
Nacho Verdugo said:
The statement of this problems is:

In diatomic molecules, the constituent atoms exert attractive forces between themselves at great distances and repulsive forces at short distances. For many molecules the Lennard-Jones law is a good approximation for the modulus of these forces.

$$ F(r)=F_0 \cdot [2(\frac{\sigma}{r})^{13}-(\frac{\sigma}{r})^7] $$

where r is the distance between the center of the nucleus of the atoms, ## \sigma ## a length parameter and ## F_0 ## a constant.

Determine the work to be performed by this force or by an external agent to approximate two atoms from infinity to the position of balance between the two.
- So, what I mean to equilibrium position is the position of balance between the two atoms, but there is no more information about that.

- The limits of integration, as I don't know the position of balance, are ## A=-\infty ## to ## B=x_0 ## (the so called position of balance).

- Didn't catch up the last advice
 
  • #14
Is there a question you wish to ask?
 
  • #15
kuruman said:
Is there a question you wish to ask?
No thanks! I just made a mistake editing this.
 

1. What is Lennard-Jones work?

Lennard-Jones work refers to the calculation of the equilibrium position of two particles interacting with each other through the Lennard-Jones potential. It is a mathematical model used in molecular dynamics simulations to describe the interaction between atoms or molecules.

2. How is the equilibrium position calculated?

The equilibrium position is calculated by finding the minimum energy point on the Lennard-Jones potential curve. This is done by solving the equation for the minimum energy point, which is given by the derivative of the potential with respect to the distance between the two particles.

3. What factors affect the equilibrium position?

The equilibrium position is affected by the strength of the interaction between the particles (determined by the parameters σ and ε in the Lennard-Jones potential), as well as the distance between the particles. Additionally, the presence of other particles in the system can also affect the equilibrium position.

4. What is the importance of calculating the equilibrium position in molecular dynamics simulations?

The equilibrium position is important as it determines the stable configuration of the particles in the system. This information is crucial in understanding the behavior and properties of the system, and can be used to make predictions about its behavior under different conditions.

5. Are there any limitations to the Lennard-Jones work method?

While the Lennard-Jones work method is widely used in molecular dynamics simulations, it does have some limitations. For example, it assumes that the particles are spherical and have a fixed interaction range. It also does not take into account quantum mechanical effects, which may be important for certain systems.

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