Sketch of an interatomic potential

In summary: Is this a reasonable explanation?In summary, the potential between 2 atoms can be expressed in the form y = y(x), with y = U/Uo and x =r/a. E2-2x = (e1-x)2, and the force between the atoms is conservative, so applying the -derivative of the potential results in the force being the same.
  • #1
CAF123
Gold Member
2,948
88

Homework Statement


Consider the interatomic potential between 2 atoms: $$U(r) = U_o \left[exp\left(\frac{-2(r-a)}{a}\right) - 2exp \left(\frac{-(r-a)}{a} \right) \right],$$ where r is the separation between the two atoms and ##U_o## and a are constants.
1)Evaluate the expressions in the square brackets for a number of values of r/a and thereby make a plot of U(r)/Uo versus r/a, plotting the two exp terms separately.
2)Sketch the total result on the same diagram.

The Attempt at a Solution



I have attached two graphs of the attractive and repulsive component of this potential. Conveniently, the potential can be expressed in the form y =y(x), with y = U/Uo and x =r/a. I don't know what a is so I am struggling to see if my graphs make any physical sense. As r/a decreases, however, I note that the repulsive term increases with limit of e2 in the repulsive case and -2e in the attractive case. To obtain these graphs, I simply plotted the two exp terms against r/a separetely, for random values of r/a. Can someone check these graphs?
Also, I am a bit unsure of how to combine them into one graph.
Many thanks.
 

Attachments

  • 001A.jpg
    001A.jpg
    13.2 KB · Views: 414
Physics news on Phys.org
  • #2
CAF123 said:

Homework Statement


Consider the interatomic potential between 2 atoms: $$U(r) = U_o \left[exp\left(\frac{-2(r-a)}{a}\right) - 2exp \left(\frac{-(r-a)}{a} \right) \right],$$ where r is the separation between the two atoms and ##U_o## and a are constants.
1)Evaluate the expressions in the square brackets for a number of values of r/a and thereby make a plot of U(r)/Uo versus r/a, plotting the two exp terms separately.
2)Sketch the total result on the same diagram.

The Attempt at a Solution



I have attached two graphs of the attractive and repulsive component of this potential. Conveniently, the potential can be expressed in the form y =y(x), with y = U/Uo and x =r/a. I don't know what a is so I am struggling to see if my graphs make any physical sense. As r/a decreases, however, I note that the repulsive term increases with limit of e2 in the repulsive case and -2e in the attractive case. To obtain these graphs, I simply plotted the two exp terms against r/a separetely, for random values of r/a. Can someone check these graphs?
Also, I am a bit unsure of how to combine them into one graph.
Many thanks.
Letting x = r/a is indeed the way to handle the parameter, a .

Both terms are basic exponential graphs. They don't flatten out for small x = r/a, unlike what you show .

[itex]\displaystyle e^{-2r/a+2}=e^2\,e^{-2r/a}\ \ [/itex] and [itex]\displaystyle \ \ 2e^{-r/a+1}=(2e)\,e^{-r/a}\ .\ [/itex]

What is each term at r=a, i.e., when x = 1 ?

At what value of x, are the two terms equal in magnitude?
 
  • #3
At r/a = x = 1, the attractive term is -2 and the repulsion term 1.
In solving for when they are equal, I get to ##e^{-2x+1} = 2e^{-x}##, which can't be solved analytically because of that 2. ( I could use Newton Raphson although I feel that is not the purpose of the problem)
What values of r/a do you suggest? Is numbers like 1,2,3.. okay?

EDIT: What I have now is two exp looking graphs. In the attractive case, I have a curve coming from ##-\infty## and tending to 0. Similarly, in the repulsive case, I have a graph starting at ##+\infty## and again decaying to zero. This looks better - thanks. How to combine them? I suppose if we are concerned only with +ve x (i.e their separation cannot be negative since this would imply one of the other particles passes the other, I draw from x ##\geq##0 i.e from y = ##-2e^2## in attractive case and y = ##e## in repulsive case, and then both tending to 0.
 
Last edited:
  • #4
CAF123 said:
##e^{-2x+1} = 2e^{-x}##, which can't be solved analytically because of that 2.
Sure it can. Just take logs both sides.
 
  • #5
haruspex said:
Sure it can. Just take logs both sides.
Hmm.. not sure how I missed that first time!
But does it help? The term I am dealing with is -2exp.. not 2exp..
 
  • #6
Does it help to note that e2-2x = (e1-x)2?
Does that suggest a way of writing the expression so that x only appears once?
 
  • #7
haruspex said:
Does it help to note that e2-2x = (e1-x)2?
Does that suggest a way of writing the expression so that x only appears once?
I am not sure I completely understand what you mean - What expression are you talking about?
 
  • #8
Ok, I got both graphs now. One quick question about something later in the problem. It asks to derive the force between these two atoms. Now I think the interatomic force(s) is/are conservative so I can apply the straightforward -derivative of potential.

One thing that I was trying to understand was why the attractive compt is modeled by the negative term in the potential. The way I tried to describe it is: The attractive component tends to lower U. Since we deal with a conservative force, T + U = const, and so T must increase. This makes sense, as the particles get closer, there is more of a attraction. Now at some point, which is the minimum of the potential, the particles gets repelled, corresponding to an increase in U and so a decrease in T, which also makes sense since if the particle is so far away, there is barely any attraction. In between these two extremes, T and U vary so as to preserve the constant. Is this a reasonable explanation?
 
  • #9
CAF123 said:
I am not sure I completely understand what you mean - What expression are you talking about?
U(r) = U(ax) = U0[e2-2x-2e1-x ]
e2-2x-2e1-x = (e1-x)2 - 2e1-x = (e1-x - 1)2 - 1
I thought that getting it down to one occurrence of x might make it easier to study.
 

1. What is a sketch of an interatomic potential?

A sketch of an interatomic potential is a visual representation of the interactions between atoms in a molecule or material. It shows the potential energy of the atoms as a function of their positions, giving insight into the stability and behavior of the system.

2. How is a sketch of an interatomic potential created?

A sketch of an interatomic potential is typically created using computational methods, such as molecular dynamics simulations. These simulations calculate the forces between atoms and use this information to generate a plot of the potential energy as a function of distance between the atoms.

3. What can a sketch of an interatomic potential tell us?

A sketch of an interatomic potential can tell us about the stability and behavior of a molecule or material. It can also provide information about the strength and type of interactions between atoms, such as covalent, ionic, or van der Waals forces.

4. How is a sketch of an interatomic potential used in research?

In research, a sketch of an interatomic potential is often used to understand and predict the properties of materials. It can also be used to guide the development of new materials with specific properties, or to investigate the behavior of molecules in different environments.

5. Can a sketch of an interatomic potential be used to accurately predict the behavior of a material?

While a sketch of an interatomic potential can provide valuable insights, it is not always able to accurately predict the behavior of a material. Other factors, such as temperature and pressure, can also affect the behavior of atoms in a material. Therefore, experimental data is often needed to validate the predictions made by a sketch of an interatomic potential.

Similar threads

  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
604
  • Introductory Physics Homework Help
Replies
1
Views
874
  • Introductory Physics Homework Help
Replies
6
Views
468
  • Introductory Physics Homework Help
Replies
10
Views
677
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
337
  • Introductory Physics Homework Help
Replies
2
Views
618
Replies
8
Views
221
  • Introductory Physics Homework Help
Replies
15
Views
340
Back
Top