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Sketch of an interatomic potential

  1. Jan 20, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    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 seperation 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.

    3. 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.
     

    Attached Files:

  2. jcsd
  3. Jan 20, 2013 #2

    SammyS

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    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?
     
  4. Jan 20, 2013 #3

    CAF123

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    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: Jan 20, 2013
  5. Jan 20, 2013 #4

    haruspex

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    Sure it can. Just take logs both sides.
     
  6. Jan 20, 2013 #5

    CAF123

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    Hmm.. not sure how I missed that first time!
    But does it help? The term I am dealing with is -2exp.. not 2exp..
     
  7. Jan 20, 2013 #6

    haruspex

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    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?
     
  8. Jan 21, 2013 #7

    CAF123

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    I am not sure I completely understand what you mean - What expression are you talking about?
     
  9. Jan 21, 2013 #8

    CAF123

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    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 modelled 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?
     
  10. Jan 21, 2013 #9

    haruspex

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    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.
     
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