
#1
Jan2013, 11:56 AM

P: 1,994

1. The problem statement, all variables and given/known data
Consider the interatomic potential between 2 atoms: $$U(r) = U_o \left[exp\left(\frac{2(ra)}{a}\right)  2exp \left(\frac{(ra)}{a} \right) \right],$$ 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)/U_{o} 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/U_{o} 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 e^{2} 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. 



#2
Jan2013, 02:39 PM

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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
Jan2013, 03:25 PM

P: 1,994

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. 



#4
Jan2013, 04:02 PM

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Sketch of an interatomic potential 



#5
Jan2013, 04:08 PM

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But does it help? The term I am dealing with is 2exp.. not 2exp.. 



#6
Jan2013, 04:21 PM

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Does it help to note that e^{22x} = (e^{1x})^{2}?
Does that suggest a way of writing the expression so that x only appears once? 



#7
Jan2113, 01:35 AM

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#8
Jan2113, 09:07 AM

P: 1,994

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? 



#9
Jan2113, 02:57 PM

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e^{22x}2e^{1x} = (e^{1x})^{2}  2e^{1x} = (e^{1x}  1)^{2}  1 I thought that getting it down to one occurrence of x might make it easier to study. 


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