Help plotting the Lennard Jones Potential

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Amanda Chun
Amanda Chun il 14 Mar 2018
Modificato: David Goodmanson il 15 Mar 2018
I have the following code to plot the Lennard-Jones potential for Xe However, when I plot the code instead of getting the expected curve, I almost just get an "L" I tried changing the axis, but it still doesn't yield the proper curve.
%%1.1: Plotting the Leonard Jones Potential
eps = 1.77; %kJ/mol
sig = 4.10; %A
r = linspace(0.01*sig,6*sig,10000);
for i = 1:length(r)
V(i) = 4*eps*((sig/r(i))^12-(sig/r(i))^6);
end
plot(V,r)
ylim([-1 1])

Risposte (2)

Image Analyst
Image Analyst il 14 Mar 2018
Try it this way:
% 1.1: Plotting the Leonard Jones Potential
eps = 1.77; %kJ/mol
sig = 4.10; %A
r = linspace(0.01*sig,6*sig,10000);
for i = 1:length(r)
V(i) = 4*eps*((sig/r(i))^12-(sig/r(i))^6);
end
semilogy(r, V, 'b-')
grid on;
% ylim([-1 1])
  2 Commenti
Amanda Chun
Amanda Chun il 14 Mar 2018
I'm not sure this is right, ideally my plot would look similar to this
Image Analyst
Image Analyst il 14 Mar 2018
Look over the formula. I know nothing of that formula or whether you entered it correctly. I just plotted what you had so you could see it better.

Accedi per commentare.


David Goodmanson
David Goodmanson il 15 Mar 2018
Modificato: David Goodmanson il 15 Mar 2018
Hi Amanda, try
plot(r,V) instead of plot(V.r)
and
ylim([-2,1])
For a more Matlablike approach you could calculate the array V all at once as a function of the array r, which you have already defined as a vector.
V = 4*eps*((sig./r).^12-(sig./r).^6);
Using ./ and .^ means all the division and power calculations are done element-by-element with the array r.

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