Help plotting the Lennard Jones Potential

15 visualizaciones (últimos 30 días)
Amanda Chun
Amanda Chun el 14 de Mzo. de 2018
Editada: David Goodmanson el 15 de Mzo. de 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])

Respuestas (2)

Image Analyst
Image Analyst el 14 de Mzo. de 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 comentarios
Amanda Chun
Amanda Chun el 14 de Mzo. de 2018
I'm not sure this is right, ideally my plot would look similar to this
Image Analyst
Image Analyst el 14 de Mzo. de 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.

Iniciar sesión para comentar.


David Goodmanson
David Goodmanson el 15 de Mzo. de 2018
Editada: David Goodmanson el 15 de Mzo. de 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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