Help plotting the Lennard Jones Potential

12 Ansichten (letzte 30 Tage)
Amanda Chun
Amanda Chun am 14 Mär. 2018
Bearbeitet: David Goodmanson am 15 Mär. 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])

Antworten (2)

Image Analyst
Image Analyst am 14 Mär. 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 Kommentare
Amanda Chun
Amanda Chun am 14 Mär. 2018
I'm not sure this is right, ideally my plot would look similar to this
Image Analyst
Image Analyst am 14 Mär. 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.

Melden Sie sich an, um zu kommentieren.


David Goodmanson
David Goodmanson am 15 Mär. 2018
Bearbeitet: David Goodmanson am 15 Mär. 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.

Kategorien

Mehr zu Oceanography and Hydrology finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by