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Topic: Lennard-Jones potential  (Read 7711 times)

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Suze

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Lennard-Jones potential
« on: May 04, 2005, 04:41:18 AM »
For water vapour, the Lennard-Jones potential has a maximum attractive energy of 1.08E-21 J at a separation of 355E-12m. From this data, how do I calculate the interaction energy at 3E-10m and 1.5E-8m?

Is the attractive energy V or E in the Lennard-Jones equation? I don't know what value of ro I use as isn't this when V=0? I'm just really confused as to what information I've been given and what the terms in the equation mean!  ??? Any help or advice would be much appreciated!  :)

Offline Dude

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Re:Lennard-Jones potential
« Reply #1 on: May 23, 2005, 04:08:43 PM »
This is probably too late for your use, however, I thought I remembered this from somewhere.  It is covered in chapter 1 in "Transport Phenomena" by Bird, Stewart and Lightfoot.  The graph in the book indicates that epsilon is directly related to the energy.  With the info you have, I would guess you would then have to take the first derivative of the LJ equation and set it equal to 0 to solve for the other constant.  That would then give you both constants to solve the equation directly for each distance.

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