How Do I Solve Part B of the Lennard-Jones Potential Problem?

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Homework Help Overview

The discussion revolves around solving part B of a problem related to the Lennard-Jones potential, focusing on the relationship between kinetic energy, potential energy, and the variables involved in the equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are considering the use of the relationship T = E - U to derive expressions for variables, questioning the role of the constant E in their calculations. Some are exploring the implications of ignoring E, while others are deriving expressions for r based on the potential's characteristics.

Discussion Status

The discussion is active, with participants sharing various approaches and expressions derived from their reasoning. There is no explicit consensus on the best method, but multiple interpretations and lines of reasoning are being explored.

Contextual Notes

Participants are working under the constraints of the problem as presented, with specific references to constants and derivatives that may influence their calculations. The nature of the potential and its minimum point is also under consideration.

Oblio
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I'm really stuck on b.) of this question.

http://www.physics.brocku.ca/Courses/2P20/problems/prob_lenardjones.pdf

I think I can use T = E - U, and solve for v from within T and integrate for x...

but what do I do with E?
 
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can i ignore E since its a constant? I got something like,
[tex]x = /frac{-6r^6A + 12B}{2r^13} t^2 [\tex][/tex]
 
The minimum is the point at which dV/dr=0
 
I'm getting the ridiculous value of :

r = [tex]\sqrt[6]{\frac{-12B}{6A}}[/tex]

by deriving the two quotients, allowing A' and B' to cancel in the process (constants) cancelling out the high denominators as much as possible... etc..
 

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