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Lennard-Jones Potential - distance of closest approach

  1. May 10, 2010 #1

    xn_

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    Here's my solution to the problem of finding the distance of closest approach for two atoms approaching each other in a Lennard-Jones potential, starting with velocities (+/-) v_0:

    They start with energy T0 = (1/2)m(2v02) which is conserved. Thus,
    E = T + V = (1/2)m(2v2) + A/r12 - B/r6 = T0. At the distance of closest approach, v=0 and we have:
    T0 = A/rt12 - B/rt6. If we let x = rt6 and rearrange:
    T0x2 + Bx - A = 0
    which has the solutions
    x = (1/2T0)(-B +/- sqrt(B2 + 4AT_0)

    My question is - what, physically, do each of these roots correspond to? They're both real, but of opposite sign - is the negative one a kind of 'virtual' turning point for the atoms passing through each other (tunnelling?) and coming out the other side?
     
  2. jcsd
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