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Find the minimum kinetic energy of two electrons in a 1D box

  1. Apr 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Problem: Consider a "crystal" consisting of two nuclei and two electrons arranged like this:
    q1 q2 q1 q2
    with a distance d betweem each. (q1=e, q2=-e)
    a) Find the potential energy as a function of d.
    b) Assuming the electrons to be restricted to a one-dimensional box of length 3d, find the minimum kinetic energy of the two electrons.
    c) Find the value of d for which the total energy is a minimum.

    2. Relevant equations
    En=n2pi2hbar2/2mL2

    And the Schrodinger equation

    3. The attempt at a solution

    The Potential energy I found to be (-7/3)(ke2/d) which is correct. (k=coulomb constant).

    I assumed the minimum Kinetic energy would be the lowest allowed energy (basically E and n=1) because Potential energy should be zero inside the box. I got as a result pi2hbar2/18md2, but the correct answer is hbar2/36md2.

    I have a factor of pi2 that I don't know how to get rid of
    I'm missing a factor of 1/2 - is that because there are two electrons and it is thus 2m instead of m?

    For c, d is supposed to equal hbar2/42mke2 and I assume it comes from the fact that Eelectric=kq2/r, but I'm not sure how to continue there. I'm guessing it has something to do with the kinetic energy I can't find.

    hbar is the reduced planck constant (h/2pi)

    Thanks!
     
  2. jcsd
  3. Apr 19, 2016 #2

    TSny

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    Perhaps they are expressing their answer in terms of h rather than hbar. Also, don't forget you have two electrons in the box.
     
  4. Dec 12, 2016 #3
    For part c, you should add the total potential energy you found in (a) to the minimum kinetic energy of the two electrons (answer to b), then differentiate the sum and solve for zero. When I solved, I got an answer of d= h^2 / (42ke^2m)
     
  5. Dec 13, 2016 #4

    PeroK

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    You're 8 months too late!
     
  6. Dec 13, 2016 #5
    Yeah I figured if someone googles the problem (like I did) then they'll find the whole problem
     
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