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## Homework Statement

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.

## Homework Equations

E

_{n}=n

^{2}pi

^{2}h

_{bar}

^{2}/2mL

^{2}

And the Schrodinger equation

## The Attempt at a Solution

[/B]

The Potential energy I found to be (-7/3)(ke

^{2}/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 pi

^{2}h

_{bar}

^{2}/18md

^{2}, but the correct answer is h

_{bar}

^{2}/36md

^{2}.

I have a factor of pi

^{2}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 h

_{bar}

^{2}/42mke

^{2}and I assume it comes from the fact that E

_{electric}=kq

^{2}/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.

h

_{bar}is the reduced planck constant (h/2pi)

Thanks!