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Introductory Physics Homework Help
Find the minimum kinetic energy of two electrons in a 1D box
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[QUOTE="danmel413, post: 5448193, member: 571853"] [h2]Homework Statement [/h2] Problem: Consider a "crystal" consisting of two nuclei and two electrons arranged like this: [I]q1 q2 q1 q2[/I] with a distance [I]d [/I]betweem each. ([I]q1=e, q2=-e[/I]) a) Find the potential energy as a function of [I]d.[/I] 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. [h2]Homework Equations[/h2] E[SUB]n[/SUB]=n[SUP]2[/SUP]pi[SUP]2[/SUP]h[SUB]bar[/SUB][SUP]2[/SUP]/2mL[SUP]2[/SUP] And the Schrodinger equation [h2]The Attempt at a Solution[/h2] [/B] The Potential energy I found to be (-7/3)(ke[SUP]2[/SUP]/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[SUP]2[/SUP]h[SUB]bar[/SUB][SUP]2[/SUP]/18md[SUP]2[/SUP], but the correct answer is h[SUB]bar[/SUB][SUP]2[/SUP]/36md[SUP]2[/SUP]. I have a factor of pi[SUP]2[/SUP] 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[SUB]bar[/SUB][SUP]2[/SUP]/42mke[SUP]2[/SUP] and I assume it comes from the fact that E[SUB]electric[/SUB]=kq[SUP]2[/SUP]/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[SUB]bar[/SUB] is the reduced Planck constant (h/2pi) Thanks! [/QUOTE]
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Introductory Physics Homework Help
Find the minimum kinetic energy of two electrons in a 1D box
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