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

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Homework Help Overview

The problem involves a system of two electrons in a one-dimensional box, with a focus on calculating the potential and kinetic energy as functions of the distance between two nuclei. The context is rooted in quantum mechanics and electrostatics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of potential energy and its relationship to kinetic energy in the context of quantum mechanics. There is a focus on the lowest allowed energy state for the electrons and the implications of having two electrons in the box.

Discussion Status

Some participants have provided calculations and expressed uncertainty regarding factors in their results, particularly concerning the correct expression for kinetic energy. There is an ongoing exploration of how to combine potential and kinetic energy to find a minimum total energy.

Contextual Notes

Participants note potential confusion regarding the factors involved in the kinetic energy calculation, especially in relation to the number of electrons and the constants used. There is also mention of a time delay in responses, indicating a potential lack of active engagement.

danmel413
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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


En=n2pi2hbar2/2mL2

And the Schrödinger equation

The Attempt at a Solution


[/B]
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!
 
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danmel413 said:
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.
Perhaps they are expressing their answer in terms of h rather than hbar. Also, don't forget you have two electrons in the box.
 
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)
 
ondryice said:
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)
You're 8 months too late!
 
PeroK said:
You're 8 months too late!
Yeah I figured if someone googles the problem (like I did) then they'll find the whole problem
 

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