Why Is My Calculation of the Particle Energy in a Rectangular Box Different?

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Homework Help Overview

The discussion revolves around calculating the energy states of a particle confined in a rectangular box with specified edge lengths. The original poster is attempting to reconcile their calculation with a provided answer for the lowest energy states.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster presents their calculation for the ground state energy and questions the discrepancy with the textbook answer. Other participants seek clarification on the energy states mentioned and the reasoning behind the original poster's calculations.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the energy states and the calculations involved. Some guidance on the correct form of the energy equation has been noted, but no consensus has been reached regarding the calculations.

Contextual Notes

The problem involves specific edge lengths for the box and requires the identification of the six lowest energy states, which may influence the calculations and assumptions made by participants.

cgw
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I am missing something.

The question is to find the 6 lowest energy states of a particle of mass m in a box with edge lengths of [tex]L_{1}=L[/tex], [tex]L_{2}=2L[/tex], [tex]L_{3}=2L[/tex].
The answer gives [tex]E_{0}=\frac{\pi^2\hbar^2}{8mL^2}[/tex].

I would have said [tex]E_{0}=\frac{3\pi^2\hbar^2}{4mL^2}[/tex] .

What am I missing?
(the answer given for the actual question is 6, 9, 9, 12, 14, 14)
 
Last edited:
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Finally figured out the latex.
 
Last edited:
Well done with the latex. Why would you say what you say ?

I don't know what you mean by "6, 9, 9, 12, 14, 14".
 
The question asks for the energy of the six lowest states [tex]\frac{E}{E_{0}}[/tex]. The textbook answer gives [tex]E_{0}=\frac{\pi^2\hbar^2}{8mL^2}[/tex]

The way I see it is:

E = E_0 at n1=1, n2=1, n3=1


[tex]E_{0}=\frac{\pi^2\hbar^2}{2m}\left{\left(\frac{n_{1}}{L}\right)^2+\left(\frac{n_{2}}{2L}\right)^2+\left(\frac{n_{3}}{2L}\right)^2\right}[/tex]

which should be [tex]E_{0}=\frac{3\pi^2\hbar^2}{4mL^2}[/tex]
 
Last edited:

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