Solving the De Broglie Problem: Calculating Length of a 1-D Box

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Homework Help Overview

The problem involves calculating the length of a one-dimensional box for an electron in the n=1 state, equating its energy to that of a photon with a wavelength of 500 nm. The subject area pertains to quantum mechanics and wave-particle duality.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the energy equations for both the electron in a box and the photon, questioning the correctness of constants and formulations used. There are suggestions to reconsider the coefficients in the equations and the use of Planck's constant versus reduced Planck's constant.

Discussion Status

The discussion is ongoing, with participants providing different interpretations of the equations involved. Some guidance has been offered regarding the correct formulation of the energy equations, but no consensus has been reached on the correct approach or solution.

Contextual Notes

There is uncertainty regarding the constants used in the equations, particularly whether to use 8 or 4 in the calculations. Participants are also questioning the arithmetic accuracy of the original poster's calculations.

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Homework Statement


What is the length of a one-dimensional box in which an electron in the n=1 state has the same energy as a photon with a wavelength of 500 nm


Homework Equations




E=h^2/8mL^2 and E=hc/lambda

making it
L=sqrt( (h*lambda)/(8cm) )


The Attempt at a Solution



I plugged in for those numbers and did not come out with the correct number. any suggestions?
 
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Are you sure that 8 should not be a 4?
 
The eightfold way is right here. 1/8=(L/2)^/2.
I think your first h should be hbar.
 
Hey
The energy if the first state of an indefinite one-dimensional box is:
[tex]E=\frac{\pi^{2}\hbar^{2}}{2mL^{2}}[/tex]
Where m is the mass of the particle and L is the length of the box.
The photon has the energy given by
[tex]E=\hbar\omega=\frac{2\pi\hbar{c}}{\lambda}[/tex]
Where [tex]\lambda[/tex] is the wave length.
And therefore the length L is [tex]L=\sqrt{\frac{\pi\hbar\lambda{c}}{4m}}[/tex]
 
Last edited:
eys_physics said:
And therefore the length L is [tex]L=\sqrt{\frac{\pi\hbar\lambda{c}}{4m}}[/tex]

Which is wrong. It should be [tex]\sqrt{\frac{\pi\hbar\lambda}{4mc}[/tex], which is identical to what the OP psingh had written correctly. Replacing [itex]h[/itex] by [itex]2\pi\hbar[/itex] won't do any good.

Perhaps the OP made some arithmetical mistake...
 

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