Electron in one dimensional infinite square well

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Homework Help Overview

The discussion revolves around an electron in a one-dimensional infinite square well, specifically regarding its excitation from the ground state when illuminated by light of a certain wavelength. Participants are exploring the energy levels associated with the electron and the photon involved in the interaction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of energy levels for the electron in the well and the energy of the photon. There is a focus on determining the correct quantum state to which the electron is excited and whether the photon energy matches the required energy difference for excitation.

Discussion Status

The conversation is ongoing, with participants questioning the calculations and assumptions made regarding energy levels. Some guidance is provided on the necessity for the photon energy to match the energy difference between states for excitation to occur, but no consensus has been reached on the calculations or the resulting quantum state.

Contextual Notes

There is a noted discrepancy between the calculated energy levels and the expected outcomes based on the photon energy. Participants are also considering the implications of the photon energy not matching any of the excited states, which raises questions about the absorption process.

whynot314
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An electron in the ground state of a one-dimensional infinite square well of width 1.10 nm is
illuminated with light of wavelength 600 nm. Into which quantum state is the electron excited?


ok so I first calculated the engery of the electron in the first ground state of the square well using
\frac{1240^{2}}{8*511000*1.10^{2}}
I get .310 eV

Then i calculate the energy of the photon
\frac{1240}{600}

and i get 2.066 eV

Then i do
2.066=\frac{1240^{2}}{8*511000*1.10^{2}}*n^{2}-.310
and solve for n i get the 3rd state. that answer says it should be in the same state n=1, I am not sure what I am doing wrong.
 
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If that fraction is 0.31 eV, how did you get n=3 as result? The solution differs significantly from 3 - indicating that the electron does not absorb this wavelength at all.
 
The .310 I calculated using the energy for a particle in a box E=\frac{h^{2} c^{2}}{8*m*c^{2}*L^{2} }*n^{2}, for n=1 for a the electron in ground state. I got n=3 by using E(photon)= (energy in a box) \frac{h^{2} c^{2}}{8*m*c^{2}*L^{2} }*n^{2}-Energy of the electron in ground state (.310). I just then solved for n.so how does the ratio compare?
 
Does the photon need to have the exact energy level to excite the electron into a different state? since the ground state is .311 eV and the 1 st excited state is 1.244 eV and the 2nd excited state is 2.799 for this length of box. And because a single photon of the light has 2.066 eV and does not match any of the other excited states it does not get absorbed?
 
whynot314 said:
Does the photon need to have the exact energy level to excite the electron into a different state?
Pretty much.
And because a single photon of the light has 2.066 eV and does not match any of the other excited states it does not get absorbed?
Exactly.
 

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