Energy uncertainty of an atom in an excited state

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Homework Help Overview

The discussion revolves around the energy uncertainty of a sodium (Na) atom in an excited state, specifically focusing on the relationship between the time spent in the excited state and the energy of the emitted photon during the transition to the ground state.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the uncertainty principle and its application to the problem, with some questioning how to utilize it effectively. There is also a mention of the specific energy of the photon emitted and its relevance to the uncertainty calculation.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the uncertainty principle and its application to the problem. Some have attempted calculations, while others express uncertainty about the relevance of certain provided values and the appropriateness of using specific equations.

Contextual Notes

Participants note constraints regarding the use of Bohr's equations due to the specifics of the sodium atom and question the necessity of the energy value given for the de-excitation process.

Ezequiel
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Homework Statement



A Na atom is in an excited state for a mean time of 1.6 \times 10^{-8}s. Then it jumps to the ground state emitting a photon with 2.105 eV of energy. Find the energy uncertainty of that excited state.

Homework Equations





The Attempt at a Solution



I don't even know where to start. Any help would be appreciated!
 
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I just don't know how to use that to solve the problem. Could you be more specific?
 
Could it be \Delta E > \frac{\hbar}{2} \frac{1}{\Delta t} = 32.96 \times 10^{-28}J?
 
Ezequiel said:
Could it be \Delta E > \frac{\hbar}{2} \frac{1}{\Delta t} = 32.96 \times 10^{-28}J?

Doesn't seem to be correct. What you did basically? And the question mentions Na atom , so we cannot use bohr equations. Also I don't know why they gave you energy of de-excitation. There is no use of it , I guess... I think you used the correct equation but did not solve it correctly.
 
Last edited:

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