Energy of Photon: Hydrogen n=4 to n=3 Transition

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Homework Help Overview

The discussion revolves around calculating the energy of a photon associated with the transition of a hydrogen atom from the n=4 to n=3 energy state, focusing on the relevant formulas and units of measurement.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the formula needed to calculate energy differences between quantum states, with some suggesting the use of the Rydberg constant. Questions arise regarding the appropriate units for energy and the validity of calculated values.

Discussion Status

Participants are actively engaging with the problem, sharing insights about the Rydberg constant and discussing unit conversions. There is an ongoing exploration of the calculations involved, with no clear consensus yet on the correct approach or final value.

Contextual Notes

Some participants question the appropriateness of using Joules versus electron volts for energy measurements, and there is mention of relativistic corrections that may or may not be considered in the calculations.

UrbanXrisis
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What is the energy of a photon when the Hydrogen atom transitions from the n=4 to n=3 energy state?
 
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Find the difference in potential between the n=3 and n=4 states.

cookiemonster
 
what formula do I use to find the energy in each state?
 
If you aren't worried about small relativistic corrections, just take the difference of R/n^2 for n=3 and n=4, where R is the Rydberg constant. Here n is the principle quantum number.
 
would energy be in Joules?
 
so to find the difference in potential... E=(R/3^2)-(R/4^2) ?
E=533264 J?
 
That's a little much... Look about 2/3 down the webpage I linked to.

eV is more commonly used, but Joules works.

cookiemonster
 
is 533264 Joules a lot for the difference in the energy states?
 
  • #10
The Rydberg constant is about 13.6 electron volts. So if you want energy in those units, it would be 13.6*(1/9 - 1/16).

The onversion from electron volt to joule is accomplished by multiplying by 1.602 x 10-19. So your number of joules is way too high.
 

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