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

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The energy of a photon during the transition of a hydrogen atom from the n=4 to n=3 state can be calculated using the formula E = R(1/n^2), where R is the Rydberg constant. The difference in energy between these states is found by calculating E = R(1/9 - 1/16). The energy is commonly expressed in electron volts (eV), with the Rydberg constant approximately equal to 13.6 eV. Converting this energy to joules involves multiplying by 1.602 x 10^-19, indicating that the initial calculation of 533264 Joules was significantly overestimated. Accurate calculations reveal that the energy difference is much lower than initially suggested.
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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?
 
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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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