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Energy of a photon

  1. May 16, 2004 #1
    What is the energy of a photon when the Hydrogen atom transitions from the n=4 to n=3 energy state?
     
  2. jcsd
  3. May 16, 2004 #2
    Find the difference in potential between the n=3 and n=4 states.

    cookiemonster
     
  4. May 16, 2004 #3
    what formula do I use to find the energy in each state?
     
  5. May 17, 2004 #4

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    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.
     
  6. May 17, 2004 #5
  7. May 17, 2004 #6
    would energy be in Joules?
     
  8. May 17, 2004 #7
    so to find the difference in potential... E=(R/3^2)-(R/4^2) ?
    E=533264 J?
     
  9. May 17, 2004 #8
    That's a little much... Look about 2/3 down the webpage I linked to.

    eV is more commonly used, but Joules works.

    cookiemonster
     
  10. May 17, 2004 #9
    is 533264 Joules a lot for the difference in the energy states?
     
  11. May 17, 2004 #10

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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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