What is the energy of a photon when the Hydrogen atom transitions from the n=4 to n=3 energy state?
Find the difference in potential between the n=3 and n=4 states.
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) ?
That's a little much... Look about 2/3 down the webpage I linked to.
eV is more commonly used, but Joules works.
is 533264 Joules a lot for the difference in the energy states?
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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