PDA

View Full Version : energy of a photon


UrbanXrisis
May16-04, 11:03 PM
What is the energy of a photon when the Hydrogen atom transitions from the n=4 to n=3 energy state?

cookiemonster
May16-04, 11:39 PM
Find the difference in potential between the n=3 and n=4 states.

cookiemonster

UrbanXrisis
May16-04, 11:47 PM
what formula do I use to find the energy in each state?

Janitor
May17-04, 12:09 AM
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.

cookiemonster
May17-04, 12:12 AM
http://scienceworld.wolfram.com/physics/BohrModel.html

cookiemonster

UrbanXrisis
May17-04, 12:13 AM
would energy be in Joules?

UrbanXrisis
May17-04, 12:18 AM
so to find the difference in potential... E=(R/3^2)-(R/4^2) ?
E=533264 J?

cookiemonster
May17-04, 12:21 AM
That's a little much... Look about 2/3 down the webpage I linked to.

eV is more commonly used, but Joules works.

cookiemonster

UrbanXrisis
May17-04, 12:39 AM
is 533264 Joules a lot for the difference in the energy states?

Janitor
May17-04, 12:53 AM
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.