It feels wierd that my first post after my long silence (which wasn't over anything important, if anyone had wondered at all; I merely had no internet for a while, was at camp, and I'd rather talk to my girlfriend than most) is to ask a homework question, but here it goes anyway;
I'm taking AP physics this year, and the summer assignment is to read a chapter of the book and answer the questions at the end. I think I got the general idea on most of them, which is all he expects, but I have no idea on the very last one. It is "Use conservation of momentum to explain why photons emitted by hydrogen atoms have slightly less energy than predicted by Eq.27-10." Eq.27-10 is Bohr's hf=Eu-El.
I really have no idea at all, and was hoping that someone here would be able to give some clue.
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I'm not at all certain, but here's my guess. Photons do not have mass (m=0), but they do have momentum. In conservation of momentum, you have several points where the calculation is based on mass * vector. Setting mass at zero for those points may give some odd results.
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Been a long time since I did physics, but think of it this way:
Assume that the momentum of the Hydrogen starting out is P (symbol changed from X in the original post).
You end up with two particles with their own momentums.
Momentum must be conserved.
Yeah, that's basically it. You can even set P = 0 so that you are in the rest frame of the hydrogen atom.
The transition of an electron into a lower energy state releases a certain amount of energy.
Most of that energy goes into the photon, but not all of it, because of conservation of momentum. Essentially, there's a tiny bit of recoil that kicks the H atom into motion in the opposite direction of the photon's motion (EDIT -- Which is necessary to keep the total momentum = 0). This energy has to come from the energy released by the electron's transition, so the resultant photon has a tiny bit less of total energy and momentum than you would predict from the formula.
Last edited by Ted_Stryker : 08-16-2007 at 02:14 PM.