Recent content by dakillerfishy

I just thought of this... Are you saying that I should solve the E max equation in terms of momentum, then use E - (pc)^2 = (mc^2)^2 to solve for energy, then solve for wavelength?

I'm sorry this is so frustrating... maybe my algebra just sucks. I already got #1, so this would be #2 So we have: E = \frac{hf}{1 + mc^{2}/2hf} E + \frac{Emc^{2}}{2hf} = hf (Eh/mc) f + Ec = \frac{h^{2}}{mc} f^{2} \frac{h^{2}}{mc}f^{2} - (Eh/mc) f - Ec = 0 f = -1.257e20 or -4.115e11...

Sorry, I wrote that wrong... the 2 is actually in the numerator and the results don't change.

So then \lambda_{2} - \lambda_{1} = h/mc(2) hc/E_{k} - \lambda_{1} = 2h/mc \lambda_{1} = hc/E_{k} - 2h/mc \lambda_{1} = (4.136e-15 * 3e8)/520000 - (2*4.136e-15)/(.511e6*3e8) = 2.386e-12 m \lambda = hc/E E = hc/\lambda E = (4.136e-15 * 3e8)/2.386e-12 = 520,000 eV So the initial...

So then for wavelength you get \lambda_{2} - \lambda_{1} = 4.86 pm, which is still the difference between lambda's. However, wouldn't \lambda_{2} be the wavelength of the photon after it has recoiled and be \lambda_{2} = hc/Ek ? Is that right, or am I still not understanding something...

Aha, that makes much more sense than what I was trying to do. Thanks! I understand why theta has to be 180 degrees, but how would you go about converting \lambda_{2} - \lambda_{1} into energy? I imagine you would add the difference in \lambda's to the original \lambda, then plug back into...

Thanks for the links, shooting star. The derivation wasn't in my book, so I looked at it in the first link and I think I have a better understanding now. I answered my first question correctly, but I'm still stuck on the second one. I tried to rearrange the max kinetic energy equation...

No, I did get what you said about the photon, but perhaps I didn't phrase my question correctly. Why are we using the electron for mc^2?

For max kinetic energy, the mc^2 term on the bottom looks like it would go to zero because the mass of the photon is negligible.

Okay, so I have two questions, both pertaining to the Compton Effect. Homework Statement Evaluate the maximum kinetic energy for a recoiling electron that is struck by a photon with momentum 0.04MeV/c Homework Equations Maximum Kinetic energy: E_{k} = \frac{hf}{1 + (mc^{2}/2hf}...

Ohhhh... I understand now. It goes back to what tiny-tim was saying about Lorentz transformations. :smile:

Got it. :smile: Thanks everyone! You've all been really helpful. :smile:

I think in this example, we're supposed to assume that he does return immediately, just for simplicity's sake. We also ignore the fact that he has to accelerate/decellerate, which will ultimately change the time dialation.

Oh I understand now. So you need to take the spacetime interval and rearrange it, so it reads: (\Deltat)^{2} = (\Deltat')^{2} - (\Deltax/c)^{2} Except you need the time dilation equation to solve for \Deltat, so you plug that into the above equation, right?

Well, I've been playing with numbers and I got this... \Deltat = \gamma\Deltat' \Deltat = \frac{1}{\sqrt{1-v^{2}/c^{2}}}\Deltat' Then plugging in numbers... 5/8 = \sqrt{1-v^{2}/c^{2}} v = .6124 c Which is apparently not right... what am I missing?