Kleppner and Kolenkow Problem 13.8 A photon of energy Eo and wavelength?o collid

Question

Kleppner and Kolenkow Problem 13.8 A photon of energy Eo and wavelength?o collides head on with a free electron of restmass mo and speed v. The photon isscattered at 90 degrees. a. Find the energyE of the scattered photon. b. The outerelectrons in a carbon atom move with speed v/c ? 6 x10-3. Using the result of part a, estimate thebroadening in wavelength of the compton scattered peak fromgraphite for ?o = 0.711 x 10-10 m and 90degree scattering. The rest mass of an electron is 0.51 MeV andh/(moc) = 2.426 x 10-12 m.Neglect the binding of the electrons. Answer to part a: E = [Eo(1+v/c)]/(1 +Eo/Ei) where Ei = moc2 / ?(1-v2/c2). I don'tknow how to get this answer through relativistic energy andmomentum conservation. I also can't figure out how to solve partb

Explanation / Answer

let p = momentum of the electron after collision, x-axis is alongthe direction of the initial photon momentum: initial photon: E0/c, initial electron:m0v, final photon: E/c, final electron: p momentum conservation: x direction: E0/c - m0v =px y direction: 0 = py + E/c p2 = (E0/c -m0v)2 + (E/c)2 p2c2 = (E0 -m0vc)2 + E2 energy of the final electron = E' E'2 = (m0c2)2 +(pc)2 = (m0c2)2 +(E0 - m0vc)2 + E2use Ei =m0c2, E'2 = (Ei/)2 +(E0 - Eiv/c)2 + E2 =Ei2/2 +E02 +Ei2v2/c2 -2E0Eiv/c + E2 energy conservation during the collision: E0 + Ei = E + E' (E0 + Ei - E)2 =E'2 E02 + Ei2 +E2 - 2E(E0 + Ei) +2E0Ei = Ei2/2 +E02 +Ei2v2/c2 -2E0Eiv/c + E2 2E(E0 + Ei) = Ei2 +2E0Ei -Ei2/2 -Ei2v2/c2 +2E0Eiv/c 2E(E0 + Ei) = Ei2(1 -Ei2/2 -Ei2v2/c2) +2E0Ei + 2E0Eiv/c 2E(E0 + Ei) = Ei2*0 +2E0Ei(1 + v/c) E(1 + E0/Ei) = E0(1 + v/c) E = E0(1 + v/c)/(1 +E0/Ei) v/c ˜ 6 x 10-3. o = 0.711x 10-10 m = 0.0711 nm, E0 = hc/o = 1240 eV/0.0711 =1.744*104 eV = 0.01744 MeV, Ei = 0.51 MeV/[1 - (v/c)2] = 0.51MeV h/(moc) = 2.426 x 10-12m. E = E0(1 + v/c)/(1 + E0/Ei) =0.01696 MeV = hc/E = 1240 eV-nm/(1.696*104 eV) = 0.0731 nm =0.731 x 10-10 m

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