Question

in the What is the momentum, in the center of mass reference frame, of a reaction +p+Kº+A if the has an energy E= 3 GeV in the laboratory reference frame? Notice that m.,- = 139.6 MeV/ca, mp = 938.3 MeV/c. (Tip: in the laboratory reference frame, the proton is at rest. In the center of mass reference frame, ộp = (P2c2 + m 4,P) and Pa- = (1pe2 +m-c, -D)) A photon with energy E = 10 MeV collides elastically with an electron at rest. What is the wave-length of the photon before the collision? What is the momentum pe' of the electron after the collision, if we observe it at an angle 0 = 2.7°?

Answer 1

Answer to the and part given, E = 10 MeV, 0 = 2.7 warelength of the photon (incident) 1240 1 240 1 = 1.24 x 10-4 nm Now, * = at t ( 1- Cost) = 1.24x10-4 + 6.626 X10 34 9. 181031 x 3 x 18 - 1.24 X10-9x109 + 2.69 44 x 10-15 = 1.266944 x 10-13 m a

& E= he и 1240 (hm) (eV) 1 2 4o 1.266944 x 10-4 (nm) I E = 9.79 x108 eV 3 9.79610°x1,6x16 19 p E' 3x108 - 2 S. 22x10-21 kg mls| Momentum of the scattered Photon