An electron of 5 GeV energy collides head-on with a photon of visible light `(E_(ph)= eV)`. Find the erergy of the scattered photon.

Answer

370 MeV.

Explanation

Prior to collision, the electron and the photon fly head-on towards each other, after collision they will move in the initial direction of the electron.s motion. The laws of conservation of energy and the momentum will assume the form
Hence it follows
`epsilon+pc=epsilon.=p.c+2hv.epsilon-pc+2hv=epsilon.-p.c`
Multiplying , we obtain
`epsilon^2-p^2c^2+2hv(epsilon+pc)=epsilon.^(2)-p.^(2)c^2+2hv.(epsilon.-p.c)`
But `epsilon^2-p^2c^2=epsilon.^(2)-p^(.2)c^2=epsilon_0^2`. Hence
`hv(epsilon+pc)=hv.(epsilon.-p.c)" or " hv(epsilon+pc)=hv.(epsilon-pc+2hv)`
Multiplying both sides of the equation by the expression `epsilon+pc` we obtain
But in the ultra - relativistic case `epsilon~~ pc` . Hence .
`hv.=(4epsilon^2hv)/(epsilon_0^2+4epsilonhv)`