InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A photon with energy exceeding eta =2.0 times the rest energy of an electron experienced a head-on collision with a stationary free electron. Find the curvature radius of the trajectory of the compton electron in a magnetic field B = 0.12 T. The compton electron is assumed to mive at right angles to the direction of the field. |
|
Answer» SOLUTION :By head on collision we understand that the electron moves on in the direction of the incident photon after the collision and the photon is scaltered backwards. Then, let us write. `cancel h omega = ETA mc^(2)` `cancel h omega' = sigma mc^(2)` `(E,p) = (epsilonc^(2), MU mc)` of the electron. The by energy momentum consrvation (cancelling factors of `mc^(2)` and `mc`) `1+eta = sigma + epsilon` `eta = mu - sigma` `epsilon^(2) = 1+mu^(2)` So eliminating `sigma& epsilon` `1+eta =- eta+ mu + sqrt(mu^(2) +)` or `(1+2eta - mu) = sqrt(mu^(2) +1)` Squaring `(1+2 eta)^(2) - 2mu (1+2eta) = 1` `4eta + 4eta^(2) = 2mu (1+2eta)` or `mu = (2eta(1+eta))/(1+2eta)` Thus the momentum of the Compton electron is `p = mu c = (2eta(1+eta)mc)/(1+2eta)`. Now in a megnetic FIELD `p = Be rho` Thus `rho = 2 eta (1+eta)//(1+2eta) (mc)/(Be)`. Substituting the values `rho = 3.412cm`. |
|