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A photon with enegry h omega = 0.15 MeV is scattered by a stationary free electron changing its wavelength by Delta lambda = 3.0 pm. Find the angle at which the compton electron moves. |
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Answer» SOLUTION :Refer to the DIAGRAM. Energy momentum conservation gives `(cancel h omega')/(c )-(cancel h omega')/(c ) cos THETA = p cos varphi` `(h omega')/(c ) SIN theta = p sin varphi` `homega + mc^(2) = h omega' +E` where `E^(2) = c^(2)p^(2)+m^(2)c^(4)`.we see `tan varphi = (omega'sin theta)/(omega-omega' cos theta) = ((1)/(lambda')sin theta)/((1)/(lambda )-(1)/(lambda')cos theta)` `= (lambda sin theta)/(lambda'- lambda cos theta) = (sin theta)/((DELTA lambda)/(lambda)+2sin^(2)((theta)/(2)))` where `Delta lambda = lambda' - lambda = 2pi lambda_(c) (1- cos theta) = 4pi lambda_(c) sin^(2)((theta)/(2))` Hence `tan varphi= (2sin((theta)/(2))cos((theta)/(2)))/((Delta lambda)/(lambda)+(Delta lambda)/(2pilambda_(e)))` But `sin theta = 2 sqrt((Delta lambda)/(4pi cancel lambda_(e)) sqrt(1-(Delta lambda)/(4pi lambda_(e))) = (Delta lambda)/(2pi cancel lambda_(e)) sqrt((4picancel lambda_(c))/(Delta lambda))-1` Thus `tan varphi = (sqrt((4pi cancel h)/(m cDelta lambda)-1))/(1+(2picancel h)/(mc lambda)) = (sqrt((4pi cancel h)/(m cDelta lambda)-1))/(1+(cancel homega)/(mc^(2))) = 31.3^(@)` 
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