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A photon with momentum p = 1.02 MeV//c, where c is the velocity of light, is scattered by a stationary free electron, changing in the process its momentum to the value p' = 0.255MeV//c. At what angle is the photon scattered? |
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Answer» Solution :We use the equation `lambda = (h)/(p) = (2pi cancel h)/(p)`. Then from COMPTON formula `(2pi cancel h)/(p') = (2pi cancel h)/(p) +2pi(cancel h)/(mc)(1-cos theta)` so `(1)/(p') = (1)/(p) + (1)/(mc).2SIN^(2) theta//2` Hence `sin^(2)((theta)/(2)) = (mc)/(2)((1)/(p')-(1)/(p))` `=(mc(p-p'))/(2pp')` or `sin((theta)/(2)) = sqrt((mc(p-p'))/(2pp'))` Substituting from the DATA `sin((theta)/(2)) = sqrt((mc^(2)(cp-cp'))/(2cp.cp')) = sqrt((0.511(1.02-0.255))/(2xx1.02xx0.255))` This GIVEN `theta = 120.2` DEGRESS. |
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