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A photon is scattered at an angle theta = 120^(@) by a stationary free electron. As a result, the electron acquires a kinetic energy T = 0.45MeV. Find the energy that the photon had prior to scattering. |
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Answer» Solution :From the COMPTON formula `lambda = lambda_(0)+(2picancel h)/(mc)(1-cos theta)` From conservation of energy `(2picancel h c)/(lambda_(0))= (2picancel h c)/(lambda)+T = (2pi cancelh c)/(lambda_(0)+(2pi cancel h)/(mc)(1-cos theta))+T` or `(4pi cancel h)/(mc)SIN^(2)((theta)/(2)) = (T)/(2pi cancel h c) lambda_(0) (lambda_(0)+(4pi cancel h)/(mc)sin^(2)((theta)/(2)))` or introducing `cancel h omega_(0) = 2pi cancelh c//lambda_(0)` `(2sin^(2)theta//2)/(mc^(2)) = (T)/(cancel h omega_(0)) ((1)/(cancel h omega_(0))+(2)/(mc^(2))sin^(2)((theta)/(2)))` Hence `((1)/(cancel omega_(0)))^(2) +2(1)/(cancelh omega_(0))sin^(2)((theta)/(2))/(mc^(2))-(2sin^(2)((theta)/(2)))/(mc^(2)T) = 0` `((1)/(cancelh omega_(0))+(sin^(2)((theta)/(2)))/(mc^(2)))^(2)=(2sin^(2)((theta)/(2)))/(mc^(2)T)+((sin^(2)((theta)/(2)))/(mc^(2)))^(2)` `(1)/(cancel h omega_(0)) = (sin^(2)((theta)/(2)))/(mc^(2)) [sqrt(1+(2MC^(2))/(TSIN^(2)theta//2))-1]` or `cancel h omega_(0) = (mc^(2)//sin^(2)theta//2)/(sqrt(1-(2mc^(2))/(Tsin^(2)((theta)/(2)))))-1` `=(T)/(2)[sqrt(1+(2mc^(2))/(Tsin^(2)theta//2))+1]` SUBSTITUTING we get `cancel h omega_(0) = 0.677MeV` |
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