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A photon with energy h omega = 250 keV is scattered at an angle theta = 120^(@) by a stationary free electron. Find the energy of the scattered photon. |
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Answer» Solution :The wave LENGTH of the incident photon is `lambda_(0) = (2pic)/(omega)` Then the wavelength of the final photon is `(2pic)/(omega) + 2pi cancel lambda_(C) (1-cos theta)` and the energy of the final photon `cancel H omega' = (2pi cancel h c)/((2pi c)/(omega) +2pi lambda_(c)(1-cos theta)) = (homega)/(1+(h omega)/(mc^(2))(1-cos theta))` `= (h omega)/(1+2((h omega)/(mc^(2)))SIN^(2)(theta//2)) = 144.1kV` |
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