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A photon of wavelength lambda = 6.0 pm is scattered at right angles by a stationary free electron. Find: (a) the frequency of the scattered photon, (b) the kinetic energy of the compton electron. |
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Answer» Solution :(a) From the Compton FROMULA `lambda' = 2PI CANCEL lambda_(C) (1- cos 90) + lambda` Thus `omega' = (2pic)/(lambda') = (2pic)/(lambda+2pi cancel lambda_(c))` where `2pi cancel lambda_(c) = (h)/(mc)`. Substituting the values. We get `omega' = 2.24 xx 10^(20)rad//sec` (b) The kinetc energy of the scattered electron (in the frame in which the initial electron was STATIONARY) is simply `T = cancelh omega - cancelh omega'` ` =(2pi cancel h c)/(lambda)-(2pi cancelh c)/(lambda+2pi cancel lambda_(c))` `= (4pi^(2)cancel h cancel lambda_(c))/(lambda(lambda+2pilambda_(c))) = (2picancel h c//lambda)/(1+lambda//2pi cancel lambda_(c)) = 59.5kV` |
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