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A narrow monochromatic X-ray beam falls on a scattering substance. The wavelengths of radiation scattered at angles theta_(1) = 60^(@) and theta_(2) = 120^(@) differ by a factor eta = 2.0. Assuming the free electrons to be responsible for the scattering, find the incident radiation wave length. |
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Answer» Solution :Let `lambda_(0) =` WAVELENGTH of the INCIDENT radiation. Then wavelength of the radiation scattened at `theta_(1) = 60^(@)` `= lambda_(1) = lambda_(0)+2pi cancel lambda_(c) (1-cos theta_(1))` where `cancel lambda_(c) = (cancelh)/(mc)`. and similarly `lambda_(2)= lambda_(0) + 2pi cancel lambda_(c) (1- cos theta_(2))` From the data `theta_(1) = 60^(@), theta_(2) = 120^(@)` and `lambda_(2) = eta lambda_(1)` Thus `(eta - 1) lambda_(0) = 2pi cancel lambda_(c) [1-cos theta_(2) - eta (1-cos theta_(1))]` `= 2pi cancel lambda_(c) [1-eta +etacos theta_(1) - cos theta_(2)]` Hence `lambda_(0) = 2pi cancel lambda_(c) [(eta cos theta_(1) - cos theta_(2))/(eta - 1)-1]` `= 4pi cancel lambda_(c) [(sin^(2)theta_(2)//2 - eta sin^(2) theta_(1)//2)/(eta - 1)] = 1.21 pm`. The expression `lambda_(0)` given in the bok contains MISPRINTS. |
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