Question

The diagram below shows two successive scatterings at locations A and B. In the first scattering, a photon of wavelength 2 hits a free electron at A resulting in a photon of wavelength in that moves at an angle e, to the incident direction. The second photon hits an electron at B and scatters at an angle e, to the direction of the second photon resulting in a photon of wavelength i. Determine the angle , (in degrees) if e, - 42.0° and the total shift in wavelength of the combined scattering is 3.18 x 10-3 nm. А relative to the incident direction

Answer 1

Total shift in ware length of combined scattering in Do=3018 8 10-3 mm = 3.18 x 10-2 AO second photan scatter of angle Oz = 4200 it in comptar scattering in the compiton shift in wavelength of photon after scattering. -> :D2 =.Ac Cl-coco). where ac = h = 0.0243 A А еxt » de = compton ware length uz plauck's constant mo z mon of electron . C=.5 peed of light Now shift in first scattering $2,- dc ci-conds) and shift in second scattering Daza de (1-2008) According to given information Dat D2z = dc (2-cond. - Condz) 3.18 x 102 A = 0.0243 A0 (2-cong, -con uz)

10.30864 = (2-cond. - 20:42) Cono, 5 0.05178, dia.cort' (0.05173) 87..0316° d= 87.0326° relative to incident direction