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A 140-KeV Photon Strikes an Electron and Scatters Through an Angle

question 98

Short Answer

A 140-keV photon strikes an electron and scatters through an angle of 120120 ^ { \circ } from its original direction. (melectron =9.11×10<sup>31</sup> kg,c=3.00×10<sup>8</sup> m/s,h=6.626×10<sup>34</sup> Js= 9.11 \times 10 <sup>- 31</sup> \mathrm {~kg} , c = 3.00 \times 10 <sup>8</sup> \mathrm {~m} / \mathrm { s } , h = 6.626 \times 10 <sup>- 34</sup> \mathrm {~J} \cdot \mathrm { s } )
(a) What is the wavelength of the photon before scattering?
(b) What is the photon wavelength after scattering?


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