assingment06-compton derivation

# assingment06-compton derivation - 37480 Tipler(Freem RIGHT...

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E 1 = hf 1 E 2 = hf 2 p 1 = h / λ 1 p 2 = h / λ 2 φ θ m p e = E 2 E 0 2 1 c Continued Fig. 3-21 The scattering of x rays can be treated as a collision of a photon of initial momentum h / l 1 and a free electron. Using conservation of momentum and energy, the momentum of the scattered photon / 2 can be related to the initial momentum, the electron mass, and the scattering angle. The resulting Compton equation for the change in the wavelength of the x ray is Equation 3-42. More More Derivation of Compton’s Equation Let 1 and 2 be the wavelengths of the incident and scattered x rays, respectively, as shown in Figure 3-21. The corresponding momenta are Eh f h 11 p 55 5 1 cc 1 and 2 p 2 c 2 using ƒ 5 c . Since Compton used the K a line of molybdenum ( 5 0.0711 nm; see Figure 3-18 b ), the energy of the incident x ray (17.4 keV) is much greater than the binding energy of the valence electrons in the carbon scattering block (about 11 eV); therefore, the carbon electron can be considered to be free. Conservation of momentum gives p 5 p 1 p 12 e or 222 p 5 p 1 p 2 2 p · p e 1 2 3-41 22 5 p 1 p 2 2 pp cos u 1 2 where

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assingment06-compton derivation - 37480 Tipler(Freem RIGHT...

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