Question

2. A 0.8 MeV photon is scattered by a free electron initially at rest. After scattering, the electron moves with a de Broglie wavelength le = 1.789 pm. (a) Determine the wavelength and scattering angle of the scattered photon. (b) Determine the kinetic energy Ec of the scattered electron, comment on whether E, satisfy the classical formula Eę = -mev? (where me is the free electron rest mass) and explain why. (25 marks)

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Answer 1

wolno? i toyles of souttering m Erhv energy of the incident photon Esmo - Damen De Brogle wavelength of the scattered electron de : 1.789 pm de Temer 2mefke : Ghale Em = Emel het (6.625x16 Exa virstie ponga, el I ne planet's statant electron *) 6.624 70-3473 me = works of elsefe 9.1810317 - 2* 9.1410194 7.5372x10-19 te: 1.789 pm 1289 4102 et 4.537281514 ew . 1.6 xor 471-07 x 103 ev 471.07 ker of elementing) : Now energy of photon ho'= d e 0.471 Mer (Kinetic energy the scattered hv - EK Cons - 0.471) mer 0.329 Mer O X 1-6x10-19 d 0.329 X 10 61626x403x108 0.526 X10-15 - 5.778 X 10*12 mm scattered of the photo coavelength This is the

hola ho = is the It hv (in osa). I angle of scattering ме Yум son of electron where hv = 0.8 mer - 28 X10! hola 0.329 mer 01526 X 10-13J 128x13 0.526 XD 1.28X10-13 It Puix 108' x 78 x rob gt (1-com) ☺ 1 + 1,562861-65X) 1+ 105628 (-65X) - 128 . . 0526 ☺ ☺ ☺ Cosa - - 0.9172 C5 x 2 5) | @ Hence 4 = cos" (0.082) : 85.25 angle of scattering = 85.250

Rest mass energy of electron - meen . 9.1x1073'x (saxloky u g 8:19X10-14 ] НЬ Хе the kinetici energy as of settered energy the kinetics eletrom CREO - 471.04 kev less than its rest mass energy, the electron is regarded as non-relativistie Ep ad me va is satisfied. particle. Hence