Question

One 0.7 MeV photon interacts with a free electron at rest. If the angle between direction of incoming photon and the outgoing photon is 45°, then calculate (a) the energy and wavelength of the outgoing photon, (b) the kinetic energy gained by the interacting electron.

Answer 1

We know that From Compton Seattening E' ho hol outgoing photone Es incident Photone Scatterd

^-^= h Loc (1-6sd). given that f= 45 Incident Ener E = 60 h = 0.7 mer Energy ho 16 pyoonew = hcü 12400 eo te he = 12480 V A = 0.7 Mev a a = 1270o - 12400x150 O, 7 M 0.1 % (06 A= 17714.28511016 m 167714 X 102 m to Incident wavele of photone wavelente > 1 Now By O ♡-13 h moc moc? a I-los p) - he, (1-6050 he noc?) (1-osa) macet (inlogy mel hc CÈTE - mace (I lost E! Where morets hoc=0.511 values mey) ( meul fest mass of é Energy put the ㅗ हो o.freu Asli Me (1-osus) ********* - t 0.7 Meu 0.29289 Oosll meu

El mo.5 Mew ooyggs6 meu @ is the Energy of scatted Photone Ehe = n'a he A = 12400 ENA Duration Now له ار کده o d=12400 tolo 0.5 106 m = 24800x10lo un 2 = 2048 tio? I scattend wave length of photone cout going photone) change in energy is gain by the Energy form. &b Now . in Kinetic So - (0.7 -0.5 K.C. = . E-E' - DE (0.7 ..somew (ks) 0.2 meve which Kinetic Energy of e is the
if you have any queries then give me a comment and give me a like thanks. .