Question

X-ray photons of wavelength 0.0268 nm are incident on a target and the Compton-scattered photons are observed at 65.0° above the photons' incident line of travel. [Use relativistic units for this problem!] (a) What is the wavelength of the scattered photons? 2.82E-110 (b) What is the momentum of the incident photons? 2.47E-231 What is the momentum of the scattered photons? 2.34E-231 (c) What is the kinetic energy of the scattered electrons? 3.6789E-1 (d) What is the momentum (magnitude and angle) of the scattered electrons? 2.21E-231 x 59.69x Submit Answer

Answer 1

Applying the concept of energy conservation and linear momentum conservation on Compton scattering we can solve this as follows...

© If rest mass of electron ke m. ane after the scattering electron moves with velocity & and relativistic mass be m.. Them from energy conservation, lot me a hit me & -frequency of incident photon d'= " of scattered " . ke - hc = (m-mo jeho » 6.681034 x3x108 [- - 1] = (m.m., c? Kinetic energy of scattered electron is (m-m.)c? so, kE = (-m.)c? = 6.6 316 34 x 3X100 [ 00268 x 159 - 21824167] J = 6,6X1634 x3*10[ 3.73x100 – 3.54x10"] J = 6,6810 34 3 x 108 (3.73-3.54) xloo J = 3,762 x 1016 J :: T KE = 3.762 X 10" ]]

(a considering the momentum conservation, we h from figure – - hy cesy + me cono - and bes' sin ø= mu sino - 2 mo es the magnitude of the scattered electrons momentum. from 0 h = h ceno + muceno > h[ + ceny ] = mv ceno - Ces 65° 7=mo Cono 2) 6.68 lo 10,0268 810 9 2.819 810 a) 6.6810 34 | 3.73x10° _ 1.49x10o ] = mo cono > 6.6x1034 x 2.24 X 10'' = mulong 2) mv ceno = 1.47 x 1023 - ③ from ② h sin 650 mv sin o 6x 10 34 x Sin 65 mv sine 2.819 X 10 > 6.66 10 34 x Sin 6s 2-819 Xio" - mv sino = 2.12 x 102 4 8 6 , we get, tano = 2012 x10 23 from 2 .12 1: 47 =1.44 1.47 x 10 o= tano (1.44) = 55.26° 10 = 55,260 = 55.269

from ein , mo can (55.26) = 1.47 x 10 a 2 3 -23 mo = = 2.57X10 1.47 x 10-23 Con (55.26) kg-m sec mo=magnitude of the momentum of scattered electrons I mo = 2.57x10 es ng-m Sec.