a) let lamda is the wavelength of the incident
photon.
use, E = h*c/lamda
lamda = h*c/E
= 6.626*10^-34*3*10^8/(200*10^3*1.6*10^-19)
= 6.21*10^-12 m
wavelength of scattered photon,
lamda' = lamda + (h/(m*c))*(1 - cos(theta))
= 6.21*10^-12 + (6.626*10^-34/(9.1*10^-31*3*10^8))*(1 -
cos(90))
= 8.64*10^-12 m
Energy of scattered photon, E' = h*c/lamda'
= 6.626*10^-34*3*10^8/(8.64*10^-12)
= 2.30*10^-14 J
= 2.30*10^-14/(1.6*10^-19)
= 143.7 keV
b) lamda = h*c/E
= 6.626*10^-34*3*10^8/(400*10^3*1.6*10^-19)
= 3.11*10^-12 m
wavelength of scattered photon,
lamda' = lamda + (h/(m*c))*(1 - cos(theta))
= 3.11*10^-12 + (6.626*10^-34/(9.1*10^-31*3*10^8))*(1 -
cos(180))
= 7.96*10^-12 m
Energy of scattered photon, E' = h*c/lamda'
= 6.626*10^-34*3*10^8/(7.96*10^-12)
= 2.49*10^-14 J
= 2.49*10^-14/(1.6*10^-19)
= 155.6 keV
<<<<<<<<<------------Answer
c) Energy of the recoil electron = 400 -
155.6
= 244.4 keV
<<<<<<<<<------------Answer
