Question

a. Compute the energy of a photon with incident energy 200 kev scattered at 90°

in a Compton event.

b. Compute the energy of the backscattered photon from a 400 kev incident

photon.

c. Compute the energy of the recoil electron.

Answer 1

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