Question

May the work please be shown

(14 pts) 5 . A photon with linear momentum of magnitude 6.50x1 0-24 kg·m/s strikes a free electron that is initially at rest and the photon scatters backwards at an angle of 180 from its original direction. a) What is the magnitude Express your answer in kg.m/s of the linear momentum of the photon after it has scattered from the electron? -2Y Ans 6.200 kog- b) What is the speed of the electron just after the photon scatters from it?

Answer 1

wavelength of incident photon, lamda = h/p

= 6.626*10^-34/(6.50*10^-24)

= 1.02*10^-10 m

a) using compton's scttering equation,

wavelgnth of scattered photon,

lamda' = lamda + (h/(m*c))*(1 - cos(theta))

= 1.02*10^-10 + (6.626*10^-34/(9.1*10^-31*3*10^8))*(1 - cos(180))

= 1.0588*10^-10 m

linear momentum of scattered the photon, p' = h/lamda'

= 6.626*10^-34/(1.0588*10^-10)

= 6.26*10^-24 kg.m/s

b) kinetic energy of the electron = loss of energy of photon

= h*c/lamda - h*c/lamda'

= 6.626*10^-34*3*10^8/(1.02*10^-10) - 6.626*10^-34*3*10^8/(1.0588*10^-10)

= 7.14*10^-17 J

now use, KE = (1/2)*m*v^2

==> v = sqrt(2*KE/m)

= sqrt(2*7.14*10^-17/(9.1*10^-31))

= 1.25*10^7 m/s