Question

An incident x-ray photon is scattered from a free electron that is initially at rest. The photon is scattered straight back at an angle of 180∘ from its initial direction. The wavelength of the scattered photon is 8.70×10⁻² nm

Part A

What is the wavelength of the incident photon?

λ =

m

Part B

What is the magnitude of the momentum of the electron after the collision?

P= _________ kg⋅m/s

Part C

What is the kinetic energy of the electron after the collision?

Ke =		J

Answer 1

here,
wavelength of scattered photon, ws = 0.0870 nm
From Compton scattering equation :
wo - wi = (1 - cosA*h/(Me*c)
wi = wo - (1 - cosA*h/(Me*c)

where,
wo is wavelength of outgoing photon
wi is wavelength of incoming photon
h is plank constant
c is speed of light
Me is mass of electron

wi = (0.0870*10^-9) - (1 - (cos180)*(6.626*10^-34)/( (9.11*10^-31) * (3*10^8) ) )

wi = 8.699*10^-11 m

Part b:
Since momentum is conserved so ;
P_electron_out + P_photon_out = P_photon_in
The momentum of a photon is given by P = h/w,
so:

|P_electron_out| = | h*(1/wi - 1/wo) |
|P_electron_out| = | (6.626*10^-34)*(1/(8.699*10^-11) - 1/(8.20*10^-11) |
|P_electron_out| = 4.635*10^-25 kg.m/s

Part c:
Energy of photon, E = c * P
Energy of photon, E = h*c * (1/wo - 1/wi)
Energy of photon, E = (6.626*10^-34)*(3*10^8)(1/(8.699*10^-11) - 1/(8.20*10^-11)
Energy of photon, E = -1.39 * 10^-16 J

so the electron must end up with this energy, and the outgoing electron has a

kinetic energy of 1.39 * 10^-16 J