Question

A photon moving in the +x-direction, scatters off a free stationary electron. The wavelength of the incident photon is 0.0230 nm. After the collision, the electron moves at an angle α below the +x-axis, while the photon moves at an angle θ = 78.3° above the +x-axis. (For the purpose of this exercise, assume that the electron is traveling slow enough that the non-relativistic relationship between momentum and velocity can be used.)

(a) What is the angle α (in degrees)? ________ ° counterclockwise from the +x-axis

(b) Determine the speed of the electron (in m/s). ____________m/s

(I posted this question earlier today, but I found the solution messy and confusing. As it's time sensative, I decided to post again.)

Answer 1

a] The momentum will be conserved before and after the scattering process in both along the direction of incident photon (x-axis) and perpendicular to this direction (y-axis).

initial momentum = p_i = E/c =

so, in the x direction,

where,

so,

=> = v_x

also, in the y direction:

=> = v_y

therefore, v = 3.847 x 10⁷ m/s

this is the speed of the electron

the angle will be:

this is the angle below the x-axis

so, the angle counter-clockwise from +ve x axis will be: .