1. Find an x-ray photon which can impart a maximum energy of 50 keV to an electron.

Ø= 180˚
h/mec = 2.43 x 10-12 m
Δλ = h/mec (1-cos Ø) = (2.43 x 10 e-12 m) (1 – cos 180˚)
= 4.86 x 10 e– 12 m
Therefore the energy of the photon can be calculated as E = hc/λ
E = hc/λ
λ   = hc/E = 1240 eV.nm/50.0
x 10e3 eV = 0.0248 nm = 2.48 x 10e-11 m
λ ‘ = λ + Δλ = 2.97 x 10 e-11 m
E’ = hc/ λ’ = 1240 eV.nm/0.0297 nm
E’ = 4.18 x 10e4 eV = 41.8 eV