Question

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

Answer 1

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