Q.54:- If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5×107 ms-1 , calculate the energy with which it is bound to the nucleus.
Energy of the incident photon= hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/(150×10-12m) = 13.25×10-16 J
Energy of the electron ejected = 1/2 mv2 = 1/2×(9.11×10-31kg)×(1.5×107ms-1)2 = 1.025×10-16 J
Energy with which the electron was bound to the nucleus = 13.25×10-16 J – 1.025×10-16 J
= 12.225×10-16 J = 12.225×10-16/1.602×10-19 eV = 7.63×103 eV