If the photon of the wavelength `150 p m` strikes an atom and one of its inner bound electrons is ejected out with a velocity of `1.5xx10^(7) ms^(-1)`, calculate the energy with which it is bound to the nucleus.

Explanation

Energy of the incident photon `= (hc)/(lamda) = ((6.626 xx 10^(-34) Js) (3.0 xx 10^(8) ms^(-1)))/((150 xx 10^(-12) m)) = 13.25 xx 10^(-16) J`
Energy of the electron ejected `= (1)/(2) mv^(2) = (1)/(2) (9.11 xx 10^(-31) kg) (1.5 xx 10^(7) ms^(-1))^(2) = 1.025 xx 10^(-16)J`
Energy with which the electron was bound to the nucleus `= 13.25 xx 10^(-16)J - 1.025 xx 10^(-16)J`
`= 12.225 xx 10^(-16)J = (12.225 xx 10^(-16))/(1.602 xx 10^(-19)) eV = 7.63 xx 10^(3) eV`