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 incident photon
`(hc)/(lambda)`
`=((6.626 xx 10^(-34) Js) xx (3.0 xx 10^(8) ms^(-1)))/((150 xx 10^(-12)m))`
`= 1.325 xx 10^(15) J`
Energy of ejected electron was bound to the nucleus
`=13.25 xx 10^(-16) - 1.025 xx 10^(-16)`
`= 12.225 xx 10^(-16)J`
or `" " = (12.225 xx 10^(-16))/(1.602 xx 10^(-19)) = 7.63 xx 10^(3) eV`.