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 xx10^(7 )ms^(–1)`, calculate the energy with which it is bound to the nucleus

Explanation

Energy of incident photon`=(hc)/lambda`
`=((6.626xx10^(-34)J s)(3.0xx10^(8) m s^(-1)))/((150xx10^(-12)m))`
`=13.25xx10^(-19)J`
Energy of the electron ejected `=1/2 mv^(2)`
`=1/2(9.11xx10^(-31)kg)(1.5xx10^(7)ms^(-1))^(2)`
`=1.025xx10^(-16)J`
Energy with which the electron was bound to the nucleus
`=13.25xx10^(-16)J-1.025xx10^(-16)J`
`=12.225xx10^(-16)J`
`=(12.225xx10^(-16))/(1.602xx10^(-19))eV=7.63xx10^(3) eV`