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 `1.5xx10^(7)m*s^(-1)`, calculate the energy with which it is bound to the nucleus.

Explanation

Energy of incident photon
`=(hc)/(lamda)=((6.626xx10^(-34)J*s)xx(3xx10^(8)m*s^(-1)))/((150xx10^(-12)m))`
`=1.3252xx10^(-15)J=13.252xx10^(-16)J`
Kinetic energy of emitted electron `((1)/(2)mv^(2))`
`=(1)/(2)xx(9.108xx10^(-31)kg)xx(1.5xx10^(7)m*s^(-1))^(2)`
So, the energy with which the electron was bound to the nucleus`=(13.252xx10^(-16)-1.025xx10^(-16))J`
`=12.227xx10^(-16)J=7.632xx10^(3)eV`