Why is first ionization enthalpy of Cu higher than that of Na ?

Explanation

`Cu= [Ar]^(18) 3d^(10)4s^(1), Na= [Ne]^(10) 3s^(1)`
Though electron to be removed in both cases in from outermost s-orbital but nuclear charge of Cu is much higher than that of Na. Hence, attraction on 4s electron in Cu is much greater than on 3s electron in Na. Moreover, removal of `1e^(-)` from 3s in Na gives a stalbe noble gas configuration whereas in case of Cu, it gives pseudo noble gas configuration.