The second ionization potential of Be is `17.98 eV` if the electron in Be is assumed to move in a spherical orbit with a centeral field of effective nuclearcharge `(Z_(6H))` consisting of the nucless and otherelectron by haw many units of charge in the nucless shicided by other electrons? (the energy of electrons in first Bohr of H is `-13.6 eV)` If the extent of shielding by the if electron of Li atom is the same as you have calculated above , find the ionisation potential of Li

Explanation

`IE = (13.6Z_(eff)^(2))/(n^(2)) = 17.98`
`Z_(eff) = 2.3`
Shieiding extent `= Z - Z_(eff) = 4 - 2.3 = 1.7 `
`Z_(eff)` for lithium `= 3 - 1.7 = 1.3 `
`:.IE of lithium = 13.6 xx (1.3)^(2) = 5.746 eV`