By what factor is the ionization energy of hydrogen-like helium greater than that of hydrogen?
1 Answer
Nov 6, 2017
Four.
The energy of hydrogen-like atoms is given by:
#E_n = -"13.61 eV" cdot Z^2/n^2# ,where
#Z# is the atomic number and#n# is the energy level,#n = 1, 2, 3, . . . # .
So, for
#E_1("He"^+) = -"13.61 eV" cdot 2^2/1^2 = -4 cdot "13.61 eV"#
For
#E_1("H") = -"13.61 eV" cdot 1^2/1^2 = -"13.61 eV"#
From Koopman's approximation theorem for single-electron orbitals, the ionization energy is the same as the energy of that one-electron orbital, with the opposite sign.
Therefore.
#"IE"_1("H") = "13.61 eV"#
#"IE"_1("He"^+) = 4 cdot "13.61 eV"#
And thus,
#color(blue)(("IE"_1("He"^+))/("IE"_1("H")) = 4)#
