Question #d1d52
1 Answer
Explanation:
For starters, the ionization energy of a hydrogen atom is equal to
So for the hydrogen atom, the energy of a given excited state, which is characterized by
#E_n = -"13.6 eV"/n^2#
For the ground state, you have
#E_ "ground state" = -"13.6 eV"#
Now, for a hydrogen-like atom like the helium cation,
#E_n = -"13.6 eV"/n^2 * Z^2#
Here
#E_ "ground state" = -"13.6 eV" * Z^2#
Since a helium atom has
#Z = 2#
you can say that its ground state will be at
#E_ "ground state He" = -"13.6 eV" * 2^2#
#E_ "ground state He" = -"54.4 eV"#
This implies that in order to remove the electron from a
#"He"^(+) + "54.4 eV" -> "He"^(2+) + "e"^(-)#