Question

What are the ionization potentials of hydrogen-like helium and lithium?

Answer 1

`"He"^(+)(g) -> "He"^(2+)(g) + e^(-)`

`DeltaH_(IP)("He"^(+)(g)) ~~ "54.42 eV"`

`"Li"^(2+)(g) -> "Li"^(3+)(g) + e^(-)`

`DeltaH_(IP)("Li"^(2+)(g)) ~~ "122.45 eV"`

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Well, since these are hydrogen-like atoms with one electron each, they ought to have similar ionization potentials that differ only by the atomic number.

In atomic units, the energy of `"H"`-like atoms is given by

`bb(E_n = -(Z^2 cdot "13.6057 eV")/(n^2))`,

where `-"13.6057 eV"` IS the ground-state energy of `"H"` atom, `Z` is the atomic number, and `n` is the principal quantum number.

In this case, only one electron is in `"He"^(+)` or `"Li"^(2+)`, so we really only need to look at `n = 1`.

The ionization potential is then the energy put into the atom to remove that single electron.

We can check each of these on NIST.

`color(blue)(DeltaH_(IP)("He"^(+)(g))) = +((2^2)("13.6057 eV"))/(1^2)`

`~~` `color(blue)(ul"54.42 eV")`

for

`"He"^(+)(g) -> "He"^(2+)(g) + e^(-)`

[
(row 2)

`color(blue)(DeltaH_(IP)("Li"^(2+)(g))) = +((3^2)("13.6057 eV"))/(1^2)`

`~~` `color(blue)(ul"122.45 eV")`

for

`"Li"^(2+)(g) -> "Li"^(3+)(g) + e^(-)`

[
(row 3)