Question

The energy of a one-electron atom or ion is given by Bohr’s equation. Which one of the following one-electron species has the lowest energy ground state? a)Li2+ b) He+ c)Be3+ d) B4+ e) H

Answer 1

The correct answer is d) B⁴⁺ has the lowest energy ground state.

Bohr's equation is

`E_n = -(Z^2R)/n^2`

For the ground state, `n = 1`, so

`E_1 = -Z^2R`

If `Z = 1`, `E = -R = "-13.6 eV"`

Since `R = "13.6 eV"`, the formula becomes

`E_1 = -13.6Z^2 "eV"`

We can see that as `Z` increases, `E_1` decreases, but let's do the calculations anyway.

• For `"H"`: `Z = 1`; `E_1 = "-13.6 eV" × 1^2 = "-13.6 eV"`
• For `"He"^+`: `Z = 2`; `E_1 = "-13.6 eV" × 2^2 = "-54.4 eV"`
• For `"Li"^(2+)`: `Z = 3`; `E_1 = "-13.6 eV" × 3^2 = "-122.4 eV"`
• For `"Be"^(3+)`: `Z = 4`; `E_1 = "-13.6 eV" × 4^2 = "-218 eV"`
• For `"B"^(4+)`: `Z = 5`; `E_1 = "-13.6 eV" × 5^2 = "-340 eV"`

The ground state energy is lowest for `Z = 5` (boron).

Answer 2

The answer is d) `B^(4+)`.

The lowest energy ground state will belong to the atom or ion that has the largest atomic number, `"Z"`.

For one-electron species, the energy of an electron that orbits the nucleus at level `"n"` is given by

`E = -Z^(2)/n^(2) * underbrace(((k_e * e^(2))^(2) * m_e)/(2bar(h)^(2)))_(R_E)`, where

`R_E` represents a constant called the Rydberg energy, `R_E = "13.6 eV"`.

The equation for energy becomes

`E = - R_E * Z^(2)/n^(2)`, where

`Z` - the atomic number;
`n` - the energy level;

Since you're interested in ground state energy, `n=1` and so

`E = -R_E * Z^(2)`

Of the species you've listed, the highest atomic number belongs to boron, `B`, for which `Z = 5`. This means that the `B^(4+)` ion will have the lowest ground state energy

`E = -"13.6 eV" * 5^(2) = color(green)(-"340 eV")`