The energy of an electron in the first Bohr orbit of the H-atom is -13.6 eV.
The magnitude of energy value of an electron in the first excited state of Be³⁺ is ________ eV (nearest integer value).

Solution

Chemistry ma’am · 4 steps

Scale hydrogen's energy by Z²/n², and remember the first excited state means n = 2.

1. One-electron ion. Be³⁺ has lost 3 of its 4 electrons, so it has only one left. Bohr's model therefore applies to it just as it does to hydrogen, with nuclear charge Z = 4.

   Eₙ = -13.6 × Z²/n² eV

2. Pick the orbit. The ground state is n = 1, so the first excited state is n = 2.
3. Put in numbers. E₂ = -13.6 × 4²/2² = -13.6 × 16/4 = -13.6 × 4 = -54.4 eV.
4. So. The magnitude is 54.4 eV, which rounds to 54.

The common slip: 218. Taking the first excited state as n = 1 gives 13.6 × 16 = 217.6, about 218, not 54.

The idea to keep. For any one-electron species (H, He⁺, Li²⁺, Be³⁺), Eₙ = -13.6 Z²/n² eV. Ground state is n = 1, first excited is n = 2.

Answer from the NTA final answer key. Solution by PrepWiser.