To find the energy of an electron in the first excited state of a Be³⁺ ion, we can use the formula for the energy of an electron in a hydrogen-like atom: E = -13.6 (Z²/n²) eV

Given: Energy of the first orbit (ground state) of the hydrogen atom:

E₁ = -13.6 eV (where Z = 1 and n = 1).

For Be³⁺: Atomic number Z = 4.

First excited state corresponds to n = 2. Calculation for Be³⁺:

The energy ratio between the hydrogen atom and the Be³⁺ ion can be written as:

E_H / E_Be³⁺ = (Z₁²/n₁²) × (n₂²/Z₂²)

Substitute the known values:

(-13.6)/E_Be³⁺ = (1²/1²) × (2²/4²)

(-13.6)/E_Be³⁺ = (1/1) × (4/16)

Solving this gives: E_Be³⁺= -13.6 × 4 = -54.4 eV

The magnitude of the energy of the electron in the first excited state of Be³⁺ is |-54.4| = 54.4 eV, which is approximately 54 eV when rounded to the nearest integer.