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:

$ \mathrm{E}_{\mathrm{T}} = -13.6 \frac{Z^2}{n^2} \text{ eV} $

Given:

Energy of the first orbit (ground state) of the hydrogen atom: $ \mathrm{E}_1 = -13.6 \text{ 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:

$ \frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{Be}^{+3}}} = \frac{Z_1^2}{n_1^2} \times \frac{n_2^2}{Z_2^2} $

Substitute the known values:

$ \frac{-13.6}{\mathrm{E}_{\mathrm{Be}^{+3}}} = \frac{1^2}{1^2} \times \frac{2^2}{4^2} $

$ \frac{-13.6}{\mathrm{E}_{\mathrm{Be}^{+3}}} = \frac{1}{1} \times \frac{4}{16} $

Solving this gives:

$ \mathrm{E}_{\mathrm{Be}^{+3}} = -13.6 \times 4 = -54.4 \text{ eV} $

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