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).

<p>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:</p> <p>$ \mathrm{E}_{\mathrm{T}} = -13.6 \frac{Z^2}{n^2} \text{ eV} $</p> <p><strong>Given:</strong></p> <p>Energy of the first orbit (ground state) of the hydrogen atom: $ \mathrm{E}_1 = -13.6 \text{ eV} $ (where $ Z = 1 $ and $ n = 1 $).</p> <p><strong>For Be³⁺:</strong></p> <p><p>Atomic number $ Z = 4 $.</p></p> <p><p>First excited state corresponds to $ n = 2 $.</p></p> <p><strong>Calculation for Be³⁺:</strong></p> <p>The energy ratio between the hydrogen atom and the Be³⁺ ion can be written as:</p> <p>$ \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} $</p> <p>Substitute the known values:</p> <p>$ \frac{-13.6}{\mathrm{E}_{\mathrm{Be}^{+3}}} = \frac{1^2}{1^2} \times \frac{2^2}{4^2} $</p> <p>$ \frac{-13.6}{\mathrm{E}_{\mathrm{Be}^{+3}}} = \frac{1}{1} \times \frac{4}{16} $</p> <p>Solving this gives:</p> <p>$ \mathrm{E}_{\mathrm{Be}^{+3}} = -13.6 \times 4 = -54.4 \text{ eV} $</p> <p>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.</p>