AIIMS 2019 Chemistry Bohr's Model MCQ Question

For a hydrogen-like atom, the energy of an electron in the nᵗʰ orbit is given by the formula $E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2}$. For the $\text{Be}^{3+}$ ion (where $Z = 4$) and $n = 2$, the energy calculation yields $E_2 = -\frac{4^2 \cdot 13.6 \, \text{eV}}{2^2} = -27.2 \, \text{eV}$, which corresponds to option B, not A. However, since the question states that the energy of their orbit is -13.6 eV, it appears to be referencing the energy of the first orbit of hydrogen, which is a common point of confusion. Options C and D are incorrect as they represent energies for higher orbits that do not apply to the given conditions.