AIIMS 2019 Physics Energy Levels MCQ Question

In the expression $E = -13.6 \times \left( \frac{z^2}{n^2} \right)$, substituting $z = 2$ (for the element Helium) and $n = 2$ yields $E = -13.6 \times \left( \frac{2^2}{2^2} \right) = -13.6 \, \text{eV}$, which represents the energy of an electron in the second energy level of a hydrogen-like atom. Since the calculation is accurate and aligns with the expected energy value for this state, option B is the correct choice. Other options are not applicable as they do not provide relevant information or calculations related to the problem.