For which atom or ion, the energy level of the second excited state is -13.6 eV? — Atomic Structure Chemistry Question

💡 Solution & Explanation

The energy of an orbit is given by E_n = -13.6 * Z^2 / n^2 eV. The second excited state corresponds to n = 3. Setting E_3 = -13.6 eV yields the equation: -13.6 = -13.6 * Z^2 / 3^2, which simplifies to Z^2 / 9 = 1 => Z^2 = 9 => Z = 3. This matches Li^2+.