The principal quantum number of initially excited level 'B' is — Atomic Structure Chemistry Question

💡 Solution & Explanation

Emission of 6 photon energies means the highest state reached is $n=4$ (since $4(3)/2 = 6$). The 2.7 eV matches a transition ending at $n=4$. For a hypothetical ionization energy (IE) of 14.4 eV, the transition from $n=2$ to $n=4$ requires $\Delta E = 14.4 (1/4 - 1/16) = 14.4 (3/16) = 2.7$ eV. Thus, level B is $n=2$. Therefore, correct answer is B.