See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Let ground state energy (in eV) be E, Then from given condition, E2n  E1 = 204 eV 1 1 2 E E 204 eV 4n   1 2 1 E 1 204 eV 4n         …(i) and E2n  En = 40.8 eV 1 1 2 2 E E