See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

For 2nd excited state to 1st excited state transition (n = 3) (n = 2) E = 7.56 eV = 13.6  2 2 2 1 1 z 2 3         z = 2 Now transition from 2nd excited state (n = 3) to ground state (n = 1) E = 13.6  z2  2 2 1 1 1 3        = 13.6  22  8 9 = 48.36 eV