Match List I with List II for a hydrogen atom. List I (Electron Transitions) List II (Energy of Emitted Photon) A n=2 n=1 I 1.89 eV B n=3 n=1 II 2.55 eV C n=4 n=2 III 10.2 eV D n=3 n=2 IV 12.09 eV Choose the correct answer from the options given below:

Step-by-step solution

The energy of an electron in the n -th orbit of a hydrogen atom is given by E_n = - 13.6 n^2 eV . Calculating the energy levels for the first four orbits: E₁ = - 13.6 1^2 = -13.6 eV E₂ = - 13.6 2^2 = -3.4 eV E₃ = - 13.6 3^2 = -1.51 eV E₄ = - 13.6 4^2 = -0.85 eV The energy of the emitted photon for a transition from n_i n_f is E = E_ n_i - E_ n_f . For A ( n=2 n=1 ): E = -3.4 - (-13.6) = 10.2 eV (Matches III) For B ( n=3 n=1 ): E = -1.51 - (-13.6) = 12.09 eV (Matches IV) For C ( n=4 n=2 ): E = -0.85 - (-3.4) = 2.55