H-atom is exposed to UV radiation having wavelength 102.8nm

the energy of photon corresponding to 102.8 nm waveegth is

E = hc/λ = 1242 eV-nm /102.8 nm

   = 12.42 eV

The energy needed to take athe H atom frm ground state to first excited state is

                    E₂-E₁ - 13.6 eV - 3.4 eV = 10.2 eV

and to  the second excited state is

                    E₃-E₁ - 13.6 eV - 1.5 eV = 12.1 eV

and to  the third excited state is

                    E₄-E₁ - 13.6 eV - 0.85 eV = 12.75 eV

thus 12.42 eV photon has large probability of being absorbed in 12.1 eV

The corresponding  waveegth is: λ = 1242/12.1

                                                               =103 nm

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