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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