The ionization energy of hydrogen in excited state is +0.85 eV. What will be the energy of the photon emitted when it returns to the ground state?


Correct option is

12.75 eV


Energy of H-atom in the ground state

            = –13.6 eV.  

Ionization energy of the hydrogen in excited state equal to +0.85 eV means  

Energy of H-atom in the excited state = –0.85 eV 

∴ Energy emitted = –0.85 – (–13.6 eV) 

                             = 12.75 eV. 



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