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?

The ionization energy of hydrogen in excited state is +0.85 eV. What w

1.	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?
Answer» SOLUTION :ENERGY of H-atom in the ground state `= - 13.6 eV` Ionization energy of 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 eV - (-1.6 eV) = 12.75 eV`