Question

the ionization energy of hydrogen atom is -13.6eV. the ionization energy of second excited stateof He+ ion will be​

Answer 1

Answer:  5.28ev

Explanation:

FORMULA FOR IONIZATION ENERGY=  13.6 x Z^2 ev

                                                                                n^2

GIVEN: IE of H atom = -13.6ev

now,  for He+ ion =  (Z=2 , n=3 because second excited state means 3rd level)    

= 13.6 x( 4 / 9 )ev

= 5.28 ev