UPSEE Solved Model Paper 4

The energy of nth state in the hydrogen state will be
En = −

13.6eV

n2

so this energy for the second excited state i.e., n = 3 will be
E3 =

−13.6eV

9

= - 1.51 ev
So ionization energy for this state will be - (- 1.51 eV) = 1.51 eV