The correct option is A 23.8 eV
We know that, in ground state, n=1
Total energy = - kinetic energy =potential energy2
⇒T.E=−K.E=PE2
⇒P.E=2[T.E]=2[−13.6 eV]=−27.2 eV
We have assumed this energy to be zero ie. potential energy is increased by 27.2 eV. Since, the kinetic energy will remain unchanged in all energy states whereas the potential energy is increased by 27.2 eV and hence the total energy in all states will increase by 27.2 eV.
⇒ for 1st excited state i.e., n=2
E2=−3.4 eV [previously]
and now,
E2=[−3.4+27.2] eV=23.8 eV