In hydrogen atom, energy of the first excited state is -3.4 eV. Then find out the K.E. of the same orbit of H-atom

In hydrogen atom, energy of the first excited state is -3.4 eV. Then f

1.	In hydrogen atom, energy of the first excited state is -3.4 eV. Then find out the K.E. of the same orbit of H-atom
Answer» `+3.4 EV` `+6.8 eV` `-13.6 eV` `+13.6 eV` SOLUTION :ENERGY in EXCITED STATE = K.E.