In Hydrogen atom, energy of first excited state is – 3.4 eV. Then find out KE of same orbit of Hydrogen atom HomeAtomic Structure MCQsIn Hydrogen atom, energy of first excited state is – 3.4 eV. Then find out KE of same orbit of Hydrogen atomQ. In Hydrogen atom, energy of first excited state is – 3.4 eV. Then find out KE of same orbit of Hydrogen atomA. 3.4 eVB. 6.8 eVC. -13.6 eVD. +13.6 eVAnswer: 3.4 eV