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In hydrogen atom, energy of first excited state is -3.4eV. Then find out KE of same orbit of hydrogen atom In hydrogen atom, energy of first excited state is -3.4eV. Then find out KE of same orbit of hydrogen atom
that means Potential Energy is -3.4 ev.Now from closest distance approach,PE/KE = -2Therfore,K.E = PE/-2 = > -3.4/-2 = 1.7 eV.
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