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Energy of electron in first excited state in Hydrogen atom is -3.4eV. Find KE and PE of electron in the ground state.

Answer» Energy of electron in n = 2 is -3.4eV
`therefore " energy in ground state " = -13.6V " " E_(n) = underset(n^(2))overset(x)"" rArr -3.4eV = underset(z^(2))overset(x)""rArr`
kE = -TE = +13.6eV energy in ground state X = -13.6eV.
PE = 2TE =`-2 xx 13.6eV = -27.2eV`


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