The electron in the hydrogen atom is in the third excited state. Its potential energy (in eV) is

Step-by-step solution

The n -th excited state corresponds to the principal quantum number n+1 . For the third excited state, n = 3 + 1 = 4 . The total energy of an electron in the n -th orbit of a hydrogen atom is given by E_n = - 13.6 n^2 eV. Substituting n = 4 , we get E₄ = - 13.6 4^2 = -0.85 eV. The potential energy U of the electron is twice its total energy E . U = 2E₄ = 2 (-0.85) = -1.70 eV. Answer: -1.70