Energy of H-atom in the ground state is -13.6 | vedclass

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(A) The ground state of a hydrogen atom corresponds to the principal quantum number $n = 1$.The second excited state corresponds to the principal quan

Vedclass · Accepted Answer

$1.51$

Vedclass · Answer

(A) The ground state of a hydrogen atom corresponds to the principal quantum number $n = 1$.The second excited state corresponds to the principal quantum number $n = 3$.The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \; eV$.For the second excited state $(n = 3)$,the energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \; eV$.The energy required to ionize the atom is the energy needed to bring the electron from the current state to the state where $E = 0$.Therefore,the ionization energy $E_{ion} = 0 - E_3 = 0 - (-1.51) = 1.51 \; eV$.

Energy of $H$-atom in the ground state is $-13.6 \; eV$. The energy needed to ionize a hydrogen atom which is in its second excited state is ....... $eV$.

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