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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 	imes \frac{Z^2}{n^2}\,eV$.For $Li^{2+}$,the ato

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$136$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 	imes \frac{Z^2}{n^2}\,eV$.For $Li^{2+}$,the atomic number $Z = 3$.The second excited state corresponds to $n = 3$.Therefore,the energy of the electron in the second excited state is $E_3 = -13.6 	imes \frac{3^2}{3^2} = -13.6\,eV$.The energy required to remove the electron (binding energy) is the magnitude of this energy,which is $13.6\,eV$.We are given that this energy is $x 	imes 10^{-1}\,eV$.So,$13.6 = x 	imes 10^{-1} \implies x = 136$.

If the binding energy of a ground state electron in a hydrogen atom is $13.6\,eV$,then the energy required to remove the electron from the second excited state of $Li^{2+}$ will be $x \times 10^{-1}\,eV$. The value of $x$ is $...........$

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