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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For a d

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$122.4$

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For a doubly ionized lithium atom $(Li^{2+})$, the atomic number $Z = 3$.In the ground state, the principal quantum number $n = 1$.Substituting these values into the formula: $E_1 = -13.6 	imes \frac{3^2}{1^2} \ eV = -13.6 	imes 9 \ eV = -122.4 \ eV$.The energy required to remove the electron (ionization energy) is the energy needed to bring the electron from the ground state to infinity $(E_{\infty} = 0)$.Therefore, Ionization Energy $= E_{\infty} - E_1 = 0 - (-122.4 \ eV) = +122.4 \ eV$.

As per the Bohr model, the minimum energy (in $eV$) required to remove an electron from the ground state of a doubly ionized $Li$ atom $(Z=3)$ is:

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