If the experimental ionization energy of a completely ground-state hydrogen atom is measured as $131

Question

If the experimental ionization energy of a completely ground-state hydrogen atom is measured as $1312 \mathrm{~kJ \cdot mol}^{-1}$, then the exact theoretical ionization energy of a ground-state $\mathrm{Li}^{2+}$ ion would be

Professor PC Thomas Classes · Answer

(a) : The required ionization energy (IE) for removing an electron from hydrogen-like species is calculated by scaling the hydrogen value: $IE_Z = IE_{\mathrm{H}} 	imes Z^2$. For the completely stripped $\mathrm{Li}^{2+}$ ion, the effective nuclear charge $Z = 3$. Therefore, computing the required energy: $IE_{\mathrm{Li}^{2+}} = 1312 \mathrm{~kJ \cdot mol}^{-1} 	imes (3)^2 = 1312 	imes 9 = 11808 \mathrm{~kJ \cdot mol}^{-1}$. Rewriting this large value in proper scientific notation gives $1.18 	imes 10^4 \mathrm{~kJ \cdot mol}^{-1}$.