Energy and radius of the ground-state Bohr orbit of \ce{He+} and \ce{Li^2+} are: [Given $R_H = 2.18 \times 10^{-18}$ J, $a_0 = 52.9$ pm]

Solution & Explanation

\textbf{Answer:} (C) $E_n(\ce{Li^2+}) = -19.62 \times 10^{-18}$ J, $r_n(\ce{Li^2+}) = 17.6$ pm; $E_n(\ce{He+}) = -8.72 \times 10^{-18}$ J, $r_n(\ce{He+}) = 26.4$ pm \textbf{Solution:} For the ground state ($n=1$): $E_n = -R_H \dfrac{Z^{2}}{n^{2}}$ and $r_n = a_0 \dfrac{n^{2}}{Z}$. For \ce{Li^2+} ($Z=3$): $E_n = -2.18\times10^{-18}\times 9 = -19.62\times10^{-18}$ J, $r_n = \dfrac{52.9}{3} = 17.6$ pm. For \ce{He+} ($Z=2$): $E_n = -2.18\times10^{-18}\times 4 = -8.72\times10^{-18}$ J, $r_n = \dfrac{52.9}{2} = 26.4$ pm.