If the radius of the second Bohr orbit of the $\mathrm{He}{}^{+}$ ion is $105.8$ pm, what is the radius of the third Bohr orbit of the $\mathrm{Li}{}^{2+}$ ion?

Solution & Explanation

\textbf{Answer:} (A) 158.7 pm \textbf{Solution:} $r\propto \frac{n^2}{Z}$. For $\mathrm{He}{}^{+}$ ($Z=2$, $n=2$): $r\propto \frac{4}{2}=2$. For $\mathrm{Li}{}^{2+}$ ($Z=3$, $n=3$): $r\propto \frac{9}{3}=3$. Ratio $=\frac{3}{2}$, so $r(\mathrm{Li}{}^{2+})=105.8\times \frac{3}{2}=158.7$ pm.