The radius of third orbit of hydrogen atom is R pm. The radius of second orbit of He⁺ ion (in pm is)

$$\begin{gathered} {\mathrm{r}}_3=0.529\times \frac{ 3^2}{1} (\mathrm{third} \mathrm{orbit} \mathrm{of} \mathrm{H})=4.761 \stackrel{0}{\mathrm{A}}=476.1 \mathrm{pm}=\mathrm{R} \\ {\mathrm{r}}_2=0.529\times \frac{ 2^2}{2} (\mathrm{second} \mathrm{orbit} \mathrm{of} \mathrm{He})=1.058 \stackrel{0}{\mathrm{A}}=105.8 \mathrm{pm} \\ \mathrm{Answer} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{R} \\ 476.1 \mathrm{pm}\times \frac{2}{9}=105.8 \mathrm{pm} \\ \mathrm{R}\times \frac{2}{9}=105.8 \mathrm{pm} \end{gathered}$$