Q45 / 180 · NEET 2025 · Chemistry
Energy and radius of first Bohr orbit of $\mathrm{He}^{+}$and $\mathrm{Li}^{2+}$ are
[Given $\mathrm{R}_{\mathrm{H}}=2.18 \times 10^{-18} \mathrm{~J}, \mathrm{a}_0=52.9 \mathrm{pm}$ ]
- Correct answerA$\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-19.62 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-8.72 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=26.4 \mathrm{pm} \end{aligned}$
- B$\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-8.72 \times 10^{-18} \mathrm{~J} \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=26.4 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-19.62 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=17.6 \mathrm{pm} \end{aligned}$
- C$\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-19.62 \times 10^{-16} \mathrm{~J} \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-8.72 \times 10^{-16} \mathrm{~J} \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=26.4 \mathrm{pm} \end{aligned}$
- D$\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-8.72 \times 10^{-16} \mathrm{~J} \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-19.62 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=17.6 \mathrm{pm} \end{aligned}$
Solution
$\begin{aligned}
& E_n=-2.18 \times 10^{-18} \times \frac{Z^2}{n^2} J \\
& \text { For } \mathrm{He}^{+} \Rightarrow \mathrm{E}_1=-2.18 \times 10^{-18} \times \frac{2^2}{1^2} \mathrm{~J} \\
& =-8.72 \times 10^{-18} \mathrm{~J} \\
& \text { For } \mathrm{Li}^{2+} \Rightarrow \mathrm{E}_1=-2.18 \times 10^{-18} \times \frac{3^2}{1^2} \mathrm{~J} \\
& =-19.62 \times 10^{-18} \mathrm{~J} \\
& \mathrm{r}_{\mathrm{n}}=52.9 \times \frac{\mathrm{n}^2}{\mathrm{Z}} \mathrm{pm} \\
& \text { For } \mathrm{He}^{+} \Rightarrow \mathrm{r}_1=52.9 \times \frac{1^2}{2} \mathrm{pm}=26.4 \mathrm{pm} \\
& \text { For } \mathrm{Li}^{2+} \Rightarrow \mathrm{r}_1=52.9 \times \frac{1^2}{3} \mathrm{pm}=17.6 \mathrm{pm}
\end{aligned}$
Hence, the correct answer is option: 'A'.