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NEET 2025
Exam
NEET
Year
2025
Subject
Chemistry
Chapter
Structure of Atom

NEET 2025 · Chemistry · Structure of Atom

Energy and radius of first Bohr orbit of He^ + and Li^ 2+ are...

Energy and radius of first Bohr orbit of He+\mathrm{He}^{+}and Li2+\mathrm{Li}^{2+} are [Given RH=2.18×10−18 J,a0=52.9pm\mathrm{R}_{\mathrm{H}}=2.18 \times 10^{-18} \mathrm{~J}, \mathrm{a}_0=52.9 \mathrm{pm} ]

Options

  1. AEn(Li2+)=−19.62×10−18 J;rn(Li2+)=17.6pmEn(He+)=−8.72×10−18 J;rn(He+)=26.4pm\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}
  2. BEn(Li2+)=−8.72×10−18 Jrn(Li2+)=26.4pmEn(He+)=−19.62×10−18 J;rn(He+)=17.6pm\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}
  3. CEn(Li2+)=−19.62×10−16 Jrn(Li2+)=17.6pmEn(He+)=−8.72×10−16 Jrn(He+)=26.4pm\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}
  4. DEn(Li2+)=−8.72×10−16 Jrn(Li2+)=17.6pmEn(He+)=−19.62×10−16 J;rn(He+)=17.6pm\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}

Correct answer

A. En(Li2+)=−19.62×10−18 J;rn(Li2+)=17.6pmEn(He+)=−8.72×10−18 J;rn(He+)=26.4pm\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}

Step-by-step solution

En=−2.18×10−18×Z2n2J For He+⇒E1=−2.18×10−18×2212J=−8.72×10−18J For Li2+⇒E1=−2.18×10−18×3212J=−19.62×10−18Jrn=52.9×n2Zpm For He+⇒r1=52.9×122pm=26.4pm For Li2+⇒r1=52.9×123pm=17.6pm\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}
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