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Structure of Atom question

2025 · Q119
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Structure of Atom question

2025 · Q119

NEETChemistryStructure of AtomMCQ+4 / −1

Energy and radius of first Bohr orbit of He+\mathrm{He}^{+}He+and Li2+\mathrm{Li}^{2+}Li2+ 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}RH​=2.18×10−18 J,a0​=52.9pm ]

  1. A
    En(Li2+)=−19.62×10−16 J;rn(Li2+)=17.6pmEn(He+)=−8.72×10−16 J;rn(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}​En​(Li2+)=−19.62×10−16 J;rn​(Li2+)=17.6pmEn​(He+)=−8.72×10−16 J;rn​(He+)=26.4pm​
  2. B
    En(Li2+)=−8.72×10−16 J;rn(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}​En​(Li2+)=−8.72×10−16 J;rn​(Li2+)=17.6pmEn​(He+)=−19.62×10−16 J;rn​(He+)=17.6pm​
  3. C
    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}​En​(Li2+)=−19.62×10−18 J;rn​(Li2+)=17.6pmEn​(He+)=−8.72×10−18 J;rn​(He+)=26.4pm​
  4. D
    En(Li2+)=−8.72×10−18 J;rn(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}​En​(Li2+)=−8.72×10−18 J;rn​(Li2+)=26.4pmEn​(He+)=−19.62×10−18 J;rn​(He+)=17.6pm​
View written solutionFree

Correct answer: C

The energy and radius of the first Bohr orbit for a hydrogen-like atom is given by:

Energy:

$ E_n = \frac{-2.18 \times 10^{-18} \times Z^2}{n^2} \, \text{J} $

Radius:

$ r_n = \frac{52.9 \times n^2}{Z} \, \text{pm} $

Where:

$ Z $ is the atomic number.

$ n $ is the orbit number (or principal quantum number).

Let's calculate these values for the ions $\text{He}^+$ and $\text{Li}^{2+}$:

For $\text{He}^+$:

The atomic number, $ Z = 2 $

Principal quantum number, $ n = 1 $

Energy, $ E_{\text{He}^+} $:

$ E_{\text{He}^+} = -2.18 \times 10^{-18} \times 2^2 = -8.72 \times 10^{-18} \, \text{J} $

Radius, $ r_{\text{He}^+} $:

$ r_{\text{He}^+} = \frac{52.9 \times 1^2}{2} = 26.45 \, \text{pm} $

For $\text{Li}^{2+}$:

The atomic number, $ Z = 3 $

Principal quantum number, $ n = 1 $

Energy, $ E_{\text{Li}^{2+}} $:

$ E_{\text{Li}^{2+}} = -2.18 \times 10^{-18} \times 3^2 = -19.62 \times 10^{-18} \, \text{J} $

Radius, $ r_{\text{Li}^{2+}} $:

$ r_{\text{Li}^{2+}} = \frac{52.9 \times 1^2}{3} = 17.63 \, \text{pm} $

These calculations show the energy and radius for the first Bohr orbit of $\text{He}^+$ and $\text{Li}^{2+}$.

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