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

Q.83) Energy and radius of first Bohr orbit of $\mathrm{He}^{+}$and $\mathrm{Li}^{2+}$ are [Given $\mathrm{R}_{\mathrm{H}}=2.18 \times 10^{-18} \mathrm{~J}, \mathrm{a}_0=52.9 \mathrm{pm}$ ]

A) $$ \begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-8.72 \times 10^{-16} \mathrm{~J} \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^2\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{\circ}\right)=-19.62 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{-}\right)=17.6 \mathrm{pm} \end{aligned} $$

B) $$ \begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-12.62 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-8.72 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=26.4 \mathrm{pm} \end{aligned} $$

C) $$ \begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)-8 \mathrm{c} 2 \times 10^{-18} \mathrm{~J} \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=26.4 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\lef

D) $$ \begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)-8 \mathrm{c} 2 \times 10^{-18} \mathrm{~J} \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=26.4 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\lef

 

Answers (1)

best_answer

Q.83) 1. Energy of nth orbit:

$$
E_n=-R_H \cdot \frac{Z^2}{n^2}
$$

2. Radius of nth orbit:

$$
r_n=a_0 \cdot \frac{n^2}{Z}
$$

- For He $^{+}(\mathrm{Z}=\mathbf{2 , n}=1)$ :
- $E=-2.18 \times 10^{-18} \cdot \frac{2^2}{1^2}=-8.72 \times 10^{-18} \mathrm{~J}$
- $r=52.9 \cdot \frac{\frac{1}{2}_2^2}{}=26.45 \mathrm{pm} \approx 26.4 \mathrm{pm}$
- For $\mathrm{Li}^{2+}(\mathrm{Z}=3, \mathrm{n}=1)$ :
- $E=-2.18 \times 10^{-18} \cdot \frac{3^2}{1^2}=-19.62 \times 10^{-18} \mathrm{~J}$
- $r=52.9 \cdot \frac{1^2}{3}=17.63 \mathrm{pm} \approx 17.6 \mathrm{pm}$
So the correct values are:
- $\mathrm{E}_n\left(\mathrm{He}^{+}\right)=-8.72 \times 10^{-18} \mathrm{~J}, r=26.4 \mathrm{pm}$
- $\mathrm{E}_n\left(\mathrm{Li}^{2+}\right)=-19.62 \times 10^{-18} \mathrm{~J}, r=17.6 \mathrm{pm}$

Hence, the correct answer is option (2).

 

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