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(D) The radius of the $n^{	ext{th}}$ orbit for a hydrogen-like atom is given by the formula: $r_n = a_0 \cdot \frac{n^2}{Z}$,where $a_0$ is the Bohr

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$\frac{2}{3}$

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(D) The radius of the $n^{	ext{th}}$ orbit for a hydrogen-like atom is given by the formula: $r_n = a_0 \cdot \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the principal quantum number,and $Z$ is the atomic number.For $Li^{2+}$,the atomic number $Z_1 = 3$ and the orbit number $n = 5$. Thus,$r_{Li^{2+}} = a_0 \cdot \frac{5^2}{3} = a_0 \cdot \frac{25}{3}$.For $He^{+}$,the atomic number $Z_2 = 2$ and the orbit number $n = 5$. Thus,$r_{He^{+}} = a_0 \cdot \frac{5^2}{2} = a_0 \cdot \frac{25}{2}$.The ratio of the radii is: $\frac{r_{Li^{2+}}}{r_{He^{+}}} = \frac{a_0 \cdot \frac{25}{3}}{a_0 \cdot \frac{25}{2}} = \frac{25}{3} \cdot \frac{2}{25} = \frac{2}{3}$.

Considering the Bohr model of hydrogen-like atoms,the ratio of the radius of the $5^{\text{th}}$ orbit of the electron in $Li^{2+}$ to that in $He^{+}$ is:

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