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The radius of the first Bohr orbit $(n = 1)$ of hydrogen atom is $53.4 pm$ . The Bohr orbit having $n = 3$ in ${Li}^{2 +}$ will be

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53.4 pm

106.8 pm

120.1 pm

160.2 pm

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detailed solution

Correct option is D

Since $r \propto n^{2} / Z$ , we have

$\frac{r_{{Li}^{2 +}}}{r_{H}} = \frac{(3^{2} / 3)}{1^{2} / 1} = 3 \Rightarrow r_{{Li}^{2 *}} = 3 r_{H} = 3 \times 53.4 pm = 160.2 pm$

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The radius of the first Bohr orbit (n=1) of hydrogen atom is 53.4 pm. The Bohr orbit having n=3 in Li2+ will be